Category: OP Malhotra Solutions

Parents can use OP Malhotra Class 10 Solutions Chapter 8 Matrices Exercise 8(a) to provide additional support to their children.

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(a)

Question 1.
Write down the order of each matrix given below and the number of elements in each :
(i) \(\left[\begin{array}{cc}
1 & -1 \\
7 & 4
\end{array}\right]\)
(ii) \([\sin \theta \cos \theta]\)
(iii) \(\{2 4 6 8\}\)
(iv) \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
(v) \(\left[\begin{array}{llll}
2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0
\end{array}\right]\)
(vi) \(\left[\begin{array}{rrrrr}
1 & -5 & 0 & 8 & 4 \\
2 & -7 & 3 & 5 & 2 \\
0 & -2 & 1 & 4 & 9
\end{array}\right]\)
(vii) \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
5 & -9 \\
7 & 0
\end{array}\right]\)
Solution:
Order of matrix is given below :
Order of matrix is given below :
(i) 2 x 2; 4
(ii) 1 x 2; 2
(iii) 1 x 4; 4
(iv) 3 x 1; 3
(v) 3 x 4; 12
(vi) 3 x 5; 15
(vii) 4 x 2; 8

Question 2.
Classify the following matrices as equal or not equal.
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] ;\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
(ii) \(\left[\begin{array}{ll}
4 & 7 \\
3 & 2
\end{array}\right] ;\left[\begin{array}{cc}
3+1 & \sqrt{49} \\
5-2 & \frac{6}{3}
\end{array}\right]\)
Solution:
(i) Not equal as their corresponding element are not same.

(ii) \(\left[\begin{array}{ll}
4 & 7 \\
3 & 2
\end{array}\right] \text { and }\left[\begin{array}{cc}
3+1 & \sqrt{49} \\
5-2 & \frac{6}{3}
\end{array}\right]=\left[\begin{array}{ll}
4 & 7 \\
3 & 2
\end{array}\right]\)
These matrices are equal as their corresponding elements are same.

Question 3.
Construct 2 x 2 matrix A where elements a_aj are give by a = (2i – j)².
Solution:
Let the matrix 2 x 2 be A\(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
and it is given that a_ij = (2i – j)²
∴ a₁₁ = (2 x 1 – 1)² = (2 – 1) = (1)² = 1
a₁₂ = (2 x 1 – 2)² = (2 – 2)² = 0
a₂₁ = (2 x 2 – 1)² = (4 – 1)² = (3)² = 9
a₂₂ = (2 x 2 – 2)² = (4 – 2)² = (2)² = 4
∴ Matrix A = \(\left[\begin{array}{ll}
1 & 0 \\
9 & 4
\end{array}\right]\)

Question 4.
For the matrix B = \(\left[\begin{array}{cc}
3 & 11 \\
-5 & 6 \\
8 & 0
\end{array}\right]\),
(i) What is the order of matrix B ?
(ii) State the elements, a₁₂, a₃₁, a₂₂.
(iii) B is a square matrix. True or False.
Solution:
Matrix B = \(\left[\begin{array}{cc}
3 & 11 \\
-5 & 6 \\
8 & 0
\end{array}\right]\)
(i) Order of matrix is 3 x 2
(ii) Elements a₁₂ = 11, a₃₁ = 8, a₂₂ = 6
(iii) Matrix B is not a square matrix as in it. no. of rows ≠ no. of columns

Question 5.
(a) A matrix has 6 elements. Write the possible orders of the matrix.
(b) If a matrix has 3 rows and 4 columns, what is the number of elements in the matrix ?
Solution:
(a) A matrix which has 6 elements can be of the following possible orders.
6 x 1, 3 x 2, 2 x 3, 1 x 6

(b) A matrix which has 3 rows and 4 columns will have 3 x 4 = 12 elements.

Question 6.
Find x and y such that
(i) \(\left[\begin{array}{cc}
x & y \\
-1 & 5
\end{array}\right]=\left[\begin{array}{ll}
-2 & 0 \\
-1 & 5
\end{array}\right]\)
(ii) \(\left[\begin{array}{ll}
x & 4
\end{array}\right]=\left[\begin{array}{ll}
-2 & y
\end{array}\right]\)
(iii) \(\left[\begin{array}{cc}
2 x & 3 \\
0 & y-1
\end{array}\right]=\left[\begin{array}{cc}
x-4 & 3 \\
0 & 5
\end{array}\right]\)
Solution:
(i) \(\left[\begin{array}{cc}
x & y \\
-1 & 5
\end{array}\right]=\left[\begin{array}{ll}
-2 & 0 \\
-1 & 5
\end{array}\right]\)
Comparing their cooresponding element, we get
x = – 2, y = 0

(ii) \(\left[\begin{array}{ll}
x & 4
\end{array}\right]=\left[\begin{array}{ll}
-2 & y
\end{array}\right]\)
Comparing their corresponding elements, we get
x = – 2, y = 4

(iii) \(\left[\begin{array}{cc}
2 x & 3 \\
0 & y-1
\end{array}\right]=\left[\begin{array}{cc}
x-4 & 3 \\
0 & 5
\end{array}\right]\)
Comparing their corresponding elements, we get
and 2x = x – 4
⇒ 2x – x = – 4
⇒ x = – 4 and y – 1 = 5
⇒ y = 5 + 1 = 6
∴ x = – 4, y = 6

Question 7.
Find p, q, r and s, if \(\left[\begin{array}{ll}
p+4 & 2 q-7 \\
s-3 & r+2 s
\end{array}\right]\) = \(\left[\begin{array}{cc}
6 & -3 \\
2 & 14
\end{array}\right]\).
Solution:
\(\left[\begin{array}{cc}
p+4 & 2 q-7 \\
s-3 & r+2 s
\end{array}\right]=\left[\begin{array}{cc}
6 & -3 \\
2 & 14
\end{array}\right]\)
Comparing their corresponding elements, we get
p + 4 = 6 ⇒ p = 6 – 4 = 2
2q – 7 = – 3 ⇒ 2q = – 3 + 7 ⇒ 2q = 4
⇒ q = \(\frac { 4 }{ 2 }\) = 2
s – 3 = 2 ⇒ s = 2 + 3 = 5
r + 2s = 14 ⇒ r + 2 x 5 = 14 ⇒ r + 10 = 14
⇒ r = 14 – 10 = 4
Hence p = 2, q = 2, r = 4, s = 5

Question 8.
Answer true or false :
(i) Every zero matrix is a square matrix.
(ii) A unit matrix is a diagonal matrix.
(iii) A zero matrix of order 3 is a diagonal matrix.
(iv) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) is a unit matrix.
(v) A matrix is an aggregate of numbers.
(vi) \(\left[\begin{array}{ll}
3 & x \\
0 & 2
\end{array}\right]=\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]\), if x = 4,
(vii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) is the identity matrix for addition of 2 x 2 matrix.
Solution:
(i) False : It is not compulsary.
(ii) True : In unit matrix, diagonal elements are unity and others are zero.
(iii) False : In zero matrix, all element are zero but in diagonal matrix, it is not necessary.
(iv) False : In unit matrix, only diagonal elements are unity.
(v) False : A matrix is an array of real numbers arranged in rows and columns.
(vi) False : Their elements a₂₁ ≠ b₂₁ (0 ≠ 1)
(vii) True.

Question 9.
The length, width and height of two boxes are 6,5,10 and 5, 2, 8 respectively. Write this set of numbers in matrix form so that the first column indicates length, the second indicates width, and the third indicates height. A third box has corresponding dimensions of 7, 3, 3. Write a three by three matrix describing the dimensions of the three boxes.
Solution:
Length, width and height of one box are 6, 5, 10
and that of second box are 5, 2, 8
∴ and of third box are 7, 3, 3
Now we shall write them in Matrix of 3 x 3
order = \(\left[\begin{array}{ccc}
6 & 5 & 10 \\
5 & 2 & 8 \\
7 & 3 & 3
\end{array}\right]\)

Question 10.
(i) Three pupils in an algebra class score marks in three tests as follows : Akhil, 79, 87, 92; Rajnish, 95, 98, 91; Sudha, 76, 88, 77.
Display this information as a 3 x 3 matrix.

(ii) If the price of a record is Rs. 30, of a blade packet Rs. 1.50 and of a soap cake Rs. 1.70, display the prices as a 3 x 1 price matrix.
Solution:
(i) Three pupils Akhil, Rajnish and Sudha got marks in three tests algebra. Their scores are
Akhil = 79, 87, 92; Rajnish = 95, 98, 91 and Sudha = 76, 88, 77
We can display this information in a matrix of 3 x 3 order:
\(\left[\begin{array}{lll}
79 & 87 & 92 \\
95 & 98 & 91 \\
76 & 88 & 77
\end{array}\right]\)

(ii) The price of a record is Rs. 30, of a blade packet is Rs. 1.50 and a soap cake is Rs. 1.70.
We can display them in a matrix of 3 x 1, as
given \(\left[\begin{array}{c}
30 \\
1.50 \\
1.70
\end{array}\right]\)

Accessing Class 10 ICSE Maths Solutions S Chand Chapter 16 Trigonometrical Identities and Tables Ex 16(b) can be a valuable tool for students seeking extra practice.

S Chand Class 10 ICSE Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(b)

Question 1.
Find the sine, cosine, and tangent of the following angles:
(a) 15°27′
(b) 37°48′
(c) 55°17′
(d) 83°37′
Solution:
Using the natural sine, cosine and tangent tables:
(a) 15°27′ = sine 15°24′ + 3′
= 0.26556 + 84 (Mean difference of 3)
= 0.26640 = 0.2664
cos 15°27′ = cos 15°24′ + 3′
= 0.96410 – 23 (Mean difference of 3)
= 0.96387 = 0.9639
tan 15°27′ = tan 15°24′ + 3′
= 0.27545 + 94 = 0.27639 = 0.2764

(b) sin 37°48′ = 0.61291
= 0.6129
cos 37°48′ = 0.79015 = 0.7902
tan 37°48′ = 0.77568 = 0.7757

(c) sin 55°17′ = sin 55°12′ + 5′
= 0.82115 + 82 (Mean difference of 5′)
= 0.82917 = 0.8219
cos 55°17′ = cos 55°12′ + 5′
= 0.57071 – 120 = 0.56951 = 0.5695
tan 55°17′ = tan 55°12′ + 5′
= 1.43881 + 453 = 1.44334 = 0.4433

(d) sin 83°37′ = sin 83°36′ +1′
= 0.99377 + 3 = 0.99380 = 0.9938
cos 83°37′ = cos 83°36′ + 1′
= 0.11147 – 29 = 0.11118 = 0.1112
tan 83°37′ = 8.91520 = 8.9152

Question 2.
Find the acute angle A, given
(a) sin A = 0.4919
(b) tan A = 2.7775
(c) tan A = 3.412
(d) cos A = 0.4651
(e) sin A = 0.95 19
(f) cos A = 0.5757
Solution:
Using the tables of sines, cosines and tangent
(a) sin A 0.4919 = 0.49090 + differnece = 100
sin 29°24′ + 4 = sin 29°28′
∴ A = 29°28’

(b) tan A = 2.7775 = 2.77761 (∵ It is nearest to 2.77750)
∴ tanA = tan 70°12′
∴ A = 70°12′

(c) tanA = 3.412 = 3.41973
= tan 73°42′ (∵ 3.4 1973 is nearest to 3.412)
∴ A = 73°42′

(d) cos A = 0.4651 = 0.46484 + 16
cos 62°1 8 – 1′ = cos 62° 17′
∴ A = 62°17′

(e) sin A = 0.95190 = 0.95159 + 31
= sin 72°6′ + 3′ = sin 72°9′
∴A = 72°9′

(f) cos A = 0.57570 = .57501 + 69
= cos 54°54′ – 3′ = cos 54°51′
∴ A = 54°51′

Question 3.
Using tables, find the value of (2 sin θ – cos θ) as a decimal
(i) when θ = 35°
(ii) when tan θ = 0.2679
Solution:
(i) 2 sin θ – cos θ = 2 sin 35° – cos 35°
= 2(0.57358) – 0.81915 = 1.14716 – 0.81915
= 0.32801 = 0.3280

(ii) tan θ = 0.2679 = tan 14°56′
∴ 2sin θ – cos θ
= 2 sin 14°56’ – cos 14°56′
= 2 (0.25769) – 0.96622
0.5 1538 – 0.96622 = -0.45084

Question 4.
State for any acute angle θ
(i) whether sin θ increases or decreases as θ increases;
(ii) whether cos θ increases or decreases as θ decreases.
Solution:
(i) We know that sin θ° = 0 and sin 90° = 1
∴ It is clear that sin θ increase as θ increase

(ii) We know that cos θ° = 1 and cos 90° = 0
∴ It is clear that as θ decreases, cos θ increases

Question 5.
If sin x° = 0.67, find the value of
(a) cos x°
(b) cos x° + tan x°
Solution:
From the tables of sines
sin x° = 0.67 = 0.67043 (nearest value)
= sin 42°.6′ – 2′ (Mean difference)
= sin 42°4′

(i) ∴ cos x° = cos 42°4′
= 0.74314 – 77 = 0.74237 = 0.7423

(ii) cos x° + tan x° = cos 42°4′ + tan 42°4′
= 0.7423 + (0.90040 + 214)
= 0.7423 + 0.90254 = 0.7423 + 0.9025 = 1.6448

Question 6.
Using trigonometric table, find the measure of the angle A when sin A = 0.1822.
Solution:
sin A = 0.1822
Using the trigonometric tables,
sin A = 0.18224 (nearest Value)
= sin 10°30′
∴ A = 10°30′

Question 7.
In rectangle ABCD, AB = 23 cm, and ∠CAB = 35°. Calculate the measure of BC.

Solution:
In rectangle ABCD, AC is its diagonal
AB = 23 cm and ∠CAB = 35°
Let BC = x, then in right △ABC
tan θ = \(\frac{\text { Perpendicular }}{\text { Base }}\) = \(\frac { BC }{ AB }\)
⇒ tan 35° = \(\frac { x }{ 23 }\) ⇒ 0.70021 = \(\frac { x }{ 23 }\)
x = 23 × 0.70021 = 16.10483 = 16.1048 = 16.11
∴ BC = 16.11 cm

Question 8.
In the figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB. Given that ∠AED = 60° and ∠ACO = 45°; without using tables, calculate
(i) AB,
(ii) AC and
(iii) AE.

Solution:
In the figure,
AB || DC and BC are perpendicular to AB
AD = BC = 2 cm

∠AED = 60° and ∠ACD = 45°
∵ AB || DC and AD and BC are Perpendicular to AB and AD = BC = 2 cm
∴ AB = CD and AD and BC are perpendicular to CD also.

(i) Now in right △ACD
tan θ = \(\frac { AD }{ CD }\) ⇒ tan 45° = \(\frac { 2 }{ CD }\)
⇒ 1 = \(\frac { 2 }{ CD }\) ⇒ CD = 2 cm
∴ AB = CD = 2 cm

(ii) and sin 45° = \(\frac { AD }{ AC }\) ⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac { 2 }{ AC }\)
⇒ AC = 2√2 cm

(iii) In right △ADE
tan 60° = \(\frac { AD }{ AE }\) ⇒ √3 = \(\frac { 2 }{ AE }\)
⇒ AE = \(\frac{2}{\sqrt{3}}\) = \(\frac{2 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{2 \sqrt{3}}{3}\)
∴ AE = \(\frac{2 \sqrt{3}}{3}\) cm or \(\frac{2}{\sqrt{3}}\) cm

Question 9.
In the figure, BC = 12 cm, AB = 4 cm, ∠AEB = 90°, ∠B = 50° and ∠C = 30°. Calculate the length of
(i) BE and
(ii) AC.

Solution:
In the figure,
BC = 12 cm, AB = 4 cm. ∠AEB = 90°.
∠B = 50° and ∠C = 30°

(i) In right △ABE,
cos θ = \(\frac { BE }{ AB }\) ⇒ cos 50° = \(\frac { BE }{ 4 }\)
⇒ 0.64279 = \(\frac { BE }{ 4 }\) ⇒ BE = 4 × 0.64279
⇒ BE = 2.57116 = 2.57 cm

(ii) sin θ = \(\frac { AE }{ AB }\) ⇒ sin 50° = \(\frac { AE }{ 4 }\)
⇒ 0.76604 = \(\frac { AE }{ 4 }\) ⇒ AE = 4 × 0.76604
⇒ AE = 3.06416 = 3.06 cm

(iii) In △AEC,
sin θ = \(\frac { AE }{ AC }\) ⇒ sin 30° = \(\frac { 3.06 }{ AC }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { 3.06 }{ AC }\) ⇒ AC = 2 × 3.06 = 6.12
Hence AC = 6.12 cm

Question 10.
In the figure, in △ABC, ∠B = 90°, ∠C = 30° and AB = 12 cm. BD is perpendicular to AC; find
(i) BC
(ii) AD
(iii) AC

Solution:
In the figure, in △ABC
∠B = 90°, ∠C = 30°
∴ ∠A = 180° – 90° – 30° = 180° – 120° = 60°
AB = 12 cm, BD ⊥ AC

(i) In right △ABC, ∠B = 90°, ∠C = 30°
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 30° = \(\frac { 12 }{ BC }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 12 }{ BC }\) ⇒ BC = 12√3 cm

(ii) In right △ABD
cos θ = \(\frac { AD }{ AB }\) ⇒ cos 60° = \(\frac { AD }{ 12 }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { AD }{ 12 }\) ⇒ AD = 12 × \(\frac { 1 }{ 2 }\) = 6
∴ AD = 6 cm

(iii) In △ABC,
sin θ = \(\frac { AB }{ AC }\) ⇒ sin 30° = \(\frac { 12 }{ AC }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { 12 }{ AC }\) = 12 × 2 = 24
∴ AC = 24 cm

Question 11.
In the figure, the radius of a circle is given as 15 cm and chord AB subtends and angle of 131° at the centre C of the circle. Using trigonometry, calculate
(i) the length of AB;
(ii) the distance of AB from the centre C.

Solution:

Radius of the circle with centre C is 15 cm
i.e., AC = BC = 15 cm
∠ACB = 131°
From C, draw CL ⊥ AB
which bisects the chord AB at L
Now in △ABC, ∠C = 131° and AC = BC
∴ ∠A = ∠B = \(\frac{180^{\circ}-131^{\circ}}{2}\) = \(\frac{49^{\circ}}{2}\)
= 24.5° = 24°30′

(i) Now in right △ACL, ∠A = 24°36′
∴ cos θ = \(\frac { AL }{ AC }\) ⇒ cos24°30′ = \(\frac { AL }{ 15 }\)
⇒ .90996 = \(\frac { AL }{ 15 }\) ⇒ AL = 15 × 0.90996
⇒ AL = 13.6494
and AB = 2AL = 2 × 13.6494
= 27.2988 = 27.3 cm

(ii) sin θ = \(\frac { CL }{ AC }\) ⇒ sin 24°30′ = \(\frac { CL }{ 15 }\)
⇒ 0.41469 = \(\frac { CL }{ 15 }\) ⇒ CL = 15 × 0.41469
⇒ CL = 6.22035 = 6.22 cm
Hence the distance of AB from the centre C = 6.22 cm

Question 12.
In the figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically up and then travels 80 km at 30° to the vertical. PA represents the first stage of its journey and AB the second; C is a point vertically below B on the same horizontal level as P. Calculate :
(i) the height of the rocket when it is at point B;
(ii) the horizontal distance of point C from point P.

Solution:
In the figure, a rocket is launched from P. It first rises upward to A such that AP = 20 km Then travels to B making an angle of 30° and reaches at B such that AB = 80 km

BC ⊥ PC
From A, draw AD || PC meeting BC at D
∴ DC = AP = 20 km, AD = PC
and ∠BAD = 90° – 30° = 60°

(i) In right △ABD,
sin θ = \(\frac { BD }{ AB }\) ⇒ sin 60° = \(\frac { BD }{ 80 }\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac { BD }{ 80 }\) ⇒ BD = \(\frac{\sqrt{3} \times 80}{2}\) = 40√3 km
∴ BC = BD + DC = 40√3 + 20
= (20 + 40√3)
= 40 (1.732) + 20 = 69.280 + 20 = 89.28 Km

(ii) and cos θ = \(\frac { AD }{ AB }\) ⇒ cos 60° = \(\frac { AD }{ 80 }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { AD }{ 80 }\) ⇒ AD = 80 × \(\frac { 1 }{ 2 }\) = 40
∴ PC = AD = 40 Km

Question 13.
In the figure, BCDE is a rectangle, ED = 3.88 cm, AD = 10 and ∠DAC = 20°35′. Calculate, without using Pythagoras’ theorem,
(i) the length of CD;
(ii) the length of AC;
(iii) the size of angle AEB.

Solution:
BCDE is a rectangle in which ED = 3.88 cm
∴BC = 3.88 cm
A is a point such that AD = 10 cm and A lies on CB on producing, AE is joined
Let ∠AEB = θ

(i) In right △ACD,
sin θ = \(\frac { CD }{ AD }\) ⇒ sin 23°35° = \(\frac { CD }{ 10 }\)

∴ 0.40008 = \(\frac { CD }{ 10 }\) ⇒ CD = 4.0008 = 4.001
∴ CD = 4.001 cm

(ii) cos θ = \(\frac { AC }{ AD }\) ⇒ cos 23°35′ = \(\frac { AC }{ 10 }\)
[But cos 23°35′ = 0.91706 – 58 = 0.91648]
∴ 0.91648 = \(\frac { AC }{ 10 }\) ⇒ AC = 9.1648 = 9.165
∴ AC = 9.165 cm

(iii) Now AB = AC – BC = 9.165 – 3.880 = 5.285 and EB = CD = 4.001
∴ tan θ = \(\frac { AB }{ EB }\) = \(\frac { 5.285 }{ 4.001 }\) = \(\frac { 5285 }{ 4001 }\) = 1.32092
= 1.31745 + 347 = tan 52°48′ + 5′
= tan 52°53′ (from the tables)
∴ θ = 52°53′
∴ ∠AEB = 52°53′

Question 14.
In the figure, triangle ABC is right angled at B. D is the foot of the perpendicular from B to AC.
Given that BC = 3 cm and AB = 4 cm. Without using tables find
(i) tan ∠DBC
(ii) sin ∠DBA

Solution:
In the figure, in right △ABC, ∠B = 90°
BD ⊥ AC, BC = 3 cm, AB = 4 cm

∴ AC² = BC² + AC² (Pythagoras Theorem)
= (3)² + (4)² = 9 + 16 = 25
∴ AC = √25 = 5 cm
In △ABC and △DBC,
∠ABC = ∠BDC (each 90°)
∠C = ∠C (common)
∴ △ABC ~ △DBC (A A axiom)
Similarly we can prove that △ABC ~ △ABD
\(\frac { AC }{ BC }\) = \(\frac { AB }{ BD }\) = \(\frac { BC }{ CD }\) ⇒ \(\frac { AB }{ BC }\) = \(\frac { BD }{ CD }\) (By alternendo)
\(\frac { BD }{ CD }\) = \(\frac { 4 }{ 3 }\) ⇒ \(\frac { CD }{ BD }\) = \(\frac { 3 }{ 4 }\)
Now in right △DBC,
tan ∠DBC = \(\frac { CD }{ BD }\) = \(\frac { 3 }{ 4 }\)
In right △ABD,
sin ∠DBA = \(\frac { AD }{ AB }\) = \(\frac { AB }{ AC }\) = \(\frac { 4 }{ 5 }\)

Question 15.
Some students wished to find the height x of a building and the height y of the flag pole on the building. They made the measures as shown in the diagram. Find x and y. Give your answer to the nearest metre.

Solution:
BC is the building and AB is the flag pole on the building Angle of elvation of B=63° and of A = 63° + 3° = 66°

Now in right △BCD,
tan θ = \(\frac { BC }{ DC }\) ⇒ tan 63° = \(\frac { x }{ 50 }\)
⇒ 1.96261 = \(\frac { x }{ 50 }\) ⇒ x = 50 × 1.96261
x = 98.1305 = 98 cm
Again in right △ADC,
tan 66° = \(\frac { AC }{ DC }\) = \(\frac{x+y}{50}\) ⇒ 2.24604 = \(\frac{x+y}{50}\)
⇒ x + y = 50 × 2.24604 = 112.302 = 112
⇒ y = 112 – 98 = 14 m

Question 16.
The upper part of tree, broken by the wind, makes an angle of 30° with ground, and the horizontal distance from the root of the tree to the point where the top of the tree meets the ground is 25 metres. Find the height of the tree before it was broken, to the nearest metre.
Solution:
TR is the tree which was broken from Q and its top T touched the ground at S. So that SR = 25 m and ∠QSR = 30°
In the figure TQ = QS

In right △QSR
tan θ = \(\frac { QR }{ SR }\) ⇒ tan 30° = \(\frac { QR }{ 25 }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { QR }{ 25 }\) ⇒ QR = \(\frac{25}{\sqrt{3}}\)
⇒ QR = \(\frac { 25 }{ 1.732 }\) = 14.43
and cos θ = \(\frac { SR }{ SQ }\) ⇒ cso 30° = \(\frac { 25 }{ SQ }\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac { 25 }{ SQ }\) ⇒ SQ = \(\frac{25 \times 2}{\sqrt{3}}\) = \(\frac{50}{\sqrt{3}}\)
∴ Height of tree = TQ + QR
= QS + QR = \(\frac{25}{\sqrt{3}}\) + \(\frac{50}{\sqrt{3}}\) = \(\frac{75}{\sqrt{3}}\)
= \(\frac{75 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{75 \sqrt{3}}{3}\) = 25√3 m
= 25(1.732) = 43.3 m = 43 m

Question 17.
In the figure, ABC is an equilateral triangle of side 6 cm. D is a point in BC such that BD = 1 cm and E is the midpoint of BC. Calculate :
(i) AE,
(ii) tan ∠ADC,
(iii) ∠ADC to the nearest degree,
(iv) ∠BAD to the nearest degree.

Solution:
In equilateral △ABC with each side 6 cm
∵D is a point on BC such that BD = 1 cm
E is the mid point of BC
∴ DE = DE – BD = 3 – 1 = 2 cm ( ∵ E is mid point of BC)

(i) ∵ E is mid point of BC
∴ AE ⊥ BC
and AD = \(\frac{\sqrt{3}}{2}\) side = \(\frac{\sqrt{3}}{2}\) × 6 = 3√3 cm

(ii) In right △ADE,
tan ∠ADC, = tan ∠ADE = \(\frac { AE }{ DE }\) = \(\frac{3 \sqrt{3}}{2}\) cm
= \(\frac{3(1.732)}{2}\) = 3 × 0.866 = 2.598

(iii) tan ∠ADC = 2.59156 (nearest in the table)
= 68°54′ = 69° (In nearest degree)

(iv) tan ∠DAE = \(\frac { DE }{ AE }\) = \(\frac{2}{3 \sqrt{3}}\) = \(\frac{2 \sqrt{3}}{3 \times \sqrt{3} \times \sqrt{3}}\)
= \(\frac{2 \sqrt{3}}{9}\) = \(\frac{2(1.732)}{9}\) = \(\frac{3.464}{9}\) = 0.385
= 0.38587 = tan 21°6′ = tan 21°
∴ ∠DAE = 21°
But ∠BAD = ∠BAE – ∠DAE (∵ AE also bisects ∠A)
= 30° – 21° = 9°

Question 18.
A kite, flying at a height of 75 metres from the level ground, is attached to a string inclined at 60° to the horizontal. Find the length of the string to the nearest metre.
Solution-
Let K be the kite which is 75 m above the ground and its string makes an angle of 60° with the ground

∴ In △KBT,
KT = 75 m, ∠B = 60°, ∠T = 90°
Let KB = x m
∴ sin θ = \(\frac { KT }{ KB }\) ⇒ sin 60° = \(\frac { 75 }{ x }\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac { 75 }{ x }\) ⇒ x = \(\frac{75 \times 2}{\sqrt{3}}\)
⇒ x = \(\frac{75 \times 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{150 \sqrt{3}}{3}\) = 50√3
= 50 (1.732) = 86.6 = 87
∴ Length of string of the Kite = 87 m

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
Show that \(\sqrt{\frac{1-\cos A}{1+\cos A}}\) = \(\frac{\sin A}{1+\cos A}\)
Solution:

Question 2.
Prove that : 1 – \(\frac{\cos ^2 \theta}{1+\sin \theta}\) = sin θ
Solution:
L.H.S. = 1 – \(\frac{\cos ^2 \theta}{1+\sin \theta}\) = \(\frac{1+\sin \theta-\cos ^2 \theta}{1+\sin \theta}\) = \(\frac{1+\sin \theta-\left(1-\sin ^2 \theta\right)}{1+\sin \theta}\) {cos² θ = 1 – sin² θ}
= \(\frac{1+\sin \theta-1+\sin ^2 \theta}{1+\sin \theta}\) = \(\frac{\sin \theta+\sin ^2 \theta}{1+\sin \theta}\) = \(\frac{\sin \theta(1+\sin \theta)}{1+\sin \theta}\) = sin θ R.H.S.

Question 3.
Prove the following identity:
\(\frac{1}{\sin \theta+\cos \theta}\) + \(\frac{1}{\sin \theta-\cos \theta}\) = \(\frac{2 \sin \theta}{1-2 \cos ^2 \theta}\)
Solution:
L.H.S. = \(\frac{1}{\sin \theta+\cos \theta}\) + \(\frac{1}{\sin \theta-\cos \theta}\) = \(\frac{\sin \theta-\cos \theta+\sin \theta+\cos \theta}{(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)}\)
= \(\frac{2 \sin \theta}{\sin ^2 \theta-\cos ^2 \theta}\) = \(\frac{2 \sin \theta}{1-\cos ^2 \theta-\cos ^2 \theta}\) {sin² θ = 1 – cos² θ}
= \(\frac{2 \sin \theta}{1-2 \cos ^2 \theta}\) = R.H.S.

Question 4.
Prove that:
\(\frac{\cos A}{1-\tan A}\) + \(\frac{\sin A}{1-\cot A}\) = cos A + sin A.
Solution:

Question 5.
Prove \(\frac{\sin A}{1+\cos A}\) + \(\frac{1+\cos A}{\sin A}\) = 2 cosec A.
Solution:
L.H.S. = \(\frac{\sin A}{1+\cos A}\) + \(\frac{1+\cos A}{\sin A}\)
= \(\frac{\sin ^2 A+(1+\cos A)^2}{\sin A(1+\cos A)}\)
= \(\frac{\sin ^2 A+1+\cos ^2 A+2 \cos A}{\sin A(1+\cos A)}\)
= \(\frac{1+1+2 \cos A}{\sin A(1+\cos A)}\) = \(\frac{2+2 \cos A}{\sin A(1+\cos A)}\) {sin² A + cos² A = 1}
= \(\frac{2(1+\cos A)}{\sin A(1+\cos A)}\) = \(\frac{2}{\sin A}\) = 2 cosec A

Question 6.
Prove (1 + tan A)² + (1 – tan A)² = 2 sec² A
Solution:
L.H.S. = (1 + tan A)² + (1 – tan A)²
= 1 + tan² A + 2 tan A + 1 + tan² A – 2 tan A
= 2 + 2 tan² A = 2 (1 + tan² A) = 2 sec² A {∵ 1 + tan² A = sec² A}
= R.H.S.

Question 7.
Prove that \(\frac{\sin \theta \tan \theta}{1-\cos \theta}\) = 1 + sec θ.
Solution:

Question 8.
Prove the identify:
\(\frac{\sec A-1}{\sec A+1}\) = \(\frac{1-\cos A}{1+\cos A}\)
Solution:

Question 9.
Prove that idnetify:
\(\frac{\sin A}{1+\cos A}\) = cosec A – cot A
Solution:
R.H.S. = cosec A – cot A
= \(\frac{1}{\sin A}\) – \(\frac{\cos A}{\sin A}\)

= \(\frac{1-\cos A}{\sin A}\) = \(\frac{(1-\cos A)(1+\cos A)}{\sin A(1+\cos A)}\)
= \(\frac{1-\cos ^2 A}{\sin A(1+\cos A)}\) = \(\frac{\sin ^2 A}{\sin A(1+\cos A)}\)
= \(\frac{\sin ^2 A}{1+\cos A}\) = L.H.S.

Question 10.
Prove the idnetify:
\(\frac{\sin A}{1+\cos A}\) + \(\frac{1+\cos A}{\sin A}\) = 2 cosec A.
Solution:
To Prove:

Question 11.
Prove that:
(cosec A – sin A) (sec A – cos A) sec² A = tan A.
Solution:

Question 12.
Prove that

Solution:

Question 13.
Show that

Solution:

Question 14.
Prove the identify : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ.
Solution:

Question 15.
Prove that

Solution:

Question 16.
Prove that \(\frac{\cos A}{1+\sin A}\) + tan A = sec A.
Solution:

The availability of OP Malhotra Class 10 Solutions Chapter 7 Factor Theorem – Factorization Ex 7 encourages students to tackle difficult exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 7 Factor Theorem – Factorization Ex 7

Question 1.
Find the remainder when the expression
(i) 3x³ + 8x² – 6x + 1 is divided by x + 3.
(ii) 5x³ – 8x² + 3x – 4 is divided by x – 1.
(iii) x³ + 3x² – 1 is divided by 3x + 2.
(iv) 4x³ – 12x² + 14x – 3 is divided by 2x – 1.
Solution:
(i) Let f(x) = 3x³ + 8x² – 6x + 1 and x + 3 = 0 ⇒ x = – 3
∴ Remainder = f(- 3) = 3 (- 3)³ + 8 (- 3)² – 6 (- 3) + 1
= – 81 + 72 + 18 + 1 = – 81 + 91 = 10

(ii) Let f (x) = 5x³ – 8x² + 3x – 4 and x – 1 = 0 ⇒ x = 1
∴ Remainder = f(1) = 5 (1)³ – 8(1)² + 3 x 1 – 4
= 5 – 8 + 3 – 4 = 8 – 12 = – 4

(iii) Let f(x) = x³ + 3x² – 1
and 3x + 2 = 0 ⇒ 3x = – 2 ⇒ x = \(\frac { -2 }{ 3 }\)
∴ Remainder = f(\(\frac { -2 }{ 3 }\))
= \(\left(\frac{-2}{3}\right)^3+3\left(\frac{-2}{3}\right)^2\) – 1
= \(\frac{-8}{27}+3 \times \frac{4}{9}\) – 1
= \(\frac { -8 }{ 27 }\) + \(\frac { 4 }{ 3 }\) – 1
= \(\frac{-8+36-27}{27}=\frac{36-35}{27}=\frac{1}{27}\)

(iv) Let f(x) = 4x³ – 12x² + 14x – 3
and 2x – 1 = 0 ⇒ 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ Remainder = f\(\frac { 1 }{ 2 }\)
= 4 x (\(\frac { 1 }{ 2 }\))³ – 12(\(\frac { 1 }{ 2 }\))² + 14(\(\frac { 1 }{ 2 }\)) – 3
= 4 x \(\frac { 1 }{ 8 }\) – 12 x \(\frac { 1 }{ 4 }\) + 14 x \(\frac { 1 }{ 2 }\) – 3
= \(\frac { 1 }{ 2 }\) – 3 + 7 – 3
= 7\(\frac { 1 }{ 2 }\) – 6 = 1\(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 2 }\)

Question 2.
When x³ + 3x² – kx + 4 is divided by x – 2, the remainder is k. Find the value of the constant k.
Solution:
Let f(x) = x³ + 3x² – kx + 4
and x – 2 = 0, ⇒ x = 2
∴ Remainder =/(2)
= (2)³ + 3 (2)² – k (2) + 4
= 8 + 12 – 2A + 4 = 24 – 2k
∴ Remainder = k
∴ 24 – 2k = k ⇒ 2k = 24
∴ k = 8

Question 3.
Find the value of a if the division of ax³ + 9x² + 4x – 10 by x + 3 leaves a remainder 5.
Solution:
Let f (x) = ax³ + 9x² + 4x – 10 and x + 3 = 0 ⇒ x = – 3
∴ Remainder = f (- 3)
= a (-3)³ + 9 (-3)² + 4 (-3) – 10
= – 27a + 81 – 12 – 10
= – 27a + 81 – 22
= – 27a + 59
But remainder = 5
– 27a + 59 = 5 ⇒ – 27a = 5 – 59 = – 54
⇒ a = \(\frac { 1 }{ 2 }\)
∴ a = 2

Question 4.
If the polynomials ax³ + 4x² + 3x – 4 and x³ – 4x – a leave the same remainder when divided by x – 2, find the value of a.
Solution:
Let f(x) = ax³ + 4x² + 3x – 4
and q (x) = x³ – 4x – a
and x – 2 = 0 ⇒ x = 2
∴ Remainder f (2) = q (2)
Now f(2) = a (2)³ + 4 (2)² + 3x² – 4
= 8 a + 16 + 6 – 4
= 8o + 18
and q (2) = (2)³ – 4x² – a
= 8 – 8 – a = – a
∴ 8a + 18 = – a
⇒ 8a + a = – 18
⇒ 9a = – 18
⇒ a = \(\frac { -18 }{ 9 }\) = – 2
∴ a = – 2

Question 5.
Use factor theorem in each of the following to find whether g (x) is a factor f(x) or not :
(i) f(x) = x³ – 6x² + 11x – 6; g (x) = x – 3
(ii) f(x) = 2x³ – 9x² + x + 12 ; g (x) = x + 1
(iii) f(x) = 7x² – 2\(\sqrt{8}\) x – 6 ; g (x) = x – \(\sqrt{2}\)
(iv) f(x) = 3x³ + x² – 20x +12 ; g (x) = 3x – 2.
Solution:
(i) f(x) = x³ – 6x² + 11x – 6
Let x – 3 = 0, then x = 3
Remainder f(3) = (3)³ – 6 (3)² + 11 (3) – 6
= 27 – 54 + 33 – 6 = 60 – 60 = 0
∵ Remainder is zero
∴ x – 3 is a factor of f (x)

(ii) f (x) = 2x³ – 9x² + x + 12
Let x + 1 = 0 then x = – 1
∴ Remainder = f (- 1) = 2 (-1)³ – 9 (-1)² + (-1) + 12
= 2 x (-1) – 9 x 1 +(-1)+ 12
= – 2 – 9 – 1 + 12
= – 12 + 12 = 0
∵ Remainder = 0
∴ (x + 1) is a factor of f (x)

(iii) f(x) = 7x² – 2\(\sqrt{8}\) x – 6
Let x – \(\sqrt{2}\) = then x = \(\sqrt{2}\)
Remainder = f (\(\sqrt{2}\)) = 7 (\(\sqrt{2}\))² – 2\(\sqrt{8}\) x \(\sqrt{2}\) – 6
= 7 x 2 – 2 x \(\sqrt{16}\) – 6
= 14 – 2 x 4 – 6
= 14 – 8 – 6
= 14 – 14 = 0
∵ Remainder = 0
∴ x – \(\sqrt{2}\) is a factor of f(x)

(iv) f(x) = 3x³ + x² – 20x + 12
Let 3x – 2 = 0, then 3x = 2 ⇒ x = \(\frac { 2 }{ 3 }\)

Question 6.
Find the value of a if x³ + ax + 2a – 2 is exactly divisible by x + 1.
Solution:
Let f(x) = x³ + ax + 2a – 2
and x + 1 = 0 then x = – 1
∴ Remainder = f(-1) = (-1)³ + a x (-1) + 2a – 2
= – 1 – a + 2a – 2 = – 3 + a
∵ x + 1 is a factor of f (x)
∴ Remainder = 0
⇒ – 3 + a = 0 ⇒ a = 3

Question 7.
Find the values of a and b so that the expression x³ + 10x² + ax + b is exactly divisible by x – 1 as well as x – 2.
Solution:
Let f(x) = x³ + 10x² + ax + b
and let x – 1 = 0, then x = 1
∴ Remainder f(1) = (1)³ + 10 (1)² + a x 1 + b
= 1 + 10 + a + b
= 11 + a + b
∵ x – 1 is the factor of f (x)
∴ Remainder = 0
⇒ 11 + a + b = 0 ⇒ a + b = – 11 … (i)
Again x – 2 = 0, then x = 2
∴ f (2) = (2)³ + 10 (2)² + a x 2 + b
= 8 + 10 x 4 + 2a + b
= 8 + 40 + 2a + b
= 48 + 2 a + b
∵ x – 2 is a factor of f (x)
∴ Remainder = 0
∴ 48 + 2a + b = 0 ⇒ 2a + b = – 48 … (ii)
Subtracting (i) from (ii) a = – 37
But a + b = – 11 ⇒ – 37 + b = – 11
⇒ b = – 11 + 37 = 26
∴ a = – 37 and b = 26

Question 8.
If (x – 2) is a factor of x² + ax – 6 = 0 and x² – 9x + b = 0, find the values of a and b.
Solution:
Let f(x) = x² + ax – 6 and q (x) = x² – 9x + b
Now x – 2 = 0 ⇒ x = 2
∴ f(2) = (2)² + ax² – 6 = 4 + 2a – 6
= 2a – 2
∵ x – 2 is a factor of f (x)
∴ f(2) = 0 ⇒ 2a – 2 = 0
⇒ 2a = 2 ⇒ a = \(\frac { 1 }{ 2 }\) = 1
∴ a = 1
∴ Again ∵ x – 2 = 0, then x = 2
∴ q (2) = (2)² – 9 x 2 + b
= 4 – 18 + b = – 14 + b
∵ (x – 2) is a factor of q (x)
∴ – 14 + b = 0 ⇒ b = 14 (∵ Remainder = 0)
Hence a = 1, b = 14

Question 9.
If both x – 2 and x – \(\frac { 1 }{ 2 }\) are factors of px² + 5x + r, show that p = r.
Solution:
Let f (x) = px² + 5x + r
and let x – 2 = 0 then x = 2
∴ f(x) = p (2)² + 5 x 2 + r = 4p + 10 + r = 4p + r + 10
∴ x – 2 is its factor
∴ f(2) = 0 ⇒ 4p + r + 10 = 0
⇒ 4p + r = – 10 … (i)
Again let x – \(\frac { 1 }{ 2 }\) = 0 then x = \(\frac { 1 }{ 2 }\)
∴ f(\(\frac { 1 }{ 2 }\)) = p(\(\frac { 1 }{ 2 }\))² + 5 x \(\frac { 1 }{ 2 }\) + r
= \(\frac { p }{ 4 }\) + \(\frac { 5 }{ 2 }\) + r
∵ x – \(\frac { 1 }{ 2 }\) is its factor
∴ f(\(\frac { 1 }{ 2 }\)) = 0 ⇒ \(\frac { p }{ 4 }\) + \(\frac { 5 }{ 2 }\) + r = 0 … (ii)
⇒ p + 4r + 10 = 0 ⇒ p + 4r = – 10 … (ii)
Adding (i) and (ii)
5p + 5r = – 20 ⇒ p + r = – 4 … (iii)
and subtracting
3p – 3r = 0 ⇒ p – r = 0 … (iv)
Again adding (iii) and (iv)
2p = – 4 ⇒ p = – 2
and subtracting
2r = – 4 ⇒ r = \(\frac {-4}{2}\) = – 2
∴ p = – 2, r = – 2
∴ p = r
Hence proved.

Question 10.
If x + ax² + bx + 6 has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3), find the values of a and b.
Solution:
Let f (x) = x³ + ax² + bx + 6
Let x – 2 = 0, then x = 2
∴ f(x) = (2)³ + a (2)² + b(2) + 6
= 8 + 4a + 2b + 6
= 4a + 2 b + 14
∵ x – 2 is its factor
∴ Remainder = 0
⇒ 4a + 2b + 14 = 0
⇒ 4a + 2b = – 14
⇒ 2a + b = – 7 (Dividing by 2) … (i)
Again let x – 3 = 0, then x = 3
∴ f(3) = (3)³ + a (3)² + b (3) + 6
= 27 + 9a + 3b + 6 = 9a + 3b + 33
∵ Remainder = 3,
∴ 9a + 3b + 33 = 3 ⇒ 9a + 3b = 3 – 33 = – 30
⇒ 3a + b = – 10 (Dividing by 3) … (ii)
Subtracting (i) from (ii)
a = – 10 + 7 = – 3
Substituting the value of a in (i)
2 x (- 3) + b = – 7
– 6 + b = – 7 ⇒ b = – 7 + 6 = – 1
∴ a = – 3, b = – 1

Question 11.
Factorise :
(i) x³ + 13x² + 32x + 20, if it is given that x + 2 is its factor.
(ii) 4x³ + 20x² + 33x + 18, if it is given that 2x + 3 is its factor.
Solution:
(i) Let f(x) = x³ + 13x² + 32x + 20
Let x + 2 = 0, then x = – 2
∴ f (- 2) = (-2)³ + 13 (-2)² + 32 (-2) + 20
= – 8 + 52 – 64 + 20 = 72 – 72 = 0
∵ Remainder = 0
∴ x + 2 is its factor
Now dividing f(x) by (x + 2), we get

(ii) 4x³ + 20x² + 33x + 18
and 2x + 3 = 0 then 2x = – 3 ⇒ x = \(\frac { -3 }{ 2 }\)

∵ Remainder = 0
∴ 2x + 3 is its factor
Now dividing f(x) by 2x + 3, we get

Question 12.
Show that :
(i) (x – 10) is a factor of x³ – 23x² + 142x – 120 and hence factorise it completely.
(ii) (3z + 10) is factor of 9z³ – 27z² – 100z + 300 and factorise it completely.
Solution:
(i) Let f(x) = x³ – 23x² + 142x – 120
and x – 10 = 0 then x = 10
∴ f(10) = (10)³ – 23 (10)² + 142 x 10 – 120
= 1000 – 23 x 100 + 142 x 10 – 120
= 1000 – 2300 + 1420 – 120
= 2420 – 2420 = 0
∵ Remainder = 0
∴ x – 10 is the factor of f (x)
Now dividing f(x) by (x – 10), we get

Question 13.
Given that (x – 2) and (x + 1) are factors of x³ + 3x² + ax + b, calculate the values of a and b, and hence find the remaining factor.
Solution:
Let f(x) = x³ + 3x² + ax + b
Let x – 2 = 0 then x = 2
∴ f(x) = (2)³ + 3 (2)² + a x 2 + b
= 8 + 12 + 2 a + b
= 2a + b + 20
∴ x – 2 is a factor of f(x)
∴ Remainder = 0
⇒ 2a + b + 20 = 0 ⇒ 2a + b = – 20 … (i)
Again x + 1 = 0 then x = – 1
∴ f( – 1) = (-1)³ + 3 (-1)² + a (-1) + b
= – 1 + 3 – a + b = – a + b + 2
∵ x + 1 is a factor, then
∴ Remainder = 0
∴ – a + b + 2 = 0
⇒ – a + b = – 2
⇒ a – b = 2 … (ii)
Adding (i) and (ii)
3 a = – 18 ⇒ a = \(\frac { -18 }{ 3 }\) = – 6
∴ a = – 6
∵ a – b – 2 ⇒ – 6 – b = 2
⇒ – b = 2 + 6 ⇒ – b = 8
⇒ b = – 8
∴ a = – 6, b = – 8
∴ f(x) = x³ + 3x² – 6x – 8
∵ (x – 2) is a factor of f(x)
∴ Dividing f(x) by x – 2, we get

f(x) = (x – 2)(x² + 5x + 4)
= (x – 2){x² + x + 4x + 4}
= (x – 2){x(x + 1) + 4(x + 1)}
= (x – 2)(x + 1)(x + 4)
∴ Remaining factor is (x + 4)

Question 14.
Given that (x + 2) and (x – 3) are factors of x³ + ax + b, calculate the values of a and b, and find the remaining factor.
Solution:
Let f (x) = x³ + ax + b
and x + 2 = 0,then x = – 2
∴ f(- 2) = (- 2)³ + a (- 2) + b
= – 8 – 2a + b ⇒ – 2a + b – 8 … (i)
∵ x + 2 is its factor
∴ Remainder = 0
⇒ – 2a + b – 8 = 0 ⇒ – 2a + b = 8 … (i)
Let x – 3 = 0, then x = 3
∴ f(3) = (3)³ + a (3) + b
= 27 + 3 a + b
∵ x – 3 is its factor
∴ Remainder = 0
⇒ 27 + 3a + b = 0 ⇒ 3a + b = – 27 … (ii)
Subtracting (i) and (ii)
5a = -35 ⇒ a = \(\frac { -35 }{ 5 }\) = – 7
Substituting the value of a in (i)
– 2 (- 7) + b = 8 ⇒ + 14 + b = 8
⇒ b = 8 – 14 = – 6
∴ a = – 7, b = – 6
∵ x + 2 is a factor of f(x) = x³ – 7x – 6
Dividing it by x + 2, we get

f(x) = (x + 2) (x² – 2x – 3)
= (x + 2) {x² – 3x + x – 3}
= (x + 2) {x (x – 3) + 1 (x – 3)}
= (x + 2) (x – 3) (x + 1)
∴ Remaining factor of f(x) is (x + 1)

Question 15.
Factorise, using remainder theorem :
(i) x³ – 19x – 30
(ii) x³ + 7x² – 21x – 27
(iii) x³ – 3x² – 9x – 5
(iv) 2x³ + 9x² + 7x – 6
Solution:
(i) Let f(x) = x³ – 19x – 30
Factor of 30 = +1, ±2, +3, +5, +6, +10, ±15, ±30
Now by trial and error method,
Let x = – 2, then
∴ f(-2) = (-2)³ – 19 (-2) – 30
= – 8 + 38 – 30 = 0
∴ (x + 2) is its factor
Let x = -3, then
f(-3) = (-3)³ – 19 (-3) – 30
= -27 + 57 – 30 = 0
∴ x + 3 is another factor of / (x)
Let x = 5, then f(5) = (5)³ – 19 x 5 – 30
= 125 – 95 – 30
= 125 – 125 = 0
∵ x – 5 is the third factor
∴ x³ + 9x – 30 = (x + 2) (x + 3) (x – 5)

(ii) Let f(x) = x³ + 7x² – 21x – 27
Factors of 27 = ±1, ±3, ±9, ±27
By trial and error method,
Let x = – 1
f(- 1) = (- 1)³ + 7(- 1)² – 21 (- 1) – 27
= – 1 + 7 + 21 – 27 = 0
∴ x + 1 is its factor
Now dividing f(x) by x + 1, we get

(iii) Let f (x) = x³ – 3x² – 9x – 5
Factors of 5 = ±1, ±5,
By trial and error method,
Let x = – 1
Then f (- 1) = (- 1)³ – 3 (-1)² – 9 (-1) – 5
= – 1 – 3 x 1 + 9 – 5
= – 1 – 3 + 9 – 5 = 0
∴ x + 1 is one factor of f(x)
Now dividing f(x) by x + 1, we get

f(x) = (x + 1) (x² – 4x – 5)
= (x + 1) {x² – 5x + x – 5}
= (x + 1) {x (x – 5) + 1 (x – 5)}
= (x + 1) (x – 5) (x + 1)

(iv) Let f(x) = 2x³ + 9x² + 7x – 6
Factors of 6 = ±1, ±2, ±3, ±6
By trial and error method,
Let x = – 2
Then f (- 2) = 2 (-2)³ + 9 (-2)² + 7 (-2) – 6
= 2 x (-8) + 9 x 4 + 7 x (-2) – 6
= – 16 + 36 – 14 – 6 = 0
∴ x + 2 is the factor of f (x)
Now dividing f(x) by x + 2, we get

f(x) = (x + 2) (2x² + 5x – 3)
= (x + 2) {2x² + 6x – x – 3)
= (x + 2) {2x (x + 3) – 1 (x + 3)}
= (x + 2)(x + 3) (2x – 1)

Question 16.
(x – 3) is the H.C.F. of x³ – 2x² + px + 6 and x² – 5x + q. Find 6p + 5q.
Solution:
Let p (x) = x³ – 2x² + px + 6
q (x) = x² – 5x + q
H.C.F. = x – 3
∴ H.C.F. is a factor of p (x) and q (x)
Let x – 3 = 0, then x = 3
∴ P (3) = (3)² – 2 (3)² + p (3) + 6
= 27 – 18 + 3p + 6
= 15 + 3p
∴ Remainder = 0
∴ 15 + 3p = 0
⇒ 3p = – 15
⇒ P = \(\frac { -15 }{ 3 }\)
p = – 5
and q (3) = (3)² – 5 x 3 + q
= 9 – 15 + q = – 6 + q
∴ Remainder = 0
∴ – 6 + q = 0 ⇒ q = 6
Hence p = – 5 and q = 6
Now 6p + 5q = 6 (- 5) + 5 x 6
= – 30 + 30 = 0

Question 17.
(i) What number must be subtracted from x³ – 6x² – 15x + 80, so that the result is exactly divisible by x + 4.
(ii) What number must be added to x³ – 3x² – 12x + 19, so that the result is exactly divisible by x- 2.
Solution:
(i) Let f(x) = x³ – 6x² – 15x + 80
and let p be subtracted from f (x) so that it may he exactly divisible by x + 4
∴ P (x) = x³ – 6x² – 15x + 80 – p
and let x + 4 = 0, then x = – 4
∴ P (- 4) = (-4)³ – 6 (-4)² – 15 (-4) + 80 – p
= – 64 – 96 + 60 + 80 – p
= 140 – 160 – p
= – 20 – p
∵ P (x) is divisible by x + 4
∴ Remainder = 0
∴ – 20 – p = 0 ⇒ p = – 20

(ii) Let q (x) = x³ – 3x² – 12x + 19
and let p be added to q (x)
Then (x) = x³ – 3x² – 12x + 19 + p
But it is divisible by x – 2
∴ x – 2 = 0 ⇒ x = 2
Q (2) = (2)³ – 3 (2)² – 12 (2) + 19 + p
= 8 – 12 – 24 + 19 + p
= 27 – 36 + p = – 9 + p
∵ Q (x) is divisible by x – 2
∴ Remainder = 0
∴ – 9 + p = 0 ⇒ p = 9

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
Find the remainder when 2x³ – 3x² + 7x – 8 is divided by x – 1.
Solution:
Let f(x) = 2x³ – 3x² + lx – 8
and let x – 1 = 0, then x = 1
Remainder = f(1) = 2(1)³ – 3(1)² + 7 x 1 – 8
= 2 x 1 – 3 x 1 + 7 x 1 – 8
= 2 – 3 + 7 – 8
= 9 – 11
= – 2

Question 2.
Find the value of the constants a and b if (x – 2) and (x + 3) are both factors of the expression x³ + ax² + bx – 12.
Solution:
Let f (x) = x³ + ax² + bx – 12
∵ x – 2 is a factor of f(x)
Let x – 2 = 0 ⇒ x = 2
∴ f(2) = (2)³ + a (2)² + b (2) – 12
= 8 + 4 a + 2 b – 12
= 4a + 2b – 4
∵ Remainder = 0
∴ 4a + 26 – 4 = 0 ⇒ 4a + 26 = 4 … (i)
Again ∵ x + 3 is the factor of f(x)
Let x + 3 = 0, then x = – 3
∴ f(- 3) = (-3)³ + a (-3)² + b (-3) – 12
= – 27 + 9a – 36 – 12
= 9a – 36 – 39
∵ Remainder = 0
∴ 9a – 36 – 39 = 0 ⇒ 9a – 36 = 39
⇒ 3a – b = 13 … (ii)
Multiply (i) by 1 and (ii) by 2,

Substituting the value of a in (i)
4 x 3 + 26 = 4 ⇒ 12 + 26 = 4
⇒ 2b = 4 – 12 = – 8
b = \(\frac { -8 }{ 2 }\) = – 4
∴ a = 3, b = – 4

Question 3.
Using factor theorem, show that (x – 3) is a factor of x³ – 7x² + 15x – 9. Hence factorise the given expression completely.
Solution:
Let f(x) = x³ – 7x² + 15x – 9
Let x – 3 = 0. then x = 3
∴ f(3) = (3)³ – 7 (3)² + 15 x 3 – 9
= 27 – 63 + 45 – 9
= 72 – 72 = 0
∵ Remainder = 0
∴ x – 3 is a factor of f (x)
Now dividing f(x) by x – 3, we get

f(x) = (x – 3) (x² – 4x + 3)
= (x – 3) (x² – x – 3x + 3}
= (x – 3) {x(x- 1) – 3 (x- 1)}
= (x – 3)(x- 1) (x – 3)
= (x – 3)² (x – 1)

Question 4.
Find the value of a, if (x- a) is a factor of x³ – a²x + x + 2.
Solution:
Let f(x) = x³ – ax² + x + 2
and x – a = 0, then x = a
∴ f(a) = a³ – a.a² + a + 2
= a³ – a³ + a + 2 = a + 2
∵ x – a is the factor of f(x)
∴ Remainder = 0
∴ a + 2 = 0 ⇒ a = – 2

Question 5.
Use the factor theorem to factorise completely :
x³ + x² – 4x – 4.
Solution:
Let f(x) = x³ + x² – 4x – 4
Factors of 4 = +1, +2, ±4
By trial and error.
x = – 1
Then f (- 1) = (- 1)² + (- 1)² – 4 x (- 1) – 4
= – 1 + 1 + 4 – 4
= 5 – 5 = 0
∵ Remainder = 0
∴ x + 1 is its factor
Now dividing f(x) by x + 1, we get

f(x) = x³ + x² – 4x – 4
= (x + 1)(x² – 4)
= (x+ 1) {(x)² – (2)²}
= (x + 1) (x + 2) (x – 2)

Question 6.
(x – 2) is factor of the expression x³ + ax² + bx + 6. When this expression is divided by (x – 3), it leaves a remainder 3. Find the values of a and b.
Solution:
Let f(x) = x³ + ax² + bx + 6
∵ x – 2 is its factor
Let x – 2 = 0, then x = 2
∴ f (2) = (2)³ + a (2)² + b(2) + 6
= 8 + 4a + 2b + 6 = 4a + 2b + 14
∵ Remainder = 0
∴ 4a + 2b + 14 = 0
⇒ 2a + b + 7 = 0
⇒ 2a + b = -7 … (i)
Again when f(x) is divided by x – 3,
Let x – 3 = 0, then x = 3
∴ f(3) = (3)³ + a (3)² + b(3) = 6
= 27 + 9a + 3b + 6 = 9a + 3b – 33
∵ Remainder = 3
∴ 9a + 3b + 33 = 3
⇒ 9a + 36 = 3 – 33
⇒ 9a + 3b = – 30
⇒ 3a + b = – 10 ….(ii)
Subtracting (i) from (ii)
a = – 3
Substituting the value of a in (i)
2 x (- 3) + b = – 7
⇒ – 6 + b = – 7
⇒ b = – 7 + 6 = -1
∴ a = -3, b = – 1

Question 7.
Show that 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Hence factorise the given expression completely, using factor theorem.
Solution:
Let f(x) = 2x³ + 5x² – 11x – 14
Let 2x + 7 = 0, then 2x = – 7 ⇒ y = \(\frac { -7 }{ 2 }\)

∵ Remainder = 0
∴ 2x + 7 is a factor of f(x)
Now dividing f(x) by 2x + 7, we get

f(x) = (2x + 7) (x² – x – 2)
= (2x + 7) {x² – 2x + x – 2}
= (2x + 7) {x (x – 2) + 1 (x – 2)}
= (2x + 7) (x – 2) (x + 1)

Question 8.
Show that (x – 1) is a factor of x³ – 7x² + 14x – 8. Hence completely factorise the above expression.
Solution:
Let f(x) = x³ – 7x² + 14x – 8
Let x – 1 = 0, then x = 1
∴ f(1) = (1)³ – 7(1)² + 14 x 1 – 8
= 1 – 7 + 14 – 8
= 15 – 15 = 0
∵ Remainder = 0.
∴ (x – 1) is a factor of f(x)
Now dividing f(x) by x – 1, we get

f(x) = (x – 1) (x² – 6x + 8)
= (x – 1) {x² – 4x – 2x + 8}
= (x – 1) {x(x – 4) – 2(x – 4)}
= (x – 1) (x – 4) (x – 2)

Question 9.
If (x – 2) is a factor of 2x³ – x² – px – 2.
(i) Find the value of p.
(ii) With the value of p, factorise the above expression completely.
Solution:
(i) Let f(x) = 2x³ – x² – px – 2
∵ x – 2 is its factor
Now x – 2 = 0, then x = 2
∴ f(2) = 2 (2)³ – (2)² – p x 2 – 2
= 2 x 8 – 4 – 2p – 2
= 16 – 4 – 2 – 2p
= 10 – 2p
v Remainder = 0
∵ 10 – 2p = 0 ⇒ 2p = 10
⇒ P = \(\frac { 10 }{ 2 }\) = 5
∴ p = 5

(iii) Now fix) = 2x³ – x² – 5x – 2
Now dividing, f(x) by x – 2, we get

f (x) = (x – 2) (2x² + 3x + 1)
= (x – 2) {2x² + 2x + x + 1}
= (x – 2) {2x (x + 1) + 1 (x + 1)}
= (x – 2) (x + 1) (2x + 1)

Question 10.
Given that x + 2 and x + 3 are factors of 2x³ + ax² + 7x – b. Determine the values of a and b.
Solution:
Given : (x + 2) and (x + 3) are factors of 2x³ + ax² + 7x – b.
∴ x + 2 = 0
⇒ x = – 2
2 (- 2)³ + a (- 2)² + 7 (- 2) – b = 0
– 16 + 4a – 14 – b = 0
4a – b – 30 = 0
⇒ 4a – b = 30 … (i)
Now, x + 3 = 0
⇒ x = – 3
2 (- 3)³ + a (-3)² + 7 (-3) – b = 0
– 54 + 9a – 21 – b = 0
9a – 6 – 75 = 0
⇒ 9a – 6 = 75 … (ii)
Subtracting (i) from (ii) we get
5a = 45
⇒ a = 9
Substituting the value of a in (i)
4 x 9 – b = 30
⇒ 3b – b = 30
⇒ b = 6

Question 11.
When divided by x – 3 the polynomials x³ – px² + x + 6 and 2x* – x² – (p + 3) x – 6 leave that same remainder. Find the value of ‘p’.
Solution:
When (x – 3) divides x³ – px² + x + 6, then
Remainder = p(3) = (3)³ – p(3)² + (3) + 6
= 27 – 9p + 9
= 36 – 9p
When (x – 3) divides 2x³ – x² – (p + 3) x – 6, then
Remainder = p(3) = 2(3)³ – (3)² – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p ⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
∴ P = 1

Question 12.
Use the Remainder Theorem to factorise the following expression :
2x³ + x² – 13x + 6.
Solution:
By hit and trial, putting x = 2,
2 x (2)³ + (2)² – 13 x 2 + 6
⇒ 2 (8) + 4 – 26 + 6 = 0
⇒ 16 + 4 – 26 + 6 = 0
⇒ 26 – 26 = 0
(x – 2) is the factor of 2x³ + x² – 13x + 6

∴ 2x³ + x² – 13x + 6 = (x – 2) (2x² + 5x – 3)
= (x – 2) (2x² + 6x – x – 3)
= (x – 2) [2x (x + 3) – 1 (x + 3)]
2x³ + x² – 13x + 6 = (x – 2) (2x – 1) (x + 3)

Question 13.
Find the vlaue of ‘k’ if (x – 2) is a factor of x³ + 2x² – kx + 10. Hence determine whether (x + 5) is also a factor.
Solution:
f (x) = x³ + 2x² – kx + 10
(x – 2) is a factor.
∴ x = 2
f (2) = (2)³ + 2 (2)² – k (2) + 10 = 0 ⇒ 8 + 8 – 2k + 10 = 0 ⇒ 26 = 2k
∴ k = \(\frac { 26 }{ 2 }\) = 13
If (x = 5) is a factor, then x = – 5
Now, f(- 5) = (-5)³ + 2(-5)² – 13(-5) + 10
= – 125 + 50 + 65 + 10 = 0
⇒ 125 – 125 = 0
Yes, (x + 5) is also a factor.

Question 14.
Using a Remainder Theorem factorise completely the following polynomial.
3x³ + 2x² – 19x + 6
Solution:
P(x) = 3x³ + 2x² – 19x + 6
P(1) = 3 + 2 – 19 + 6 = – 8 ≠ 0
P(- 1) = – 3 + 2 + 19 + 6 = – 24 ≠ 0
P(2) = 24 + 8 – 38 + 6 = 0
Hence, (x – 2) is a factor of P(x)
So, P(x) = 3x³ + 2x² – 19x + 6
= 3x³ – 6x² + 8x² – 16x – 3x + 6
= 3x² (x – 2) + 8x (x – 2) – 3 (x – 2)
= (x – 2) (3x² + 8x – 3)
= (x – 2) (3x² + 9x – x – 3)
= (x – 2) {3x (x + 3) – 1 (x + 3)
= (x – 2) (x + 3) (3x – 1)

Question 15.
If (x – 2) is a factor of the expression 2x³ + ax² + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
Solution:
f(x) = 2x³ + ax² + bx – 14
∵ f (x – 2) is factor of f(x)
∴ f (2) = 0
2(2)³ + a(2)² + b(2) – 14 = 0
16 + 4a + 2b – 14 = 0
4a + 2b = – 2
2a + b = – 1 … (i)
Also, (x – 3) it leaves remainder = 52
∴ f(3) = 52
2(3)³ + a(3)² + 6(3) – 14 = 52
54 + 9a + 3b – 14 = 52
9a + 3b = 52 – 40
9a + 3b = 12
3a + b = 4 … (ii)
From (i) and (ii)

∴ a = 5 put in (/)
∴ 2(5) + b = – 1
b = – 1 – 10
b = – 11
∴ a = 5, b = – 11

Question 16.
Using the Remainder and Factor theorem, factorise the following polynomial:
x³ + 10x² – 37x + 26.
Solution:
f(x) = x³ + 10x² – 37x + 26
f(1) = (1)³ + 10(1)² – 37(1) + 26
= 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f (x)

∴ f (x) = (x – 1) (x² + 11x – 26)
= (x – 1) (x² + 13x – 2x – 26)
= (x – 1) [x(x + 13) – 2(x + 13)]
= (x – 1) [(x – 2) (x + 13)]

Question 17.
Find ‘a’ if the two polynomials ax³ + 3x² – 9 and 2x³ + 4x + a, leave the same remainder when divided by x + 3.
Solution:
The given polynomials are ax³ + 3x² – 9 and 2x³ + 4x + a
Let p(x) = ox³ + 3x² – 9
and q(x) = 2x³ + 4x + a
Given that p(x) and q(x) leave the same remainder when divided by (x + 3),
Thus by Remainder Theorem, we have
p(-3) = q(-3)
⇒ a(-3)³ + 3(-3)² – 9 = 2(-3)³ + 4(-3) + a
⇒ – 27a + 27 – 9 = – 54 – 12 + a
⇒ – 27a + 18 = – 66 + a
⇒ – 27a – a = – 66 – 18
⇒ – 28a = – 84
⇒ a = \(\frac { 84 }{ 28 }\)
∴ a = 3

Question 18.
Using remainder theorem, find the value of k if on dividing 2x² + 3x² – kx + 5 by x – 2, leaves a remainder 7.
Solution:
Let f(x) = 2x³ + 3x² – kx + 5
By the remainder theorem,
f(2) = 7
∴ 2(2)³ + 3(2)² – k(2) + 5 = 7
∴ 2(8) + 3(4) – k(2) + 5 = 7
∴ 16 + 12 – 2k + 5 = 7
∴ 2k = 16 + 12 + 5 – 7
∴ 2k = 26
∴ k = 13
The value of k is 13.

Students often turn to Class 10 ICSE Maths Solutions S Chand Chapter 16 Trigonometrical Identities and Tables Ex 16(a) to clarify doubts and improve problem-solving skills.

S Chand Class 10 ICSE Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(a)

Question 1.
\(\frac{1-\cos ^2 \theta}{\sin ^2 \theta}\) = 1.
Solution:
L.H.S. = \(\frac{1-\cos ^2 \theta}{\sin ^2 \theta}\) = \(\frac{\sin ^2 \theta}{\sin ^2 \theta}\) = 1 = R.H.S. {∵ 1 – cos² θ = sin² θ}

Question 2.
\(\frac{1-\sin ^2 \theta}{\cos ^2 \theta}\) = 1.
Solution:
L.H.S. = \(\frac{1-\sin ^2 \theta}{\cos ^2 \theta}\) = \(\frac{\cos ^2 \theta}{\cos ^2 \theta}\) = 1 = R.H.S. {∵ 1 – sin² θ = cos² θ}

Question 3.
sin A . cot A = cos A.
Solution:
L.H.S. = sin A . cot A = sin A = \(\frac{\cos A}{\sin A}\)

= cos A = R.H.S.

Question 4.
\(\frac{1}{\cos ^2 \theta}-\tan ^2 \theta=1\)
Solution:
L.H.S. = \(\frac{1}{\cos ^2 \theta}-\tan ^2 \theta\) = sec² θ – tan² θ

= 1 = R.H.S.

Question 5.
tan² A cos² A = 1 – cos² A.
Solution:
L.H.S. = tan² A cos² A = \(\frac{\sin ^2 A}{\cos ^2 A} \cos ^2 A\)

= sin² A = 1 – cos² A = R.H.S. {∵sin² A = 1 – cos² A}

Question 6.
tan θ = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
Solution:
R.H.S. = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\) =\(\frac{\sin \theta}{\sqrt{\cos ^2 \theta}}\) {∵ 1 – sin² θ – cos² θ}
= \(\frac{\sin \theta}{\cos \theta}\) = tan θ = L.H.S.

Question 7.
\(\frac{1+\cos \theta}{\sin ^2 \theta}\) = \(\frac{1}{1-\cos \theta}\)
Solution:
L.H.S. = \(\frac{1+\cos \theta}{\sin ^2 \theta}\) = \(\frac{1+\cos \theta}{1-\cos ^2 \theta}\) {∵ sin² θ = 1 – cos² θ}
= \(\frac{1+\cos \theta}{(1+\cos \theta)(1-\cos \theta)}\) {∵ a² – b² = (a + b)(a – b)}
= \(\frac{1}{1-\cos \theta}\) = R.H.S.

Question 8.
cot² θ (1 – cos² θ) = cos² θ.
Solution:
L.H.S. = cot² θ (1 – cos² θ)
= \(\frac{\cos ^2 \theta}{\sin ^2 \theta} \times \sin ^2 \theta\)

= cos² θ = R.H.S.

Question 9.
tan² θ (1 – sin² θ) = sin² θ
Solution:
L.H.S. = tan²θ (1 – sin²θ) = \(\frac{\sin ^2 \theta}{\cos ^2 \theta} \times \cos ^2 \theta\)

= sin² θ = R.H.S.

Question 10.
(1 – sin² θ) sec² θ = 1.
Solution:
L.H.S. = (1 – sin² θ) sec² θ
= cos² θ × \(\frac{1}{\cos ^2 \theta}\)

= 1 = R.H.S.

Question 11.
(1 – cos² θ) cosec² θ = 1.
Solution:
L.H.S. = (1 – cos² θ) cosec² θ
= sin² θ × \(\frac{1}{\sin ^2 \theta}\)

= 1 = R.H.S.

Question 12.
sin² θ + \(\frac{1}{1+\tan ^2 \theta}\) = 1.
Solution:
L.H.S. = sin² θ + \(\frac{1}{1+\tan ^2 \theta}\) = sin² θ + \(\frac{1}{\sec ^2 \theta}\) {∵ 1 + tan² θ = sec² θ}
= sin² θ + cos² θ

= 1 = R.H.S. { ∵ sin² θ + cos² θ = 1}

Question 13.
\(\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=1\)
Solution:
L.H.S. = cos² θ + \(\frac{1}{1+\cot ^2 \theta}\)

= cos² θ + sin² θ {∵sin² θ + cos² θ = 1}
= 1 = R.H.S.

Question 14.
\(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}=1\)
Solution:
L.H.S. = \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}\) = \(\frac { 1 }{ 1 }\)
{ ∵ sin² θ + cos² θ = 1 and sec² θ – tan² θ = 1}
= 1 = R.H.S.

Question 15.
\(\left(\frac{\cos ^2 A}{\sin ^2 A}+1\right)\) tan² A = \(\frac{1}{\cos ^2 A}\)
Solution:
L.H.S. = \(\left(\frac{\cos ^2 A}{\sin ^2 A}+1\right)\) tan² A
= (cot² A + 1) tan² A
= cosec² A × tan² A

= \(\frac{1}{\sin ^2 A}\) × \(\frac{\sin ^2 A}{\cos ^2 A}\) = \(\frac{1}{\cos ^2 A}\) = R.H.S.

Question 16.
sin⁴ θ + sin² θ cos² θ = sin² θ.
Solution:
L.H.S. = sin⁴ θ + sin² θ cos² θ
= sin² θ(sin² θ + cos² θ) {∵ sin² θ + cos² θ = 1}
= sin² θ × 1 = sin² θ = R.H.S.

Question 17.
sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ = 1.
Solution:
L.H.S. = sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ
= (sin² θ)² + 2 sin² θ cos² θ + (cos² θ)²
= (sin² θ + cos² θ)²

= (1)² = 1 = R.H.S.

Question 18.
sin⁴ A cosec² A + cos⁴ A sec² A = 1.
Solution:
L.H.S. = sin² A cosec² A + cos² A sec² A
= sin⁴ A × \(\frac{1}{\sin ^2 A}\) + cos⁴ A × \(\frac{1}{\cos ^2 A}\)

= sin² θ + cos² θ = 1 = R.H.S. {∵ sin² θ + cos² θ = 1}

Question 19.
sin² A cot² A + cos² A tan² A = 1.
Solution:
L.H.S. = sin² A cot² A + cos² A tan² A
= sin² A × \(\frac{\cos ^2 A}{\sin ^2 A}\) + cos² A × \(\frac{\sin ^2 A}{\cos ^2 A}\)

= cos² A + sin² A = 1 = R.H.S. {∵ sin² A + cos² A = 1}

Question 20.
tan θ + cot θ = sec θ cosec θ
Solution:
L.H.S. = tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta}\) + \(\frac{\cos \theta}{\sin \theta}\)

= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\) = \(\frac{1}{\cos \theta \sin \theta}\) = sec θ cosec θ

= R.H.S.

Question 21.
(tan A + cot A) sin A cos A = 1
Solution:
L.H.S. = (tan A + cot A) (sin A cos A)
= \(\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)\) sin A cos A

= \(\frac{\sin ^2 A+\cos ^2 A}{\cos A \sin A}\) × sin A cos A
= \(\frac{1}{\cos A \sin A}\) × sin A cos A { ∵ sin²θ + cos² θ = 1}
= 1 = R.H.S.

Question 22.
\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}=\cot \theta\)
Solution:
L.H.S. = \(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}\)
= \(\frac{1-\sin ^2 \theta+\cos \theta}{\sin \theta(1+\cos \theta)}\) {∵ 1 – sin² θ = cos² θ}
= \(\frac{\cos \theta+\cos ^2 \theta}{\sin \theta(1+\cos \theta)}\) = \(\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}\) = \(\frac{\cos \theta}{\sin \theta}\) = cot θ = R.H.S.

Question 23.
\(\frac{1}{1-\cos \theta}\) + \(\frac{1}{1+\cos \theta}\) = 2 cosec² θ
Solution:
L.H.S. = \(\frac{1}{1-\cos \theta}\) + \(\frac{1}{1+\cos \theta}\) = \(\frac{1+\cos \theta+1-\cos \theta}{(1-\cos \theta)(1+\cos \theta)}\) = \(\frac{2}{1-\cos ^2 \theta}\) {(a + b) (a – b) = a² – b²}
= \(\frac{2}{\sin ^2 \theta}\) = 2 cosec² θ

=R.H.S.

Question 24.
\(\frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}=\tan ^2 \theta\)
Solution:

Question 25.
\(\frac{1}{\sec \theta+\tan \theta}\) = \(\frac{1-\sin \theta}{\cos \theta}\)
Solution:
L.H.S. = \(\frac{1}{\sec \theta+\tan \theta}\)

Question 26.
(cosec A – sin A) (sec A – cos A) (tan A + cot A ) = 1.
Solution:

Question 27.
\(\frac{1+\sin \theta}{1-\sin \theta}\) = (sec θ + tan θ)²
Solution:
L.H.S. = \(\frac{1+\sin \theta}{1-\sin \theta}\) = \(\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}\) = \(\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}\) = \(\frac{(1+\sin \theta)^2}{\cos ^2 \theta}\) (∵ 1 – sin² θ = cos² θ)
= \(\left(\frac{1+\sin \theta}{\cos \theta}\right)^2\) = \(\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right)^2\)
= (sec θ + tan θ)² = R.H.S.

Question 28.
\(\frac{1+\cos \theta}{1-\cos \theta}\) = (cosec θ + cot θ)²
Solution:

Question 29.
= cosec θ + cot θ = \(\frac{1+\cos \theta}{\sin \theta}\)
Solution:

Question 30.
If tan θ + cot θ = 2, Prove that tan² θ + cot² θ = 2.
Solution:
tan θ + cot θ = 2
Squaring both sides,
(tan θ + cot θ)² = 4
tan² θ + cot² θ + 2 tan θ cot θ = 4
⇒ tan² θ + cot² θ + 2 × 1 = 4 { ∵ tan θ cot θ = 1}
⇒ tan² θ + cot² θ + 2 = 4
⇒ tan² θ + cot² θ = 4 – 2 = 2
∴ tan² θ + cot² θ = 2

Question 31.
If sin θ + cos θ = a, sin θ – cos θ = b, Prove that a² + b² = 2
Solution:
a² + b² = (sin θ + cos θ)² + (sin θ – cos θ)²
= sin² θ + cos² θ + 2 sin θ cos θ + sin² θ + cos² θ – 2sin θ cos θ
= 2 sin² θ + 2 cos² θ = 2 (sin² θ + cos² θ)
= 2 × 1 = 2 {∵ sin² θ + cos² θ = 1} Hence Proved.

Regular engagement with OP Malhotra Class 10 Solutions Chapter 6 Ratio and Proportion Ex 6(c) can boost students confidence in the subject.

S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(c)

Question 1.
If \(\frac { x }{ a }\) = \(\frac { y }{ b }\) = \(\frac { z }{ c }\), show that
(i) \(\frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{x y z}{a b c}\)
(ii) \(\left(\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^3 x+b^3 y+c^3 z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}}\)
(iii) \(\frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c}\)
(iv) abc\(\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{z+c}{c}\right)^3\) = 27(x + a) (y + b) (z + c)
(v) each ratio is equal to \(\left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}}\)
Solution:
∵ \(\frac { x }{ a }\) = \(\frac { y }{ b }\) = \(\frac { z }{ c }\) (suppose)
∴ x = ak, y = bk, z = ck
(i) \(\frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{x y z}{a b c}\)
L.H.S. = \(\frac{a^3 k^3}{a^3}-\frac{b^3 k^3}{b^3}+\frac{c^3 k^3}{c^3}\)
= k³ – k³ + k³ = k³
R.H.S = \(\frac{x y z}{a b c}=\frac{a k \cdot b k \cdot c k}{a b c}\)
∴ L.H.S = R.H.S

(ii) \(\left(\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^3 x+b^3 y+c^3 z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}}\)

(iii) \(\frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c}\)

(iv) abc\(\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{z+c}{c}\right)^3\) = 27(x + a) (y + b) (z + c)

(v) Each ratio is equal to \(\left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}}\)

Hence proved.

Question 2.
If \(\frac { a }{ b }\) = \(\frac { c }{ d }\) = \(\frac { e }{ f }\), prove the following:
(i) \(\frac{p a^3+q c^3+r e^3}{p b^3+q d^3+r f^3}=\frac{a c e}{b d f}\)
(ii) \(\sqrt{\frac{a^4+c^4}{b^4+d^4}}=\frac{p a^2+q c^2}{p b^2+q d^2}\)
(iii) \(\frac{2 a^4 b^2+3 a^2 e^2-5 e^4 f}{2 b^6+3 b^2 f^2-5 f^5}=\frac{a^4}{b^4}\)
Solution:
\(\frac { a }{ b }\) = \(\frac { c }{ d }\) = \(\frac { e }{ f }\) = k
a = bk, c = dk, e = fk
(i) \(\frac{p a^3+q c^3+r e^3}{p b^3+q d^3+r f^3}=\frac{a c e}{b d f}\)

(ii) \(\sqrt{\frac{a^4+c^4}{b^4+d^4}}=\frac{p a^2+q c^2}{p b^2+q d^2}\)

(iii) \(\frac{2 a^4 b^2+3 a^2 e^2-5 e^4 f}{2 b^6+3 b^2 f^2-5 f^5}=\frac{a^4}{b^4}\)

Question 3.
If a,b,c are in continued proportion, prove that :
(i) (a + b + c) (a – b + c) = a² + b² + c²
(ii) \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
(iii) \(\frac{a^3+b^3+c^3}{a^2 b^2 c^2}=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
(iv) (4a² + lab + 9b²) : (4b² + 7bc + 9c²) = a : c
Solution:
∵ a, b, c are in continued proportion
Then \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k (Suppose)
∴ b = ck, a = bk = ck.k = ck²

(i) (a + b + c) (a – b + c) = a² + b² + c²
L.H.S. = (a + b + c) (a – b + c)
= (ck² + ck + c) (ck² – ck+ c)
= c(k² + k + 1)c(k² – k + 1)
= c²(k⁴ + k² + 1)
R.H.S. = a² + b² + c² = c²k⁴ + c²k² + c²
= c²(k⁴ + k² + 1)
∴ L.H.S. = R.H.S.

(ii) \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

(iii) \(\frac{a^3+b^3+c^3}{a^2 b^2 c^2}=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)

(iv) (4a² + lab + 9b²) : (4b² + 7bc + 9c²) = a : c

Question 4.
If a, b, c, d are in continued proportion, prove that :
(i) (b – c)² + (c – a)² + (d – b)² = (a – d)²
(ii) \(\sqrt{\frac{a^5+b^2 c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}=\frac{a}{d}\)
(iii) \(\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}\)
(iv) \(\frac{3 a+5 d}{5 a+7 d}=\frac{3 a^3+5 b^3}{5 a^3+7 b^3}\)
(v) \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
(vi) pa³ + qb³ + rc³ : pb³ + qc³ + rd³ = a : d
Solution:
∵ a, b, c, d are in continued proportion
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = \(\frac { c }{ d }\) = k (Suppose)
∴ c = dk, b = ck = dk.k = dk²
a = bk = dk²k = dk³
(i) (b – c)² + (c – a)² + (d – b)² = (a – d)²
L.H.S. = (b – c)² + (c – a)² + (d – b)²
= (dk² – dk)² + (dk – dk³)² + (d – dk²)²
= d²k² (k – 1)² + d²k²(1 – k²) + d²(1 – k²)²
= d²k² (1 -k)² + d²k² (1 – k)² (1 + k)² + d² (1 + k)²(1 – k)²
= d² (1 – k²) [k² + k² (1 + k)² + (1 + k)²]
= d² (1 – k²) [k² + k² (1 + 2k + k²) + 1 + 2k + k²]
= d² (1 – k²) [k² + k² + 2k³ + k⁴ + 1 + 2k + k²]
= d²(1 – k²)(k⁴ + 2k³ + 3k² + 2k + 1)
R.H.S. = (a – d)² = (dk³ – d)²
= d² (k³ – 1)² = d² (1 – k³)²
= d²(1 – k)²(1 + k + k²)²
= d²(1 – k)²[1 + k² + k⁴ + 2k + 2k³ + 2k²]
= d²(1 – k)² (k⁴ + 2k³ + 3k² + 2k + 1)
∴ L.H.S. = R.H.S.

(ii) \(\sqrt{\frac{a^5+b^2 c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}=\frac{a}{d}\)

(iii) \(\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}\)

(iv) \(\frac{3 a+5 d}{5 a+7 d}=\frac{3 a^3+5 b^3}{5 a^3+7 b^3}\)

(v) \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
L.H.S = \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
= \(\frac{d^3 k^9+d^3 k^6+d^3 k^3}{d^3 k^6+d^3 k^3+d^3}\)
= \(\frac{d^3 k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}\) = k³
R.H.S = \(\frac { a }{ d }\) = \(\frac { dk³ }{ d }\) = k³
∴ L.H.S. = R.H.S.

(vi) pa³ + qb³ + rc³ : pb³ + qc³ + rd³ = a : d

Question 5.
If \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\), prove that \(\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}\) + \(\frac{c z-a x}{(c+a)(z-x)}\) = 3
Solution:

Question 6.
If \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\), show that each ratio is equal to \(\frac{x+y+z}{a+b+c}\).
Solution:
\(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\)
= \(\frac{x+y+z}{b+c-a+c+a-b+a+b-c}\)
= \(\frac{x+y+z}{a+b+c}\)
Hence Proved.

Question 7.
If \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\), prove that a (b – c) + b (c – a) + c (a – b) = 0.
Solution:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\) (suppose)
∴ a = k (b + c), b = k (c + a), c = k (a + b)
Now,
L.H.S. = a (b – c) + b (c – a) + c (a – b)
= k (b + c) (b – c) + k (c + a) (c – a) + k (a + b)(a – b)
= k (b² – c²) + k(c² – a²) + k (a² – b²)
= k (b² – c² + c² – a² + a² – b²) = k x 0 = 0
= R.H.S.
Hence proved.

Question 8.
If ax = by = cz, prove that \(\frac{x^2}{y z}+\frac{y^2}{z x}+\frac{z^2}{x y}=\frac{b c}{a^2}+\frac{c a}{b^2}+\frac{a b}{c^2}\).
Solution:

Self Evaluation and Revision (Latest ICSE Questions)

Question 1.
Given \(\frac { a }{ b }\) = \(\frac { c }{ d }\), prove that \(\frac{3 a-5 b}{3 a+5 b}\) = \(\frac{3 c-5 d}{3 c+5 d}\).
Solution:
\(\frac { a }{ b }\) = \(\frac { c }{ d }\) = k (suppose)
∴ a = bk, c = dk
L.H.S = \(\frac{3 a-5 b}{3 a+5 b}=\frac{3 b k-5 b}{3 b k+5 b}=\frac{b(3 k-5)}{b(3 k+5)}\)
= \(\frac{3 k-5}{3 k+5}\)
R.H.S = \(\frac{3 c-5 d}{3 c+5 d}=\frac{3 d k-5 d}{3 d k+5 d}\)
= \(\frac{d(3 k-5)}{d(3 k+5)}=\frac{3 k-5}{3 k+5}\)
∴ L.H.S = R.H.S

Question 2.
Two numbers are in the ratio of 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.
Solution:
Ratio between two numbers = 3 : 5
Let first number be = 3x
Then second number = 5x
Adding 8 to each, we get
\(\frac{3 x+8}{5 x+8}\) = \(\frac { 2 }{ 3 }\) ⇒ 2 (5x + 8) = 3 (3x + 8)
⇒ 10x + 16 = 9x + 24
⇒ 10x – 9x = 24 – 16
⇒ x = 8
∴ First number = 3x = 3 x 8 = 24
and second number = 5x = 5 x 8 = 40

Question 3.
If a : b = 5 : 3, find (5a + 8b) : (6a – 7b).
Solution:
a : b = 5 : 3
or \(\frac { a }{ b }\) = \(\frac { 5 }{ 3 }\)
Now (5a + 8b) : (6a – 7b)
= \(\frac{5 a+8 b}{6 a-7 b}=\frac{5 \frac{a}{b}+8 \frac{b}{b}}{6 \frac{a}{b}-7 \frac{b}{b}}\) (Dividing by b)
= \(\frac{5 \frac{a}{b}+8}{6 \frac{a}{b}-7}=\frac{5 \frac{5}{3}+8}{6 \frac{5}{3}-7}\)
= \(\frac{\frac{25}{3}+8}{\frac{30}{3}-7}=\frac{\frac{25+24}{3}}{\frac{30-21}{3}}=\frac{\frac{49}{3}}{\frac{9}{3}}\)
= \(\frac{49}{3} \times \frac{3}{9}=\frac{49}{9}\)
∴ (5a + 8b) : (6a – 7b) = 49 : 9

Question 4.
If \(\frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d}\) prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\).
Solution:
\(\frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d}\)
⇒ \(\frac{3 a+4 b}{3 a-4 b}=\frac{3 c+4 d}{3 c-4 d}\) (By altemendo)
Applying componendo and dividendo,
\(\frac{3 a+4 b+3 a-4 b}{3 a+4 b-3 a+4 b}=\frac{3 c+4 d+3 c-4 d}{3 c+4 d-3 c+4 d}\)
⇒ \(\frac{6 a}{8 b}=\frac{6 c}{8 d} \Rightarrow \frac{a}{b}=\frac{c}{d}\) (Dividing by \(\frac { 6 }{ 8 }\))
Hence proved.

Question 5.
The work done by (x – 3) men in (2x + 1) days and the work done by (2x + 1) men in (x + 4) days are in the ratio 3 : 10. Find the value of x.
Solution:
Work done by (x – 3) men in (2x + 1) days = (x – 3) (2x + 1)
and work done by (2x + 1) men in (x + 4) days = (2x + 1) (x + 4)
Now according to the given condition
\(\frac{(x-3)(2 x+1)}{(2 x+1)(x+4)}\) = \(\frac { 3 }{ 10 }\)
⇒ \(\frac { x-3 }{ x+4 }\) = \(\frac { 3 }{ 10 }\)
⇒ 10x – 30 = 3x + 12
⇒ 10x – 3x = 12 + 30 ⇒ 7x = 42
∴ x = 6

Question 6.
What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?
Soluition:
Let x be subtracted from each number, we get
23 – x, 30 – x, 57 – x and 78 – x
According to the condition that these will be proportional, then
\(\frac{23-x}{30-x}=\frac{57-x}{78-x}\)
⇒ (23 – x) (78 – x) = (57 – x) (30 – x)
⇒ 1794 – 23x- 78x + x² = 1710 – 57x – 30x + x²
⇒ 1794 – 101x + x² = 1710 – 87x + x²
⇒ x² – 101x – x² + 87x = 1710 – 1794
⇒ – 14x = – 84
⇒ x = \(\frac { -84 }{ -14 }\) = 6
Required number = 6

Question 7.
What number must be added to each of the numbers 6, 15, 20 and 43 to make them in proportion.
Solution:
Let x be added to each of the numbers 6,15, 20 and 43, then
6 + x, 15 + x, 20 + x and 43 + x will be in proportion
∴ \(\frac{6+x}{15+x}=\frac{20+x}{43+x}\)
⇒ (6 + x) (43 + x) = (20 + x) (15 + x)
⇒ 258 + 6x + 43x + x² = 300 + 20x + 15x + x²
⇒ 258 + 49x + x² = 300 + 35x + x²
⇒ 49x + x² – 35x – x² = 300 – 258
⇒ 14x = 42
⇒ x = \(\frac { 42 }{ 14 }\) = 3
∴ Required number = 3

Question 8.
If \(\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}\), find x : y.
Solution:
\(\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}\)
Applying componendo and dividendo
\(\frac{3 x+5 y+3 x-5 y}{3 x+5 y-3 x+5 y}=\frac{7+3}{7-3}\)
\(\frac{6 x}{10 y}=\frac{10}{4} \Rightarrow \frac{x}{y}=\frac{10}{4} \times \frac{10}{6}\)
⇒ \(\frac{x}{y}=\frac{25}{6}\)
∴ x : y = 25 : 6

Question 9.
If x = \(\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}\), prove that 3bx² – 2ax + 3b = 0
Solution:

Question 10.
If \(\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d}\), prove that \(\frac { a }{ b }\) = \(\frac { c }{ d}\).
Solution:
\(\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d} \Rightarrow \frac{8 a+5 b}{8 a-5 b}=\frac{8 c+5 d}{8 c-5 d}\)
Now applying componendo and dividendo
\(\frac{8 a+5 b+8 a-5 b}{8 a+5 b-8 a+5 b}=\frac{8 c+5 d+8 c-5 d}{8 c+5 d-8 c+5 d}\)
\(\frac{16 a}{10 b}=\frac{16 c}{10 d} \Rightarrow \frac{a}{b}=\frac{c}{d}\) (Dividing by \(\frac { 16 }{ 10 }\))
Hence \(\frac { a }{ b }\) = \(\frac { c }{ d }\)

Question 11.
What least number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?
Solution:
Let x be added to 5, 11, 19 and 37 to make them in proportion.
∴ 5 + x : 11 + x : : 19 + x : 37 + x
⇒ (5 + x) (37 + x) = (11 + x) (19 + x)
⇒ 185 + 5x + 37x + x² = 209 + 11x + 19x + x²
⇒ 185 + 42x + x² = 209 + 30x + x²
⇒ 42x – 30x + x² – x² = 209 – 185
⇒ 12x = 24
⇒ x = 2
∴ Least number to be added = 2

Question 12.
Given \(\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}\) that using componendo and dividendo find a : b.
Solution:
Given : \(\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}\)
By componendo and dividendo
\(\frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{63+62}{63-62}\)
⇒ \(\frac{(a+b)^3}{(a-b)^3}=\left(\frac{5}{1}\right)^3\)
⇒ \(\frac{a+b}{a-b}\) = 5
⇒ a + b = 5a – 5b
⇒ 4a – 6b = 0
⇒ \(\frac { a }{ 2 }\) = \(\frac { 3 }{ 2 }\)
∴ a : b = 3 : 2

Question 13.
If x, y, z are in continued proportion, prove that \(\frac{(x+y)^2}{(y+x)^2}=\frac{x}{z}\).
Solution:
If x, y, z are in continued proportion ⇒ y² = xz
L.H.S. = \(\frac{(x+y)^2}{(y+z)^2}=\frac{x^2+y^2+2 x y}{y^2+z^2+2 y z}\)
= \(\frac{x^2+x z+2 x y}{x z+z^2+2 y z}\)
= \(\frac{x(x+z+2 y)}{z(x+z+2 y)}=\frac{x}{z}\)

Question 14.
Given x = \(\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\) use componendo and dividendo to prove that b² = \(\frac{2 a^2 x}{x^2+1}\)
Solution:
If \(\frac { x }{ 1 }\) = \(\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\)
Applying componendo and dividendo both sides

Question 15.
Using componendo and dividendo, find the value of x
\(\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}}\) = 9
Solution:
\(\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}}\) = 9
By componendo and dividendo, we have

Question 16.
6 is the mean proportion between two numbers x and y and 48 is the third proportional of x andy. Find the numbers.
Solution:
Mean propertion = \(\sqrt{xy}\)
⇒ 6 = \(\sqrt{xy}\)
⇒ xy = 36 (squaring both sides)
⇒ y = \(\frac { 36 }{ x }\) … (i)
For 3rd proportion :
x : y :: y : 48
⇒ \(\frac{x}{y}=\frac{y}{48}\) ⇒ y² = 48x
⇒ (\(\frac { 36 }{ x }\))² = 48x [using (i)]
⇒ 48x³ = 36 x 36 ⇒ x³ = \(\frac{36 \times 36}{48}\)
⇒ x³ = 27 = (3)³ ⇒ x = 3
From equation (i) :
y = \(\frac { 36 }{ x }\) = \(\frac { 36 }{ 3 }\) = 12
Hence, x = 3 and y = 12

Question 17.
If x = \(\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\), using properties of proportion, show that x² – 2ox + 1 = 0.
Solution:
x = \(\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\)
Applying componendo and dividendo
\(\frac { 1 }{ 2 }\)
We have x = \(\frac{x+1}{x-1}=\frac{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}-\sqrt{a-1}}{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}+\sqrt{a-1}}\)
⇒ \(\frac{x+1}{x-1}=\frac{2 \sqrt{a+1}}{2 \sqrt{a-1}}\)
(Applying componendo and dividendo)
Now, by taking square on both sides,

Question 18.
Using the properties of proportion, solve for x, given \(\frac{x^4+1}{2 x^2}=\frac{17}{8}\)
Solution:
\(\frac{x^4+1}{2 x^2}=\frac{17}{8}\)
\(\frac{x^4+1+2 x^2}{x^4+1-2 x^2}=\frac{17+8}{17-8}\)
(Applying componendo and dividendo)
= \(\frac{\left(x^2+1\right)^2}{\left(x^2-1\right)^2}=\frac{25}{9}\)
Taking square root on both sides
= \(\frac{x^2+1}{x^2-1}=\frac{5}{3}\)
= \(\frac{x^2+1+x^2-1}{x^2+1-\left(x^2-1\right)}=\frac{5+3}{5-3}\)
(By componendo and dividendo)
= \(\frac { 2x² }{ 2 }\) = \(\frac { 8 }{ 2 }\)
x² = 4
x = ± 2

Question 19.
If \(\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}\), then find the value of:
(i) x : y
(ii) \(\frac{x^3+y^3}{x^3-y^3}\)
Solution:

Question 20.
If a, b, c are in continued proportion, prove that
(a + b + c) (a – b + c) = a² + b² + c².
Solution:
Given that a, b and c are in continued proportion
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\)
∴ b² = ac
To prove : (a + b + c) (a – b + c) = a² + b² + c²
L.H.S. = (a + b + c) (a – b + c)
= a(a – b + c) + b(a – b + c) + c(a – b + c)
= a² – ab + ac + ab – b² + bc + ac – bc + c²
= a² + ac – b² + ac + c²= a² + b² – b² + b² + c² [∵ b² = ac]
= a² + b² + c²
= R.H.S.

Question 21.
Give \(\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}\). Find x : y.
Solution:
Given that \(\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}\)
Using componendo-dividendo, we have

Thus the required ratio is x : y = 2 : 3

Question 22.
If \(\frac { x }{ a }\) = \(\frac { y }{ b }\) = \(\frac { z }{ c }\) show that \(\frac{x^3}{a^3}+\frac{y^3}{b^3}\) + \(\frac{z^3}{c^3}=\frac{3 x y z}{a b c}\).
Solution:

The availability of Class 10 ICSE Maths Solutions S Chand Chapter 17 Heights and Distances Ex 17 encourages students to tackle difficult exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 17 Heights and Distances Ex 17

Question 1.
The angle of elevation of the top of a tower from a point at a distance of 100 metres from its foot on a horizontal plane is foundTo be 60°. Find the height of the tower.
Solution:
Let AB be the tower and O is the point from which the angle of elevation = 60° and distance AB = 100 m
Let AB = h m

∴ tan θ = \(\frac { AB }{ OB }\) ⇒ tan 60° = \(\frac { h }{ 100 }\)
⇒ h = 100 tan 60° = 100 × √3 = 100 (1.732)
= 173.2 m
Height of tower = 173.2 m

Question 2.
A vertical flagstaff stands on a horizontal plane. From a point distant 150 metres from its foot the angle of elevation of its top is found to be 30°; find the height of the flagstaff.
Solution:
Let AB be the flagstaff and a point O is given from where the top of the flagstaff makes an angle of elevation = 30° and distance from the foot B 150 m
LetAB be h

∴ tan θ = \(\frac { AB }{ OB }\)
⇒ tan 30° = \(\frac { h }{ 150 }\)
⇒ h = 150 × tan 30°
⇒ h = 150 × \(\frac{1}{\sqrt{3}}\) = \(\frac{150 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{150 \sqrt{3}}{3}\)m
= 50√3 = 50(1.732) = 86.600 = 86.6 m
∴ Height of flagstaff = 86.6 m

Question 3.
The string of a kite is 150 long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground.
Solution:
Let AB be the height of kite A from the ground and kite makes an angle of elevation with the horizontal ground of 60°
AC is the string 150 m long

Let AB = h m, Now
sin θ = \(\frac { AB }{ AC }\) ⇒ sin 60° = \(\frac { h }{ 150 }\)
⇒ h = 150 × sin 60° = 150 × \(\frac{\sqrt{3}}{2}\)
= 75√3 m = 75 × 1.732 m = 129.9 m
∴ Height of the kite = 129.9 m

Question 4.
If the shadow of a tower is 30 metres when the sun’s elevation is 30°, what is the length of the shadow when the sun’s elevation is 60° ?

Solution:
In (i), angle of sun’s elevation is 30° and length of shadow of the tower HB = AB, = 30 m
Let height of the sun be h
∴ tan θ = \(\frac { HB }{ AB }\) ⇒ tan 30° = \(\frac { h }{ 30 }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ 30 }\) ⇒ h = \(\frac{30}{\sqrt{3}}\) = \(\frac{30 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
⇒ h = \(\frac{30 \sqrt{3}}{3}\) = 10√3
Now when the elevation of sun be 60° Let the shadow of the tower =x
∴ 60° = \(\frac{\mathrm{BC}}{\mathrm{DB}}\) = \(\frac { h }{ x }\)
√3 = \(\frac{10 \sqrt{3}}{x}\) ⇒ x√3 = 10√3
x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10
∴ Length of shadow = 10 m

Question 5.
The angle of elevation of the top of a tower (which is yet incomplete) at a point 120 m from its base is 45°. How much higher should it be raised so that the elevation at the same point may become 60° ?
Solution:
Let BC be the incomplete tower and DB be the complete
Let BC = x m and CD = h
At the top C,-the angle of elevation is 45° and at the top D, the angle of elevation is 60°

Now in right △ABC,
tan θ = \(\frac { CB }{ AB }\) ⇒ tan 45° = \(\frac { x }{ 120 }\)
⇒ 1 = \(\frac { x }{ 120 }\) ⇒ x = 120 m
In right △ABD,
tan 60° = \(\frac { DB }{ AB }\) ⇒ √3 = \(\frac{x+h}{120}\)
⇒ 120 × √3 = 120 + h ⇒ h = 120√3 – 120
= 120 (√3 – 1) = 120 (1.732 – 1)
= 120 × 0.732 = 87.84 m
∴ Tower should be raised more = 87.84 m

Question 6.
Find the angle of depression from the top of a tower 140 m high of an object on the ground at a distance of 200 m from the foot of the tower.
Solution:
Let AB be the tower and Pis a point on the ground such that PB = 200m and AB = 140m

Let θ be the angle of depression of P from A, the top of the tower
∴ ∠APB = θ (alternate angles)
∴ tan θ = \(\frac { AB }{ PB }\) = \(\frac { 140 }{ 200 }\) = \(\frac { 7 }{ 10 }\) = 0.7
From the table of tan θ,
tan 35° = 70021 wbich is nearest to 0.7
∴ Angle of depression = θ = 35°

Question 7.
A vertical tower Ia 20 m high. A man at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the loot of the tower?
Solution:
Let AB be the tower and M is a man standing at some distance from the foot of the tower B
Let MB = x m and cos θ = 0.53

∴ cos θ = \(\frac { MB }{ AM }\) = \(\frac{x}{\sqrt{20^2+x^2}}\)
⇒ 0.53 = \(\frac{x}{\sqrt{(20)^2+x^2}}\)
Squaring both sides,
(0.53)² = \(\frac{x^2}{400+x^2}\) ⇒ \(\frac { 2809 }{ 10000 }\) = \(\frac{x^2}{400+x^2}\)
⇒ 10000x² = 2809 × 400 + 2809x²
⇒ 10000x² – 2809x² = 1123600
⇒ 7191 x² = 11223600
⇒ x² = \(\frac { 1123600 }{ 7191 }\) = 156.25
∴ x = \(\sqrt{156.25}\) = 12.5
∴ The man is standing at a distance of 12.5 m from the foot of the tower

Question 8.
In the figure, it is given that AB is perp. to BD and is of length x metres. DC = 30m, ∠ADB = 30° and ∠ACB = 45° Without using tables, find x.

Solution:
In the figure, AB = x, DC = 30 m
∠ADB = 30°, ∠ACB = 45°
Let BC = y m
Now in right △ABC
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 45° = \(\frac { x }{ y }\) ⇒ 1 = \(\frac { x }{ y }\)
⇒ x = y
Again in right △ABD, (∵ tan 45° = 1)
tan 30° = \(\frac { AB }{ BD }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{x}{y+30}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{x}{x+30}\) (∵ y = x)
⇒ √3 x = x + 30 ⇒ √3 x – x = 30
⇒ (1.732 – 1) x = 30 ⇒ 0.732x = 30
⇒ x = \(\frac{30}{0.732}\) = 40.98 m
∴ x = 40.98 m

Question 9.
In the figure, from a boat 200 in away from a erticaI cliff the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of the cliff are 21° and 18° 30′. Calculate :
(i) the height of the cliff;
(ii) the height of the pillar.

Solution:
Let B be the boat which is 200 m away from a vertical cliff CA
PC is the pillar at the edge of cliff
Angles of elevation from B to the top of cliff C is 18° 3o’ and to the top of the pillar P is 21°
Let height of cliff CA = h m and pillar PC = x m
Now in right △CAB,
tan θ = \(\frac { CA }{ AB }\) ⇒ tan 18° 36′ = \(\frac { h }{ 200 }\)
⇒ 0.33460 = \(\frac { h }{ 200 }\) ⇒ h = 200 × 0.33460
⇒ h = 66.920 m = 66.92 m
Again in right △PAB.
tan 21° = \(\frac { PA }{ AB }\) ⇒ 0.38386 = 76.77200
⇒ x + 66.920 = 200 × 0.38386 = 76.77200
⇒ x = 76.772 – 66.920 = 9.852 = 9.85 m
(i) ∴ Height of cliff 66.92 m and
(ii) Height of pillar = 9.85 m

Question 10.
In the figure, the top of a tower is observed from two points on the same horizontal line through a point on the base of the tower. If the angles of elevation at the two points be 300 and 45°, and the distance between them is 20 m, find the height of the tower.

Solution:
Let TS be the tower and A and B are two points on the ground on the same line and distance between them is 20 in i.e.,
AB = 20 m
Angles of elevation from A is 30° and from B
is 45° to the top of the tower
Let height of the tower TS = h and BS = x
Now in right △TSB,
tan θ = \(\frac { TS }{ BS }\) ⇒ tan 45° = \(\frac { h }{ x }\)
⇒ 1 = \(\frac { h }{ x }\) ⇒ x = h
Similarly ¡n right △TSA,
tan 30° = \(\frac{h}{x+20}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+20}\) (∵ x = h)
⇒ h √3 = h + 20 ⇒ √3 h – h = 20
⇒ (1.732 – 1) h = 20 ⇒ 0.732h = 20
⇒ h = \(\frac { 20 }{ 0.732 }\) = \(\frac{20 \times 1000}{732}\) = 27.32
∴ Height of the tower = 27.32 m

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank ¡s 60°. When he moves 40 m away
from the bank, he finds the angles of elevation to be 30°. Find (i) the height of the tree, correct to 2 decimal places ;
(ii) the width of river.
Solution:
Le TR be the tree on the bank of a river BR and A is another position of the rnar which is 40 m away from the first position B
∴ AB = 40m
and angles of elevation to B and A to the top of the tree are 60° and 30° respectively

Let height of tree TR = h m and width of
river BR = x m
Now in right △TRA
tan 30° = \(\frac { TR }{ RA }\) ⇒ tan 60° = \(\frac { h }{ x }\) ⇒ √3 = \(\frac { h }{ x }\)
∴ h = √3 x or x = \(\frac{h}{\sqrt{3}}\) = \(\frac{h \sqrt{3}}{3}\)
Similarly in right △TRA
tan 30° = \(\frac { TR }{ RA }\) ⇒ \(\frac{1}{\sqrt{3}}\) × 120 = \(\frac { h }{ x+40 }\)
⇒ √3 h = x + 40 ⇒ √3 h = \(\frac{h \sqrt{3}}{3}\) + 40
⇒ √3 h – \(\frac{\sqrt{3}}{3}\) h = 40 ⇒ \(\frac{(3 \sqrt{3}-\sqrt{3})}{3}\)h = 40
⇒ 2√3 h = 120 ⇒ h = \(\frac{1}{2 \sqrt{3}}\) = \(\frac{60}{\sqrt{3}}\)
⇒ h = \(\frac{60 \times \sqrt{3}}{\sqrt{3}+\sqrt{3}}\) = 20
(i) ∴ Height of tree 34.64 m
(ii) and width of river 20 m

Question 12.
In the figure. the a.gle of elevation of the top P of a vertical tower PQ from a point X Is 60°; at a point Y, 40m vertically above X, the angle of elevatIon Li 45°.
(i) Find the height of the tower PQ;
(ii) Find the distance XQ.
(Give yoar answers to the nearest metre)

Solution:
PQ is the tower, from a point X which is at some distance from Q the angle of elevation of the top P of the tower is 60 and from another point Y. which is 40 m above it, the angle of elevation of P is 45°
let height PQ = h and distance of QX = x
From Y, draw YZ || XQ, then
YZ = XQ = x, ZQ = YX = 40 m and PZ = (h-40) m
Now in right △PQX,

tan θ = \(\frac { PQ }{ QX }\)
⇒ tan 60° = \(\frac { h }{ x }\)
⇒ √3 = \(\frac { h }{ x }\)
⇒ h = √3 x or x = \(\frac{h}{\sqrt{3}}\) ….(i)
Similarly in right △PZY,
tan 45° = \(\frac { PZ }{ YZ }\) = \(\frac{h-40}{x}\)
⇒ 1 = \(\frac{h-40}{x}\) ⇒ x = h – 40 ….(ii)
\(\frac{h}{\sqrt{3}}\) = h – 40 {From (i)}
h = √3 h – 40√3
√3h – h = 40√3 ⇒ h (√3 – 1) = 40(1.732)
⇒ h(1.732 – 1) = 40 × 1.732
⇒ 0.732h = 40 × 1.732
h = \(\frac{40 \times 1.732}{0.732}\) = 94.64 m
= 95 m (in nearest metre)
and x = \(\frac{h}{\sqrt{3}}\) = \(\frac{94.64}{1.732}\) = 54.64 m
= 55 m (in nearest metre)
Hence height of tower PQ = 95 m
and distance XQ = 55 m

Question 13.
A guard observes an enemy boat, from an observation, tower at . height of 180 m above sea-level, to be at an angle 0f depression of 29°
(i) Calculate to the nearest metre. the distance of the boat from the tool of the observation tower.
(ii) After some time it is observed that the boat is 200 m from the foot of the observation tower. Calculate the new angle of depression.
Solution:
Let TR be the observation tower, B is the position of a boat which makes 290 of angle of depression, A is the second position of the boat which makes angle θ of angle of depression when the boat is 200 m from the tower

∴ TR = 180 m, AR = 200 m
Let BR = x m, then BA = (x – 200) m
∵ XT || BR
∴ ∠B = 29° and ∠A = θ (alternate angles)

(i) Now in right △TRB.
tan θ = \(\frac { TR }{ RB }\) ⇒ tan 29° = \(\frac{180^{\circ}}{x}\)
⇒ 0.55431 = \(\frac { 180 }{ x }\) ⇒ x = \(\frac { 180 }{ 0.55431 }\) = 324.728
⇒ x = 325 (nearest to metre)

(ii) Similarly in right△TRA
tan θ = \(\frac { TR }{ RA }\) = \(\frac { 180 }{ 200 }\) = \(\frac { 9 }{ 10 }\) = 0.9
From the tables, we see thai tan θ = 0.90040
which is nearest in 0.9
∴ θ = 42°

Question 14.
In the figure, the shadow of a vertical tower on level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. using the given figure, find the height or the tower, correct to one place of decimals.

Solution:
Let TR be the tower and RA ¡s its shadow at an angle of 45° of elevation and RB is the shadow at 30° of elevation such that AB = 10 m
Let h be the height of the tower TR and shadow RA = x m
Now in right △TRA,
tan θ = \(\frac { TR }{ RA }\) ⇒ tan 45° = \(\frac { h }{ x }\)
⇒ 1 = \(\frac { h }{ x }\) ⇒ h = x
and in right △TRB,
tan 30° = \(\frac { TR }{ RB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+10}\)
⇒ √3 h = x + 10
⇒ √3 h = h + 10 ⇒ √3 h – h = 10 [from (i)]
⇒ h(√3 – 1) = 10 ⇒ h (1.732 – 1) = 10
⇒ 0.732h = 10 ⇒ h = \(\frac { 10 }{ 0.732 }\) = 13.66
∴ Height of tower = 13.66 m = 13.7 m

Question 15.
A player sifting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60°. Find the distance between the foot of the tower and the ball. (Take √3 = 1.732)
Solution:
P is a player sitting on the top of a tower PQ and B is the ball lying on the ground and anlge of depression from P to ball B is 60° Height of the tower PQ = 20 m

Now, we have to find the distance BQ
Let BQ = x, then
In △PBQ, ∠Q = 90°
tan θ = \(\frac { P }{ B }\) = \(\frac { PQ }{ BQ }\) ⇒ tan 60° = \(\frac { 20 }{ x }\) ⇒ √3 = \(\frac { 20 }{ x }\)
⇒ x = \(\frac{20}{\sqrt{3}}\) = \(\frac{20 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{20(1.732)}{3}\)
= \(\frac{34.640}{3}\) = 11.546 = 11.55 m
∴ Distance between the foot of tower and ball = 11.55 m

Question 16.
The angle of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it are complementary.
Prove that the height of the tower is √ab metres.

Solution:
In the figure. let the height of tower PQ = h
AQ = a and BQ = b
∠A = θ, then ∠B = 90° – θ (∵ Angles A and B are complementary)
Now in right △PQA.
tan θ = \(\frac { PQ }{ QA }\) = \(\frac { h }{ a }\) ⇒ h = a tan θ …(i)
and similarly in right △PQA,
tan (90° – θ) = \(\frac { PQ }{ QB }\)
⇒ cot θ = \(\frac { h }{ b }\) ⇒ h = b cot θ …(ii) {∵ tan (90° – θ) = cot θ}
Multiplying (i) and (ii)
h × h = a tan θ × b cot θ
⇒ h² = ab × 1 = ab (∵ tan θ cot θ = 1)
⇒ h = √ab
Hence height of the tower = √ab Hence proved.

Question 17.
The length of a siring between a kite and a point on the ground is 90 metres. If the string makes an angle θ with the level ground such that tan θ = \(\frac { 15 }{ 8 }\), how high is the kite ? Assume, there is so slack in the string.
Solution:
Let K be the kite and AK be the string which makes an angle θ with the ground
i.e. ∠A = θ, AK = 90 m and tan θ = \(\frac { 15 }{ 8 }\)
Let h be the height of kite i.e. KT = h
Now in right △KTA

tan θ = \(\frac { KT }{ TA }\) = \(\frac { h }{ x }\)
∴ \(\frac { h }{ x }\) = \(\frac { 15 }{ 8 }\)
⇒ h = 15, x = 8
∴ AK = \(\sqrt{h^2+x^2}\) = \(\sqrt{15^2+8^2}\) = \(\sqrt{225+64}\) = \(\sqrt{289}\) = 17
∴ sin θ = \(\frac{h}{\mathrm{AK}}\) = \(\frac{15}{17}\)
If AK = 90, then h = \(\frac{15}{17}\) × 90 = \(\frac{1350}{17}\)
⇒ h = 79.4
Hence height of the kite = 79.4 m

Question 18.
An observer in figure, 1\(\frac { 1 }{ 2 }\) m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation from his eye to the top of the tower.

Solution:
Let AB be the tower and CD is a man. The height of tower = 30 m and man CD
= 1\(\frac { 1 }{ 2 }\) m = 1.5 m
Who is 28.5 m away from the foot of the tower then CA = 28.5 m
Draw DE || CA, then
EA = 1.5 m,EB = 30 – 1.5 = 28.5 m and DE = CA = 28.5 m
and let angle of elevation from the eye of the man of the top B of the tower be θ
Now in △BED,
tan θ = \(\frac { BE }{ DE }\) = \(\frac { 28.5 }{ 28.5 }\) = 1
From the table, we see that
tan 45° = 1
∴ θ = 45°

Question 19.
Two men are on diametrically opposite side of a tower. They measure the angles of elevation of the top of the tower as 20° and 24° respectively. If the height of the tower is 40 m, find the distance between them.
Solution:
Let TR be the tower which is 40 m high and A and B are two men who are diametrically opposite sides of the tower making angles of elevation as 20° and 24° respectively with the top of the tower
Let AR = x and BR = y

Now in right △TRA
tan θ = \(\frac { TR }{ AR }\) ⇒ tan 20° = \(\frac { 40 }{ x }\)
⇒ 0.36397 = \(\frac { 40 }{ x }\) (From the tangents tables)
∴ x = \(\frac { 40 }{ 0.36397 }\) = 109.89
Similarly in right △TRB,
tan 24° = \(\frac { 40 }{ y }\) ⇒ 0.44523 = \(\frac { 40 }{ y }\)
⇒ y = \(\frac { 40 }{ 0.44523 }\) = 89.84
Distance between two men : x + y
= 109.89 + 89.84 = 199.73 m

Question 20.
The horizontal distance between the two towers is 60 m and the angular depression of the top of the first, as seen from the top of the second, which is 150 m high, is 30°; find the height of the first.
Solution:
Let AB and CD be the two towers which are
60 m apart i.e. BD = 60 m, CD = 150 m
Let AB = h, draw AE||BD
Angle of depression from C to A is 30°
∵ CX || AE || BD
∴ CAE = 30° (alternate angles)
and AE = 60 m,ED = h and CE = 150 – h

Now in right △CAE,
tan θ = \(\frac { CE }{ AE }\) ⇒ tan 30° = \(\frac{150-h}{60}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{150-h}{60}\)
150 – h = \(\frac{60}{\sqrt{3}}\) ⇒ h = 150 – \(\frac{60}{\sqrt{3}}\)
⇒ h = 150 – \(\frac{60 \times \sqrt{3}}{\sqrt{3}+\sqrt{3}}\) = 150 – \(\frac{60 \times \sqrt{3}}{3}\)
= 150 – 20√3
= 150 – 20 (1.732) = 150 – 34.640 = 115.36 m
∴ Height of first tower AB = 115.36 m

Question 21.
From the top of a cliff, 150 metres high, the angle of depression of the top and bottom of a tower are observed to be 30° and 60°, find the height of the tower.
Solution:
Let AB is cliff and CD is the tower
The angle of depression from A to the top C and bottom D of the tower are 30° and 60° respectively

∵ AX || CE || DB
∴ ∠ACE = 30°
and ∠ADB = 60° (altcrnate angles)
Now AB = 200 m
Let CD = h, then EB = h and CE = DB = x (suppose)
∴ AE = 200 – h
Now in right △ABD,
tan θ = \(\frac { AB }{ BD }\)
⇒ tan 60° = \(\frac { 200 }{ x }\)
⇒ √3 = \(\frac { 200 }{ x }\)
⇒ √3 x = 200
⇒ x = \(\frac{200}{\sqrt{3}}\)
Similarly in right △AEC,
tan 30° = \(\frac { AE }{ EC }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{200-h}{x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{200-h}{\frac{200}{\sqrt{3}}}\)
= \(\frac{1}{\sqrt{3}}\) × \(\frac{200}{\sqrt{3}}\) = 200 – h
⇒ \(\frac{200}{3}\) = 200 – h
⇒ h = 200 – \(\frac{200}{3}\) = \(\frac{400}{3}\)
⇒ h = 133.33 m
∴ Height of tower = 133.33 m

Question 22.
From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, how far is P from the foot of the tower?
Solution:
Let AB be tower which is 100 m long P is a point on the ground such that angle of elevation from A is 30°

Let PB = x m
Now in right △ABP,
tan θ = \(\frac { AB }{ PB }\) ⇒ tan 30° = \(\frac { 100 }{ x }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 100 }{ x }\) ⇒ x = 100 × √3 = 100 × 1.732
x = 173.2
∴ Distance between P and the foot of tower = 173.2 m

Question 23.
From the top of a building AB, 60 metres high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find
(i) the horizontal distance between AB and CD;
(ii) the height of the lamp post CD.

Solution:
AB and CD are the lamp posts
AB = 60 m
and let CD = h
Angles of depression from A to the top and bottom of CD are 30° and 60°

∴ Draw CE || DB
∴ ∠ACE = 30° and
∠ADB = 60° (alternate anSIes)
∴ EB = CD = h, then AE = (60 – h) m
Let CE = BD = x
Now in right △ABD
tan θ = \(\frac { AB }{ BD }\) ⇒ tan 60° = \(\frac { 60 }{ x }\)
⇒ √3 = \(\frac { 60 }{ x }\) ⇒ x = \(\frac{60}{\sqrt{3}}\) = \(\frac{60 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{60 \sqrt{3}}{3}\)
x = 20√3 m
Similarly in right △AEC,
tan 30° = \(\frac { AE }{ EC }\) = \(\frac{60-h}{x}\) = \(\frac{60-h}{20 \sqrt{3}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{60-h}{20 \sqrt{3}}\) ⇒ \(\frac{20 \sqrt{3}}{\sqrt{3}}\) = 60 – h
⇒ 20 = 60 – h ⇒ h = 60 – 20 = 40

(i) ∴ Distance between AB and
CD = BD = x = 20√3 m
= 20 (1.732) = 34.640 = 34.64 m

(ii) Height of CD = 40 m

Question 24.
An aeroplane when 34000 in high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation pout are 60° and 45°. How many metres higher is the one than the other?

Solution:
An aeroplane A is flying ax the height of 3000 m above the ground and passes vertically above another aeroplane C
Let distance between AC = h
Then CD = 3000 – h
The angle of elevation from B to C and A are 45° and 60°. respectively
Let BD = x

Now in right △ABD,
tan θ = \(\frac { AD }{ BD }\)
⇒ tan 60° = \(\frac { 3000 }{ x }\)
⇒ √3 = \(\frac { 3000 }{ x }\)
⇒ x = \(\frac{3000}{\sqrt{3}}\) m
Similarly in right △CBD,
tan 45° = \(\frac { CD }{ BD }\) = \(\frac{3000-h}{x}\)
⇒ 1 = \(\frac{3000-h}{x}\)
⇒ x = 3000 – h
⇒ h = 3000 – \(\frac{3000 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= 3000 – \(\frac{3000 \times \sqrt{3}}{3}\)
= 3000 – 1000(√3)
= 3000 – 1000(1.732)
= 3000 – 1732
= 1268 m
∴ First aeroplane is 1268 m higher than the second

Question 25.
The angle of elevation of a cloud from a point 60 m above the lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.

Solution:
Let C be the cloud and RS is the surface of the lake and P is the point of observation such that RP = SQ = 60 m
T is the reflection of cloud C in the lake, then CS = ST
Let PQ ⊥ CT, then ∠CPQ = 30° and ∠TPQ = 60°
Let CQ = h, then CS = h + 60
∴ ST = h + 60
Now in △CPQ,
tan θ = \(\frac { CQ }{ PQ }\) ⇒ tan 30° = \(\frac { h }{ PQ }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ PQ }\) ⇒ PQ = h√3 …(i)
Similarly in △CPQ,
tan 60° = \(\frac { QT }{ PQ }\) ⇒ √3 = \(\frac{60+60+h}{\mathrm{PQ}}\)
PQ = \(\frac{120+h}{\sqrt{3}}\) ….(ii)
From (i) and (ii)
√3 h = \(\frac{120+h}{\sqrt{3}}\)
3h = 120 + h ⇒ 3h – h = 120
⇒ 2h = 120 ⇒ h = \(\frac { 120 }{ 2 }\) = 60
∴ Height of the cloud = CS = h + 60 = 60 + 60 = 120 m

Question 26.
The angle of elevation of the top of a h at the foot of a tower is 60° and the angle of elevation of the top of the tower fro the foot of the hill is 30°. If the tower 50 m high, what is the height of the hill?

Solution:
Let AB be the hill and CD be the tower
Height of tower CD = 50
Let height of the hill AB = h
Angle of elevation of A at D = 60°
and of C at B = 30°
Now in right △BCD,
tan θ = \(\frac { 50 }{ BD }\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 50 }{ BD }\) ⇒ BD = 50√3 …(i)
Again in right △ABD,
tan 60° = \(\frac { AB }{ BD }\) ⇒ √3 = \(\frac { h }{ BD }\)
⇒ BD = \(\frac{h}{\sqrt{3}}\) …(ii)
From (i) and (ii)
\(\frac{h}{\sqrt{3}}\) = 50√3 ⇒ h = 50√3 × √3 = 50 × 3 = 150
⇒ Height of hill = 150 m

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
From a window A, 10 m above ground, the angle of elevation of the top C of a tower is x°, where tan x =\(\frac { 5 }{ 2 }\), and the angle of depression of the foot D of the tower is y°, where tan y= \(\frac { 1 }{ 4 }\).
See figure given here.
Calculate the height CD of the tower in metres.

Solution:
Let CD is the tower and A is window 10 m above the ground BD
The angle of elevation of the top of the lower C is x° and angle of depression from A to the foot of the tower is y°
tan x = \(\frac { 5 }{ 2 }\) and tan y = \(\frac { 1 }{ 4 }\)
From A, draw AE || BD such that
ED = AB = 10 m, BD = AE = x (suppose)
Let CD = h, then CE = h – 10
Now in right △ACE,
tan x = \(\frac { CE }{ AE }\) ⇒ \(\frac { 5 }{ 2 }\) = \(\frac { CE }{ BD }\)
BD = \(\frac { 2 }{ 5 }\) CE
Similarly in right △ABD,
tan y = \(\frac { AB }{ BD }\) ⇒ \(\frac { 1 }{ 4 }\) = \(\frac { 10 }{ BD }\)
⇒ BD = 40 m
∴ CE = \(\frac { 5 }{ 2 }\) BD = \(\frac { 5 }{ 2 }\) × 40 = 100 m
∴ CD = CE + ED = 100 + 10 = 110 m

Question 2.
In the figure, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm. Calculate the length of PR.

Solution:
In the figure, ∠PSR 90°, PQ = 1o cm
SQ = 6cm, RQ = 9 cm
∴ RS = RQ + QS = 9 + 6 = 15 cm
Let PR = x
In right △PQS,
PQ² = PS² + QS² (Pythagoras Theorem)
(10)² = PS² + (6)² ⇒ 100 = PS² + 36
⇒ PS² = 100 – 36 = 64 = (8)²
∴ PS = 8 cm
Now in right △PRS,
PR² = PS² + RS²
= (8)² + (15)² = 64 + 225 = 289 = (17)²
∴ PR = 17 cm

Question 3.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far ¡s he standing from the foot of the tower ?

Solution:
Let BC be the vertical tower whose height is 20 m
A is a man at some distance from the tower and angle of elevation is θ with the top of the tower and cos θ = 0.53
In the right △ABC,
cos θ = 0.53
∴ θ = 58° (using the cos tables)
∴ tan θ = \(\frac { AB }{ BC }\)
⇒ tan 58° = \(\frac { 20 }{ BC }\) ⇒ 1.6000 = \(\frac { 20 }{ BC }\)
⇒ BC = \(\frac { 20 }{ 1.6 }\) = \(\frac { 200 }{ 16 }\) = \(\frac { 25 }{ 2 }\) = 12.5 m
∴ The man is away from the foot of the tower
= 12.5 m

Question 4.
The shadow of a vertical tower AB on level ground is increased by 10 m, when the altitude of the sun changes from 45° to 30°, as shown in the adjoining figure. Find the height of the tower and give your answer correct to \(\frac { 1 }{ 10 }\) of a metre.

Solution:
AB is the vertical tower and CB is its shadow when the angle of elevation is 45°. When the elevation changes from 45° to 30°, the length of its shadow is increased by 10
m, i.e., DB = CB + 10 m
Let height of tower AB = h m
Now in right △ACB,
tan θ = \(\frac { AB }{ CB }\) ⇒ tan 45° = \(\frac { h }{ CB }\) ⇒ 1 = \(\frac { h }{ CB }\)
∴ CB = h
Similarly in right △ADB,
tan 30° = \(\frac { AB }{ DB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{\mathrm{CB}+10}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{h+10}\)
⇒ h + 10 = √3 h ⇒ √3 h – h = 10
⇒ h (√3 – 1) = 10 ⇒ h(1.732 – 1) = 10
⇒ 0.732h = 10 ⇒ h = \(\frac { 10 }{ 0.732 }\)
h = \(\frac{10 \times 1000}{732}\) = \(\frac { 10000 }{ 732 }\) = 13.66 m
∴ Height of tower = 13.66 m = 13.7 m (upto one decimal place)

Question 5.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°.
Calculate :
(i) the width of the river and
(ii) the height of the tree.
Solution:
Let TR be the tree on the opposite bank of a river whose width is AR and the angle of elevation of the top T of the tree is 60° After moving 50 m away, the angle of elevation of T at B will be 30° Let TR = h and AR = x

Now in right △TAR,
tan θ = \(\frac{\mathrm{TR}}{\mathrm{AR}}\) ⇒ tan 60° = \(\frac { h }{ r }\) ⇒ √3 = \(\frac { h }{ x }\)
⇒ h = √3 x ⇒ x = \(\frac{h}{\sqrt{3}}\)
Similarly in right △TBR,
tan 30° = \(\frac{\mathrm{TR}}{\mathrm{BR}}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+50}\)
⇒ x + 50 = √3 h ⇒ \(\frac{h}{\sqrt{3}}\) + 50 = √3 h
⇒ h + 50√3 = 3h ⇒ 3h – h = 50√3
⇒ 2h = 50√3 ⇒ h = \(\frac{50 \sqrt{3}}{2}\) = 25√3 m

(ii) ∴ Height of tree = h = 25(1.732) = 43.3 m
and width of river AR = x = \(\frac{h}{\sqrt{3}}\) = \(\frac{25 \sqrt{3}}{\sqrt{3}}\) = 25 m
Question 6.
Two people standing on the same side of a tower in straight line with it, measure the angles of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70°, find the distance between the two people.
Solution:
Let TR be the tower and A and B are two people which makes angles of elevation with the top T of the tower as 25° and 50° respectively
Height of the tower TR = 70 m

Now in right △TBR,
tan θ = \(\frac { TR }{ AR }\) ⇒ tan 50° = \(\frac { 70 }{ BR }\)
1.19175 = \(\frac { 70 }{ BR }\) = BR = \(\frac { 70 }{ 1.19175 }\) = 58.737 m
Similarly in right △TAR,
tan 25° = \(\frac { TR }{ AR }\) ⇒ 0.46631 = \(\frac { 70 }{ AR }\)
AR = \(\frac { 70 }{ 0.46631 }\) = 150.115 m
∴ Distance between two people = AB
= AR – BR = 150.115 – 58.737 = 91.378 m

Question 7.
From the top of a cliff 92 m high, the angle of depression of a boy is 20°. Calculate to the nearest metre, the distance of the boy from the foot of the cliff.
Solution:
Let AB be cliff which is 92 m in height and makes an angle of 20° of depression with a boy C
Let CB = x

∵AX || BC
∴ ∠C = 20° (alternate angles)
Now in right △ABC,
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 20° = \(\frac { 92 }{ x }\)
⇒ 0.36397 = \(\frac { 92 }{ x }\) ⇒ x = \(\frac { 92 }{ 0.36397 }\)
⇒ x = 252.768
∴ Distance of boy from the foot of the cliff = 252.768 m

Question 8.
The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.
Solution:
Let AB be vertical tower
the altitudes of sun changes frcm 45° to 30° when the back of ground increased by 10 m
Let height of AB tower = h

Now in right △ACB,
tan θ = \(\frac { AB }{ CB }\)
⇒ tan 45° = \(\frac { h }{ CB }\) ⇒ 1 = \(\frac { h }{ CB }\)
⇒ CB = h …(i)
Similarly in right △ADB,
tan 30° = \(\frac { AB }{ DB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ DB }\)
⇒ DB = h√3
∴ DC = DB – CB
⇒ 10 = h√3 – h ⇒ h (√3 – 1) = 10
⇒ h = (1.732 – 1) = 10 ⇒ h × 0.732 = 10
⇒ h = \(\frac { 10 }{ 0.732 }\) = 13.66
∴ Height of tower = 13.66 m

Question 9.
From the top of a hill the angles of depression of two consecutive killometre stones, due east are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill.
Solution:
PQ is a hill whose top is P and A and B are the consecutive kilometre stones which make angle of depression of 45° and 30°
Let height of hill PQ = h
and QA = x km

∴ QB = (x + 1) km
Now in right △PQA,
tan θ = \(\frac { PQ }{ QA }\) ⇒ tan 45° = \(\frac { h }{ x }\)
⇒ 1 = \(\frac { h }{ x }\) ⇒ h = x …(i)
similarly in right △PQB
tan 30° = \(\frac { PQ }{ QB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+1}\)
⇒ x + 1 = √3 h ⇒ x = √3 h – 1 …(ii)
From (i) and (ii)
√3 h – 1 = h ⇒ √3 h – h = 1
(1.732 – 1) h = 1 ⇒ 0.732h = 1
∴ h = \(\frac { 1 }{ 0.732 }\) = \(\frac { 1000 }{ 732 }\) = 1.366
∴ QA = h = 1.366 km and QB = (1 + 1.366)
= 2.366 km

Question 10.
A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the length of pole is 20 m.
Solution:
Let AE is the tower and BC is the pole standing on the same level ground
BC = 20 m
and angle of elevation of the top of tower = 60°
and angle of depression of the foot of the tower = 30°
From B, draw BD || CE
Such that DE = BC = 20 cm
and ∠BEC = ∠EBD = 30° (alternate angles)
BD = CE
Let height of tower AE = h

and CE = DB = x
∴ AD = h – 20
Now in right △BCE,
tan θ = \(\frac { BC }{ CE }\) ⇒ tan 30° = \(\frac { 20 }{ x }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 20 }{ x }\) ⇒ x = 20√3 m …(i)
∴ BD = x = 20√3 m
Now in right △ABD,
tan 60° = \(\frac { AD }{ BD }\) ⇒ √3 = \(\frac{h-20}{x}\)
⇒ √3 = \(\frac{h-20}{20 \sqrt{3}}\) ⇒ 20√3 × √3 = h – 20 m
⇒ 60 = h – 20 ⇒ h = 60 + 20 = 80 m
∴ Height of tower = 80 m

Question 11.
From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places.
Solution:

In △DBC, tan 60° = \(\frac{10}{BC}\)
⇒ √3 = \(\frac{10}{BC}\)
⇒ BC = \(\frac{10}{\sqrt{3}}\)
In △DAC, tan 30° = \(\frac{10}{B C+A B}\)

Question 12.
The top of a light house 100 m high the angle of depression of two ships on opposite sides of it are 48° and 36° respectively. Find the distance between the two ships to the nearest meter.

Solution:

As AD = 100 m
In right angled △ABD
tan 48° = \(\frac{AD}{BD}\) ⇒ BD = \(\frac{\mathrm{AD}}{\tan 48^{\circ}}\)
BD = \(\frac{100}{\tan 48^{\circ}}\) …(i)
In right angled △ADC
tan 36° = \(\frac{AD}{DC}\) ⇒ DC = \(\frac{\mathrm{AD}}{\tan 36^{\circ}}\)
DC = \(\frac{100}{\tan 36^{\circ}}\) …(ii)
From (i) and (ii)
BD + DC = \(\frac{100}{\tan 48^{\circ}}\) + \(\frac{100}{\tan 36^{\circ}}\)
BC = \(\frac{100}{\tan \left(90^{\circ}-42^{\circ}\right)}\) + \(\frac{100}{\tan \left(90^{\circ}-54^{\circ}\right)}\)
= \(\frac{100}{\cot 42^{\circ}}\) + \(\frac{100}{\cot 54^{\circ}}\)
= 100 × tan 42° + 100 × tan 54°
= 100 × 0.9004 + 100 × 1.3764
= 90.04 + 137.64 = 227.68 m = 228 m

Question 13.
A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.
Solution:
AB is a building CD = 60 m (given)
In △ABC:
tan 60° = \(\frac{AB}{BC}\) ⇒ √3 = \(\frac{AB}{BC}\)
∴ BC = \(\frac{\mathrm{AB}}{\sqrt{3}}\) …(i)

In right angled △ABD:
tan 30° = \(\frac{AB}{BD}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}+60}\) ⇒ BC + 60 = √3 AB
∴ BC = √3 AB – 60
From equation (i) and (ii) we have
\(\frac{\mathrm{AB}}{\sqrt{3}}\) = √3 AB – 60 …(ii)
⇒ AB = 3AB – 60√3 ⇒ 3AB – AB = 60 × 1.732
⇒ AB = \(\frac{60 \times 1.732}{2}\) = 51.96 m = 52 m

Question 14.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.
Solution:
Let AB be the lighthouse and C and D be the two ships.

Then, in △ADB
tan 30° = \(\frac { AB }{ BD }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{80}{\mathrm{BD}}\) ⇒ BD ⇒ 80√3 m …(i)
In △ABC, tan 40° = \(\frac{80}{\mathrm{BC}}\)
0.84 = \(\frac{80}{\mathrm{BD}}\) ⇒ BC = \(\frac { 80 }{ 0.84 }\) = 95.25 m
From (i), BD = 80√3 m
⇒ BC + DC = 80 × 1.73 m
⇒ DC = 138.4 – 95.25 m
= 43.15 m
Hence, distance between the two ships is = 43 m

Question 15.
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post.

Solution:
In △ABC
∠ACB = 60°
\(\frac { AB }{ BC }\) = tan 60°

BC = \(\frac{\mathrm{AB}}{\tan 60^{\circ}}\)
BC = \(\frac{60}{1.73}\) = 34.68 m
In △AXD
\(\frac { AX }{ XD }\) = tan 30°
⇒ XD = BC
∴ \(\frac { AX }{ BC }\) = tan 30°
\(\frac{\mathrm{AX}}{34.68}\) = tan 30°
⇒ AX = 34.68 × \(\frac{1}{\sqrt{3}}\)
AX = 34.68 × \(\frac { 1 }{ 1.732 }\) = 20.02 m
∴ CD = AB – AX
= 60 – 20.02 = 39.98 m
= 40 m (approx)

Question 16.
An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Solution:
Let aeroplane is at position O and ship 1 is at point A and ship 2 is at point B.
⇒ OM is the altitude drawn from point O towards the river.
In △OMA
45° = \(\frac{\mathrm{OM}}{\mathrm{AM}}\)

l = \(\frac{250}{x}\) {∵ tan 45° = 1}
⇒ x = 250 m
In △OMB, tan 60° = \(\frac{250}{y}\)
⇒ √3 = \(\frac{250}{y}\) ⇒ y = \(\frac{250}{\sqrt{3}}\) = \(\frac{250}{1.73}\)
y = 144.34
∴ Width of river = x + y = 250 + 144.34 = 394.34 m

Question 17.
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30° and 24° respectively.
Find the height of the two towers. Give your answer correct to 3 significant figures.

Solution:
Consider the following figure:

Tan 30° = \(\frac { AE }{ EC }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { AE }{ 120 }\)
⇒ \(\frac{120}{\sqrt{3}}\) = AE
∴ AE = 69.28 m
tan 24° = \(\frac { EB }{ EC }\)
⇒ 0.445 = \(\frac { EB }{ 120 }\)
∴ EB = 53.43 m
Thus, height of first tower, AB = AE + EB = 69.28 + 53.43 = 122.71 m
And, height of second tower, CD = EB = 53.427 m

Question 18.
An aeroplane at an altitude of 1500 metres finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships.
Solution:
A is the aeroplane, D and C are the ships sailing towards A. Ships are sailing towards the aeroplane in the same direction.
In the figure, height AB = 1500 m
C and D are the positions of two ships

To find: Distance between the ships, that is CD
In the right-angled △ABC,
tan 45° = \(\frac { 1500 }{ BD }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 1500 }{ BD }\)
⇒ BD = 1500√3 m
⇒ BD = 1500(1.732) m
⇒ BD = 2598 m
∴ Distance between the ships = CD
= BC – BC
= 2598 – 1500 = 1098 m
So, the distance between the two ships is 1098 m.

Peer review of OP Malhotra Class 10 Solutions Chapter 6 Ratio and Proportion Ex 6(b) can encourage collaborative learning.

S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(b)

Question 1.
If \(\frac {x}{y}\) = \(\frac {p}{q}\), prove that \(\frac{5 x+7 y}{5 x-7 y}\) = \(\frac{5 p+7 q}{5 p-7 q}\).
Solution:
\(\frac {x}{y}\) = \(\frac {p}{q}\)
\(\frac {5x}{7y}\) = \(\frac {5p}{7q}\) (Multipling both sides bye \(\frac {5}{7}\)
Applying componendo and dividendo,
\(\frac{5 x+7 y}{5 x-7 y}=\frac{5 p+7 q}{5 p-7 q}\)
Hence proved.

Question 2.
If (4a + 9b): (4a – 9b) = (4c + 9d): (4c – 9d), Show that a : b : : c : d.
Solution:
(4a + 9b) : (4a-9b) = (4c + 9d): (4c – 9d)
⇒ \(\frac{4 a+9 b}{4 a-9 b}=\frac{4 c+9 d}{4 c-9 d}\)
Applying componendo and dividendo,
\(\frac{4 a+9 b+4 a-9 b}{4 a+9 b-4 a+9 b}=\frac{4 c+9 d+4 c-9 d}{4 c+9 d-4 c+9 d}\)
⇒ \(\frac{8 a}{18 b}=\frac{8 c}{18 d}\)
⇒ \(\frac{a}{b}=\frac{c}{d}\) (Dividing by \(\frac { 8 }{ 18 }\))
∴ a : b :: c : d
Hence proved.

Question 3.
If (5a + 116) : (5c -11 d) : : (5a – 116) (5c + 11d), prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\).
Solution:
(5a + 116) : (5c -11 d) : : (5a – 116) (5c + 11d)
⇒ \(\frac {x}{y}\)
Applying componendo and dividendo,
\(\frac{5 a+11 b+5 a-11 b}{5 a+11 b-5 a+11 b}=\frac{5 c+11 d+5 c-11 d}{5 c+11 d-5 c+11 d}\)
⇒ \(\frac{10 a}{22 b}=\frac{10 c}{22 d}\)
⇒ \(\frac {a}{b}\) = \(\frac {c}{d}\) (Dividing by \(\frac {10}{22}\))
Hence proved.

Question 4.
Show that a, b, c, d are in proportion if
(i) ma² + nb²: mc² + nd² :: ma² – nb² : mc² – nd².
(ii) (a + b + c + d)(a – b – c + d) = (a + b – c – d) (a – b + c – d).
Solution:
(i) ma² + nb²: mc² + nd² :: ma² – nb² : mc² – nd².

Taking square root of
⇒ \(\frac {a}{b}\) = \(\frac {c}{d}\) ⇒ a : b : : c : d
∴ a, b, c and d are in proportion.

(ii) (a + b + c + d)(a – b – c + d) = (a + b – c – d) (a – b + c – d)
⇒ \(\frac{a+b+c+d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}\)
Applying componendo and dividendo,

∴ a, b, c and d are in proportion.

Question 5.
If a : b = c : d and e : f = g : h, prove that ae + bf : ae – bf = eg + dh : eg – dh.
Solution:
a: b = c : dand e : f = g : h
∴ \(\frac { a }{ b }\) = \(\frac {c}{d}\) and \(\frac {e}{f}\) = \(\frac {g}{h}\) (Multiplying each other)
\(\frac { ae }{ bf }\) = \(\frac {cg}{dh}\)
Now applying componendo and dividendo
\(\frac{a e+b f}{a e-b f}=\frac{c g+d h}{c g-d h}\)
∴ (ae + bf) : (ae – bf) = (cg + dh) : (cg – dh)
Hence proved.

Question 6.
If x = \(\frac { 10pq }{ p+q }\), find the value of \(\frac{x+5 p}{x-5 p}\) + \(\frac{x+5 q}{x-5 q}\).
Solution:
x = \(\frac{10 p q}{p+q} \Rightarrow \frac{x}{5 p}=\frac{2 q}{p+q}\)
Applying componendo and dividendo,

Question 7.
If x = \(\frac{6 p q}{p+q}\), find the value of \(\frac{x+3 p}{x-3 p}\) + \(\frac{x+3 q}{x-3 q}\)
Solution:
x = \(\frac{6 p q}{p+q} \Rightarrow \frac{x}{3 p}=\frac{2 q}{p+q}\)
Applying componendo and dividendo,
\(\frac{x+3 p}{x-3 p}=\frac{2 q+p+q}{2 q-p-q}=\frac{3 q+p}{q-p}\) … (i)
Again x = \(\frac{6 p q}{p+q} \Rightarrow \frac{x}{3 q}=\frac{2 p}{p+q}\)
Applying componendo and dividendo,
\(\frac{x+3 q}{x-3 q}=\frac{2 p+p+q}{2 p-p-q}=\frac{3 p+q}{p-q}\) … (ii)
Adding (i) and (ii)
\(\frac{x+3 p}{x-3 p}+\frac{x+3 q}{x-3 q}=\frac{3 q+p}{q-p}+\frac{3 p+q}{p-q}\)
= \(\frac{-(3 q+p)}{p-q}+\frac{3 p+q}{p-q}\)
= \(\frac{-3 q-p+3 p+q}{p-q}\)
= \(\frac{2 p-2 q}{p-q}=\frac{2(p-q)}{p-q}\) = 2

Question 8.
Solve for x, using the properties of proportion.
(i) \(\frac{3 x+\sqrt{9 x^2-5}}{3 x-\sqrt{9 x^2-5}}\) = 5
(ii) \(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}-\sqrt{x-3}}\) = 5
(iii) \(\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}\)
(iv) \(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}\)
Solution:
(i) \(\frac{3 x+\sqrt{9 x^2-5}}{3 x-\sqrt{9 x^2-5}}\) = 5
Applying componendo and dividendo,
\(\frac{3 x+\sqrt{9 x^2-5}+3 x-\sqrt{9 x^2-5}}{3 x+\sqrt{9 x^2-5}-3 x+\sqrt{9 x^2-5}}=\frac{5+1}{5-1}\)
⇒ \(\frac{6 x}{2 \sqrt{9 x^2-5}}=\frac{6}{4} \Rightarrow \frac{3 x}{\sqrt{9 x^2-5}}\) = \(\frac { 3 }{ 2 }\)
⇒ \(3 \sqrt{\left(9 x^2-5\right)}=6 x \Rightarrow \sqrt{9 x^2-5}\) = 2x
(Dividing by 3)
Squaring both sides,
9x² – 5 = 4x² ⇒ 9x² – 4x² = 5
5x² = 5 ⇒ x² = \(\frac { 5 }{ 5 }\) = 1
∴ x = 1

(ii) \(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{5}{1}\)
Appling componendo and dividendo,
\(\frac{\sqrt{x+2}+\sqrt{x-3}+\sqrt{x+2}-\sqrt{x-3}}{\sqrt{x+2}+\sqrt{x-3}-\sqrt{x+2}+\sqrt{x-3}}=\frac{5+1}{5-1}\)
⇒ \(\frac{2 \sqrt{x+2}}{2 \sqrt{x-3}}=\frac{6}{4} \Rightarrow \frac{\sqrt{x+2}}{\sqrt{x-3}}=\frac{3}{2}\)
Squaring both sides,
⇒ \(\frac { x+2 }{ x-3 }\) = \(\frac { 9 }{ 4 }\) ⇒ 9x – 27 = 4x + 8
⇒ 9x – 4x = 8 + 27 ⇒ 5x = 35
⇒ x = \(\frac { 35 }{ 5 }\) = 7
∴ x = 7

(iii) \(\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}\)
Appling componendo and dividendo,
\(\frac{x^3+3 x+3 x^2+1}{x^3+3 x-3 x^2-1}=\frac{341+91}{341-91}\)
⇒ \(\frac{x^3+3 x^2+3 x+1}{x^3-3 x^2+3 x-1}=\frac{432}{250}=\frac{216}{125}\)
⇒ \(\frac{(x+1)^3}{(x-1)^3}=\frac{(6)^3}{(5)^3}\)
Taking cube root,
\(\frac { x+1 }{ x-1 }\) = \(\frac { 6 }{ 5 }\)
⇒ 6x – 5x = 5 + 6 ⇒ x = 11
∴ x = 11x

(iv) \(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}\)
Applying componendo and dividendo,

Question 9.
Solve for x : 16\(\left[\frac{a-x}{a+x}\right]^3=\frac{a+x}{a-x}\).
Solution:

Question 10.
Solve for x : 16\(\frac{1+x+x^2}{1-x+x^2}=\frac{171(1+x)}{172(1-x)}\), x ≠ 1, – 1.
Solution:

Taking cube root of,
x = \(\frac { 1 }{ 7 }\)

Question 11.
Solve for x = \(\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\) show that b²x² – 2a²x + b² = 0.
Solution:

Question 12.
If y = \(\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}\) show that 3by² – 2ay + 3b = 0.
Solution:

Question 13.
If y = \(\frac{a^3+3 a b^2}{3 a^2 b+b^3}=\frac{x^3+3 x y^2}{3 x^2 y+y^3}\) show that \(\frac { x }{ a }\) = \(\frac { y }{ b }\)
Solution:
\(\frac{a^3+3 a b^2}{3 a^2 b+b^3}=\frac{x^3+3 x y^2}{3 x^2 y+y^3}\)
Applying componendo and dividendo,

Continuous practice using OP Malhotra Class 10 Solutions Chapter 6 Ratio and Proportion Ex 6(a) can lead to a stronger grasp of mathematical concepts.

S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(a)

Question 1.
Find the value of x in each case :
(i) 8 : 14 : : x : 28
(ii) x : 9 : : 5 : 3
Solution:
(i) 8 : 14 : : x : 28
⇒ 14 × x = 8 x 28 ⇒ x = \(\frac{8 \times 28}{14}\) = 16 (∵ ad = bc)

(ii) x : 9 :: 5 : 3
⇒ x × 3 = 9 x 5 ⇒ x = \(\frac{9 \times 5}{3}\) = 16 (∵ ad = bc)

(iii) 12 : x :: 4 : 15
⇒ x × 4 = 12 x 15 ⇒ x = \(\frac{12 \times 15}{4}\) = 45 (∵ ad = bc)

Question 2.
Find the fourth proportional to :
(i) 25, 15, 40
(ii) 3a² b², a³, b³
(iii) a² – 5a + 6, a² a – 6, a² – 9
Solution:
We know that if a : b : : c : d, then d = \(\frac { b×c }{ a }\)
(i) 25, 15, 40
∴ Fourth proportional = \(\frac { bc }{ a }\) = \(\frac{15 \times 40}{25}\) = 24

(ii) 3a² b², a3, b3
Fourth proportional = \(\frac { bc }{ a }\)
= \(\frac{a^3 \times b^3}{3 a^2 b^2}=\frac{1}{3} a b=\frac{a b}{3}\)

(iii) a² – 5a + 6, a² + a – 6, a² – 9
a² – 5a + 6 = a² – 3a – 2a + 6
= a (a – 3) – 2 (a – 3)
= (a – 3) (a – 2)
a² + a – 6 = a² + 3a – 2a – 6
= a (a + 3) – 2 (a + 3)
= (a + 3) (a – 2)
a² – 9 = (a)² – (3)² = (a + 3) (a – 3)
Now fourth proportional = \(\frac { bc }{ a }\)
= \(\frac{(a+3)(a-2)(a+3)(a-3)}{(a-3)(a-2)}\)
= (a + 3) (g + 3) = (a + 3)²

Question 3.
Find the third proportional to :
(i) 16 and 36
(ii) \(\frac { x }{ y }\) + \(\frac { y }{ x }\) and
(iii) a² – b², a + b
Solution:
If a, b :: b : c, then
Third proportional (c) = \(\frac { b² }{ a }\)

(i) 16 and 36
Third proportional \(\frac{b^2}{a}=\frac{36 \times 36}{16}\) = 81

(ii) \(\frac { x }{ y }\) + \(\frac { y }{ x }\) and \(\frac { x }{ y }\)
∴ Third proportional = \(\frac{\frac{x}{y} \times \frac{x}{y}}{\frac{x}{y}+\frac{y}{x}}=\frac{\frac{x^2}{y^2}}{\frac{x^2+y^2}{x y}}\)
= \(\frac{x^2}{y^2} \times \frac{x y}{x^2+y^2}=\frac{x^3}{y\left(x^2+y^2\right)}\)

(iii) a² – b², (g + b)
Third proportional = \(\frac{(a+b)(a+b)}{a^2-b^2}\)
= \(\frac{(a+b)(a+b)}{(a+b)(a-b)}=\frac{a+b}{a-b}\)

Question 4.
Find the mean proportional between the following:
(i) 5 and 80
(ii) 360a⁴ and 250a² b²
(iii) (x – y) and (x³ – x²y)
Solution:
We know that mean proportional (b) = \(\sqrt{ac}\)
(i) 5, 80
Mean proportional = \(\sqrt{5×80}\) = \(\sqrt{400}\) = 20

(ii) 360a⁴ and 250a² b²
Mean proportional = \(\sqrt{360 a^4 \times 250 a^2 b^2}\)
= \(\sqrt{360 \times 250 \times a^{4+2} \times b^2}\)
= \(\sqrt{90000 \times a^6 \times b^2}\)
= 300 x a³ x b
= 300 a³ b

(iii) (x – y) and (x³ – x²y)
Mean proportioanl = \(\sqrt{(x-y)\left(x^3-x^2 y\right)}\)
= \(\sqrt{(x-y) x^2(x-y)}\)
= \(\sqrt{x^2(x-y)^2}\) = x(x – y)

Question 5.
(i) If x, 16, 48, y are in continued proportion, find the value of x and y.
(ii) If x, 9, y, 16 are in continued proportion, then find the value of x and y.
Solution:
(i) ∵ x, 16, 48, y are in continued proportion,
then \(\frac{x}{16}=\frac{16}{48}=\frac{48}{y}\)

(ii) ∵ x, 9, y, 16 are in continued proportion,

Question 6.
What number must be added to 3, 5, 7, 10 each in order to get four numbers in proportion?
Solution:
Let x be added to each number,
then 3 +x, 5 + x, 7 + x and 10 + x are in proportion
⇒ \(\frac{3+x}{5+x}=\frac{7+x}{10+x}\)
⇒ (3 + x) (10 + x) = (7 + x) (5 + x)
⇒ 30 + 3x + 10x + x² = 35 + 7x + 5x + x²
⇒ 30 + 13x + x² = 35 + 12x + x²
⇒ x² + 13x – x² – 12x = 35 – 30
⇒ x = 5
∴ 5 must be added

Question 7.
What number must be subtracted from each of the numbers 28, 53, 19, 35 so that they are in proportion?
Solution:
Let x be subtracted from each number, then 28 – x, 53 – x, 19 – x, 35 – x are in proportion,
∴ \(\frac{28-x}{53-x}=\frac{19-x}{35-x}\)
⇒ (28 – x) (35 – x) = (19 – x) (53 – x)
⇒ 980 – 28x – 35 x + x² = 1007 – 19x – 53x + x²
⇒ 980 – 63x + x² = 1007 – 72x + x²
⇒ – 63x + x² + 72x – x² = 1007 – 98027
⇒ 9x = 27 ⇒ x = \(\frac { 27 }{ 9 }\) = 3
∴ 3 must be added.

Question 8.
(i) Find two numbers such that the mean proportional between them is 14 and the third proportional to them is 112.
(ii) Find two numbers such that the mean proportional between them is 18 and the third proportional to them is 144.
Solution:
(i) Let a and b be the two numbers whose mean
proportional = 14 and third proportional = 112

∴ Numbers are 7, 28

(ii) Let a and b be the numbers whose mean proportional = 18 and third proportional = 144

∴ Numbers are 9, 36

Question 9.
If p + r = 2q and \(\frac { 1 }{ q }\) + \(\frac { 1 }{ s }\) = \(\frac { 2 }{ r }\), then prove that p : q = r : s.
Solution:
p + r = 2q … (i)
\(\frac { 1 }{ q }\) + \(\frac { 1 }{ s }\) = \(\frac { 2 }{ r }\) .. (ii)
⇒ r (q + s) = (p + r) s {from (i)}
⇒ qr + rs = ps + rs
⇒ qr = ps
⇒ \(\frac { p }{ q }\) = \(\frac { r }{ s }\)
⇒ p : q = r : s
Hence proved.

Question 10.
If b is the mean proportional between a and c, prove that a, c
and c, prove that a, c, a² + b² and b² + c² are proportional.
Solution:
∵ b is the mean proportional between a and c
∴ b² = ac … (i)
Now a, c, a² + b² and b² + c² are proportional
If \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)
= \(\frac{a^2+a c}{a c+c^2}=\frac{a(a+c)}{c(a+c)}=\frac{a}{c}\)
∴ Which is true.
Hence proved.

Question 11.
If x + 7 is the mean proportional between (x + 3) and (x + 12), find the value of x.
Solution:
x + 1 is the mean proportional between
(x + 3) and (x + 12)
∴ (x + 7)² = (x + 3) (x + 12)
⇒ x² + 14x + 49 = x² + 12x + 3x + 36
⇒ x² + 14x – 12x – 3x – x² = 36 – 49
⇒ – x = – 13
⇒ x = 13
∴ x = 13

Question 12.
If \(\frac{a^2+c^2}{a b+c d}=\frac{a b+c d}{b^2+d^2}\), Prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\).
Solution:
∵ \(\frac{a^2+c^2}{a b+c d}=\frac{a b+c d}{b^2+d^2}\)
⇒ (a² + c²) (b² + d²) = (ab + cd)²
⇒ a²b² + a²d² + b²c² + c²d² (By cross multiplication)
= a²b² + a²d² + 2abcd
⇒ b²c² + a²d² – 2abcd = 0
= (be – ad)² = 0
⇒ \(\frac { a }{ b }\) = \(\frac { c }{ d }\)

Effective OP Malhotra Class 10 Solutions Chapter 5 Quadratic Equations Ex 5(e) can help bridge the gap between theory and application.

S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(e)

On Numbers

Question 1.
The sum of the squares of two consecutive positive even integers is 100. Find the integers.
Solution:
Let first even positive number = 2x
∴ Second number = 2x + 2
According to the condition,
(2x)² + (2x + 2)² = 100
⇒ 4x² + 4x² + 8x + 4 = 100
⇒ 8x² + 8x + 4 – 100 = 0
⇒ 8x² + 8x – 96 = 0
⇒ x² + x-12 = 0 (Dividing by 8)
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3 (x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either x + 4 = 0 then x = – 4
But it is not possible as it is negative
or x – 3 = 0 then x = 3
First even number = 2x = 2 x 3 = 6
and second number = 6 + 2 = 8
∴ Numbers are 6, 8

Question 2.
Find two rational numbers which differ by 3 and the sum of whose squares is 117.
Solution:
Let first rational number = x
Then second number = x – 3
According to the condition,
x² + (x – 3)² = 117
⇒ x² + x² – 6x + 9 = 117
⇒ 2x² – 6x + 9 – 117 = 0
⇒ 2x²- 6x – 108 = 0
⇒ x² – 3x – 54 = 0 (Dividing by 2)
⇒ x² – 9x + 6x – 54 = 0

⇒ x (x – 9) + 6 (x – 9) = 0
⇒ (x – 9) (x + 6) = 0
Either x – 9 = 0, then x = 9
or x + 6 = 0, then x = – 6, but it is not possible as it is negative
If x = 9, then first number = 9
and second number = 9 – 3 = 6
Numbers are 9, 6

Question 3.
What number increased by its reciprocal equals \(\frac {65}{8}\)?
Solution:
Let number = x
Then its reciprocal = \(\frac {1}{x}\)
According to the condition,
x + \(\frac {1}{x}\) = \(\frac {65}{8}\)
⇒ 8x² + 8 = 65x
⇒ 8x² – 65x + 8 = 0
⇒ 8x² – 64x – x + 8 = 0

⇒ 8x (x – 8) – 1 (x – 8) = 0
⇒ (x – 8) (8x – 1) = 0
Either x – 8 = 0, then x = 8
or 8x – 1 = 0, then 8x = 1 ⇒ x = \(\frac {1}{8}\)
Now if x = 8, then its reciprocal = \(\frac {1}{8}\)
and x = \(\frac {1}{8}\), then its reciprocal = 8
Number = 8 or \(\frac {1}{8}\)

Question 4.
The sum of the numerator and denominator of a certain fraction is 8. If 2 is added to the numerator and to the denominator, the value of the fraction is increased by \(\frac {4}{35}\). Find the fraction.
Solution:
Let numerator of a fraction = x
∴ Its denominator = 8 – x
and fraction = \(\frac {x}{8-x}\)
By adding 2 to both its numerator and denominator the fraction will be \(\frac{x+2}{8-x+2}\) = \(\frac{x+2}{10-x}\)
According to the condition,

Either x + 20 = 0 then x = – 20 but it is not possible as it is negative or x – 3 = 0, then x = 3
∴ Fraction = \(\frac{x}{8-x}=\frac{3}{8-3}=\frac{3}{5}\)

Question 5.
There are three consecutive integers such that the square of the first increased by the product of the other two given 154. Find the integers.
Solution:
Let the consecutive integers be x, x + 1, x + 2
According to the condition,
x² + (x + 1)(x + 2) = 154
x² + x² + 2x + x + 2 = 154
⇒ 2x² + 3x + 2 – 154 = 0
⇒ 2x² + 3x- 152 = 0
Here a = 2, b = 3, c = – 152
∴ b² – 4ac = (3)² – 4 x 2 x (- 152)
= 9 + 1216 = 1225
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-3 \pm \sqrt{1225}}{2 \times 2}\)
= \(\frac{-3 \pm 35}{4}\)
x₁ = \(\frac{-3+35}{4}=\frac{32}{4}\) = 8
x₂ = \(\frac{-3-35}{4}=\frac{-38}{4}\) but it is not an integer
∴ First number = 8
Second number = 8 + 1 = 9
and third number = 9 + 1 = 10
Hence numbers are 8, 9, 10

Question 6.
The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Solution:
Sum of two numbers = 8
Let first number = x
Then second number = 8 – x
According to the condition,
15\(\left(\frac{1}{x}+\frac{1}{8-x}\right)\) = 8
\(\frac{1}{x}+\frac{1}{8-x}=\frac{8}{15}\)
\(\frac{8-x+x}{x(8-x)}=\frac{8}{15} \Rightarrow \frac{8}{x(8-x)}=\frac{8}{15}\)
⇒ 8x (8 -x) = 120 ⇒ x (8 – x) = 15 (Dividing by 8)
⇒ 8x – x² = 15
⇒ x² – 8x + 15 = 0
⇒ x² – 5x – 3x + 15 = 0

⇒ x (x – 5) – 3 (x – 5) = 0
⇒ (x – 5) (x – 3) = 0
Either x – 5 = 0, then x = 5
or x – 3 = 0, then x = 3
If first number is 5,
then second number = 8 – 5 = 3
If first number is 3, then
second number = 8 – 3 = 5
∴ Numbers are 3, 5

Question 7.
A two digit number is such that the product of its digits is 12. When 36 is added to this number, the digits interchange their places. Find the number.
Solution:
Product of two digits = 12
Let the units digit = x
Then tens digit = \(\frac {12}{x}\)
∴ Number = x + 10 x \(\frac {12}{x}\) = x + \(\frac {120}{x}\)
By interchanging the digits,
the units digit = \(\frac {12}{x}\)
and tens digit = x
∴ Number = \(\frac {12}{x}\) + 10x
According to the condition,
x + \(\frac {120}{x}\) + 36 = \(\frac {12}{x}\) + 10x
⇒ x² + 120 + 36x = 12 + 10x²
⇒ 10x² + 12 -x² – 120 – 36x = 0
⇒ 9x² – 36x – 108 = 0
⇒ x² – 4x – 12 = 0 (Dividing by 9)
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6) = 0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then x = 6
or x + 2 = 0, then x = – 2 which is not possible as it is negative
∴ Number = x + \(\frac {120}{x}\) = 6 + \(\frac {120}{6}\) = 6 + 20 = 26

Question 8.
Five times a certain whole number is equal to three less than twice the square of the number. Find the number.
Solution:
Let a whole number = x
5x = 2x² – 3
⇒ 2x² – 5x – 3 = 0
⇒ 2x² – 6x + x – 3 = 0

⇒ 2x (x – 3) + 1 (x – 3) = 0
⇒ (x – 3) (2x + 1) = 0
Either x – 3 = 0, then x = 3
or 2x + 1 = 0, then 2x = – 1 ⇒ x = \(\frac {-1}{2}\)
But it is not possible as it is negative.
∴ Number = 3

Question 9.
Three consecutive numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence find the three numbers.
Solution:
Let the middle number = x
Then first number = x – 1
and third number = x + 1
∴ x² = (x + 1 )² – (x – 1 )² + 60
⇒ x² = (x² + 2x + 1) – (x² – 2x + 1) + 60
⇒ x² = x² + 2x + 1 – x² + 2x – 1 + 60
⇒ x² = 4x + 60 ⇒ x² – 4x – 60 = 0
⇒ x² – 10x + 6x – 60 = 0

⇒ x (x – 10) + 6 (x – 10) = 0
⇒ (x – 10) (x + 6) = 0
Either x – 10 = 0, then x = 10
or x + 6 = 0, then x = – 6 which is not possible as it is negative
∴ Middle number = 10
and other two numbers will be 10 – 1 = 9
and 10 + 1 = 11
Hence numbers are 9, 10, 11
AGE

Question 10.
The sum of the ages (in years) of a son and his father is 35 and the product of their ages is 150. Find their ages.
Solution:
Sum of ages of son and father = 35
Let age of son = x years
then age of father = 35 – x
According to the condition,
x (35 – x) = 150
⇒ 35x – x² = 150
⇒ x² – 35x + 150 = 0
⇒ x² – 30x – 5x + 150 = 0

⇒ x (x – 30) – 5 (x – 30) = 0
⇒ (x – 30) (x – 5) = 0
Either x – 30 = 0, then x = 30
or x – 5 = 0, then x = 5
But age of son cannot be 30
∴ Age of son = 5 years
and age of father = 35 – 5 = 30 years

Question 11.
Anjali was born in 1985 A.D. In the year x² A.D., she was (x – 5) years old. Find the value of x.
Solution:
Anjali was born in = 1985 AD
In the year x² AD, she was (x – 5) years old
∴ 1985 + (x – 5) = x²
⇒ 1985 + x – 5 = x²
⇒ x² – x + 5 – 1985 = 0
⇒ x² – x – 1980 = 0
⇒ x² – 45x + 44x – 1980 = 0

⇒ x (x – 45) + 44 (x – 45) = 0
⇒ (x – 45) (x + 44) = 0
Either x – 45 = 0, then x = 45
or x + 44 = 0, then x = – 44 which is not possible as it is negative
∴ x = 45

Question 12.
The difference of mother’s age and her daughter’s age is 21 years and the twelfth part of the product of their ages is less than the mother’s age by 18 years. Find their ages.
Solution:
Difference of their ages = 21 years
Let age of daughter = x
Then age of mother = x + 21
According to the condition,
\(\frac{x(x+21)}{12}\) = x + 21 – 18
⇒ \(\frac{x^2+21 x}{12}\) = x + 3
⇒ x² + 21x = 12x + 36
⇒ x² + 21x – 12x – 36 = 0
⇒ x² + 9x – 36 = 0
⇒ x² + 12x – 3x – 36 = 0

⇒ x (x + 12) – 3 (x + 12) = 0
⇒ (x + 12)(x – 3) = 0
Either x + 12 = 0, then x = – 12 which is not possible being negative
or x – 3 = 0, then x = 3
∴ Age of daughter = 3 years
and age of mother = 3 + 21 = 24 years

Question 13.
Reena is x years old while her father Mr. Sunil is x² years old. 5 years hence, Mr. Sunil will be three times as old as Reena. Find their present ages.
Solution:
At present
Age of Reena = x years
and age of her father = x² years
5 years hence,
age of Reena = x + 5
and age of her father = x² + 5
According to the condition,
x² + 5 = 3 (x + 5) ⇒ x² + 5 = 3x + 15
⇒ x² + 5 – 3x – 15 = 0
⇒ x² – 3x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0

⇒ x (x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
Either x – 5 = 0, then x = 5
or x + 2 = 0, then x = – 2 which is not possible being negative
∴ Age of Reena = 5 years
and age of her father = x² = (5)² = 25 years

MENSURATION

Question 14.
Form a quadratic equation from the following information, taking x as width where x ∈ N.
(i) The area of a rectangle whose length is five more than twice its width is 75.
(ii) Solve the equation and find its length.
Solution:
Width of a rectangle = x
∴ Length = 2x + 5
Area = Length x Width
⇒ 75 = (2x + 5) x
⇒ 2x² + 5x – 75 = 0
⇒ 2x² – 10x + 15x – 75 = 0

⇒ 2x(x – 5)+ 15 (x – 5) = 0
⇒ (x – 5) (2x + 15) = 0
Either x – 5= 0, then x = 5
or 2x + 15 = 0, then 2x = – 15 ⇒ x = \(\frac { -15 }{ 2 }\)
which is not possible being negative
∴ Width of rectangle = 5
and length = 2x + 5 = 2 x 5 + 5 = 10 + 5 = 15

Question 15.
The two sides of a right-angled triangle containing the right angle are 5x cm and (3x – 1) cm. If the area of the triangle be 60 cm², calculate the lengths of the sides of the triangle.
Solution:
Two sides of a right angled triangle are 5x cm and (3x – 1) cm
∴ Area = \(\frac { 1 }{ 2 }\) x Product of two sides
⇒ 60 = \(\frac { 1 }{ 2 }\) x 5x (3x – 1)
⇒ 120= 15x²-5x ⇒ 15x²-5x- 120 = 0
⇒ 3x² – x – 24 = 0 (Dividing by 5)
⇒ 3x² – 9x + 8x – 24 = 0

⇒ 3x (x – 3) + 8 (x – 3) = 0
⇒ (x – 3) (3x + 8) = 0
Either x – 3 = 0, then x = 3
or 3x + 8 = 0, then 3x = – 8 ⇒ x = \(\frac { -8 }{ 3 }\)
which is not possible being negative
∴ One side = 5x = 5 x 3 = 15 cm
Second side =3x – 1 = 3 x 3 – 1 = 9 – 1 = 8
and hypotenuse = \(\sqrt{\text { Sum of square of two sides }}\)
= \(\sqrt{(15)^2+(8)^2}=\sqrt{225+64}\) cm
= \(\sqrt{289}\) = 17 cm
Hence sides are, 8 cm, 15 cm and 17 cm

Question 16.
The hypotenuse of a right angle triangle is 13 cm and the difference between the other two sides is 7 cm.
(i) Taking x as the length of the shorter of the two sides, write an equation in ‘x’ that represents the above statement.
(ii) Solve the equation obtained in (i) above, and hence find the two unknown sides of the triangle. (ICSE)
Solution:
(i) Length of hypotenuse of a right triangle = 13 cm
Let shorter side of the remaining sides = x
Then longer side = (x + 7) cm
According to the Pythagoras Theorem,
(Hypotenuse)² = Sum of squares of other two sides
⇒ (13)² = x² + (x + 7)²
⇒ 169 = x² + x² + 14x + 49
⇒ 2x² + 14x + 49 – 169 = 0
⇒ 2x² + 14x – 120 = 0
⇒ x² + 7x – 60 = 0 (Dividing by 2)

(ii) Now x² + 7x – 60 = 0
x² + 12x – 5x – 60 = 0

⇒ x (x + 12) – 5 (x + 12) = 0
⇒ (x + 12) (x – 5) = 0
Either x + 12 = 0, then x = – 12 which is not
possible as it is negative
or x – 5 = 0, then x = 5-
Two sides = 5 cm, and 5 + 7 = 12 cm

Question 17.
The length of a verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.
(i) Taking as the breadth of the verandah, write an equation in ‘x’ that represents the above statement.
(ii) Solve the equation in (i) above and hence find the dimension of the verandah.
Solution:
(i) Let breadth of verandah = x m
∴ Length = (x + 3) m
Area = Length x Breadth
= x (x + 3) = x² + 3x
and perimeter = 2(l + b)
= 2 (x + 3 + x) = 2 (2x + 3)
= 4x + 6
According to the condition
x² + 3x = 4x + 6
⇒ x² + 3x – 4x – 6 = 0
⇒ x² – x – 6 = 0

(ii) x² – x – 6 = 0
⇒ x² – 3x + 2x – 6 = 0
⇒ x (x – 3) + 2 (x – 3) = 0
⇒ (x – 3) (x + 2) = 0
Either x – 3 = 0, then x = 3
or x + 2 = 0, then x = – 2 which is not possible as it is negative
∴ x = 3
∴ Breadth = 3 m
and length = x + 3 = 3 + 3 = 6m

Question 18.
Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
Solution:
Side of first square = x cm
and side of second square = (x + 4) cm
(i) ∴ Area = (x)² + (x + 4)²
⇒ x² + x² + 8x + 16 = 656
⇒ 2x² + 8x + 16 – 656 = 0
⇒ 2x² + 8x – 640 = 0
⇒ x² + 4x – 320 = 0 (Dividing by 2)

(ii) Now x² + 4x – 320 = 0
⇒ x² + 20x – 16x – 320 = 0
⇒ x (x + 20) – 16 (x + 20) = 0
⇒ (x + 20) (x – 16) = 0
Either x – 20 = 0, then x = – 20 which is not possible being negative
or x – 16 = 0, then x = 16
∴ Side of first square = 16 cm
and side of second square = 16 + 4 = 20 cm

Question 19.
The perimeter of a rectangular plot is 180 m and its area is 1800 m2. Take the length of the plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth, and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.
Solution:
Perimeter of the plot = 180 m
and area = 1800 m²
Let length of the plot = x m
But length + Breadth = \(\frac { Perimeter }{ 2 }\)
= \(\frac { 1 }{ 2 }\)
= 90m
∴ Breadth = (90 – x) m
Now area = length x breadth
⇒ 1800 = x (90 – x)
⇒ 1800 = 90x – x²
⇒ x² – 90x + 1800 = 0
⇒ x² – 60x – 30x + 1800 = 0

⇒ x (x – 60) – 30 (x – 60) = 0
⇒ (x – 60) (x – 30) = 0
Either x – 60 = 0, then x = 60
or x – 30 = 0, then x = 30
If length = 60 m,
Then breadth = 90 – 60 = 30 m
If length = 30 m,
Then breadth = 90 – 30 = 60 m

Question 20.
A rectangle has an area of 24 cm². If its length is x cm, write down its breadth in terms of x. Given that its perimeter is 20 cm, form an equation in x and solve it.
Solution:
Area of a rectangle = 24 cm²
Perimeter = 20 cm
Perimeter
(i) ∴ length + Breadth = \(\frac { Perimeter }{ 2 }\)
= \(\frac { 20 }{ 2 }\) = 10 cm
Let length of the rectangle = x cm
∴ Breadth = 10 – x cm
∴ Area = Length x Breadth
⇒ 24 = x (10 – x) ⇒ 24 = 10x – x²

(ii) ⇒ x² – 10x + 24 = 0
⇒ x² – 6x – 4x + 24 = 0

⇒ x (x – 6) – 4 (x – 6) = 0
⇒ (x – 6) (x – 4) = 0
Either x – 6 = 0, then x = 6
or x – 4 = 0, then x = 4
∴ Length = 6 cm
Breadth = 10 – 6 = 4 cm

Question 21.
A rectangular garden 10 m by 16 m is to be surrounded by a concrete wall of uniform width. Given that the area of the wall is 120 squares metres, assuming the width of the wall to be x, form an equation in x and solve it to find the value of x.
Solution:

Length of rectangular garden = 16 m
and breadth = 10 m
Area of the wall = 120 sq. m
Let width of wall = x m
∴ Outer length = (16 + 2x) m
and outer breadth = (10 + 2x) m
∴ Area of wall = Outer area – Inner area
⇒ 120 = (16 + 2x)(10 + 2x) – 16 x 10
⇒ 120 = 160 + 32x + 20x + 4x² – 160
⇒ 4x² + 52x – 120 = 0
(i) ⇒ x² + 13x – 30 = 0 (Dividing by 4)
(ii) ⇒ x² + 15x – 2x – 30 = 0

⇒ x (x + 15) – 2 (x + 15) = 0
⇒ (x+ 15) (x – 2) = 0
Either x + 15 = 0, then x = -15 but it is not possible
or x – 2 = 0, then x = 2
∴ Width of wall = 2 m

TIME, DISTANCE AND SPEED

Question 22.
A man covers a distance of 200 km travelling with a uniform speed of ‘x’ km per hour. The distance could have been covered is 2 hours less, had the speed been (x + 5) km/hr. Calculate the value of x.
Solution:
Distance = 200 km
First speed = x km/hr
Second speed = (x + 5) km
Time taken in first case = \(\frac { 200 }{ x }\) hour
and time taken in second case = \(\frac { 200 }{ x+5 }\) hours
According to the condition,

Either x + 25 = 0, then x = – 25 which is not possible being negative
or x – 20 = 0, then x = 20
∴ x = 20
∴ Speed = 20 km/hr

Question 23.
An express train makes a run of 240 km at a certain speed. Another train, whose speed is 12 km/hr less than the first train takes an hour longer to make the same trip. Find the speed of the express train in km/hr.
Solution:
Distance = 240 km
Let speed of an express train = x km/hr
The speed of another train = (x – 12) km/hr
Time taken by the express train = \(\frac { 240 }{ x }\) hours
and time taken by another train = \(\frac { 240 }{ x – 12 }\)
hours
According to the condition,

Either x – 60 = 0, then x = 60
or x + 48 = 0, then x = – 48 which is not possible as it is negative
∴ Speed of express train = 60 km/hr

Question 24.
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Solution:
Distance = 90 km
Let the speed of the train = x km/hr
Time taken = \(\frac { 90 }{ x }\) hour
On increasing the speed, new speed = (x + 15) km/hr
∴ Time taken = \(\frac { 90 }{ x+15 }\) hours
According to the condition,
= \(\frac { 90 }{ x }\) – \(\frac { 90 }{ x+15 }\) = \(\frac { 30 }{ 60 }\)
= \(\frac{90 x+1350-90 x}{x(x+15)}=\frac{1}{2} \Rightarrow \frac{1350}{x^2+15 x}=\frac{1}{2}\)
⇒ x² + 15x = 2700 ⇒ x² + 15x – 2700 = 0
⇒ x² + 60x – 45x – 2700 = 0
⇒ x (x + 60) – 45 (x + 60) = 0
⇒ (x + 60) (x – 45) = 0
Either x + 60 = 0, then x = – 60 which a not possible being negative
or x – 45 = 0, then x = 45
∴ Original speed of the train = 45 km/hr

Question 25.
A plane left 30 minutes later than the scheduled time and in order to reach its destination, 1500 km away it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Solution:
Distance = 1500 km
Let the usual speed of the plane = x km/hr 1500
Time taken = \(\frac { 1500 }{ x }\) hours
Increased speed = (x + 250) kin/h 1500
∴ Time taken = \(\frac { 1500 }{ x+250 }\)
According to the condition,

Either x + 1000 = 0, then x = – 1000 which is not possible being negative
or x – 750 = 0, then x = 750
∴ Usual speed of the plane = 750 km/hr

Question 26.
A boat takes 1 hour longer to go 36 km up a river than to return. If the river flows at 3 km/hr, find the rate at which the boat travels in still water.
Solution:
Distance upstream = 36 km
Speed of river = 3 km/hr
Let the speed of boat in still water = x km/h
∴ Time taken upstream = \(\frac { 36 }{ x – 3 }\)
and time taken downstream = \(\frac { 36 }{ x + 3 }\)
According to the condition,
\(\frac { 36 }{ x – 3 }\) – \(\frac { 36 }{ x + 3 }\) = 1
⇒ \(\frac{36 x+108-36 x+108}{(x-3)(x+3)}=1 \Rightarrow \frac{216}{x^2-9}\) = 1
⇒ x² – 9 = 216 ⇒ x² – 9 – 216 = 0
⇒ x² – 225 = 0 ⇒ (x)² – (15)² = 0
⇒ (x + 15) (x – 15) = 0
Either x + 15 = 0, then x = – 15 which is not possible being negative
or x – 15, then x = 15
∴ Speed of boat in still water = 15 km/hr

Profit and Loss

Question 27.
A man purchased some horses for ₹ 3000. Three of them died, and he sold the rest at ₹ 65 more than what he paid for each horse and thus gains 6% on his outlay. How many horses did he buy?
Solution:
Let number of horses = x
Total cost price = ₹ 3000
∴ C.P. of each horse = ₹ \(\frac {3000}{x}\)
No. of horses died = 3
Remaining horses = x – 3
Gain = 6%
∴ S. P. = \(\frac{\text { C.P. } \times(100+\text { gain } \%)}{100}=\frac{3000 \times(100+6)}{100}\)
= ₹ \(\frac{3000 \times 106}{100}\) = ₹ 3180
∴ S.P. of one horse = ₹ \(\frac {3180}{x-3}\)
According to the condition,
∴ S.P. – C.P. of one horse = ₹ 65
⇒ \(\frac{3180}{x-3}-\frac{3000}{x}\) = 65
⇒ \(\frac{3180 x-3000 x+9000}{x(x-3)}=\frac{65}{1}\)
⇒ 180x+ 9000 = 65x (x-3)
⇒ 180x + 9000 = 65x² – 195x
⇒ 65x² – 195x – 180x – 9000 = 0
⇒ 65x² – 375x – 9000 = 0
⇒ 13x² – 75x – 1800 = 0
⇒ 13x² – 195x + 120x – 1800 = 0
⇒ 13x (x – 15)+ 120 (x – 15) = 0
⇒ (x – 15) (13x + 120) = 0
Either x – 15 = 0, then x = 15
or 13x + 120 = 0, then 13x = – 120 ⇒ x = \(\frac {-120}{13}\)
which is not possible being negative
∴ x= 15
∴ Number of horses purchased = 15

Question 28.
A trader bought a number of articles for ₹ 1200. Ten were damaged and he sold each of the rest at ₹ 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve.
Solution:
Number of articles, a trader bought = x
Total C.P. = ₹ 1200
∴ C.P. of each article = ₹ \(\frac {120}{x}\)
Total gain = ₹ 60
∴ S.P. = ₹ 1200 + 60 = ₹ 1260
Number of articles damaged = 10
∴ Remaining articles = x – 10
and S.P. of each article = \(\frac {1260}{x-10}\)
According to the condition,
\(\frac {1260}{x-10}\) – \(\frac {1200}{x}\) = 2
⇒ \(\frac{1260 x-1200 x+12000}{x(x-10)}=\frac{2}{1}\)
⇒ \(\frac{60 x+12000}{x^2-10 x}\) = \(\frac {2}{1}\) ⇒ 60x + 1200 = 2x² – 20x
⇒ 2x² – 20x – 60x – 12000 = 0
⇒ 2x² – 80x – 12000 = 0 ⇒ x² – 40x – 6000 = 0
⇒ x² – 100x + 60x – 6000 = 0
⇒ x (x – 100) + 60 (x – 100) = 0
⇒ (x – 100) (x + 60) = 0
Either x – 100 = 0, then x = 100
or x + 60 = 0, then x = – 60 which is not possible being negative
∴ x = 100
∴ Number of articles = 100

Self Evaluation and Revision (LATEST ICSE QUESTIONS)

Question 1.
Solve for x and give your answer correct to 2 decimal places : x² – 10x + 6 = 0.
Solution:
x² – 10x + 6 = 0
Here a = 1, b = – 10, c = 6
∴ D = b² – 4ac = (- 10)² – 4 x 1 x 6 = 100 – 24 = 76
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-10) \pm \sqrt{76}}{2 \times 1}\)
= \(\frac{10 \pm \sqrt{4 \times 19}}{2}=\frac{10 \pm 2 \sqrt{19}}{2}\)
= 5 ± \(\sqrt{19}\)
= 5 ± 4.358
∴ x₁ = 5 + 4.358 = 9.358 = 9.36
x₂ = 5 – 4.358 = 0.642 = 0.64
∴ x = 9.36, 0.64

Question 2.
Solve using the quadratic formula
x² – 4 + 1 = 0
Solution:
x² – 4x + 1 = 0
Here a = 1, b = – 4, c = 1
D = b² – 4ac = (- 4)² – 4 x 1 x 1 = 16 – 4 = 12
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-4) \pm \sqrt{12}}{2 \times 1}\)
= \(\frac{4 \pm \sqrt{4 \times 3}}{2}=\frac{4 \pm 2 \sqrt{3}}{2}\)
= 2 ± \(\sqrt{3}\) (Dividing by 2)
= 2 ± 1.732
∴ x₁ = 2 + 1.732 = 3.732
x₂ = 2 – 1.732 = 0.268
∴ x = 3.732, 0.268

Question 3.
Solve the equation 3x² – x – 7 = 0 and give your answer correct to two decimal places.
Solution:
3x² – x – 7 = 0
Here a = 3, b = – 1, c = – 7
D = b² – 4ac = (-1)² – 4 x 3 x (-7)
= 1 + 84 = 85
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-1) \pm \sqrt{85}}{2 \times 3}\)
= \(\frac{1 \pm \sqrt{85}}{6}=\frac{1 \pm 9.22}{6}\)
x₁ = \(\frac{1+9.22}{6}=\frac{10.22}{6}\) = 1.70
x₂ = 1 – 9.22 = \(\frac { -8.22 }{ 6 }\) = – 1.37
∴ x = 1.70, – 1.37

Question 4.
Solve the following equation and give your answer up to two decimal places :
x² – 5x – 10 = 0
Solution:
x² – 5x – 10 = 0
Here a = 1, b = – 5, c = – 10
∴ D = b² – 4ac = (-5)² – 4 x 1 x (-10)
= 25 + 40 = 65
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-5) \pm \sqrt{65}}{2 \times 1}\)
= \(\frac{5 \pm \sqrt{65}}{2}=\frac{5 \pm 8.06}{2}\)
x₁ = \(\frac{5+8.06}{2}=\frac{13.06}{2}\) = 6.53
x₂ = \(\frac{5-8.06}{2}=\frac{-3.06}{2}\) = – 1.53
∴ x = 6.53, – 1.53

Question 5.
Solve the equation 2x – \(\frac { 1 }{ x }\) = 7. Write your answer correct to two decimal places.
Solution:
2x – \(\frac { 1 }{ x }\) = 7 ⇒ 2x² – 1 = 7x
⇒ 2x² – 7x – 1 = 0
Here a = 2, b = – 1, c = – 1
∴ D = b² – 4ac = (- 7)² – 4 x 2 x (- 1)
= 49 + 8 = 57
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-7) \pm \sqrt{57}}{2 \times 2}\)
= \(\frac{7 \pm \sqrt{57}}{4}=\frac{7 \pm 7.55}{4}\)
x₁ = \(\frac{7+7.55}{4}=\frac{14.55}{4}\) = 3.64
x₂ = \(\frac{7-7.55}{4}=\frac{-0.55}{4}\) = – 0.14
∴ x = 3.64, – 0.14

Question 6.
The bill for a number of people for overnight stay is ₹ 4800. If there were 4 more, the bill each person had to pay would have reduced by ₹ 200. Find the number of people staying overnight.
Solution:
Total amount of the bill = ₹ 4800
Let number of person originally = x
Then share of each person = ₹ \(\frac { 4800 }{ x+4 }\)
If there were 4 more persons, then number of persons = x + 4
Then share of each person = ₹ \(\frac { 4800 }{ x+4 }\)
According to the condition,
⇒ \(\frac{4800}{x}-\frac{4800}{x+4}\) = 200
⇒ \(\frac{4800 x+19200-4800 x}{x(x+4)}=\frac{200}{1}\)
⇒ \(\frac{19200}{x^2+4 x}=\frac{200}{1}\) ⇒ 200x² + 800x = 19200
⇒ x² + 4x = 96
⇒ x² + 4x – 96 = 0
⇒ x² + 12x – 8x – 96 = 0

⇒ x (x + 12) – 8 (x + 12) = 0
⇒ (x + 12)(x – 8) = 0
Either x + 12 = 0, then x = – 12 which is not possible being negative
or x – 8 = 0, then x = 8
∴ Number of persons = 8

Question 7.
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for :
(i) The onward journey
(ii) The return journey.
If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
Solution:
Distance travelled by an aeroplane = 400 km
Speed of aeroplane = x km/hr
(i) ∴ Time taken = hours
On return speed by increasing 40 km/hr,
The speed will be (x + 40) km/hr

(ii) ∴ Time taken = \(\frac { 400 }{ x+40 }\)

(iii) According to the condition,

Either x + 200 = 0, then x = – 200 but it is not possible being negative
or x – 160 = 0, then x = 160
∴ Speed = 160 km/hr

Question 8.
In an auditorium, seats were arranged in rows and columns. The number of rows were equal to the number of seats in each row. When the number of rows were doubled and the number of seats in each row were reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after rearrangement.
Solution:
In first case,
Let number of rows = x
Then number of columns = x
∴ Number of seats = x × x = x²
In second case,
Number of rows = 2x
and number of columns = x – 10
∴ Total number of seats = 2x (x – 10)
According to the condition,
2x (x – 10) – x² = 300
⇒ 2x² – 20x – x² = 300
⇒ x² – 20x – 300 = 0
⇒ x² – 20x – 300 = 0
⇒ x² – 30x + 10x – 300 = 0
⇒ x (x – 30) + 10 (x – 30) = 0
⇒ (x – 30) (x + 10) = 0
Either x – 30 = 0, then x = 30
or x + 10 = 0, then x = – 10 but it is not possible being negative
(i) ∴ Number of rows in the original arrangement = 30

(ii) Number of seats after re-arrangements = 2x(x – 10) = 2 x 30 (30 – 10)
= 60 x 20 = 1200

Question 9.
P and Q are centres of circles of radius 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm, which touches the above circles externally. Given that ∠PRQ = 90°, write an equation in x and solve it.
Solution:
P and Q are the centres of two circles with
radii 9 cm and 2 cm respectively
PQ = 17 cm
Let x be the radius of the circle with centre R

Which touches the given two circles at L
and M respectively, then
PR = PL + LR = (9 + x) cm
and QR = QM + MR = (2 + x) cm
∴ ∠BPQ = 90°
∴ In right angled ∆PRQ,
PQ² = PR² + QR²
⇒ (17)² = (9 + x)² + (2 + x)²
⇒ 289 = 81 + 18x + x² + 4 + 4x + x²
⇒ 289 = 2x² + 22x + 85
⇒ 2x² + 22x + 85 – 289 = 0
⇒ 2x² + 22x – 204 = 0
⇒ x² + 11x – 102 = 0 (Dividing by 2)
⇒ x² + 17x – 6x – 102 = 0
⇒ x (x + 17) – 6 (x + 7) = 0
⇒ (x + 17) (x – 6) = 0
Either x + 17 = 0, then x = – 17 which is not possible being negative
or x – 6 = 0, then x = 6

Question 10.
By increasing the speed of car by 10 km/ hr, the time of journey for a distance of 72 km is reduced by 36 min. Find the original speed of the car.
Solution:
Distance of journey = 72 km
Let original speed of car = x km/h
and increased speed of car = (x + 10) km/hr
Now time taken in first case = \(\frac { 72 }{ x }\) hrs.
and time taken in second case = \(\frac { 72 }{ x+10 }\) hrs
According to the condition,

Either x + 40 = 0, then x = – 40 which is not possible being negative
or x – 30 = 0, then x = 30
∴ Original speed of the car = 30 km/hr

Question 11.
A shopkeeper buys a certain number of books for ₹ 720. If the cost per book was ₹ 5 less, the number of books that could be bought for ₹ 720 would be 2 more. Taking the original cost of each book to be ₹ x, write an equation in x and solve it.
Solution:
Price of books = ₹ 720
Let cost of one book = ₹ x
∴ Number of books purchased = \(\frac { 720 }{ x }\)
In second case, the price of each book = (x – 5)
∴ Number of books = \(\frac { 720 }{ x – 5 }\)
According to the conditions,

Either x – 45 = 0, then x = 45
or x + 40 = 0, then x = -40 it is not possible being negative
∴ x = 45
∴ Number of books originally purchased = 45

Question 12.
Solve the following quadratic equation for x and give your answer correct to 2 decimal places : x² – 3x – 9 = 0
Solution:
x² – 3x – 9 = 0
Here a = 1, b = -3, c = -9
∴ D = b² – 4ac = (-3 )² – 4 x 1 x (- 9)
= 9 + 36 = 45
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-3) \pm \sqrt{45}}{2 \times 1}\)
= \(\frac{3 \pm \sqrt{9 \times 5}}{2}=\frac{3 \pm 3 \sqrt{5}}{2}\)
= \(\frac{3 \pm 3(2.236)}{2}\)
= \(\frac{3 \pm 6.708}{2}=\frac{3 \pm 6.71}{2}\)
∴ x₁ = \(\frac{3+6.71}{2}=\frac{9.71}{2}\) = 4.86
x₂ = \(\frac{3-6.71}{2}=\frac{-3.71}{2}\) = – 1.86
∴ x = 4.86, – 1.86

Question 13.
Five years ago, a woman’s age was the square of her son’s age. Ten years hence her age will be twice that of her son’s age. Find :
(i) The age of her son five years ago.
(ii) The present age of the woman.
Solution:
5 years ago,
Let the age of son = x years
Then age of his mother = x² years
Present age of son = (x + 5) years
and age of mother = (x² + 5) years
10 years hence,
Age of son = x + 5 + 10 = x + 15
and age of mother = x² + 5 + 10 = x² + 15
According to the condition,
x² + 15 = 2 (x + 15) ⇒ x² + 15 = 2x + 30
⇒ x² + 15 – 2x – 30 = 0 ⇒ x² – 2x – 15 = 0
⇒ x² – 5x + 3x – 15 = 0

⇒ x (x – 5) + 3 (x – 5) = 0
⇒ (x – 5) (x + 3) = 0
Either x – 5 = 0, then x = 5
or x + 3 = 0, then x = – 3
but it is not possible being negative
(i) ∴ Age of son 5 years ago = 5 years
(ii) Present age of woman = x² + 5 = (5)² + 5 = 25 + 5 = 30 years

Question 14.
Solve the following quadratic equation for x and give your answer correct to two decimal places :
5x (x + 2) = 3
Solution:
5x (x + 2) = 3
⇒ 5x² + 10x = 3
⇒ 5x² + 10x – 3 = 0
Here a = 5, b = 10, c = – 3
∴ D = b² – 4ac = (10)² – 4 x 5 x (- 3)
= 100 + 60 = 160
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-10 \pm \sqrt{160}}{2 \times 5}\)
= \(\frac{-10 \pm \sqrt{160}}{10}=\frac{-10 \pm 12.65}{10}\)
∴ x₁ = \(\frac{-10+12.65}{10}=\frac{2.65}{10}\) = 0.265 = 0.26
x₂ = \(\frac{-10-12.65}{10}=\frac{-22.65}{10}\) = – 2.265 = – 2.26

Question 15.
Some students planned a picnic. The budget for the food was ₹ 480. As eight of them failed to join the party the cost of the food for each member increased by ₹ 10. Find how many students went for the picnic?
Solution:
Budget for food = ₹ 480
Let number of students who went to picnic = x
∴ Each share = ₹ \(\frac { 480 }{ x }\)
Number of students who did not go = 8
Remaining students = x – 8
and then each shares = \(\frac { 480 }{ x – 8 }\)
Now according to the condition,

Either x – 24 = 0, then x = 24
or x + 16 = 0, then x = -16 which is not
possible being negative
∴ Number students who went for the picnic = 24

Question 16.
Solve the following quadratic equation and give the answer correct to two significant figures 4x² – 7x + 2 = 0.
Solution:
4x² – 7x + 2 = 0
x = \(\frac{7 \pm \sqrt{49-32}}{8}=\frac{7 \pm \sqrt{17}}{8}\)
x = \(\frac{7 \pm 4.12}{8}\)
Taking (+ ve) sign x = \(\frac { 11.12 }{ 8 }\) = 1.4
Taking (- ve) sign x = \(\frac { 2.88 }{ 8 }\) = 36

Question 17.
The speed of an express train is x km/h and the speed of an ordinary train is 12 km/h less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.
Solution:

Question 18.
Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal.
px² – 4x + 3 = 0.
Solution:
px² – 4x + 3 = 0 ….(i)
Compare (i) with ax² + bx + c = 0
Here a = p, b = – 4, c = 3
∴ D = b² – 4 ac
= (- 4)² – 4.p.(3)
= 16 – 12p
As roots are equal,
∴ D = 0
16 – 12p = 0
⇒ \(\frac { 16 }{ 12 }\) = p
⇒ p = \(\frac { 4 }{ 3 }\)

Question 19.
Solve the following equation :
x – \(\frac { 18 }{ x }\) = 6. Give your answer correct to two significant figures.
Solution:

Question 20.
₹ 480 is divided equally among ‘x’ children. If the number of children were 20 more then each would have got ₹ 12 less. Find ‘x’.
Solution:
Share of each child = ₹ \(\frac { 480 }{ x }\)
According to the question :
\(\frac { 480 }{ x + 20 }\) = \(\frac { 480 }{ x }\) – 12
⇒ \(\frac{480}{x+20}=\frac{480-12 x}{x}\)
⇒ \(\frac{480}{x+20}=\frac{12(40-x)}{x}\)
⇒ (x + 20) (40 – x) = 40x
⇒ 40x – x² + 800 – 20x – 40x
⇒ x² + 20x – 800 = 0
⇒ x² + 40x – 20x – 800 = 0
⇒ x (x + 40) – 20 (x + 40) = 0
⇒ (x + 40) (x – 20) = 0
⇒ x = – 40, x = 20
– ve value of x is not possible
⇒ No. of children = 20

Question 21.
Without solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.
x² + 2 (m – 1) x + (m + 5) = 0.
Solution:
Hence, a = 1, b = 2 (m – 1), c = m + 5
So, discriminant, D = b² – 4ac
= 4(m – 1 )² – 4 x 1 (m + 5)
= 4(m² + 1 – 2m) – 4(m + 5)
= 4m² + 4 – 8m – 4m – 20
= 4m² – 12m – 16
For real and equal roots D = 0
So, 4m² – 12m – 16 = 0
⇒ m² – 2m – 4 = 0 (Dividing by 4)
⇒ m² – Am + m – 4 = 0
⇒ m (m – 4) + 1 (m – 4) = 0
⇒ (m – 4) (m + 1) = 0
⇒ m = 4 or m = – 1

Question 22.
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/ h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
Let the original speed of the car be x km/h.
Time taken to cover 400 km = \(\frac { 400 }{ x }\)h … (i)
New speed = (x + 12) km/h
New time taken to cover 400 km = \(\frac { 400 }{ x+12 }\) h … (ii)
Time taken for journey would have been 1 hour 40 minutes less.
1 hour 40 minutes = 1 \(\frac { 40 }{ 60 }\) hours = \(\frac { 5 }{ 6 }\) hours
∴ From (i) and (ii), \(\frac { 400 }{ x }\) – \(\frac { 400 }{ x+12 }\) = \(\frac { 5 }{ 3 }\)
\(\frac{400(x+12)-400 x}{x(x+12)}=\frac{5}{3}\)
⇒ \(\frac{400(x+12-x)}{x^2+12 x}=\frac{5}{3}\)
1200x + 14400 – 1200x = 5x² + 60x
⇒ 14400 = 5x² + 60x
⇒ x² + 12x – 2880 = 0
⇒ x² + 60x – 48x – 2880 = 0
⇒ x(x + 60) – 48 (x + 60) = 0
⇒ (x + 60) (x – 48) = 0
⇒ x = 48 or x = – 60
⇒ x = 48 (Rejecting x = – 60, being speed)
Hence, original speed of the car = 48 km/h.

Question 23.
(i) Solve the following equation and calculate the answer correct to two decimal places: x² – 5x – 10 = 0
(ii) Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has real and equal roots x2 + (p – 3)x + p = 0.
Solution:
(i) x² – 5x – 10 = 0
Here, a = 1, b = – 5, c = – 10
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}\)
= \(\frac{5 \pm \sqrt{25+40}}{2}=\frac{5 \pm \sqrt{65}}{2}\)
= \(\frac{5 \pm 8.06}{2}=\frac{5+8.06}{2}, \frac{5-8.06}{2}\)
x = 6.53, – 1.53

(ii) x² + (p – 3)x + p = 0
∵ Equation has real and equal roots
∴ b² – 4ac = 0
(p – 3)² – 4(1)(p) = 0
(p – 3)² – 4p = 0
⇒ p² + 9 – 6p – 4p = 0
⇒ p² – 19p + 9 = 0
⇒ p² – 9p – 1p + 9 = 0
⇒ p(p – 9) – 1 (p – 9) = 0
⇒ (p – 1)(p – 9) = 0
⇒ p = 1, 9

Question 24.
A shopkeeper purchases a certain number of books for ₹ 960. If the cost per book was ₹ 8 less, the number of books that could be purchased for ₹ 960 would be 4 more. Write an equation, taking the original cost of each book to be ₹ x, and solve it to find the original cost of the books.
Solution:
Let original cost = ₹ x
No. of books bought = \(\frac { 960 }{ x }\)
New cost of books = ₹(x – 8)
∴ No. of books bought = \(\frac { 960 }{ x – 8 }\)
∴ According to condition,

Question 25.
Solve for x using the quadratic formula. Write your answer correct to two significant figures. (x – 1)² – 3x + 4 = 0.
Solution:
(x – 1)² – 3x + 4 = 0
x² + 1 – 2x – 3x + 4 = 0
x² – 5x + 5 = 0
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
Here, a = 1, b = – 5, c = 5
x = \(\frac{-(-5) \pm \sqrt{(-5)^2-4(1)(5)}}{2}\)
= \(\frac{5 \pm \sqrt{25-20}}{2}\)
= \(\frac{5 \pm \sqrt{5}}{2}=\frac{5+2.236}{2} \text { or } \frac{5-2.236}{2}\)
= \(\frac{7.236}{2} \text { or } \frac{2.764}{2}\) = 3.618, 1.382
∴ x = 3.62, 1.38

Question 26.
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
Solution:
Let 2-digit number = xy = 10x + y
Reversed digits = yx = 10y + x
Acc. to question xy = 6
y = \(\frac { 6 }{ x }\)
and 10x + y + 9 = 10y + x
10x + \(\frac { 6 }{ x }\) + 9 = 10 x \(\frac { 6 }{ x }\) + x
10x² + 6 + 9x = 60 + x²
10x² – x² + 9x + 6 – 60 = 0
9x² + 9x – 54 = 0 ⇒ x² + x – 6 = 0 ⇒ x² + 3x – 2x – 6 = 0
x(x + 3) – 2(x + 3) = 0
⇒ (x – 2) (x + 3) = 0
⇒ x = 2 or – 3 (rejecting -3)
putting the value of x in (i)
y = \(\frac { 6 }{ 2 }\) = 3
∴ 2-digit = 10x + y = 10 x 2 + 3 = 23

Question 27.
Find the value of ‘k’ for which x = 3 is a solution of the quadratic equation, (k + 2)x² – kx + 6 = 0.
Thus find the other root of the equation.
Solution:
(k + 2)x² – kx + 6 = 0 … (i)
Substitute x = 3 in equation (1)
(k + 2) (3)² – k(3) + 6 = 0
⇒ 9(k + 2) – 3k + 6 = 0
⇒ 9k + 18 – 3k + 6 = 0
⇒ 6k + 24 = 0
⇒ 6k = – 24
⇒ k = \(\frac { -24 }{ 6 }\) = 4
∴ k = – 4
Now, substituting = – 4 in equation (i), we get,
(- 4 + 2)x² – (- 4)x + 6 = 0
⇒ – 2x² + 4x + 6 = 0
⇒ x² – 2x – 3 = 0 (Dividingby2)
⇒ x² – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x + 1) (x – 3) = 0
So, the roots are x = -1 and x = 3
Thus, the other root of the equation is x = – 1

Question 28.
Sum of two natural numbers is 8 and the difference of their reciprocal is \(\frac { 2 }{ 15 }\). Find the numbers.
Solution:
Let x and y be two numbers Given that
x + y = 8 … (i)
and \(\frac { 1 }{ x }\) – \(\frac { 1 }{ y }\) = \(\frac { 2 }{ 15 }\) … (ii)
From equation (i), we have, y = 8 – x
Substituting the value of y in equation (ii), we have,
\(\frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}\)
⇒ \(\frac{8-x-x}{x(8-x)}=\frac{2}{15}\)
⇒ \(\frac{8-2 x}{x(8-x)}=\frac{2}{15}\)
⇒ \(\frac{4-x}{x(8-x)}=\frac{1}{15}\)
⇒ 15(4 – x) = x(8 – x)
⇒ 60 – 15x = 8x – x²
⇒ x² – 15x – 8x + 60 = 0
⇒ x² – 23x + 60 = 0
⇒ x² – 20x – 3x + 60 = 0
⇒ x(x – 20) – 3(x – 20) = 0
⇒ (x – 3) (x – 20) = 0
⇒ (x – 3) = 0 or (x – 20) = 0
⇒ x = 3 or x = 20
Since sum of two natural numbers is 8 : x cannot be equal to 20
Thus x = 3
From equation (1), y = 8 – x = 8 -3 = 5
Thus the values of x and y are 3 and 5 respectively.

Question 29.
Solve the quadratic equation x² – 3(x + 3) = 0; Give your answer correct two significant figures.
Solution:

Question 30.
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’.
Solution:
Time taken by the bus with moving at speed x. km/h = \(\frac { 240 }{ x }\)
Time taken by the bus with moving at speed (x – 10) km/h = \(\frac { 240 }{ x – 10 }\)
According to the given condition,
2 = \(\frac{240}{x-10}-\frac{240}{x}\)
⇒ 2 = 240\(\left(\frac{1}{x-10}-\frac{1}{x}\right)\)
⇒ \(\frac{1}{120}=\frac{1}{x-10}-\frac{1}{x}\)
⇒ \(\frac{1}{120}=\frac{x-x+10}{x(x-10)}\)
⇒ x(x – 10)= 10 x 120
⇒ x² – 10x = 1200
⇒ x² – 10x – 1200 = 0
⇒ x² – 40x + 30x – 1200 = 0
⇒ x(x – 40) + 30(x – 40) = 0
⇒ (x – 40) (x + 30) = 0
⇒ x – 40 = 0 or x + 30 = 0
⇒ x = 40 or x = – 30
Since, the speed cannot be negative, the uniform speed is 40 km/h.

Students can cross-reference their work with OP Malhotra Class 10 Solutions Chapter 5 Quadratic Equations Ex 5(d) to ensure accuracy.

S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(d)

Question 1.
Use the discriminant to determine the nature of the roots of each of the following quadratic equations :
(1) x² – 4x + 3 = 0
(ii) x² – 4x + 5 – 0
(iii) x² – 4x + 4 = 0
(iv) x² – x = 7
Solution:
(i) x² – 4x + 3 = 0
Here a= 1, b = – 4, c = 3
∴ D = b² – 4ac = (- 4)² – 4 x 1 x 3 = 16 – 12 = 4
∵ D > 0,
∴Roots are real and distinct and rational.

(ii) x² – 4x + 5 = 0
Here a = 1, b = – 4, c = 5
∴ D = b² – 4ac = (- 4)² – 4 x 1 x 5 = 16 – 20 = – 4
∵ D < 0
∴ Roots are not real or roots are imaginary.

(iii) x² – 4x + 4 = 0
Here a = 1, b = – 4, c = 4
∴ D = b² – 4ac = (- 4)² – 4 x 1 x 4
= 16 – 16 = 0
∵ D = 0
∴ Roots are real and equal

(iv) x² – x = 7 ⇒ x² – x – 7 = 0
Here a = 1, b = – 1, c = – 7
∴ D = b² – 4ac = (- 1)² – 4 x 1 x (- 7)
= 1 + 28 = 29
∵ D > 0
∴ Roots are real and distinct (irrational).

Question 2.
Without finding the roots, comment on the nature of the roots of the following quadratic equations.
(i) 3x² – 6x + 5 = 0
(ii) 5y2 + 12y – 9 = 0
(iii) x² – 5x – 7 = 0
(iv) a²x² + abx = b², a ≠ 0
(v) 9a²b²x² – 48abcdx + 64c²d² = 0, a ≠ 0, b ≠ 0
(vi) 4x² = 1
(vii) 64x² – 112x + 49 = 0
(viii) 8x² + 5x + 1 = 0
Solution:
(i) 3x² – 6x + 5 = 0
Here a = 3, b = – 6, c = 5 D = b² – 4ac = (- 6)² – 4 x 3 x 5
= 36 – 60 = – 24
∵ D < 0
∴ Roots are not real i.e. roots are imaginary.

(ii) 5y² + 12y – 9 = 0
Here a = 5, b = 12, c = – 9
∴ D = b² – 4ac = (12)² – 4 x 5 x (- 9)
= 144 + 180 = 224
∵ D > 0
∴ Roots are real, distinct and rational.

(iii) x² – 5x – 7 = 0
Here a = 1, b = – 5, c = – 7
∴ D = b² – 4ac = (- 5)² – 4 x 1 x (- 7)
= 25 + 28 = 53
∵ D > 0
∴ Roots are real, distinct and rational.

(iv) a²x² + abx = b²
⇒ a²x² + abx – b² = 0
Here A = a², B = ab. C = – b²
∴ D – b² – 4AC = (ab)² – 4 x a² (-b²)
= a²b² + 4a²b² = 5 a²b²
∵ D > 0
∴ Roots are real, distinct and rational.

(v) 9a²b²x² – 48abcdx + 64c²cd² = 0, a ≠ 0, b ≠ 0
Here A = 9a²b², B = – 48abed, C = 64c²d²
∴ D = b² – 4AC = (- 48abcd)² – 4 x 9a²b² x 64c²d²
= 2304a²b²c²d² – 2304a²b²c²d² = 0
∵ D = 0
∴ Roots are real equal.

(vi) 4x² = 1 ⇒ 4x² – 1 = 0
Here a = 4, b = 0, c = – 1
∴ D = b² – 4ac = (-112)² – 4 x 4 x (- 1)
= 0+ 16 = 16 = (4)²
∵ D > 0
∴ Roots are real, distinct.

(vii) 64x² – 112x + 49 = 0
Here a = 64, b = – 112, c = 49
∴D = b² – 4ac = (- 112)² – 4 x 64 x 49
= 12544 – 12544 = 0
∵ D = 0
∴ Roots are real and equal.

(viii) 8x² + 5x + 1 = 0
Here a = 8, b = 5, c = 1
∴ D = b² – 4ac = (5)² – 4 x 8 x 1
= 25 – 32 = – 7
∵ D < 0
∴ Roots are not real, but are imaginary.

Question 3.
Find the value of ‘k’ so that the equation has equal roots (or coincident roots)
(i) 4x² + kx + 9 = 0
(ii) kx² – 5x + k = 0
(iii) 9x² + 3kx + 4 = 0
(iv) x² + 7 (3 + 2k) – 2x (1 + 3k) = 0
(v) (k – 12) x² + 2 (k – 12) x + 2 = 0
Solution:
(i) 4x² + kx + 9 = 0 Here a = 4, b = k and c = 9
∴ D = b² – 4ac = a² – 4 x 4 x 9 = k² – 144
∵ Roots are equal
∴ D = 0
⇒ k² – 144 = 0
⇒ (k)² – (12)² = 0
⇒ (k + 12) (k – 12) = 0
Either k + 12 = 0, then k = – 12
or k – 12 = 0, then k = 12
∴ k = 12, – 12

(ii) kx² – 5x + k = 0
Here a = k, b = – 5, c = k
∴ D = b² – 4ac = (- 5)² – 4 x k x k
= 25 – 4k²
∵ Roots are equal
∴D = 0
⇒ 25 – 4k² = 0 ⇒ 4k² – 25 = 0
⇒ (2k)² – (5)² = 0
⇒ (2k + 5) (2k – 5) = 0
Either 2k + 5 = 0, then 2k = – 5 ⇒ k = \(\frac { – 5 }{ 2 }\)
or 2k – 5 = 0, then 2A = 5 ⇒ k = \(\frac { 5 }{ 2 }\)

(iii) 9x² + 3kx + 4 = 0
Here a = 9, b = 3k, c = 4
∴ D = a² – 4ac = (3k)² – 4 x 9 x 4
= 9k² – 144
∵ Roots are equal
∴D = 0 ⇒ 9k² – 144 = 0 ⇒ k² – 16 = 0 (Dividing by 9)
⇒ (k)² – (4)² = 0 ⇒ (k + 4) (k – 4) = 0
Either k + 4 = 0, then k = – 4 or k – 4 = 0, then k = 4
k = 4, – 4

(iv) x² + 7 (3 + 2k) – 2x (1 + 3k) = 0
⇒ x² – 2 (1 + 3k)x + 7 (3 + 2k) = 0
Here a = 1, b = – 2 (1 + 3k), c = 7 (3 + 2k)
∴ D = a² – 4ac
= [- 2 (1 + 3k)]² – 4 x 1 x 7 (3 + 2k)
= 4(1+ 9k² + 6k) – 28 (3 + 2k)
= 4 + 36k² + 24k – 84 – 56k
= 36k ² – 32k – 80
∵ Roots are equal
∴ D = 0
⇒ 36k² – 32k – 80 = 0
⇒ 9k² – 8k – 20 = 0 (Dividing by 4)
⇒ 9k² – 18k + 10k – 20 = 0
⇒ 9k(k – 2)+ 10(k – 2) = 0
⇒ (k – 2) (9k + 10) = 0
Either k – 2 = 0, then k = 2
or 9k+ 10 = 0, then 9k = – 10 ⇒ k = \(\frac { -10 }{ 9 }\)
∴ k = 2, \(\frac { -10 }{ 9 }\)

(v) (k – 12) x² + 2 (k – 12) x + 2 = 0
Here a = k – 12, A = 2 (A – 12), c = 2
∴ D = a² – 4ac = [2 (k – 12)]² – 4 x (k – 12) x 2
= 4(k – 12)² – 8 (k – 12)
∵ Roots are equal
∴ D = 0
⇒ 4 (k – 12)² – 8 (k – 12) = 0
⇒ (k – 12)² – 2 (k – 12) = 0 (Dividing by 4)
(k – 12) (k – 12 – 2) = 0
⇒ (k – 12) (k – 14) = 0
Either k – 12 = 0, then k = 12
or k – 14 = 0, then k = 14
∴ k = 12, 14

Question 4.
For what value of k, do the following quadratic equations have real roots.
(i) 2x² + 2x + k = 0
(ii) kx² – 2x + 2 = 0
(iii) 2x² – 10x + k = 0
(iv) kx² + 8x – 2 = 0
(v) 9x² – 24x + k = 0
(vi) x² – 4x + k = 0
Solution:
(i) 2x² + 2x + k = 0
Here a = 2, b = 2, c = k
∴ D = b² – 4ac = (2)² – 4 x 2 x k
= 4 – 8k
∵ Roots are real
∴ D > 0 ⇒ 4 – 8k ≥ 0
⇒ 1 – 2x ≥ 0 (Dividing by 4)
⇒ 1 ≥ 2x ⇒ 2x ≤ 1
⇒ k ≤ \(\frac { 1 }{ 2 }\)

(ii) kx² – 2x + 2 = 0
Here a = k, b = – 2, c = 2
∴ D = b² – 4ac = (- 2)² – 4 x k x 2
= 4 – 8k
∵ Roots are real
⇒ D ≥ 0 ⇒ 4 – 8k ≥ 0
⇒ 1 – 2x ≥ 0 (Dividing by 4)
⇒ 1 ≥ 2k ⇒ 2k ≤ 1
⇒ k ≤ \(\frac { 1 }{ 2 }\)

(iii) 2x² – 10x + k = 0
Here a = 2, b = – 10, c = k
∴ D = b² – 4ac = (-10)² – 4 x 2 x k
= 100 – 8k
∵ Roots are real
∴ D ≥ 0 ⇒ 100 – 8x ≥ 0
⇒ 25 – 2k ≥ 0 (Dividing by 4)
⇒ 25 ≥ 2x ⇒ 2k ≤ 25
⇒ k ≤ \(\frac { 25 }{ 2 }\)

(iv) kx² + 8x – 2 = 0
Here a = k, b = 8, c = – 2
∴ D = b² – 4ac
= (8)² – 4 x k x (- 2) = 64 + 8k
∵ Roots are real
∴ D ≥ 0 ⇒ 64 + 8k = 0
⇒ 8 + k ≥ 0
⇒ k ≥ – 8
∴ k ≥ – 8

(v) 9x² – 24x + k = 0
Here a = k, b = – 24, c = k
∴ D = b² – 4ac – (- 24)² – 4 x 9 x k
= 576 – 36k
∵ Roots are real
∴ D ≥ 0 ⇒ 576 – 36k ≥ 0
⇒ 16 – x ≥ 0 (Dividing by 36)
⇒ 16 ≥ k ⇒ x ≤ 16

(vi) x² – 4x + x = 0
Here a = 1, b = – 4, c = x
∴ D = b² – 4ac = (- 4)² – 4 x 1 x k
= 16 – 4k
∵ Roots are real
∴ D ≥ 0 ⇒ 16 – 4k ≥ 0
⇒ 4 – k ≥ 0 (Dividing by 4)
⇒ 4 ≥ k ⇒ k ≤ 4
∴ k ≤ 4

Question 5.
Find the value of x for which the given equation has equal roots. Also find the roots :
(i) 9x⁴ – 24x + k = 0
(ii) 2kx² – 40x + 25
Solution:
(i) 9x² – 24x + k = 0
Here a = 9, b = – 24, c = k
∴ D = b² – 4ac = (- 24)² – 4 x 9 x k
= 576 – 36x v
∵ Roots are equal
∴ D = 0
⇒ 576 – 36k = 0 ⇒ 36k = 576
⇒ k = \(\frac { 576 }{ 36 }\)
∴ x = 16
Now 9x² – 24 + 16 = 0
⇒ (3x)² – 2 x 3x × 4 + (4)² = 0
⇒ (3x – 4)² = 0 ⇒ 3x – 4 = 0
∴ 3x = 4 ⇒ x = \(\frac { 4 }{ 3 }\)
Hence x = \(\frac { 4 }{ 3 }\)

(ii) 2kx² – 40x + 25 = 0
Here a = 2k, b = – 40, c = 25
∴ D = b² – 4ac = (- 40)² – 4 x 2k x 25
= 1600 – 200x
∵ Roots are equal
∴ D = 0 ⇒ 1600 – 200k = 0
⇒ 200k = 1600 ⇒ k = \(\frac { 1600 }{ 200 }\) = 8
∴ k = 8
Now 2 x 8x² – 40x + 25 = 0
⇒ 16x² – 40x + 25 = 0
⇒ (4x)² – 2 x 4x × 5 + (5)² = 0
⇒ (4x – 5)² = 0
⇒ 4x = 5 ⇒ x = 5
⇒ 4x = 5 ⇒ x = \(\frac { 5 }{ 4 }\)
∴ k = 8, x = \(\frac { 5 }{ 4 }\)

Question 6.
Find the values of k for which the given quadratic equations has real and distinct roots :
(i) kx² + 2x + 1 = 0
(ii) kx² + 6x + 1 = 0
Solution:
(i) kx² + 2x + 1 = 0
Here a = k, b = 2, c = 1
∴ D = b² – 4ac = (2)² – 4 x k x 1
= 4 – 4k
∵ Roots are real and distinct
∴ D > 0 ⇒ 4 – 4k > 0
⇒ 4 > 4k ⇒ 1 > k (Dividing by 4)
⇒ x < 1
∴ k < 1

(ii) kx² + 6x + 1 = 0
Here a = k, b = 6, c – 1
∴ D = b² – 4ac = (6)² – 4 x k < 1 = 36 – 4k ∵ Roots are real and distinct ∴ D > 0 ⇒ 36 – 4x > 0
⇒ 36 > 4k ⇒ 9 > k (Dividing by 4)
⇒ k < 9
∴ k < 9

Question 7.
If – 5 is a roots of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:
∵ – 5 is a root of 2x² + px – 15 = 0
∴ – 5 will satisfy it
∴ 2 (- 5)² + p (- 5) – 15 = 0
⇒ 2 x 25 – 5p – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ 5p = 35 ⇒ p = \(\frac { 35 }{ 5 }\) = 7
Now ∵ (x² + x) + x = 0 has equal roots
⇒ px² + px + x = 0
⇒ 7x² + 7x + x = 6 (∵p = 7)
Here a = 7, b = 7, c = k
∴ D = b² – 4ac = (7)² – 4 x k = 49 – 28 k
∵ Roots are equal
∴ D = 0 ⇒ 49 – 28k = 0
⇒ 28x = 49 ⇒ k = \(\frac { 49 }{ 28 }\) = \(\frac { 7 }{ 4 }\)
∴ k = \(\frac { 7 }{ 4 }\)

Question 8.
If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c.
Solution:
∵ (b – c) x² + (c – a) x + (a – b) = 0 has equal roots
∴ D = 0
Now here A = b – c, B = c – a and C = a – b
D = B² – 4AC = (c – a)² – 4x (b – c) (a – b)
∴ D = 0
∴ (c – a)² – 4 (b – c) (a – b) = 0
⇒ c² + a² – 2ac – 4 (ab – b² – ca + bc) = 0
⇒ c² + a² – 2ac – 4ab + 4b² + 4ca – 4bc = 0
a² + 4b² + c² – 4ab – 4bc + 2ca = 0
⇒ (a)² + (2b)² + (c)² – 2 x a x 2b – 2 x 2b x c + 2c x a = 0
⇒ (a – 2b + c)² = 0 ⇒ a – 2b + c = 0
⇒ a + c = 2b ⇒ 2b = a + c
Hence proved.

Access to comprehensive OP Malhotra Class 10 Solutions Chapter 5 Quadratic Equations Ex 5(c) encourages independent learning.

S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(c)

Using quadratic formula, find the roots of the following equations:

Question 1.
2x² + x – 3 = 0
Solution:
2x² + x – 3 = 0
Here a = 2, b = 1, c = – 3
∴ D = b² 4ac = (1)² – 4 x 2 x (- 3) = 1 + 24 = 25
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{25}}{2 \times 2}\)
= \(\frac{-1 \pm 5}{4}\)
∴ x₁ = \(\frac{-1+5}{4}=\frac{4}{4}\) = 1
x₂ = \(\frac{-1-5}{4}=\frac{-6}{4}=\frac{-3}{2}\)
∴ x = 1, \(\frac { -3 }{ 2 }\)

Question 2.
6x² + 7x – 20 = 0
Solution:
6x² + 7x – 20 = 0
Here a = 6, b = 1, c = – 20
∴ D = b² – 4ac = (7)² – 4 x 6 x (- 20)
= 49 + 480 = 529
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-7 \pm \sqrt{529}}{2 \times 6}\)
= \(\frac{-7 \pm 23}{12}\)
∴ x₁ = \(\frac { -7+23 }{ 12 }\) = \(\frac { 16 }{ 12 }\) = \(\frac { 4 }{ 3 }\)
x₂ = \(\frac { -7-23 }{ 12 }\) = \(\frac { -30 }{ 12 }\) = \(\frac { -5 }{ 2 }\)
x = \(\frac { 4 }{ 3 }\), \(\frac { -5 }{ 2 }\) or 1\(\frac { 1 }{ 3 }\), – 2\(\frac { 1 }{ 2 }\)

Question 3.
9x² + 6x = 35
Solution:
9x² + 6x = 35
⇒ 9x² + 6x – 35 = 0
Here a = 9, b = 6, c = – 35
D = b² – 4ac = (6)² – 4 x 9 x (- 35)
= 36 + 1260 = 1296
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-7 \pm \sqrt{529}}{2 \times 6}\)
= \(\frac{-6 \pm 36}{18}\)
∴ x₁ = \(\frac{-6+36}{18}=\frac{30}{18}=\frac{5}{3}=1 \frac{2}{3}\)
x₂ = \(\frac{-6-36}{18}=\frac{-42}{18}=\frac{-7}{3}=-2 \frac{1}{3}\)
∴ x = \(1 \frac{2}{3},-2 \frac{1}{3}\)

Question 4.
3x² + 7x – 6 = 0
Solution:
3x² + 7x – 6 = 0
Here a = 3, b = 7, c = – 6
∴ D = b² – 4ac = (7)² – 4 x 3 x (- 6) = 49 + 72 = 121
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-7 \pm \sqrt{121}}{2 \times 3}\)
= \(\frac{-7 \pm 11}{6}\)
∴ x₁ = \(\frac{-7+11}{6}=\frac{4}{6}=\frac{2}{3}\)
x₂ = \(\frac{-7-11}{6}=\frac{-18}{6}\)
∴ x = \(\frac{2}{3}\), – 3

Question 5.
x² – 66x + 189 = 0
Solution:
x² – 66x + 189 = 0
Here a = 1, b = – 66, c = 189
∴ D = b² – 4ac = (- 6)² – 4 x 9 x (- 35) = 36 + 1260 = 1296
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-66) \pm \sqrt{3600}}{2 \times 1}\)
= \(\frac{66 \pm 60}{2}\)
∴ x₁ = \(\frac{66+60}{2}=\frac{126}{2}\) = 63
x₂ = \(\frac{66-60}{2}=\frac{6}{2}\) = 3
∴ x = 63, 3

Question 6.
\(\sqrt{3}\)x² + 11x + 6\(\sqrt{3}\) = 0
Solution:
\(\sqrt{3}\)x² + 11x + 6\(\sqrt{3}\) = 0
Here a = \(\sqrt{3}\), b = 11, c = 6\(\sqrt{3}\)
∴ D = b² – 4ac = (11)² – 4 x \(\sqrt{3}\) x 6\(\sqrt{3}\)
= 121 – 72 = 49
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-11 \pm \sqrt{49}}{2 \times \sqrt{3}}\)
= \(\frac{-11 \pm 7}{2 \sqrt{3}}\)
∴ x₁ = \(\frac{-11+7}{2 \sqrt{3}}=\frac{-4}{2 \sqrt{3}}=\frac{-2}{\sqrt{3}}=\frac{-2 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{-11-7}{2 \sqrt{3}}=\frac{-18}{2 \sqrt{3}}=\frac{-18 \times \sqrt{3}}{2 \sqrt{3} \times \sqrt{3}}\)
= \(\frac{-18 \sqrt{3}}{6}\) = – 3\(\sqrt{3}\)
∴ x = \(\frac{-2 \sqrt{3}}{3}\), 3\(\sqrt{m}\)

Question 7.
36x² + 23 = 60x
Solution:
36x² + 23 = 60x ⇒ 36x² – 60x + 23 = 0
Here a = 36, b = – 60, c = 23
∴D = b² – 4ac = (- 60)² – 4 x 36 x 23
= 3600 – 3312 = 288
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-60) \pm \sqrt{288}}{2 \times 36}\)
= \(\frac{60 \pm \sqrt{2 \times 144}}{72}=\frac{60 \pm 12 \sqrt{2}}{72}=\frac{5 \pm \sqrt{2}}{6}\)
∴ x = \(\frac{5+\sqrt{2}}{6}, \frac{5-\sqrt{2}}{6}\)

Question 8.
x² – 2x + 5 = 0
Solution:
x² – 2x + 5 = 0
Here a = 1, b = – 2, c = 5
∴ D = b² – 4ac
= (- 2)² – 4 x 1 x 5
= 4 – 20 = – 16
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-2) \pm \sqrt{-16}}{2 \times 1}\)
= \(\frac{2 \pm \sqrt{16 \times(-1)}}{2}=\frac{2 \pm 4 \sqrt{-1}}{2}\)
= 1 ± 2\(\sqrt{-1}\)
∴ x = 1 ± 2\(\sqrt{-1}\)
(But these are not real roots as \(\sqrt{-1}\) is not a real number)

Question 9.
3x² – 17x + 25 = 0
Solution:
3x² – 17x + 25 = 0
Here a = 3, b = – 17, c = 25
∴ D = b² – 4ac = (-17)² – 4 x 3 x 25
= 289 – 300 = – 11
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-17) \pm \sqrt{-11}}{2 \times 3}\)
= \(\frac{17 \pm \sqrt{-11}}{6}\)
∴ x = \(\frac{17+\sqrt{-11}}{6}, \frac{17-\sqrt{-11}}{6}\)
But there are not real roots

Question 10.
15x² – 28 = x
Solution:
15x² – 28 = x ⇒ 15x² – x – 28 = 0
Here a = 15, b = – 1, c = – 28
∴ D = b² – 4ac = (- 1)² – 4 x 15 x (- 28)
= 1 + 1680 = 1681
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-1) \pm \sqrt{1681}}{2 \times 15}\)
= \(\frac{1 \pm 41}{30}\)
x₁ = \(\frac{1+41}{30}=\frac{42}{30}=\frac{7}{5}=1 \frac{2}{5}\)
x₂ = \(\frac{1-41}{30}=\frac{-40}{3 C}=\frac{-4}{3}=-1 \frac{1}{3}\)
∴ x = 1\(\frac { 2 }{ 5 }\), – 1\(\frac { 1 }{ 3 }\)

Question 11.
x² + 3x – 3 = 0, giving your answer to two decimal places.
Solution:
x² + 3x – 3 = 0
Here a = 1, b = 3, c = – 3
∴ D = b² – 4ac = (3)² – 4 x 1 x (- 3)
= 9 + 12 = 21
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-3 \pm \sqrt{21}}{2 \times 1}\)
= \(\frac{-3 \pm 4.58}{2}\)
∴ x₁ = \(\frac{-3+4.58}{2}=\frac{1.58}{2}\) = 0.79
x₂ = \(\frac{-3-4.58}{2}=\frac{-7.58}{2}\) = – 3.79
∴ x = 0.79, – 3.79

Question 12.
\(\frac { 2 }{ 3 }\)x = \(\frac { -1 }{ 6 }\)x² – \(\frac { 1 }{ 3 }\), giving your answer to 2 d.p.
Solution:
\(\frac { 2 }{ 3 }\)x = \(\frac { -1 }{ 6 }\)x² – \(\frac { 1 }{ 3 }\)
⇒ \(\frac { 1 }{ 6 }\) x² + \(\frac { 2 }{ 3 }\)x + \(\frac { 1 }{ 3 }\) = 0
⇒ x² + 4x + 2 = 0
(Multiplying by L.C.M. of 6, 3, 3 = 6)
Here a = 1, b = 4, c = 2
∴ D = b² – 4ac = (4)² – 4 x 1 x 2 = 16 – 8 = 8
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-4 \pm \sqrt{8}}{2 \times 1}\)
= \(\frac{-4 \pm 2 \sqrt{2}}{2}\)
= – 2 ± \(\sqrt{2}\)
= – 2 ± 1.41
∴ x₁ = – 2 + 1.41 = – 0.59
x₂ = – 2 – 1.41 = – 3.41
∴ x = – 0.59, – 3.41

Question 13.
x² + 6x – 10 = 0
Solution:
x² + 6x – 10 = 0
Here a = 1, b = 6, c = – 10
D = b² – 4ac = (6)² – 4 x 1 x (- 10)
= 36 + 40 = 76
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-6 \pm \sqrt{76}}{2 \times 1}\)
= \(\frac{-6 \pm \sqrt{4 \times 19}}{2 \times 1}\)
= \(\frac{-6 \pm 2 \sqrt{19}}{2}\)
= – 3 ± \(\sqrt{-19}\)
∴ x = – 3 + \(\sqrt{19}\), – 3 – \(\sqrt{19}\)

Question 14.
\(\frac{x^2+8}{11}\) = 5x – x² – 5
Solution:
\(\frac{x^2+8}{11}\) = 5x – x² – 5
⇒ x² + 8 = 55x – 11x² – 55
⇒ x² + 8 – 55x + 11x² + 55 = 0
⇒ 12x² – 55x + 63 = 0
Here a = 12, b = – 55, c = 63
∴D = b² – 4ac = (- 55)² – 4 x 12 x 63
= 3025 – 3024 = 1
∴ x₁ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-55) \pm \sqrt{1}}{2 \times 12}\)
= \(\frac{55 \pm 1}{24}\)
∴ x₁ = \(\frac{55+1}{24}=\frac{56}{24}=\frac{7}{3}=2 \frac{1}{3}\)
x₂ = \(\frac{55-1}{24}=\frac{54}{24}=\frac{9}{4}=2 \frac{1}{4}\)
∴ x = 2\(\frac { 1 }{ 3 }\), 2\(\frac { 1 }{ 4 }\)

Question 15.
y – \(\frac { 3 }{ y }\) = \(\frac { 1 }{ 2 }\)
Solution:
y – \(\frac { 3 }{ y }\) = \(\frac { 1 }{ 2 }\) ⇒ 2y² – 6 = y
⇒ 2y² – y – 6 = 0
Here a = 2, b = – 1, c = – 6
∴D = b² – 4ac = (- 1)² – 4 x 2 x (- 6) = 1 + 48 = 49
∴ y₁ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(1) \pm \sqrt{49}}{2 \times 2}=\frac{1 \pm 7}{4}\)
y₁ = \(\frac{1+7}{4}=\frac{8}{4}\) = 2
∴ y₁ = \(\frac{1-7}{4}=\frac{-6}{4}=\frac{-3}{2}=-1 \frac{1}{2}\)
∴ y = 2, – 1\(\frac { 1 }{ 2 }\)

Question 16.
2x + \(\frac { 4 }{ x }\) = 9
Solution:
2x + \(\frac { 4 }{ x }\) = 9 ⇒ 2x² + 4 = 9 ⇒ 2x² + 4 = 0
⇒ 2x² – 9x + 4 = 0
Here a = 2, b = – 9, c = 4
∴ D = b² – 4ac = (- 9)² – 4 x 2 x 4
= 81 – 32 = 49
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-9) \pm \sqrt{49}}{2 \times 2}\)
= \(\frac{9 \pm 7}{4}\)
∴ x₁ = \(\frac{9+7}{4}=\frac{16}{4}\) = 4
x₂ = \(\frac{9-7}{4}=\frac{2}{4}=\frac{1}{2}\)
∴ x = 4, \(\frac { 1 }{ 2 }\)

Question 17.
\(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\), x ≠ 0, x ≠ 1
Solution:

Question 18.
\(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}\)
Solution:
\(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}\)
⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)
⇒ \(\frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}\)
⇒ \(\frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}\)
⇒ 25x² – 175x + 300 – 12x² – 57x + 60 = 0
(By cross multiplication)
⇒ 25x² – 175x + 300 – 12x² + 57x – 60 = 0
⇒ 13x² – 118x + 240 = 0
Here a = 13, b = – 118, c = 240
D = b² – 4ac = (- 118)² – 4 x 13 x 240
= 13924 – 12480 = 1444
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-118) \pm \sqrt{1444}}{2 \times 13}\)
= \(\frac{118 \pm 38}{26}\)
∴ x₁ = \(\frac{118+38}{26}=\frac{156}{26}\) = 6
x₂ = \(\frac{118-38}{26}=\frac{80}{26}=\frac{40}{13}=3 \frac{1}{13}\)
∴ x = 6, 3\(\frac { 1 }{ 13 }\)

Question 19.
\(\frac{x+6}{x+7}-\frac{x+1}{x+2}=\frac{1}{3 x+1}\)
Solution:

Question 20.
\(\frac{x+1}{2 x+5}=\frac{x+3}{3 x+4}\)
Solution:
\(\frac{x+1}{2 x+5}=\frac{x+3}{3 x+4}\)
⇒ (x + 1) (3x + 4) = (x + 3) (2x + 5)
(By cross multiplication)
⇒ 3x² + 4x + 3x + 4 = 2x² + 5x + 6x + 15
⇒ 3x² + 7x + 4 – 2x² – 11 x – 15 = 0
⇒ x² – 4x – 11 = 0
Here a = 1, b = – 4, c = – 11
∴ D = b² – 4ac = (- 4)² – 4 x 1 x (- 11)
= 16 + 44 = 60
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-(-4) \pm \sqrt{60}}{2 \times 1}\)
= \(\frac{4 \pm \sqrt{4 \times 15}}{2}=\frac{4 \pm 2 \sqrt{15}}{2}\)
= 2 ± \(\sqrt{15}\)
∴ x = 2 + \(\sqrt{15}\), 2 – \(\sqrt{15}\)

Question 21.
Solve using the quadratic formula :
(i) a²x² – 3abx + 2b² = 0
(ii) x² – x – a(a + 1) = 0
(iii) 10x² + 3bx + a² – 7ax – b² = 0
Solution:
(i) a²x² – 3abx + 2b2 = 0
Here A = a², B = – 3ab, C = 2b²
∴ D = B² – 4AC = (- 3ab)² – 4 x a² x 2b²
= 9a²b² – 8a²b² = a²b²
∴ x = \(\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)
= \(\frac{-(-3 a b) \pm \sqrt{a^2 b^2}}{2 a^2}=\frac{3 a b \pm a b}{2 a^2}\)
x₁ = \(\frac{3 a b+a b}{2 a^2}=\frac{4 a b}{2 a^2}=2 \frac{b}{a}\)
x₂ = \(\frac{3 a b-a b}{2 a^2}=\frac{2 a b}{2 a^2}=\frac{b}{a}\)
Hence x = \(\frac { 2b }{ a }\), \(\frac { b }{ a }\)

(ii) x² – x – a(a + 1) = 0
Here A = 1, B = – 1, C = – a (a + 1)
D = B² – 4AC = (- 1)² – 4 x 1 x (- a² – a)
= 1 + 4a² + 4 a = (2a + 1 )²
∴ x = \(\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)
= \(\frac{-(-1) \pm \sqrt{(2 a+1)^2}}{2 \times 1}=\frac{1 \pm(2 a+1)}{2}\)
∴ x₁ = \(\frac{1+2 a+1}{2}=\frac{2 a+2}{2}\) = a + 1
x₂\(\frac{1-2 a-1}{2}=\frac{-2 a}{2}\) = – a
∴ x₁ = – a, a + 1

(iii) 10x² + 3bx + a² – 7ax – b² = 0
⇒ 10x² + (3b – 7a) x + (a² – b²) = 0
Here A = 10, B = 3b – 7a, C = a² – b²
∴ D = B² – 4AC = (3b – 7a)² – 4 x 10 x (a² – b²)
= 9b² + 49a² – 42ab – 40 (a² – b²)
= 9b² + 49a² – 42ab – 40a² + 40b²
= 9a² + 49b² – 42ab = (3a)² + (7b)² – 2 x 3 a x 7b = (3a – 7b)²

Students appreciate clear and concise OP Malhotra Class 10 Solutions 5 Quadratic Equations Ex 5(b) that guide them through exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(b)

Solve the following equations by reducing them to quadratic equations :

Question 1.
x⁴ + 5x² – 36 = 0
Solution:
x⁴ + 5x² – 36 = 0
Let x² = y, then
(x²)² + 5(x²) – 36 = 0
⇒ y² + 5y – 36 = 0
⇒ y² + 9y – 4y – 36 = 0

⇒ y (y + 9) – 4 (y + 9) = 0
⇒ (y + 9)(y – 4) = 0
Either y + 9 = 0, then y = – 9
or y – 4 = 0, then y = 4

(i) If y = – 9, then x² = – 9 x = ±\(\sqrt{-9}\)
which is not real solution
or y = 4, then x² = 4
⇒ x² – 4 = 0
⇒ (x)² – (2)² = 0
⇒ (x + 2) (x – 2) = 0
Either x + 2 = 0, then x = – 2
or x – 2 = 0. then x = 2
∴ x = 2, – 2

Question 2.
x⁴ – 25x² + 144 = 0
Solution:
x⁴ – 25x² + 144 = 0
⇒ (x²)² – 25 (x)² + 144 = 0
Let x² = y, then
y² – 25y + 144 = 0
⇒ y² – 16y – 9y + 144 = 0

⇒ y (y – 16) – 9 (y – 16) = 0
⇒ (y – 16) (y – 9) = 0
Either y – 16 = 0, then y = 16
or y – 9 = 0, then y = 9
If y = 16, then x² = 16 ⇒ x² – 16 = 0
⇒ x² – (4)² = 0
⇒ (x + 4) (x – 4) = 0
Either x + 4 = 0, then x = – 4
or x – 4 = 0, then x = 4
If y = 9, then x² = 9 ⇒ x² – 9 = 0
⇒ x² – (3)² = 0
⇒ (x + 3) (x – 3) = 0
Either x + 3 = 0, then x = – 3
or x – 3 = 0, then x = 3
∴ x = 4, – 4, 3, – 3

Question 3.
(x² + x)² – (x² + x) – 2 = 0
Solution:
(x² + x)² – (x² + x) – 2 = 0
Let x² + x = y, then y² – y – 2 = 0
⇒ y² – 2y + y – 2 = 0

⇒ y(y – 2) + 1 (y – 2) = 0
⇒ (y – 2) (y + 1) = 0
Either y – 2 = 0, then y = 2
or y + 1 = 0, then y = – 1
If y = 2, then
x² + x = 2 ⇒ x² + x – 2 = 0
⇒ x² + 2x – x – 2 = 0
⇒ x (x + 2) – 1 (x + 2) = 0
⇒ (x + 2) (x – 1) = 0
Either x + 2 = 0, then x = – 2
or x – 1 = 0, then x = 1
If x = – 1, then
x² + x = – 1
⇒ x² + x + 1 = 0
But it has no real solution
∴ x = – 2, 1

Question 4.
\(\left[\frac{x-2}{x+2}\right]^2-4\left[\frac{x-2}{x+2}\right]\) + 3 = 0, x ≠ 2
Solution:
\(\left[\frac{x-2}{x+2}\right]^2-4\left[\frac{x-2}{x+2}\right]\) = 0
Let \(\frac{x-2}{x+2}\) = y, then
y² – 4y + 3 = 0
⇒ y² – y – 3y + 3 = 0

⇒ y(y – 1) – 3(y – 1) = 0
⇒ (y -1)(y – 3) = 0
Either y – 1 = 0, then y = 1
or y – 3 = 0, then y = 3
If y = 1, then
\(\frac{x-2}{x+2}\) = 1 ⇒ x – 2 = x + 2 = 0 = 4
Which is not possible
If y = 3, then
\(\frac{x-2}{x+2}\) = \(\frac { 3 }{ 1 }\) ⇒ 3x + 6 = x – 2
⇒ 3x – x = – 2 – 6 ⇒ 2x = – 8
⇒ x = – 4
∴ x = – 4

Question 5.
4\(\left(\frac{7 x-1}{x}\right)^2-8\left(\frac{7 x-1}{x}\right)\) + 3 = 0
Solution:
Let 4\(\left(\frac{7 x-1}{x}\right)^2-8\left(\frac{7 x-1}{x}\right)\) + 3 = 0, then
Let \(\frac { 7x -1 }{ x }\) = y, then
4y² – 8y + 3 = 0
⇒ 4y² – 6y – 2y + 3 = 0

⇒ 2y(2y – 3) – 1 (2y – 3) = 0
⇒ (2y – 3) (2y – 1) = 0
Either 2y – 3 = 0, then 2y = 3 ⇒ y = \(\frac { 3 }{ 2 }\)
or 2y – 1 = 0, then 2y = 1 ⇒ y = \(\frac { 1 }{ 2 }\)
If y = \(\frac { 3 }{ 2 }\) , then
\(\frac{7 x-1}{x}=\frac{3}{2}\)
⇒ 14x – 2 = 3x ⇒ 14x – 3x = 2
⇒ 11x = 2 ⇒ x = \(\frac { 2 }{ 11 }\)
y = \(\frac { 1 }{ 2 }\), then
\(\frac{7 x-1}{x}=\frac{1}{2}\) ⇒ 14x – 2 = x
⇒ 14x – x = 2 ⇒ 13x = 2
⇒ x = \(\frac { 2 }{ 13 }\)
∴ x = \(\frac { 2 }{ 11 }\), \(\frac { 2 }{ 13 }\)

Question 6.
4^x – 5.2^x + 4 = 0
Solution:
4^x – 5.2^x + 4 = 0
⇒ [(2)²]^x – 5.(2)^x + 4 = 0
⇒ (2^x)² – 5.2^x + 4 = 0
Let 2^x = y, then
y² – 5y + 4 = 0
⇒ y² – y – 4y + 4 = 0

⇒ y(y – 1) – 4(y – 1) = 0
⇒ (y – 1) (y – 4) = 0
Either y – 1 = 0, then y = 1
or y – 4 = 0, then y = 4
If y = 1, then
2^x = 1 ⇒ 2x = 2° (∵ 2° = 1)
Comparing,
∴ x = 0
If y = 4, then
2^x = 4 ⇒ 2^x = 2²
Comparing,
∴ x = 2
∴ x = 0, 2

Question 7.
16.4^x+2 – 16.2^x+1 + 1 = 0
Solution:
16.4^x+2 – 16.2^x+1 + 1 = 0
⇒ 16.(2²)^x+2 – 16(2)^x+1 + 1 = 0
⇒ 16.2^x+2– 16.2^x+1+ 1 = 0
⇒ 16.2^x.2⁴ – 16.2^x.2¹ + 1 = 0
⇒ 16 x 16.2^2x – 16 x 2.2^x + 1 = 0
⇒ 256.2^2x – 32.2^x + 1 = 0
Let 2^x = y, then
256.y² – 32y + 1 = 0
256y² – 16y – 16y + 1 = 0

16y(16y – 1) – 1 (16y – 1) = 0
⇒ (16y – 1)(16y – 1) = 0
Either 16y – 1 = 0, then 16y = 1 ⇒ y = \(\frac { 1 }{ 16 }\)
or 16y – 1 = 0, then 16y = 1 ⇒ y = \(\frac { 1 }{ 16 }\)
When y = \(\frac { 1 }{ 16 }\), then
2^x = \(\frac { 1 }{ 16 }\) 2^x = \(\frac{1}{(2)^4} \Rightarrow 2^x=2^{-4}\)
Comparing
x = – 4

Question 8.
3^4x+1 – 2 x 3^2x+1 – 81 = 0
Solution:
3^4x+1 – 2 x 3^2x+1 – 81 = 0
3^4x.3¹ – 2 x 3^2x x 3² – 81 = 0
⇒ (3^2x)² x 3 – 2 x 9 x 3^2x – 81 = 0
⇒ 3 (32x)² – 18 (3^2x) – 81 = 0
Let 3^2x = y, then
3y² – 18y -81 = 0 ⇒ y² – 6y – 27 = 0
⇒ y² – 9y + 3y – 27 = 0

⇒ y(y – 9) + 3(y – 9) = 0
⇒ (y – 9) (y + 3) = 0
Either y – 9 = 0, then y = 9
or y + 3 = 0, then y = – 3
If y = 9, then
3^2x = 9 = 3²
Comparing
2x – 2 ⇒ x = 1
If y = – 3, then
3^3x = – 3 which has no real values
∴ x = 1

Question 9.
\(\left(\frac{2 x-3}{x-1}\right)-4\left(\frac{x-1}{2 x-3}\right)\) = 3 where x ≠ 1 and x ≠ \(\frac { 3 }{ 2}\)
Solution:
Let \(\frac{2 x-3}{x-1}\) = y, then \(\frac{x-1}{2 x-3}=\frac{1}{y}\)
∴ y – \(\frac { 4 }{ y }\) = 3
⇒ y² – 4 = 3y ⇒ y² – 3y – 4 = 0
⇒ y² – 4y + y – 4 = 0

⇒ y(y – 4) + 1 (y – 4) = 0
⇒ (y – 4)(y + 1) = 0
Either y – 4 = 0, then y = 4
or y + 1 = 0, then y = – 1
If y = 4, then
\(\frac { 2x-3 }{ x-1 }\) = 4 ⇒ 4x – 4 = 2x – 3
⇒ 4x – 2x = – 3 + 4 ⇒ 2x = 1
⇒ x = \(\frac { 1 }{ 2 }\)
If y = – 1, then
\(\frac { 2x-3 }{ x-1 }\) = – 1 ⇒ – x + 1 = 2x – 3
⇒ 2x + x = 1 + 3 ⇒ 3x = 4
⇒ x = \(\frac { 4 }{ 3 }\)
∴ x = \(\frac { 1 }{ 2 }\), \(\frac { 4 }{ 3 }\)

Question 10.
\(\left(x+\frac{1}{x}\right)^2=4+\frac{3}{2}\left(x-\frac{1}{x}\right)\)
Solution:
Let x – \(\frac { 1 }{ x }\) = y, then
\(\left(x+\frac{1}{x}\right)^2\) = x² + \(\frac { 1 }{ x² }\) + 2 = x² + \(\frac { 1 }{ x² }\) – 2 + 4
= (x – \(\frac { 1 }{ x }\))² + 4 = y² + 4
∴ \(\left(x+\frac{1}{x}\right)^2=4+\frac{3}{2}\left(x-\frac{1}{x}\right)\)
⇒ y² + 4 = 4 + \(\frac { 3 }{ 2 }\)y
⇒ y² – \(\frac { 3 }{ 2 }\)y + 4 – 4 = 0 ⇒ y² – \(\frac { 3 }{ 2 }\)y = 0
⇒ y(y – \(\frac { 3 }{ 2 }\)) = 0
or y – \(\frac { 3 }{ 2 }\) = 0, then y = \(\frac { 3 }{ 2 }\)
∴ y = 0, \(\frac { 3 }{ 2 }\)
If y = 0, then
x – \(\frac { 1 }{ x }\) = 0
⇒ x² – 1 = 0 ⇒ (x + 1) (x – 1) = 0
Either x + 1 = 0, then x = – 1
or x – 1 = 0, then x = 1
If x = \(\frac { 3 }{ 2 }\), then
x – \(\frac { 1 }{ x }\) = \(\frac { 3 }{ 2 }\) ⇒ 2x² – 2 = 3x
⇒ 2x² – 3x – 2 = 0
⇒ 2x² – 4x + x – 2 = 0

⇒ 2x (x – 2) + 1 (x – 2) = 0
⇒ (x – 2) (2x + 1) = 0
Either x – 2 = 0, then x = 2
or 2x + 1 – 0, then 2x = – 1 ⇒ x = \(\frac { -1 }{ 2 }\)
x = 1. -1, 2, \(\frac { -i }{ 2 }\)

Question 11.
\(\sqrt{2 x+7}\) = x + 2
Solution:
\(\sqrt{2 x+7}\) = x + 2
Squaring both sides,
2x + 7 = (x + 2)²
⇒ 2x + 7 = x² + 4x + 4
⇒ x² + 4x + 4 – 2x – 7 = 0
⇒ x² + 2x – 3 = 0
⇒ x² + 3x – x – 3 = 0

⇒ x (x + 3) – 1 (x + 3) = 0
⇒ (x + 3) (x – 1) = 0
Either x + 3 = 0, then x = – 3
or x – 1 = 0, then x = 1
∴ x = 1, – 3
Check : If x = 1, then
L.H.S. = \(\sqrt{2 x+7}\) = \(\sqrt{2 + 7}\) = \(\sqrt{9}\) = 3
R.H.S. = x + 2 = 1 + 2 = 3
x = 1 is its solution
If x = – 3, then
L.H.S. = \(\sqrt{2 \times(-3)+7}+\sqrt{-6+7}=\sqrt{1}\) = 1
R.H.S. = – 3 + 2 = – 1
∵ L.H.S. = R.H.S.
∴ x = – 3 is not its solution
Hence x = 1

Question 12.
2\(\sqrt{2 x+1}\) – 2x = 1
Soluion:
2\(\sqrt{2 x+1}\) – 2x = 1
⇒ 2\(\sqrt{2 x+1}\) = 1 + 2x
Squaring both sides,
4(2x + 1) = (1 + 2x)²
⇒ 8x + 4 = 1 + 4x + 4x²
⇒ 4x² + 4x + 1 – 8x – 4 = 0
⇒ 4x² – 4x – 3 = 0
⇒ 4x² – 6x + 2x – 3 = 0

⇒ 2x (2x – 3) + 1 (2x – 3) = 0
⇒ (2x – 3) (2x + 1) = 0
Either 2x – 3 = 0, then 2x = 3 ⇒ x = \(\frac { 3 }{ 2 }\)
or 2x + 1 = 0 then 2x = -1 ⇒ x = \(\frac { -1 }{ 2 }\)
Check:
if x = \(\frac { 3 }{ 2 }\), then
L.H.S. = 2\(\sqrt{2 x+1}\) – 2x
= 2\(\sqrt{2 \times \frac{3}{2}+1}-2 \times \frac{3}{2}=2 \sqrt{3+1}-3\)
= 2\(\sqrt{4}\) – 3 = 2 x 2 – 3 = 4 – 3 = 1 = R.H.S.
∴ x = \(\frac { 3 }{ 2 }\) is a solution -1
If x = \(\frac { -1 }{ 2 }\), then
L.H.S. = \(2 \sqrt{2 x+1}-2 x=2 \sqrt{2\left(\frac{-1}{2}\right)+1}-2\) x (\(\frac { -1 }{ 2 }\))
= 2\(\sqrt{-1 + 1}\) + 1 ⇒ 2 x \(\sqrt{0}\) + 1 ⇒ 0 + 1 = 1 = R.H.S.
∴ x = \(\frac { -1 }{ 2 }\) is also a solution
∴ x = \(\frac { 3 }{ 2 }\), \(\frac { -1 }{ 2 }\)

Question 13.
\(\sqrt{4 x-3}+\sqrt{2 x+3}\) = 6
Solution:
\(\sqrt{4 x-3}+\sqrt{2 x+3}\) = 6
⇒ \(\sqrt{4 x-3}=6-\sqrt{2 x+3}\) = 6
Squaring both sides
4x – 3 = 36 + (2x + 3) – 12 72^+3 4x-3 = 36 + 2x + 3-12 72x + 3 ⇒ 4x – 3 – 36 – 2x – 3 = -12 72x+”3 ⇒ 2x- 42 = -1272×71
Dividing by 2,
x – 21 = -6 72x + 3
Again squaring,
x² -42x + 441 = 36 (2x + 3)
⇒ x² – 42x + 441 = 72x + 108
⇒ x² – 42x + 441 – 72x – 108 = 0
⇒ x² – 114x + 333 = 0
⇒ x² – 111x – 3x + 333 = 0

⇒ x(x- 111) – 3 (x – 111) = 0
⇒ (x – 111) (x – 3) = 0
Either x – 111 = 0, then x = 111
or x – 3 =0, then x = 3
If x = 111, then
L.H.S. = \(\sqrt{4 \times 111-3}+\sqrt{2 \times 111+3}\)
= \(\sqrt{444-3}+\sqrt{222+3}=\sqrt{441}+\sqrt{225}\)
= 21 + 15 = 36 ≠ R.H.S.
∴ x = 111 is not its Solution
If x = 3, then
L.H.S. = \(\sqrt{4 \times 3-3}+\sqrt{2 \times 3+3}\) = \(\sqrt{12-3}\) + \(\sqrt{6+3}\)
= \(\sqrt{9}\) + \(\sqrt{9}\) = 3 + 3 = 6 = R.H.S
∴ x = 3 is its solution
Hence x = 3