OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Ex 5(e)

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.