## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test.

Question 1.
Find the equation of a line whose inclination is 60° and y-intercept is – 4.
Solution:
Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
=> y – √3x – 4 Ans.

Question 2.
Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Solution:
Slope of the line 3y + 2x = 12
=> 3y = 12 – 2x
=> 3y = – 2x + 12

Question 3.
If the equation of a line is y – √3x + 1, find its inclination.
Solution:
In the line
y = √3 x + 1
Slope = √3 => tan θ = √3
θ = 60° (∵ tan 60° = √3)

Question 4.
If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.
Solution:
The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points – 4 = 2 m + c …(i)
and 1 = – 3 m + c …(ii)
Subtracting we get,

Question 5.
If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.
Solution:
Let the points to be A (1, 4), B (3, – 2) and C (p, – 5) are collinear and let B (3, – 2)
divides AC in the ratio of m1 : m2

Question 6.
Find the inclination of the line joining the points P (4, 0) and Q (7, 3).
Solution:
Slope of the line joining the points P (4, 0) and Q (7, 3)

Question 7.
Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to $$– \frac { 3 }{ 7 }$$
Solution:
Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5
Adding we get,

Question 8.
If the point A is reflected in the y-axis, the co-ordinates of its image A1, are (4, – 3),
(i) Find the co-ordinates of A
(ii) Find the co-ordinates of A2, A3 the images of the points A, A1, Respectively under reflection in the line x = – 2
Solution:
(i) ∵ A is reflected in the y-axis and its image is A1 (4, – 3)
Co-ordinates of A will be ( – 4, – 3)

Question 9.
If the lines $$\frac { x }{ 3 } +\frac { y }{ 4 } =7$$ and 3x + ky = 11 are perpendicular to each other, find the value of k.
Solution:
Given
Equation of lines are

Question 10.
Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).
Solution:
The equation of the line is x – 2y + 8 = 0
=> 2y = x + 8

Question 11.
Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.
Solution:
Equations of the line is
3y = 5 – x
=> 3y = – x + 5

Question 12.
Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.
Solution:
Let slope of the line joining the points A (1, 2) and B (6, 7) be m1

Question 13.
The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.
Solution:
Slope of line AC (m1)

Question 14.
A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:

(i) the co-ordinates of A and B
(ii) the equation of the line AB
Solution:
A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y) , and P (2, 1) divides BA in the ratio 3 : 1.

Question 15.
A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.
Solution:
Let the line make intercept a and b with the x-axis and y-axis respectively then the line passes through

Question 16.
If the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.
Solution:
Co-ordinates of A are (3, – 2).

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test.

Question 1.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B
Solution:
Base BC of an equilateral ∆ABC lies on y-axis co-ordinates of point C are (0, – 3), origin (0, 0) is the mid-point of BC.

Question 2.
A and B have co-ordinates (4, 3) and (0, 1), Find
(i) the image A’ of A under reflection in the y – axis.
(ii) the image of B’ of B under reflection in the lineAA’.
(iii) the length of A’B’.
Solution:

(i) Co-ordinates of A’, the image of A (4, 3) reflected in y-axis will be ( – 4, 3).
(ii) Co-ordinates of B’ the image of B (0, 1) reflected in the line AA’ will be (0, 5).
(iii) Length A’B’

Question 3.
Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.
Solution:
Let R be the point whose co-ordinates are (x, y) which divides PQ in the ratio of 3:1.

Question 4.
Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Co-ordinates of A (3, 1) and B ( – 2, 5)
P lies on AB such that

Question 5.
P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Solution:
Given
AP = PQ = QB

Question 6.
The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).
Solution:
Let A (2, – 2), B (8, – 2) and centre of the circle be
O (α + 2, α – 5)

Question 7.
The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:
Given
(3, 5) is the mid-point of A (2, p) and B (q, 4)

Question 8.
The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.
Solution:
Let AB be the diameter where co-ordinates of A are (3, 0) and of B are ( – 5, 6).
Co-ordinates of its origin O will be

Question 9.
In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?
Solution:
Let the point ( – 4, 6) divides the line segment joining the points
A ( – 6, 10) and B (3, – 8), in the ratio m : n

Question 10.
Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.
Solution:
Let P ( – 3, p) divides AB in the ratio of m1 : m2 coordinates of
A ( – 5, – 4) and B ( – 2, 3)

Question 11.
In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.
Solution:
Let the point P which is on x-axis, divides the line segment joining the points A (4, 2) and B (3, – 5) in the ratio of m1 : m2.
and let co-ordinates of P be (x, 0)

Question 12.
If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.
Solution:
Let co-ordinates of A be ( – 4, 3) and of B (6, 3) and of P be (2, y)
Let the ratio in which the P divides AB be m1 : m2

Question 13.
Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.
Solution:
Points are given A (2, – 2), B (3, 7)
and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2 at P and let co-ordinates of

Question 14.
The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.
(i) Write the coordinates of A’ and A”.
(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.
Solution:
A’ is the reflection of A(2, – 3) in the x-axis
(i) ∴ Co-ordinates of A’ will be (2, 3)
Draw a line x = 4 which is parallel to y-axis
A” is the reflection of A’ (2, 3)
∴Co-ordinates OA” will be (6, 3)
(ii) Join AA” which intersects x-axis at P whose
co-ordinate are (4, 0)
Let P divide AA” in the ratio in m1 : m2

Hence P(4, 0) divides AA” in the ratio 1 : 1

Question 15.
ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.
Solution:
Let the co-ordinates of C be (x, y) and other three vertices of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)
∴ ABCD is a parallelogram
Its diagonals bisect each other.
Let AC and BD intersect each other at O.
∴O is mid-points of BD
∴ Co-ordinates of O will be

Question 16.
ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.
Solution:
ABCD is a || gm , m which co-ordinates of A are (2, – 3) and B (-1, -1)
Its diagonals AC and BD bisect each other at M (1, – 4)
∴ M is mid point of AC and BD Let co-ordinates of C be (x1, y1) and of D be (x2, y2) when M is mid point of AC then

Question 17.
In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.

Solution:
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y) and P (3, 1) divides it in the ratio of 2 : 3.

Question 18.
Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.
Find : (i) the co-ordinates of Q.
(ii)the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.

Solution:
(i) Let co-ordinates of Q be (x’, y’) and of R (x”,y”)
Point P (1, 2) divides OQ in the ratio of 2 : 3

Question 19.
If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Solution:
A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of ∆ABC
D, E and F are the midpoints of sides BC, CA and AB respectively and G is the centroid of the ∆ABC

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Matrices Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Matrices Chapter Test.

Question 1.
Find the values of a and below
$$\begin{bmatrix} a+3 & { b }^{ 2 }+2 \\ 0 & -6 \end{bmatrix}=\begin{bmatrix} 2a+1 & 3b \\ 0 & { b }^{ 2 }-5b \end{bmatrix}$$
Solution:
$$\begin{bmatrix} a+3 & { b }^{ 2 }+2 \\ 0 & -6 \end{bmatrix}=\begin{bmatrix} 2a+1 & 3b \\ 0 & { b }^{ 2 }-5b \end{bmatrix}$$
comparing the corresponding elements
a + 3 = 2a + 1
=> 2a – a =3 – 1
=> a = 2
b² + 2 = 3b
=>b² – 3b + 2 = 0
=> b² – b – 2b + 2 = 0
=> b (b – 1) – 2 (b – 1) = 0
=> (b – 1) (b – 2) = 0.
Either b – 1 = 0, then b = 1 or b – 2 = 0,
then b = 2
Hence a = 2, 5 = 2 or 1 Ans.

Question 2.
Find a, b, c and d if $$3\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 4 & a+b \\ c+d & 3 \end{bmatrix}+\begin{bmatrix} a & 6 \\ -1 & 2d \end{bmatrix}$$
Solution:
Given
$$3\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 4 & a+b \\ c+d & 3 \end{bmatrix}+\begin{bmatrix} a & 6 \\ -1 & 2d \end{bmatrix}$$

Question 3.
Find X if Y = $$\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$$ and 2X + Y = $$\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$$
Solution:
Given
2X + Y = $$\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$$
=> 2X = 2X + Y = $$\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$$ – Y

Question 4.
Determine the matrices A and B when
A + 2B = $$\begin{bmatrix} 1 & 2 \\ 6 & -3 \end{bmatrix}$$ and 2A – B = $$\begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix}$$
Solution:
A + 2B = $$\begin{bmatrix} 1 & 2 \\ 6 & -3 \end{bmatrix}$$…..(i)
2A – B = $$\begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix}$$…….(ii)
Multiplying (i) by 1 and (ii) by 2

Question 5.
(i) Find the matrix B if A = $$\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$$ and A² = A + 2B
(ii) If A = $$\begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix}$$, B = $$\begin{bmatrix} 0 & 1 \\ -2 & 5 \end{bmatrix}$$
and C = $$\begin{bmatrix} -2 & 0 \\ -1 & 1 \end{bmatrix}$$ find A(4B – 3C)
Solution:
A = $$\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$$
let B = $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

Question 6.
If A = $$\begin{bmatrix} 1 & 4 \\ 1 & 0 \end{bmatrix}$$, B = $$\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$$ and C = $$\begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix}$$ compute (AB)C = (CB)A ?
Solution:
Given
A = $$\begin{bmatrix} 1 & 4 \\ 1 & 0 \end{bmatrix}$$,
B = $$\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$$ and
C = $$\begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix}$$

Question 7.
If A = $$\begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix}$$ and B = $$\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix}$$ find the each of the following and state it they are equal:
(i) (A + B)(A – B)
(ii)A² – B²
Solution:
Given
A = $$\begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix}$$ and
B = $$\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix}$$

Question 8.
If A = $$\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$$ find A² – 5A – 14I
Where I is unit matrix of order 2 x 2
Solution:
Given
A = $$\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$$

Question 9.
If A = $$\begin{bmatrix} 3 & 3 \\ p & q \end{bmatrix}$$ and A² = 0 find p and q
Solution:
Given
A = $$\begin{bmatrix} 3 & 3 \\ p & q \end{bmatrix}$$

Question 10.
If A = $$\begin{bmatrix} \frac { 3 }{ 5 } & \frac { 2 }{ 5 } \\ x & y \end{bmatrix}$$ and A² = I, find x,y
Solution:
Given
A = $$\begin{bmatrix} \frac { 3 }{ 5 } & \frac { 2 }{ 5 } \\ x & y \end{bmatrix}$$

Question 11.
If $$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ find a,b,c and d
Solution:
Given
$$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$

Question 12.
Find a and b if
$$\begin{bmatrix} a-b & b-4 \\ b+4 & a-2 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} -2 & -2 \\ 14 & 0 \end{bmatrix}$$
Solution:
Given
$$\begin{bmatrix} a-b & b-4 \\ b+4 & a-2 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} -2 & -2 \\ 14 & 0 \end{bmatrix}$$

Question 13.
If A = $$\begin{bmatrix} { sec60 }^{ o } & { cos90 }^{ o } \\ { -3tan45 }^{ o } & { sin90 }^{ o } \end{bmatrix}$$ and B = $$\begin{bmatrix} 0 & { cos45 }^{ o } \\ -2 & { 3sin90 }^{ o } \end{bmatrix}$$
Find (i) 2A – 3B (ii) A² (iii) BA
Solution:
Given
A = $$\begin{bmatrix} { sec60 }^{ o } & { cos90 }^{ o } \\ { -3tan45 }^{ o } & { sin90 }^{ o } \end{bmatrix}$$ and
B = $$\begin{bmatrix} 0 & { cos45 }^{ o } \\ -2 & { 3sin90 }^{ o } \end{bmatrix}$$

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Matrices Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Chapter Test.

Question 1.
Find the compound ratio of:
(a + b)2 : (a – b )2 ,
(a2 – b2) : (a2 + b2),
(a4 – b4) : (a + b)4
Solution:
(a + b)2 : (a – b )2 ,
(a2 – b2) : (a2 + b2),
(a4 – b4) : (a + b)4

Question 2.
If (7 p + 3 q) : (3 p – 2 q) = 43 : 2 find p : q
Solution:
(7p + 3q) : (3p – 2q) = 43 : 2
=> $$\frac { 7p+3q }{ 3p-2q } =\frac { 43 }{ 2 }$$

Question 3.
If a : b = 3 : 5, find (3a + 5b): (7a – 2b).
Solution:
a : b = 3 : 5
=> $$\frac { a }{ b } =\frac { 3 }{ 5 }$$
3a + 5n : 7a – 2b
Dividing each term by b

Question 4.
The ratio of the shorter sides of a right angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.
Solution:
Let the two shorter sides of a right angled triangle be 5x and 12x.
Third (longest side)

Question 5.
The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged ?
Solution:
Let the savings of Lokesh and his sister are 5x and 6x.
and the Lokesh should save Rs y more Now, according to the problem,
=> $$\frac { 5x+y }{ 6x+30 } =\frac { 5 }{ 6 }$$
=> 30x + 6y = 30x + 150

Question 6.
In an examination, the number of those who passed and the number of those who failed were in the ratio of 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.
Solution:
Let number of passed = 3 x
and failed = x
Total candidates appeared = 3x + x = 4x. In second case
No. of candidates appeared = 4 x + 8
and No. of passed = 3 x – 6
and failed = 4x + 8 – 3x + 6 = x + 14
then ratio will be = 2 : 1
Now according to the condition

Question 7.
What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional ?
Solution:
Let x be added to each number, then numbers will be 15 + x, 17 + x, 34 + x, and 38 + x.
Now according to the condition

Question 8.
If (a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion, prove that b is the mean proportional between a and c.
Solution:
(a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion
=> $$\frac { a+2b+c }{ a-c } =\frac { a-c }{ a-2b+c }$$

Question 9.
If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.
Solution:
2, 6, p, 54 and q are in continued proportional then
=> $$\frac { 2 }{ 6 } =\frac { 6 }{ p } =\frac { p }{ 54 } =\frac { 54 }{ q }$$

Question 10.
If a, b, c, d, e are in continued proportion, prove that: a : e = a4 : b4.
Solution:
a, b, c, d, e are in continued proportion
=> $$\frac { a }{ b } =\frac { b }{ c } =\frac { c }{ d } =\frac { d }{ e }$$ = k (say)

Question 11.
Find two numbers whose mean proportional is 16 and the third proportional is 128.
Solution:
Let x and y be two numbers
Their mean proportion = 16
and third proportion = 128

Question 12.
If q is the mean proportional between p and r, prove that:
$${ p }^{ 2 }-{ 3q }^{ 2 }+{ r }^{ 2 }={ q }^{ 4 }\left( \frac { 1 }{ { p }^{ 2 } } -\frac { 3 }{ { q }^{ 2 } } +\frac { 1 }{ { r }^{ 2 } } \right)$$
Solution:
q is mean proportional between p and r
q² = pr

Question 13.
If $$\frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f }$$, prove that each ratio is
(i) $$\sqrt { \frac { { 3a }^{ 2 }-{ 5c }^{ 2 }+{ 7e }^{ 2 } }{ { 3b }^{ 2 }-{ 5d }^{ 2 }+{ 7f }^{ 2 } } }$$
(ii) $${ \left[ \frac { { 2a }^{ 3 }+{ 5c }^{ 3 }+{ 7e }^{ 3 } }{ { 2b }^{ 3 }+{ 5d }^{ 3 }+{ 7f }^{ 3 } } \right] }^{ \frac { 1 }{ 3 } }$$
Solution:
$$\frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f }$$ = k(say)
∴ a = k, c = dk, e = fk

Question 14.
If $$\frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c }$$, prove that
$$\frac { { 3x }^{ 3 }-{ 5y }^{ 3 }+{ 4z }^{ 3 } }{ { 3a }^{ 3 }-{ 5b }^{ 3 }+{ 4c }^{ 3 } } ={ \left( \frac { 3x-5y+4z }{ 3a-5b+4c } \right) }^{ 3 }$$
Solution:
$$\frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c }$$ = k (say)
x = ak,y = bk,z = ck

Question 15.
If x : a = y : b, prove that
$$\frac { { x }^{ 4 }+{ a }^{ 4 } }{ { x }^{ 3 }+{ a }^{ 3 } } +\frac { { y }^{ 4 }+{ b }^{ 4 } }{ { y }^{ 3 }+{ b }^{ 3 } } =\frac { { \left( x+y \right) }^{ 4 }+{ \left( a+b \right) }^{ 4 } }{ { \left( x+y \right) }^{ 3 }+{ \left( a+b \right) }^{ 3 } }$$
Solution:
$$\frac { x }{ a } = \frac { y }{ b }$$ = k (say)
x = ak, y = bk

Question 16.
If $$\frac { x }{ b+c-a } =\frac { y }{ c+a-b } =\frac { z }{ a+b-c }$$ prove that each ratio’s equal to :
$$\frac { x+y+z }{ a+b+c }$$
Solution:
$$\frac { x }{ b+c-a } =\frac { y }{ c+a-b } =\frac { z }{ a+b-c }$$ = k(say)
x = k(b + c – a),
y = k(c + a – b),
z = k(a + b – c)

Question 17.
If a : b = 9 : 10, find the value of
(i) $$\frac { 5a+3b }{ 5a-3b }$$
(ii) $$\frac { { 2a }^{ 2 }-{ 3b }^{ 2 } }{ { 2a }^{ 2 }+{ 3b }^{ 2 } }$$
Solution:
a : b = 9 : 10
=> $$\frac { a }{ b } = \frac { 9 }{ 10 }$$

Question 18.
If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of $$\frac { { 3x }^{ 4 }+{ 25y }^{ 4 } }{ { 3x }^{ 4 }-{ 25y }^{ 4 } }$$ ;
Solution:
$$\frac { { 3x }^{ 4 }+{ 25y }^{ 4 } }{ { 3x }^{ 4 }-{ 25y }^{ 4 } } =\frac { 11 }{ 9 }$$
Applying componendo and dividendo

Question 19.
If $$x=\frac { 2mab }{ a+b }$$ , find the value of
$$\frac { x+ma }{ x-ma } +\frac { x+mb }{ x-mb }$$
Solution:
$$x=\frac { 2mab }{ a+b }$$
=> $$\frac { x }{ ma } +\frac { 2b }{ a+b }$$
Applying componendo and dividendo

Question 20.
If $$x=\frac { pab }{ a+b }$$ ,prove that $$\frac { x+pa }{ x-pa } -\frac { x+pb }{ x-pb } =\frac { 2\left( { a }^{ 2 }-{ b }^{ 2 } \right) }{ ab }$$
Solution:
$$x=\frac { pab }{ a+b }$$
=> $$\frac { x }{ pa } +\frac { b }{ a+b }$$
Applying componendo and dividendo

Question 21.
Find x from the equation $$\frac { a+x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ a+x-\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =\frac { b }{ x }$$
Solution:
$$\frac { a+x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ a+x-\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =\frac { b }{ x }$$
Applying componendo and dividendo,

Question 22.
If $$x=\frac { \sqrt [ 3 ]{ a+1 } +\sqrt [ 3 ]{ a-1 } }{ \sqrt [ 3 ]{ a+1 } -\sqrt [ 3 ]{ a-1 } }$$, prove that :
x³ – 3ax² + 3x – a = 0
Solution:
$$x=\frac { \sqrt [ 3 ]{ a+1 } +\sqrt [ 3 ]{ a-1 } }{ \sqrt [ 3 ]{ a+1 } -\sqrt [ 3 ]{ a-1 } }$$
Applying componendo and dividendo,

Question 23.
If $$\frac { by+cz }{ b^{ 2 }+{ c }^{ 2 } } =\frac { cz+ax }{ { c }^{ 2 }+{ a }^{ 2 } } =\frac { ax+by }{ { a }^{ 2 }+{ b }^{ 2 } }$$, prove that each of these ratio is equal to $$\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c }$$
Solution:
$$\frac { by+cz }{ b^{ 2 }+{ c }^{ 2 } } =\frac { cz+ax }{ { c }^{ 2 }+{ a }^{ 2 } } =\frac { ax+by }{ { a }^{ 2 }+{ b }^{ 2 } }$$

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Factorization Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Factorization Chapter Test.

Question 1.
Find the remainder when 2x3 – 3x2 + 4x + 7 is divided by
(i) x – 2
(ii) x + 3
(iii) 2x + 1
Solution:
f(x) = 2x3 – 3x2 + 4x + 7
(i) Let x – 2 = 0, then x = 2
Substituting value of x in f(x)
f(2) = 2 (2)3 – 3 (2)2 + 4 (2) + 7
= 2 × 8 – 3 × 4 + 4 × 2 + 7
= 16 – 12 + 8 + 7 = 19
Remainder = 19 Ans.
(ii) Let x + 3 = 0, then x = – 3
Substituting the value of x in f(x)

Question 2.
When 2x3 – 9x2 + 10x – p is divided by (x + 1), the remainder is – 24.Find the value of p.
Solution:
Let x + 1 = 0 then x = – 1,
Substituting the value of x in f(x)
f(x) = 2x3 – 9x2 + 10x – p

Question 3.
If (2x – 3) is a factor of 6x2 + x + a, find the value of a. With this value of a, factorise the given expression.
Solution:
Let 2 x – 3 = 0 then 2x = 3
=>x = $$\\ \frac { 3 }{ 2 }$$
Substituting the value of x in f(x)

Question 4.
When 3x2 – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also factorise the polynomial 3x2 – 5x + p – 3.
Solution:
f(x) = 3x2 – 5x+ p
Let (x – 2) = 0, then x = 2
f(2) = 3 (2)2 – 5(2) + p
= 3 x 4 – 10 + p
= 12 – 10 + p
= 2 + p

Question 5.
Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorise the given polynomial completely.
Solution:
f(x) = 5x3 + 4x2 – 5x – 4
Let 5x + 4 = 0, then 5x = – 4
=> x = $$\\ \frac { -4 }{ 2 }$$

Question 6.
Use factor theorem to factorise the following polynomials completely:
(i) 4x3 + 4x2 – 9x – 9
(ii) x3 – 19x – 30
Solution:
(i) f(x) = 4x3 + 4x2 – 9x – 9
Let x = – 1,then
f( – 1) = 4 ( – 1)3 + 4 ( – 1)2 – 9 ( – 1) – 9

Question 7.
If x3 – 2x2 + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorise the given polynomial completely.
Solution:
f(x) = x3 – 2x2 + px + q
(x + 2) is a factor
f( – 2) = ( – 2)3 – 2( – 2)2 + p ( – 2) + q

Question 8.
If (x + 3) and (x – 4) are factors of x3 + ax2 – bx + 24, find the values of a and b: With these values of a and b, factorise the given expression.
Solution:
f(x) = x3 + ax2 – bx + 24
Let x + 3 = 0, then x = – 3
Substituting the value of x in f(x)

Question 9.
If 2x3 + ax2 – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorise the given expression.
Solution:
f(x) = 2x3 + ax2 – 11 x + b
Let x – 2 = 0, then x = 2,
Substituting the vaue of x in f(x)

Question 10.
If (2x + 1) is a factor of both the expressions 2x2 – 5x + p and 2x2 + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.
Solution:
Let 2x + 1 = 0, then 2x = – 1
x = $$– \frac { 1 }{ 2 }$$
Substituting the value of x in

Question 11.
When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is,, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1) (x – 2).
Solution:
When f(x) is divided by (x – 1),
Remainder = 5
Let r – 1 = 0 => x = 1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Factorization Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test

Solve the following equations (1 to 4) by factorisation :

Question 1.
(i) x² + 6x – 16 = 0
(ii) 3x² + 11x + 10 = 0
Solution:
x² + 6x – 16 = 0
=> x² + 8x – 2x – 16 = 0
x (x + 8) – 2 (x + 8) = 0

Question 2.
(i) 2x² + ax – a² = 0
(ii) √3x² + 10x + 7√3 = 0
Solution:
(i) 2x² + ax – a² = 0
=> 2x² + 2ax – ax – a² = 0

Question 3.
(i) x(x + 1) + (x + 2)(x + 3) = 42
(ii) $$\frac { 6 }{ x } -\frac { 2 }{ x-1 } =\frac { 1 }{ x-2 }$$
Solution:
(i) x(x + 1) + (x + 2)(x + 3) = 42
2x² + 6x + 6 – 42 = 0

Question 4.
(i)$$\sqrt { x+15 } =x+3$$
(ii)$$\sqrt { { 3x }^{ 2 }-2x-1 } =2x-2$$
Solution:
(i)$$\sqrt { x+15 } =x+3$$
Squaring on both sides
x + 15 = (x + 3)²

Solve the following equations (5 to 8) by using formula :

Question 5.
(i) 2x² – 3x – 1 = 0
(ii) $$x\left( 3x+\frac { 1 }{ 2 } \right) =6$$
Solution:
(i) 2x² – 3x – 1 = 0
Here a = 2, b = – 3, c = – 1

Question 6.
(i) $$\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 }$$
(ii) $$\frac { 2 }{ x+2 } -\frac { 1 }{ x+1 } =\frac { 4 }{ x+4 } -\frac { 3 }{ x+3 }$$
Solution:
(i) $$\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 }$$
(2x + 5)(x + 3) = (x + 1)(3x + 4)

Question 7.
(i) $$\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 }$$
(ii) $$\frac { 4 }{ x } -3=\frac { 5 }{ 2x+3 } ,x\neq 0,-\frac { 3 }{ 2 }$$
Solution:
(i) $$\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 }$$
let $$\frac { 3x-4 }{ 7 }$$ = y,then

Question 8.
(i)x² + (4 – 3a)x – 12a = 0
(ii)10ax² – 6x + 15ax – 9 = 0,a≠0
Solution:
(i)x² + (4 – 3a)x – 12a = 0
Here a = 1,b = 4 – 3a,c = – 12a

Question 9.
Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)² – 3x + 4 = 0. (2014)
Solution:
(x – 1)² – 3x + 4 = 0
x² + 1 – 2x – 3x + 4 = 0

Question 10.
Discuss the nature of the roots of the following equations:
(i) 3x² – 7x + 8 = 0
(ii) x² – $$\\ \frac { 1 }{ 2 } x$$ – 4 = 0
(iii) 5x² – 6√5x + 9 = 0
(iv) √3x² – 2x – √3 = 0
Solution:
(i) 3x² – 7x + 8 = 0
Here a = 3, b = – 7,c = 8

Question 11.
Find the values of k so that the quadratic equation (4 – k) x² + 2 (k + 2) x + (8k + 1) = 0 has equal roots.
Solution:
(4 – k) x² + 2 (k + 2) x + (8k + 1) = 0
Here a = (4 – k), b = 2 (k + 2), c = 8k + 1

or k – 3 = 0, then k= 3
k = 0, 3 Ans.

Question 12.
Find the values of m so that the quadratic equation 3x² – 5x – 2m = 0 has two distinct real roots.
Solution:
3x² – 5x – 2m = 0
Here a = 3, b = – 5, c = – 2m

Question 13.
Find the value(s) of k for which each of the following quadratic equation has equal roots:
(i)3kx² = 4(kx – 1)
(ii)(k + 4)x² + (k + 1)x + 1 =0
Also, find the roots for that value (s) of k in each case.
Solution:
(i)3kx² = 4(kx – 1)
=> 3kx² = 4kx – 4
=> 3kx² – 4kx + 4 = 0

Question 14.
Find two natural numbers which differ by 3 and whose squares have the sum 117.
Solution:
Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)² = 117

Question 15.
Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:
Let larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition

Question 16.
Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
Solution:
Ratio in two natural numbers = 3 : 4
Let the numbers be 3x and 4x
According to the condition,

Question 17.
Two squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.
Solution:
Side of first square = x cm .
and side of second square = (x + 4) cm
Now according to the condition,

or x – 16 = 0 then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 – 4
= 20 cm Ans.

Question 18.
The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.
Solution:
Let breadth = x m
then length = (x + 12) m
Area = l × b = x (x + 12) m²
and perimeter = 2 (l + b)
= 2(x + 12 + x) = 2 (2x + 12) m
According to the condition.

Question 19.
A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.
Solution:
Area of rectangular garden = 100 cm²
Length of barbed wire = 30 m
Let the length of side opposite to wall = x

Question 20.
The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.
Solution:
Let the length of shortest side = x m
Length of hypotenuse = 2x – 1
and third side = x + 1
Now according to the condition,

Question 21.
A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.
Solution:
Perimeter of a right angled triangle = 112 cm
Hypotenuse = 50 cm
∴ Sum of other two sides = 112 – 50 = 62 cm
Let the length of first side = x
and length of other side = 62 – x

Question 22.
Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.
(i) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.
(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.
Solution:
Distance travelled by car A in one litre = x km
and distance travelled by car B in one litre = (x + 5) km
(i) Consumption of car A in covering 400 km

Question 23.
The speed of a boat in still water is 11 km/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream
Solution:
Speed of boat in still water =11 km/hr
Let the speed of stream = x km/hr.
Distance covered = 12 km.
Time taken = 2 hours 45 minutes .

Question 24.
By selling an article for Rs. 21, a trader loses as much percent as the cost price of the article. Find the cost price.
Solution:
S.R of an article = Rs. 21
Let cost price = Rs. x
Then loss = x%

Question 25.
A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.
Solution:
Amount spent = Rs. 2800
Price of each plant = Rs. x
Reduced price = Rs. (x – 1)

Question 26.
Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.
Solution:
Let Partap’s present age = x years
40 years hence his age = x + 40
and 32 years ago his age = x – 32
According to the condition

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test.

Question 1.
Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.
Solution:
5x – 2 < 3(3 – x)
=> 5x – 2 ≤ 9 – 3x
=> 5x + 3x ≤ 9 + 2

Question 2.
Solve the inequations :
6x – 5 < 3x + 4, x ∈ I.
Solution:
6x – 5 < 3x + 4
6x – 3x < 4 + 5
=> 3x <9
=> x < 3
x∈I
Solution Set = { – 1, – 2, 2, 1, 0….. }

Question 3.
Find the solution set of the inequation
x + 5 < 2 x + 3 ; x ∈ R
Graph the solution set on the number line.
Solution:
x + 5 ≤ 2x + 3
x – 2 x ≤ 3 – 5
=> – x ≤ – 2
=> x ≥ 2

Question 4.
If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.
Solution:
– 1 < 3 – 2x ≤ 7
– 1 < 3 – 2x and 3 – 2x ≤ 7
2 x < 3 + 1 and – 2x ≤ 7 – 3
2 x < 4 and – 2 x ≤ 4
x < 2 and – x ≤ 2
and x ≥ – 2 or – 2 ≤ x
x∈R
Solution set – 2 ≤ x < 2
Solution set on number line

Question 5.
Solve the inequation :
$$\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } ,x\in R$$
Solution:
$$\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 }$$
$$\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le \frac { 8 }{ 5 } +\frac { 3x-1 }{ 7 }$$

Question 6.
Find the range of values of a, which satisfy 7 ≤ – 4x + 2 < 12, x ∈ R. Graph these values of a on the real number line.
Solution:
7 < – 4x + 2 < 12
7 < – 4x + 2 and – 4x + 2 < 12

Question 7.
If x∈R, solve $$2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x$$
Solution:
$$2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x$$
$$2x-3\ge x+\frac { 1-x }{ 3 }$$ and $$x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x$$

Question 8.
Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.
Solution:
Let the positive integer = x
According to the problem,
5a – 6 < 4x
5a – 4x < 6 => x < 6
Solution set = {x : x < 6}
= { 1, 2, 3, 4, 5, 6} Ans.

Question 9.
Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is atleast 3.
Solution:
Let first least natural number = x
then second number = x + 1
and third number = x + 2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Shares and Dividends Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Shares and Dividends Chapter Test.

Question 1.
If a man received ₹1080 as dividend from 9% ₹20 shares, find the number of shares purchased by him.
Solution:
Income on one share = $$\\ \frac { 9 }{ 100 }$$ x 20
= Rs $$\\ \frac { 9 }{ 5 }$$
.’. No. of shares = 1080 x $$\\ \frac { 5 }{ 9 }$$
= 120 x 5 = 600 Ans.

Question 2.
Find the percentage interest on capital invested in 18% shares when a Rs 10 share costs Rs 12.
Solution:
Dividend on one share = 18% of Rs 10
= $$\\ \frac { 18\times 10 }{ 100 }$$
= Rs $$\\ \frac { 9 }{ 5 }$$

Question 3.
Rohit Kulkami invests Rs 10000 in 10% Rs 100 shares of a company. If his annual dividend is Rs 800, find :
(i) The market value of each share.
(ii) The rate percent which he earns on his investment.
Solution:
Investment = Rs 10000
Face value of each share = Rs 100
Rate of dividend = 10%
Annual dividend = Rs 800

Question 4.
At what price should a 9% Rs 100 share be quoted when the money is worth 6% ?
Solution:
If interest is 6 then investment = Rs 100
and if interest is 9, then investment
= Rs $$\\ \frac { 100\times 9 }{ 6 }$$
= Rs 150
Market value of each share = Rs 150 Ans

Question 5.
By selling at Rs 92, some 2.5% Rs 100 shares and investing the proceeds in 5% Rs 100 shares at Rs 115, a person increased his annual income by Rs 90. Find:
(i) the number of shares sold.
(ii) the number of shares purchased.
(iii) the new income.
(iv) the rate percent which he earns on his investment.
Solution:
Rate of dividend = 2.5% and market price = Rs 92
Let number of shares purchased = x.
Selling price of x shares = 92 x
Income from investing

Question 6.
A man has some shares of Rs. 100 par value paying 6% dividend. He sells half of these at a discount of 10% and invests the proceeds in 7% Rs. 50 shares at a premium of Rs. 10. This transaction decreases his income from dividends by Rs. 120. Calculate:
(i) the number of shares before the transaction.
(ii) the number of shares he sold.
(iii) his initial annual income from shares.
Solution:
Let no. of shares = x
Value of x shares = x × 100 = 100 x

Question 7.
Divide Rs. 101520 into two parts such that if one part is invested in 8% Rs. 100 shares at 8% discount and the other in 9% Rs. 50 shares at 8% premium, the annual incomes are equal.
Solution:
Total investment = Rs. 101520
Let investment in first part = x
and in second part = (101520 – x)
Market value of first kind of shares = Rs. 100 – Rs. 8
= Rs. 92
and rate of dividend = 8%

Question 8.
A man buys Rs. 40 shares of a company which pays 10% dividend. He buys the shares at such a price that his profit is 16% on his investment. At what price did he buy each share ?
Solution:
Face value of each share = Rs. 40
Dividend = 10%
Gain on investment = 10%

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Shares and Dividends Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Banking Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Banking Chapter Test

Question 1.
Mr. Dhruv deposits Rs 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.
Solution:
Deposit per month = Rs 600
Rate of interest = 10% p.a.
Period (n) = 5 years 60 months.
Total principal for one month

Question 2.
Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a $$2 \frac { 1 }{ 2 }$$ years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?
Solution:
In case of Ankita,
Deposit per month = Rs 400
Period (n) = 3 years = 36 months
Rate of interest = 10%
Total principal for one month

Question 3.
Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find
(i) the rate of simple interest,
(ii) the total interest earned by Shilpa
Solution:
Deposit per month (P) = Rs 800
Amount of maturity = Rs 48200

Question 4.
Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for $$4 \frac { 1 }{ 2 }$$ years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.
Solution:
Let each monthly instalment = Rs x
Rate of interest = 11 %
Period (n) = $$4 \frac { 1 }{ 2 }$$ years or 54 months,
Total principal for one month

Question 5.
Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.
Solution:
Deposit during the month (P) = Rs 600
Rate of interest = 7% p.a.
Amount of maturity = Rs 15450
Let time = n months

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Banking Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. aplustopper try to provide online math tutoring for you.

## ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes

### ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.1

Solution 01:
It gives accurate measurement and avoids error due to thickness of ruler or positioning of eye (due to angular viewing)

Solution 02:
By measuring the lengths of the given figure

(i) AB = CD
(ii) BC < AB
(iii) AC = BD
(iv) CD < BD

Solution 03:
Given that AC = 10 CM, AB = 6 CM and BC = 4 CM
By constructing line segment by the given data, the model drawn as below.

Point B lies in between A and C

Solution 04:

By measuring the Lengths of line segments in the above figure
AB = 3 CM
BC = 1.5 CM
(i) It can be observed that AC = AB + BC [i.e. 4.5 CM = 3 CM + 1.5 CM]
(ii) AC – BC = AB 4.5 CM – 1.5 CM = 3 CM by measurement fount that AB = 3 CM, so AC – BC = AB.

Solution 05:
By measuring the lengths of the given figure.

Given data
AB = 1.9 CM
BC = 0.7 CM
CD = 1.9 CM
AD = 4.5 CM
(i) AC + BD = 2.6 CM + 2.6 CM = 5.2 CM AD + BC = 4.5 CM + 0.7 CM = 5.2 CM Hence, AC + BD = AD +BC.
(ii) AB + CD = 1.9 CM + 1.9 CM = 3.8 CM AD – BC = 4.5 CM – 0.7 CM = 3.8 CM Hence, AC + BD = AD +BC.

Solution 06:
By measuring the lengths of the given triangle ABC as below

AB = 2.6 CM, AC = 3.8 CM and BC = 3.8 CM.
(i) AB + BC = 2.6 CM + 3.8 CM = 6.4 CM AC = 3.8 CM Hence, AB + BC > AC.
(ii) BC + AC = 3.8 CM + 3.8 CM = 7.6 CM AB = 2.6 CM Hence, AB > BC + AC.
(iii) AC + AB = 3.8 CM + 2.6 CM = 5.4 CM BC = 3.8 CM Hence, AC + AB > BC.

### ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.2

Solution 01:
(i) When the hour hand moves from 4 to 10 clockwise, fraction of revolution turned = ½ Number of right angles turned = 2.

(ii) When the hour hand moves from 2 to 5 clockwise, fraction of revolution turned = ¼. Number of right angles turned = 1.

(iii) When the hour hand moves from 7 to 10 clockwise, fraction of revolution turned = ¼. Number of right angles turned = 1.

(iv) When the hour hand moves from 8 to 5 clockwise, fraction of revolution turned = ¾. Number of right angles turned = 3.

(v) When the hour hand moves from 11 to 5 clockwise, fraction of revolution turned = ½ Number of right angles turned = 2.

(vi) When the hour hand moves from 6 to 3 clockwise, Fraction of revolution turned = ¾. Number of right angles turned = 3.

Solution 02:
(i) When the hour hand moves from 10 and makes half revolution, clockwise it will stop at 4.

(ii) When the hour hand moves from 4 and makes 1/4 revolution, clockwise it will stop at 7.

(iii) When the hour hand moves from 4 and makes 3/4 revolution, clockwise it will stop at 1.

Solution 03:
(i) When the hour hand moves from 6 and turns through 1 right angle, clockwise it will stop at 9.

(ii) When the hour hand moves from 8 and turns through 2 right angles, clockwise it will stop at 2.

(iii) When the hour hand moves from 10 and turns through 3 right angles, clockwise it will stop at 7.

(iv) When the hour hand moves from 7 and turns through 2 straight angles, clockwise it will stop at 7.

Solution 04:
(i) While turning from north to south Fraction of a revolution = ¾. Number of right angles = 3.

(ii) While turning from south to east Fraction of a revolution = 1/4. Number of right angles = 1.

(iii) While turning from east to west (clockwise). Fraction of a revolution = 1/2. Number of right angles = 2.

Solution 05:
(i) Straight angle – (c) Half of a revolution
(ii) Right angle – (d) One fourth of a revolution
(iii) Complete angle – (f) One complete revolution
(iv) Acute angle – (b) Less than one fourth of a revolution
(v) Obtuse angle – (e) Between ¼ and ½ of a revolution
(vi) Reflex angle – (a) More than half of a revolution

Solution 06:
(i) Acute angle
(ii) Obtuse angle
(iii) Right Angle
(iv) Straight angle
(v) Reflex angle
(vi) Reflex angle
(vii) Acute angle
(viii) Obtuse angle

Solution 07:
(i) Angle a and Angle c are acute, Angle b is obtuse
(ii) Angle x and Angle z are Obtuse, Angle y is acute
(iii) Angle p is obtuse, Angle q and Angle s are acute and Angle r is reflex.

Solution 08:

By measuring the protractor marked angles are as follows
(i) 62o
(ii) 116o
(iii) 121o

Solution 09:

By measuring the protractor marked angles are as follows
(i) 315o
(ii) 235o

Solution 10:
In the clock the angle between every numeric is 30o i.e. angle between 1 and 2 is 30o, 2 and 3 is 30o and 4 and 6 is 30o x 2 = 60o

Similarly,
(i) Angle between the hands of the clock – 60o

(ii) Angle between the hands of the clock – 30o

(iii) Angle between the hands of the clock – 150o

Solution 011:

Smaller angle formed by the hour and minutes hands of a clock at 7’O clock is 150o [30o x 5 = 150o] (Type of Angle – Obtuse angle) as shown in the above model
Other Angle = 360o – 150o = 210o (Type of Angle – Reflex angle)

Solution 12:
One is a 30o – 60o – 90o set square; the other is a 45o – 45o -90o set square. The angle of measure 90o is common between them.

### ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.3

Solution 01:
Two straight line are called perpendicular lines if they intersect at right angles.
In the given models (i), (iii) and (iv) are perpendicular lines.

Solution 02:
(i) Yes, CE = EG; E is the midpoint of CG
(ii) Yes, PF Line bisect segment BH – E is the midpoint of BH and Line P bisects line segment BH.
(iii) Line segment DF, Line segment BH
(iv) All are true (AC > FG, CD = GH and BC < EG)

### ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.4

Solution 01:

(i) Two sides are equal – Isosceles triangle
(ii) Three sides are different – Scalene triangle
(iii) Three sides are equal – Equilateral triangle

Solution 02:

(i) Angle is 90o – Right angled triangle
(ii) Angle is more than 90o – Obtuse angled triangle
(iii) Angle is less than 90o – acute angled triangle

Solution 03:

(i) Angle is less than 90o – acute angled triangle and two sides are in equal in length- Isosceles triangle.
(ii) Angle is 90o – right angled triangle and three sides are in not equal in length- scalene triangle.
(iii) Angle is more than 90o – Obtuse angled triangle and two sides are in equal in length- Isosceles triangle.
(iv) Angle is 90o – right angled triangle and two sides are equal in length- Isosceles triangle.
(v) Angle is less than 90o – acute angled triangle and three sides are in equal in length- Equilateral triangle.
(vi) Angle is more than 90o – Obtuse angled triangle and three sides are in not equal in length- scalene triangle.

Solution 04:
(i) Three sides of equal length – (e) Equilateral
(ii) 2 Sides of length – (g) Isosceles
(iii) All sides of different length – (a) Scalene
(iv) 3 acute angles – (f) Acute angled
(v) 1 right angle – (d) Right Angled
(vi) 1 Obtuse angle- C) Obtuse Angled
(vii) 1 Right angle with two sides of equal length – (b)Right angled Isosceles

Solution 05:
(i) False
(ii) True
(iii) True
(iv) False
(v) False
(vi) False
(vii) True
(viii) False

### ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.5

Solution 01:
(i) True
(ii) True
(iii) True
(iv) True
(v) False
(vi) False
(vii) False
Solution 02:
(i) Not a polygon, because it is not a closed curve
(ii) Polygon, because it is a simple closed curve made up entirely of line segments
(iii) Not a polygon, because it is not a simple curve
(iv) Not a polygon, because it is not made up of entirely line segments.

Solution 03:
(i) Pentagon

(ii) Quadrilateral

(iii) Hexagon

(iv) Octagon

Solution 04:
ABCDE is a regular pentagon and diagonals as in the below figure.

Solution 05:

Let ABCDEF be regular hexagon then
(i) Triangle ABC is an Isosceles triangle.
(ii) Triangle CEF is a right angled triangle.

Solution 06:
ABCD is a regular quadrilateral – Square.

### ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.6

Solution 01:
(i) Cuboid
(ii) Cuboid
(iii) Cuboid
(iv) Cylinder
(v) Cube
(vi) Sphere

Solution 02:
(i) Cone

(ii) Sphere

(iii) Cube

(iv) Pyramid

(v) Cylinder

(vi) Cuboid

Solution 03:
(i) A cube has 6 square faces, 12 edges and 8 vertices.
(ii) A triangular prism has 2 triangular faces, 3 rectangular faces, 9 edges and 6 vertices.
(iii) A triangular pyramid has 4 faces, 6 edges and 4 vertices.

## ML Aggarwal ICSE Solutions for Class 9 Maths Chapter 5 Simultaneous Linear Equations

Understanding ICSE Mathematics Class 9 ML Aggarwal Solutions Pdf Download Chapter 5 Simultaneous Linear Equations

## ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 14 Mensuration.

Understanding ICSE Mathematics Class 6 ML Aggarwal Solutions Pdf Download Chapter 14 Mensuration