Author Archives: Priyanka

Exercise MCQ

Question 1

The equation of the x-axis is

(a) x = 0

(b) y = 0

(c) x = y

(d) x + y = 0Solution 1

Correct option: (b)

The equation of the x-axis is y = 0.Question 2

The equation of the y-axis is

(a) x = 0

(b) y = 0

(c) x = y

(d) x + y = 0Solution 2

Correct option: (a)

The equation of the y-axis is x = 0. Question 3

The point of the form (a,a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y = x

(d) the line x + y = 0Solution 3

Question 4

The point of the form (a,-a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y-x=0

(d) the line x + y = 0Solution 4

Question 5

The linear equation 3x – 5y = 15 has

(a) a unique solution

(b) two solutions

(c) infinitely many solutions

(d) no solutionSolution 5

Question 6

The equation 2x + 5y = 7 has a unique solution, if x and y are

(a) natural numbers

(b) rational numbers

(c) positive real numbers

(d) real numbersSolution 6

Correct option: (a)

The equation 2x + 5y = 7 has a unique solution, if x and y are natural numbers.

If we take x = 1 and y = 1, the given equation is satisfied. Question 7

The graph of y = 5 is a line

(a) making an intercept 5 on the x-axis

(b) making an intercept 5 on the y-axis

(c) parallel to the x-axis at a distance of 5 units from the origin

(d) parallel to the y-axis at a distance of 5 units from the originSolution 7

Correct option: (c)

The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin. Question 8

The graph of x = 4 is a line

(a) making an intercept 4 on the x-axis

(b) making an intercept 4 on the y-axis

(c) parallel to the x-axis at a distance of 4 units from the origin

(d) parallel to the y-axis at a distance of 4 units from the originSolution 8

Correct option: (d)

The graph of x = 4 is a line parallel to the y-axis at a distance of 4 units from the origin. Question 9

The graph of x + 3 = 0 is a line

(a) making an intercept -3 on the x-axis

(b) making an intercept -3 on the y-axis

(c) parallel to the y-axis at a distance of 3 units to the left of y-axis

(d) parallel to the x-axis at a distance of 3 units below the x-axisSolution 9

Correct option: (c)

The graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis. Question 10

The graph of y + 2 = 0 is a line

(a) making an intercept -2 on the x-axis

(b) making an intercept -2 on the y-axis

(c) parallel to the x-axis at a distance of 2 units below the x-axis

(d) parallel to the y-axis at a distance of 2 units to the left of y-axisSolution 10

Correct option: (c)

The graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis. Question 11

The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point

(a) (2, 0)

(b) (3, 0)

(c) (0, 2)

(d) (0, 3)Solution 11

Correct option: (c)

When a graph meets the y-axis, the x coordinate is zero.

Thus, substituting x = 0 in the given equation, we get

2(0) + 3y = 6

⇒ 3y = 6

⇒ y = 2

Hence, the required point is (0, 2).Question 12

The graph of the linear equation 2x + 5y = 10 meets the x-axis at the point

(a) (0, 2)

(b) (2, 0)

(c) (5, 0)

(d) (0, 5)Solution 12

Correct option: (c)

When a graph meets the x-axis, the y coordinate is zero.

Thus, substituting y = 0 in the given equation, we get

2x + 5(0) = 10

⇒ 2x = 10

⇒ x = 5

Hence, the required point is (5, 0). Question 13

The graph of the line x = 3 passes through the point

(a) (0,3)

(b) (2,3)

(c) (3,2)

(d) None of theseSolution 13

Question 14

The graph of the line y = 3 passes though the point

(a) (3, 0)

(b) (3, 2)

(c) (2, 3)

(d) none of theseSolution 14

Correct option: (c)

Since, the y coordinate is 3, the graph of the line y = 3 passes through the point (2, 3).Question 15

The graph of the line y = -3 does not pass through the point

(a) (2,-3)

(b) (3,-3)

(c) (0,-3)

(d) (-3,2)Solution 15

Question 16

The graph of the linear equation x-y=0 passes through the point

Solution 16

Question 17

If each of (-2,2), (0,0) and (2,-2) is a solution of a linear equation in x and y, then the equation is

(a) x-y=0

(b) x+y=0

(c) -x+2y=0

(d) x – 2y=0Solution 17

Question 18

How many linear equations can be satisfied by x = 2 and y = 3?

(a) only one

(b) only two

(c) only three

(d) Infinitely manySolution 18

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2 and y = 3. Question 19

A linear equation in two variable x and y is of the form ax+by+c=0, where

(a) a≠0, b≠0

(b) a≠0, b=0

(c) a=0, b≠0

(d) a= 0, c=0Solution 19

Question 20

If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is

(a) 6

(b) 5

(c) 2

(d) 4Solution 20

Correct option: (d)

Since, (2, 0) is a solution of the linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given equation, we have

2(2) + 3(0) = k

⇒ 4 + 0 = k

⇒ k = 4 Question 21

Any point on x-axis is of the form:

(a) (x,y), where x ≠0 and y ≠0

(b) (0,y), where y ≠0

(c) (x,0), where x ≠0

(d) (y,y), where y ≠0Solution 21

Question 22

Any point on y-axis is of the form

(a) (x,0), where x ≠ 0

(b) (0,y), where y ≠ 0

(c) (x,x), where x ≠ 0

(d) None of theseSolution 22

Question 23

x = 5, y = 2 is a solution of the linear equation

(a) x + 2y = 7

(b) 5x + 2y = 7

(c) x + y = 7

(d) 5x + y = 7Solution 23

Correct option: (c)

Substituting x = 5 and y = 2 in L.H.S. of equation x + y = 7, we get

L.H.S. = 5 + 2 = 7 = R.H.S.

Hence, x = 5 and y = 2 is a solution of the linear equation x + y = 7. Question 24

If the point (3, 4) lies on the graph of 3y = ax + 7 then the value of a is

(a) 

(b) 

(c) 

(d) Solution 24

Correct option: (b)

Since the point (3, 4) lies on the graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we get

3(4) = a(3) + 7

⇒ 12 = 3a + 7

⇒ 3a = 5

Exercise Ex. 4B

Question 1(vii)

Draw the graph of each of the following equation.

y + 5 = 0 Solution 1(vii)

y + 5 = 0

⇒ y = -5, which is a line parallel to the X-axis, at a distance of 5 units from it, below the X-axis.

Question 1(viii)

Draw the graph of each of the following equation.

y = 4Solution 1(viii)

y = 4 is a line parallel to the X-axis, at a distance of 4 units from it, above the X-axis.

Question 1(i)

Draw the graph of each of the following equation.

x = 4Solution 1(i)

x = 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to its right.

Question 1(ii)

Draw the graph of each of the following equation.

x + 4 = 0Solution 1(ii)

x + 4 = 0

⇒ x = -4, which is a line parallel to the Y-axis, at a distance of 4 units from it, to its left.

Question 1(iii)

Draw the graph of each of the following equation.

y = 3Solution 1(iii)

y = 3 is a line parallel to the X-axis, at a distance of 3 units from it, above the X-axis.

Question 1(iv)

Draw the graph of each of the following equation.

y = -3Solution 1(iv)

y = -3 is a line parallel to the X-axis, at a distance of 3 units from it, below the X-axis.

Question 1(v)

Draw the graph of each of the following equation.

x = -2Solution 1(v)

x = -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to its left.

Question 1(vi)

Draw the graph of each of the following equation.

x = 5Solution 1(vi)

x = 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to its right.

Question 2(i)

Draw the graph of the equation y = 3x.

From your graph, find the value of y when x = 2.Solution 2(i)

y = 3x

When x = 1, then y = 3(1) = 3

When x = -1, then y = 3(-1) = -3

Thus, we have the following table:

x	1	-1
y	3	-3

Now, plot the points A(1, 3) and B(-1, -3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of y = 3x.

Reading the graph

Given: x = 2. Take a point M on the X-axis such that OM = 2.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 6

Thus, when x = 2, then y = 6.Question 2(ii)

Draw the graph of the equation y = 3x. From your graph, find the value of y when x = -2.Solution 2(ii)

The given equation is y = 3x.

Putting x = 1, y = 3  1 = 3

Putting x = 2, y = 3  2 = 6

Thus, we have the following table:

x	1	2
y	3	6

Plot points (1,3) and (2,6) on a graph paper and join them to get the required graph.

Take a point P on the left of y-axis such that the distance of point P from the y-axis is 2 units.

Draw PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to x-axis meeting y-axis at N.

So, y = ON = -6.Question 3(ii)

Draw the graph of the equation x + 2y – 3 = 0.

From your graph, find the value of y when x = -5Solution 3(ii)

x + 2y – 3 = 0

⇒ 2y = 3 – x

When x = -1, then  

When x = 1, then 

Thus, we have the following table:

x	-1	1
y	2	1

Now, plot the points A(-1, 2) and B(1, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + 2y – 3 = 0.

Reading the graph

Given: x = -5. Take a point M on the X-axis such that OM = -5.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 4

Thus, when x = -5, then y = 4. Question 3(i)

Draw the graph of the equation x + 2y – 3 = 0. From your graph, find the value of y when x = 5.Solution 3(i)

The given equation is,

x + 2y – 3 = 0

x = 3 – 2y

Putting y = 1,x = 3 – (2 1) = 1

Putting y = 0,x = 3 – (2 0) = 3

Thus, we have the following table:

x	1	3
y	1	0

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P.

Through P, draw PM parallel to x-axis cutting y-axis at M.

So, y = OM = -1.Question 4

Draw the graph of the equation 2x – 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x when y = 3.Solution 4

The given equation is, 2x – 3y = 5

Now, if x = 4, then

And, if x = -2, then

Thus, we have the following table:

x	4	-2
y	1	-3

Plot points (4,1) and (-2,-3) on a graph paper and join them to get the required graph.

(i) When x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to its right cutting the line at Q and through Q draw a line parallel to x-axis cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

(ii) When y = 3, draw a line parallel to x-axis at a distance of 3 units from x-axis and above it, cutting the line at point P. Through P, draw a line parallel to y-axis meeting x-axis at a point which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.Question 5

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.Solution 5

The given equation is 2x + y = 6

 y = 6 – 2x

Now, if x = 1, then y = 6 – 2  1 = 4

And, if x = 2, then y = 6 – 2  2 = 2

Thus, we have the following table:

x	1	2
y	4	2

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).Question 6

Draw the graph of the equation 3x + 2y = 6. Find the coordinates of the point, where the graph cuts the y-axis.Solution 6

The given equation is 3x + 2y = 6

 2y = 6 – 3x

Now, if x = 2, then

And, if x = 4, then

Thus, we have the following table:

x	2	4
y	0	-3

Plot points (2, 0) and (4,-3) on a graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6 cuts the y-axis at a point P which is 3 units above the x-axis.

So, co-ordinates of P are (0,3).Question 7

Draw the graphs of the equations 3x – 2y = 4 and x + y – 3 = 0. On the same graph paper, find the coordinates of the point where the two graph lines intersect.Solution 7

Graph of the equation 3x – 2y = 4

⇒ 2y = 3x – 4

When x = 2, then  

When x = -2, then 

Thus, we have the following table:

x	2	-2
y	1	-5

Now, plot the points A(2, 1) and B(-2, -5) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 3x – 2y = 4.

Graph of the equation x + y – 3 = 0

⇒ y = 3 – x

When x = 1, then y = 3 – 1 = 2 

When x = -1, then y = 3 – (-1) = 4

Thus, we have the following table:

x	1	-1
y	2	4

Now, plot the points C(1, 2) and D(-1, 4) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x + y – 3 = 0.

The two graph lines intersect at point A(2, 1). Question 8(i)

Draw the graph of the line 4x + 3y = 24.

Write the coordinates of the points where this line intersects the x-axis and the y-axis.Solution 8(i)

4x + 3y = 24

⇒ 3y = 24 – 4x

When x = 0, then  

When x = 3, then 

Thus, we have the following table:

x	0	3
y	8	4

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

Reading the graph

The graph of line 4x + 3y = 24 intersects the X-axis at point C(6, 0) and the Y-axis at point A(0, 8). Question 8(ii)

Draw the graph of the line 4x + 3y = 24.

Use this graph to find the area of the triangle formed by the graph line and the coordinate axes.Solution 8(ii)

4x + 3y = 24

⇒ 3y = 24 – 4x

When x = 0, then  

When x = 3, then 

Thus, we have the following table:

x	0	3
y	8	4

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

Reading the graph

Required area = Area of ΔAOC

Question 9

Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these two lines and the x-axis. Find the area of the shaded region.Solution 9

Graph of the equation 2x + y = 6

⇒ y = 6 – 2x

When x = 1, then y = 6 – 2(1) = 6 – 2 = 4 

When x = 2, then y = 6 – 2(2) = 6 – 4 = 2

Thus, we have the following table:

x	1	2
y	4	2

Now, plot the points A(1, 4) and B(2, 2) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 2x + y = 6.

Graph of the equation 2x – y + 2 = 0

⇒ y = 2x + 2

When x = -1, then y = 2(-1) + 2 = -2 + 2 = 0 

When x = 2, then y = 2(2) + 2 = 4 + 2 = 6

Thus, we have the following table:

x	-1	2
y	0	6

Now, plot the points C(-1, 0) and D(2, 6) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x – y + 2 = 0.

The two graph lines intersect at point A(1, 4).

The area enclosed by the lines and X-axis is shown in the graph.

Draw AM perpendicular from A on X-axis.

PM = y-coordinate of point A(1, 4) = 4

And, CP = 4

Area of shaded region = Area of ΔACP

Question 10

Draw the graphs of the lines x – y = 1 and 2x + y = 8. Shade the area formed by these two lines and the y-axis. Also, find this area.Solution 10

Graph of the equation x – y = 1

⇒ y = x – 1

When x = 1, then y = 1 – 1 = 0 

When x = 2, then y = 2 – 1 = 1

Thus, we have the following table:

x	1	2
y	0	1

Now, plot the points A(1, 0) and B(2, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x – y = 1.

Graph of the equation 2x + y = 8

⇒ y = 8 – 2x

When x = 2, then y = 8 – 2(2) = 8 – 4 = 4 

When x = 3, then y = 8 – 2(3) = 8 – 6 = 2 

Thus, we have the following table:

x	2	3
y	4	2

Now, plot the points C(2, 4) and D(3, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x + y = 8.

The two graph lines intersect at point D(3, 2).

The area enclosed by the lines and Y-axis is shown in the graph.

Draw DM perpendicular from D on Y-axis.

DM = x-coordinate of point D(3, 2) = 3

And, EF = 9

Area of shaded region = Area of ΔDEF

Question 11

Draw the graph for each of the equations x + y = 6 and x – y = 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect.

*Back answer incorrect.Solution 11

Graph of the equation x + y = 6

⇒ y = 6 – x

When x = 2, then y = 6 – 2 = 4 

When x = 3, then y = 6 – 3 = 3

Thus, we have the following table:

x	2	3
y	4	3

Now, plot the points A(2, 4) and B(3, 3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 6.

Graph of the equation x – y = 2

⇒ y = x – 2

When x = 3, then y = 3 – 2 = 1 

When x = 4, then y = 4 – 2 = 2 

Thus, we have the following table:

x	3	4
y	1	2

Now, plot the points C(3, 1) and D(4, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x – y = 2.

The two graph lines intersect at point D(4, 2).Question 12

Two students A and B contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation to satisfy the above data and draw its graph.Solution 12

Let the amount contributed by students A and B be Rs. x and Rs. y respectively.

Total contribution = 100

⇒ x + y = 100

⇒ y = 100 – x

When x = 25, then y = 100 – 25 = 75

When x = 50, then y = 100 – 50 = 50

Thus, we have the following table:

x	25	50
y	75	50

Now, plot the points A(25, 75) and B(50, 50) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 100.

Exercise Ex. 4A

Question 1(i)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3x + 5y = 7.5 Solution 1(i)

We have,

3x + 5y = 7.5

⇒ 3x + 5y – 7.5 = 0

⇒ 6x + 10y – 15 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = 10 and c = -15 Question 1(ii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(ii)

On comparing this equation with ax + by + c = 0, we obtain

a = 10, b = -1 and c = 30 Question 1(iii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3y – 2x = 6Solution 1(iii)

We have,

3y – 2x = 6

⇒ -2x + 3y – 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = -2, b = 3 and c = -6 Question 1(iv)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

4x = 5ySolution 1(iv)

We have,

4x = 5y

⇒ 4x – 5y = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 4, b = -5 and c = 0 Question 1(v)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(v)

⇒ 6x – 5y = 30

⇒ 6x – 5y – 30 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = -5 and c = -30 Question 1(vi)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(vi)

On comparing this equation with ax + by + c = 0, we obtain

a = , b =  and c = -5 Question 2(i)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

x = 6Solution 2(i)

We have,

x = 6

⇒ x – 6 = 0

⇒ 1x + 0y – 6 = 0

⇒ x + 0y – 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 0 and c = -6 Question 2(ii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3x – y = x – 1Solution 2(ii)

We have,

3x – y = x – 1

⇒ 3x – x – y + 1 = 0

⇒ 2x – y + 1 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = -1 and c = 1 Question 2(iii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

2x + 9 = 0Solution 2(iii)

We have,

2x + 9 = 0

⇒ 2x + 0y + 9 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = 0 and c = 9 Question 2(iv)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

4y = 7Solution 2(iv)

We have,

4y = 7

⇒ 0x + 4y – 7 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 0, b = 4 and c = -7 Question 2(v)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

x + y = 4Solution 2(v)

We have,

x + y = 4

⇒ x + y – 4 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 1 and c = -4 Question 2(vi)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 2(vi)

We have,

⇒ 3x – 8y – 1 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 3, b = -8 and c = -1 Question 3(i)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(4, 0)Solution 3(i)

Given equation is 5x – 4y = 20

Substituting x = 4 and y = 0 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(4) – 4(0)

= 20 – 0

= 20

= R.H.S.

Hence, (4, 0) is the solution of the given equation.Question 3(ii)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(0, 5)Solution 3(ii)

Given equation is 5x – 4y = 20

Substituting x = 0 and y = 5 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(0) – 4(5)

= 0 – 20

= -20

≠ R.H.S.

Hence, (0, 5) is not the solution of the given equation. Question 3(iii)

Check which of the following are the solutions of the equation 5x – 4y = 20.

Solution 3(iii)

Given equation is 5x – 4y = 20

Substituting x = -2 and y =  in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(-2) – 4

= -10 – 10

= -20

≠ R.H.S.

Hence,  is not the solution of the given equation. Question 3(iv)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(0, -5)Solution 3(iv)

Given equation is 5x – 4y = 20

Substituting x = 0 and y = -5 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(0) – 4(-5)

= 0 + 20

= 20

= R.H.S.

Hence, (0, -5) is the solution of the given equation. Question 3(v)

Check which of the following are the solutions of the equation 5x – 4y = 20.

Solution 3(v)

Given equation is 5x – 4y = 20

Substituting x = 2 and y =  in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(2) – 4

= 10 + 10

= 20

= R.H.S.

Hence,  is the solution of the given equation. Question 4(a)

Find five different solutions of each of the following equations:

2x – 3y = 6Solution 4(a)

Given equation is 2x – 3y = 6

Substituting x = 0 in the given equation, we get

2(0) – 3y = 6

⇒ 0 – 3y = 6

⇒ 3y = -6

⇒ y = -2

So, (0, -2) is the solution of the given equation.

Substituting y = 0 in the given equation, we get

2x – 3(0) = 6

⇒ 2x – 0 = 6

⇒ 2x = 6

⇒ x = 3

So, (3, 0) is the solution of the given equation.

Substituting x = 6 in the given equation, we get

2(6) – 3y = 6

⇒ 12 – 3y = 6

⇒ 3y = 6

⇒ y = 2

So, (6, 2) is the solution of the given equation.

Substituting y = 4 in the given equation, we get

2x – 3(4) = 6

⇒ 2x – 12 = 6

⇒ 2x = 18

⇒ x = 9

So, (9, 4) is the solution of the given equation.

Substituting x = -3 in the given equation, we get

2(-3) – 3y = 6

⇒ -6 – 3y = 6

⇒ 3y = -12

⇒ y = -4

So, (-3, -4) is the solution of the given equation.Question 4(b)

Find five different solutions of each of the following equations:

Solution 4(b)

Given equation is  

Substituting x = 0 in (i), we get

4(0) + 3y = 30

⇒ 3y = 30

⇒ y = 10

So, (0, 10) is the solution of the given equation.

Substituting x = 3 in (i), we get

4(3) + 3y = 30

⇒ 12 + 3y = 30

⇒ 3y = 18

⇒ y = 6

So, (3, 6) is the solution of the given equation.

Substituting x = -3 in (i), we get

4(-3) + 3y = 30

⇒ -12 + 3y = 30

⇒ 3y = 42

⇒ y = 14

So, (-3, 14) is the solution of the given equation.

Substituting y = 2 in (i), we get

4x + 3(2) = 30

⇒ 4x + 6 = 30

⇒ 4x = 24

⇒ x = 6

So, (6, 2) is the solution of the given equation.

Substituting y = -2 in (i), we get

4x + 3(-2) = 30

⇒ 4x – 6 = 30

⇒ 4x = 36

⇒ x = 9

So, (9, -2) is the solution of the given equation.Question 4(c)

Find five different solutions of each of the following equations:

3y = 4xSolution 4(c)

Given equation is 3y = 4x

Substituting x = 3 in the given equation, we get

3y = 4(3)

⇒ 3y = 12

⇒ y = 4

So, (3, 4) is the solution of the given equation.

Substituting x = -3 in the given equation, we get

3y = 4(-3)

⇒ 3y = -12

⇒ y = -4

So, (-3, -4) is the solution of the given equation.

Substituting x = 9 in the given equation, we get

3y = 4(9)

⇒ 3y = 36

⇒ y = 12

So, (9, 12) is the solution of the given equation.

Substituting y = 8 in the given equation, we get

3(8) = 4x

⇒ 4x = 24

⇒ x = 6

So, (6, 8) is the solution of the given equation.

Substituting y = -8 in the given equation, we get

3(-8) = 4x

⇒ 4x = -24

⇒ x = -6

So, (-6, -8) is the solution of the given equation.Question 5

If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.Solution 5

Since x = 3 and y = 4 is a solution of the equation 5x – 3y = k, substituting x = 3 and y = 4 in equation 5x – 3y = k, we get

5(3) – 3(4) = k

⇒ 15 – 12 = k

⇒ k = 3 Question 6

If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.Solution 6

Since x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, substituting these values in equation, we get

4(3k + 2) – 3(2k – 1) + 1 = 0

⇒ 12k + 8 – 6k + 3 + 1 = 0

⇒ 6k + 12 = 0

⇒ 6k = -12

⇒ k = -2 Question 7

The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y).Solution 7

Let the cost of one pencil be Rs. x and that of one ballpoint be Rs. y.

Then,

Cost of 5 pencils = Rs. 5x

Cost of 2 ballpoints = Rs. 2y

According to given statement, we have

5x = 2y

⇒ 5x – 2y = 0

Exercise Ex. 18A

Question 1

If the mean of 5 observation x, x + 2, x + 4, x + 6 and x  + 8 is 11, find the value of x.Solution 1

Question 2

If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?Solution 2

Question 3

Compute the mean of the following data:

Class	1-3	3-5	5-7	7-9
Frequency	12	22	27	19

Solution 3

Question 4

_{Find the mean, using direct method:}

Class	Frequency
0 – 1010 – 2020- 3030 – 4040 – 5050 – 60	7561282

Solution 4

We have

Class	Frequency	Mid Value
0-1010-2020-3030-4040-5050-60	7561282	51525354555	3575150420360110

Mean Question 5

_{Find the mean, using direct method:}

Class	Frequency
25 – 3535 – 4545 – 5555 – 6565 – 75	6108124

Solution 5

We have

Class	Frequency	Mid – value
25 – 3535 – 4545 – 5555 – 6565 – 75	6108124	3040506070	180400400720280

 Mean, Question 6

_{Find the mean, using direct method:}

Class	Frequency
0 – 100100 – 200200 – 300300 – 400400 – 500	6915128

Solution 6

We have

Class	Frequency	Mid Value
0 – 100100 – 200200 – 300300 – 400400 – 500	6915128	50150250350450	3001350375042003600
	= 50

 Mean, Question 7

Using an appropriate method, find the mean of the following frequency distribution:

Class interval	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	8	10	16	23	12	11

Which method did you use, and why?Solution 7

Question 8

If the mean of the following frequency distribution is 24, find the value of p.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	3	4	p	3	2

Solution 8

Question 9

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is Rs.18, find the missing frequency f.

Daily pocket allowance (in )	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

Solution 9

Question 10

If the mean of the following frequency distribution is 54, find the value of p.

Class	0-20	20-40	40-60	60-80	80-100
Frequency	7	p	10	9	13

Solution 10

Question 11

The mean of the following data is 42. Find the missing frequencies x and y if the sum of frequencies is 100.

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	10	x	13	y	10	14	9

Solution 11

Question 12

The daily expenditure of 100 families are given below. Calculate f₁ and f₂ if the mean daily expenditure is Rs.188.

Expenditure (in Rs.)	140-160	160-180	180-200	200-220	220-240
Number of families	5	25	f1	f2	5

Solution 12

Question 13

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50.

Class	Frequency
0 – 20	7
20 – 40
40 – 60	12
60 – 80
80 – 100	8
100 – 120	5

Find and .Solution 13

We have

Class	Frequency	Mid Value
0 – 20	7	10	70
20 – 40		30	30
40 – 60	12	50	600
60 – 80	=18 –	70	1260 – 70
80 – 100	8	90	720
100 – 120	5	110	550
	= 50

Question 14

During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarized as follows:

Number of heart-beats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of patients	2	4	3	8	7	4	2

Find the mean heartbeats per minute for these patients, choosing a suitable method.Solution 14

Question 15

Find the mean, using assumed mean method:

Marks	No, of students
0 – 1010 – 2020 -3030 – 4040 – 5050 – 60	12182720176

Solution 15

We have, Let A = 25 be the assumed mean

Marks	Frequency	Mid value	Deviation
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60	12182720176	51525 = A354555	-20-100102030	-240-1800200340180
	= 100			= 300

Hence mean = 28Question 16

Find the mean, using assumed mean method:

Class	Frequency
100 – 120120 – 140140 – 160160 – 180180 – 200	102030155

Solution 16

Let the assumed mean be 150, h = 20

Marks	Frequency	Mid value	Deviationdi =  – 150	di
100 – 120120 – 140140 – 160160 – 180180 – 200	102030155	110130150=A170190	-40-2002040	-400-4000300200
	= 80			di=-300

Hence, Mean = 146.25
Question 17

Find the mean, using assumed mean method:

Class	Frequency
0 – 20	20
20 – 40	35
40 – 60	52
60 – 80	44
80 – 100	38
100 – 120	31

Solution 17

Let A = 50 be the assumed mean, we have

Marks	Frequency	Mid value	Deviation
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120	203552443831	103050 = A 7090110	-40-200204060	-800-700088015201860
	= 220

Question 18

The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.

Literacy rate (%)	45-55	55-65	65-75	75-85	85-95
Number of cities	4	11	12	9	4

Solution 18

Question 19

Find the mean of the following frequency distribution using step-deviation method.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	10	15	8	10

Solution 19

Question 20

Find the mean of the following data, using step-deviation method:

Class	5-15	15-25	25-35	35-45	45-55	55-65	65-75
Frequency	6	10	16	15	24	8	7

Solution 20

Question 21

The weights of tea in 70 packets are shown in the following table:

Weight (in grams)	200-201	201-202	202-203	203-204	204-205	205-206
Number of packets	13	27	18	10	1	1

Find the mean weight of packets using step-deviation method.Solution 21

Question 22

Find the mean of the following frequency distribution using a suitable method:

Class	20-30	30-40	40-50	50-60	60-70
Frequency	25	40	42	33	10

Solution 22

Question 23

In a annual examination marks (out of 90) obtained by students of class X in mathematics are given below:

Marks obtained	0-15	15-30	30-45	45-60	60-75	75-90
Number of students	2	4	5	20	9	10

Find the mean marks.Solution 23

Question 24

Find the arithmetic mean of the following frequency distribution using step-deviation method:

Age (in year)	18-24	24-30	30-36	36-42	42-48	48-54
Number of workers	6	8	12	8	4	2

Solution 24

Question 25

Find the arithmetic mean of each of the following frequency distribution using step-deviation method:

Class	Frequency
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620	1495435

Solution 25

Let h = 20 and assume mean = 550, we prepare the table given below:

Age	Frequency	Mid value
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620	1495435	510530550 = A570590610	-2-10123	-27-904615
	= 40

Thus, A = 550, h = 20, and  = 40, 

Hence the mean of the frequency distribution is 544Question 26

Find the mean age from the following frequency distribution:

Age(in years)	No. of persons
25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59	4142216653

Hint: change the given series to the exclusive seriesSolution 26

The given series is an inclusive series, making it an exclusive series, we have

Class	Frequency	Mid value
24.5 – 29.529.5 – 34.534.5 – 39.539.5 – 44.544.5 – 49.549.5 – 54.554.5 – 59.5	4142216653	27323742 = A475257	-3-2-10123	-12-28-2206109
	= 70

Thus, A = 42, h = 5,  = 70 and 

Hence, Mean = 39.36 years
Question 27

The following table shows the age distribution of patients of malaria in a village during a particular month:

Age(in years)	No. of cases
5 – 1415 – 2424 – 3435 – 4445 – 5455 – 64	6112123145

Find the average age of the patients.Solution 27

The given series is an inclusive series making it an exclusive series,we get

class	Frequency	Mid value
4.5 – 14.514.5 – 24.524.5 – 34.534.5 – 44.544.5 – 54.554.5 – 64.5	6112123145	9.519.529.5=A39.549.559.5	-2-10123	-12-110232815
	= 80

Thus, A = 29.5, h = 10,  = 80 and 

Hence, Mean = 34.87 years
Question 28

Weight of 60 eggs were recorded as given below:

Weight (in grams)	75-79	80-84	85-89	90-94	95-99	100-104	105-109
Number of eggs	4	9	13	17	12	3	2

Calculate their mean weight to the nearest gram.Solution 28

Question 29

The following table shows the marks scored by 80 students in an examination:

Marks	Less than 5	Less than 10	Less than 15	Less than 20	Less than 25	Less than 30	Less than 35	Less than 40
Number of students	3	10	25	49	65	73	78	80

Calculate the mean marks correct to 2 decimal places.Solution 29

Exercise Ex. 18B

Question 1

In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.

Age (in years)	0-15	15-30	30-45	45-60	60-75
Number of patients	5	20	40	50	25

Solution 1

Question 2

Compute the median from the following data:

Marks	No. of students
0 – 77 – 1414 – 2121 – 2828 – 3535 – 4242 – 49	347110169

Solution 2

We prepare the frequency table, given below

Marks	No. of students	C.F.
0 – 77 – 1414 – 2121 – 2828 – 3535 – 4242 – 49	347110169	371425254150
	N =  = 50

Now, 

The cumulative frequency is 25 and corresponding class is 21 – 28.

Thus, the median class is 21 – 28

l = 21, h = 7, f = 11, c = C.F. preceding class 21 – 28 is 14 and  = 25

Hence the median is 28.Question 3

The following table shows the daily wages of workers in a factory:

Daily wages	No. of workers
0 – 100100 – 200200 – 300300 – 400400 – 500	403248228

Find the median daily wage income of the workers.Solution 3

We prepare the frequency table given below:

Daily wages	Frequency	C.F.
0 – 100100 – 200200 – 300300 – 400400 – 500	403248228	4072120142150
	N =  = 150

Now, N = 150, therefore 

The cumulative frequency just greater than 75 is 120 and corresponding class is 200 – 300.

Thus, the median class is 200 – 300

l = 200, h = 100, f = 48

c = C.F. preceding median class = 72 and 

Hence the median of daily wages is Rs. 206.25.Hence the median is 28.Question 4

Calculate the median from the following frequency distribution:

Class	Frequency
5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535- 4040 – 45	5615105422

Solution 4

We prepare the frequency table, given below:

Class	Frequency	C.F
5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535- 4040 – 45	5615105422	511263641454749
	= 49

Now, N = 49 

The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 – 20.

Thus, the median class is 15 – 20

 l = 15, h = 5, f = 15

c = CF preceding median class = 11 and 

Median of frequency distribution is 19.5Question 5

Given below is the number of units of electricity consumed in a week in a certain locality:

Consumption(in units)	No. of consumers
65 – 8585 – 105105 – 125125 – 145145 – 165165 – 185185 – 205	4513201474

Calculate the median.Solution 5

We prepare the cumulative frequency table as given below:

Consumption	Frequency	C.F
65 – 8585 – 105105 – 125125 – 145145 – 165165 – 185185 – 205	4513201474	492242566367
	N =  = 67

Now, N = 67 

The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 – 145.

Thus, the median class is 125 – 145

 l = 125, h = 20,  and c = CF preceding the median class = 22,  = 33.5

Hence median of electricity consumed is 136.5Question 6

Calculate the median from the following data:

Height(in cm)	No. of boys
135 – 140140 – 145145 – 150150 – 155155 – 160160 – 165165 – 170170 – 175	6101822201563

Solution 6

Frequency table is given below:

Height	Frequency	C.F
135 – 140140 – 145145 – 150150 – 155155 – 160160 – 165165 – 170170 – 175	6101822201563	6163456769197100
	N =  =100

N = 100, 

The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155

Thus, the median class is 150 – 155

l = 150, h = 5, f = 22, c = C.F.preceding median class = 34

Hence, Median = 153.64
Question 7

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Class	Frequency
0 – 1010 – 2020 – 3030 – 4040 – 50	525×187

Solution 7

The frequency table is given below. Let the missing frequency be x.

Class	Frequency	C.F
0 – 1010 – 2020 – 3030 – 4040 – 50	525×187	53030 + x48 + x55 + x

Median = 24  Median class is 20 – 30

l = 20, h = 10, f = x, c = C.F. preceding median class = 30

Hence, the missing frequency is 25.Question 8

The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

Class	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40
Frequencies	12	a	12	15	b	6	6	4

Solution 8

Question 9

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.

Runs scored	2500-3500	3500-4500	4500-5500	5500-6500	6500-7500	7500-8500
Number of batsmen	5	x	y	12	6	2

Solution 9

Question 10

If the median of the following frequency distribution is 32.5, find the value of .

Solution 10

Let be the frequencies of class intervals 0 – 10 and 40 – 50

Median is 32.5 which lies in 30 – 40, so the median class is 30 – 40

l = 30, h = 10, f = 12, N = 40 and 

Question 11

Calculate the median for the following data:

Age(in years)	Frequency
19 – 2526 – 3233 – 3940 – 4647 – 5354 – 60	359668102354

Solution 11

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

Class	Frequency	C.F
18.5 – 25.525.5 – 32.532.5 – 39.539.5 – 46.546.5 – 53.553.5 – 60.5	359668102354	35131199301336340
	fi = N = 340

The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.

Median class is 32.5 – 39.5

l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131

Hence median is 36.5 yearsQuestion 12

Find the median wages for the following frequencies distribution:

Wages per day(in Rs)	Frequency
61 – 7071 – 8081 – 9091 – 100101 – 110111 – 120	5152030208

Solution 12

Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get

Wages per day(in Rs)	Frequency	C.F
60.5 – 70.570.5 – 80.580.5 – 90.590.5 – 100.5100.5 – 110.5110.5 – 120.5	5152030208	52040709098
	fi = N =98

The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 – 100.5.

median class is 90.5 – 100.5

l = 90.5, h = 10, f = 30, c = CF preceding median class = 40

Hence, Median = Rs 93.50Question 13

Find the median from the following table:

Class	Frequency
1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3535 – 4040 – 45	710163224161152

Solution 13

The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get

Marks	Frequency	C.F
0.5 – 5.55.5 – 10.510.5 – 15.515.5 – 20.520.5 – 25.525.5 – 30.530.5 – 35.535.5 – 40.540.5 – 45.5	710163224161152	717336589105116121123
	fi = N =123

The cumulative frequency just greater than 61.5 is 65.

The corresponding median class is 15.5 – 20.5.

Then the median class is 15.5 – 20.5

l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33

Hence, Median = 19.95
Question 14

Find the median from the following data:

Marks	No. of students
Below 10Below 20Below 30Below 40Below 50Below 60Below 70Below 80	1232578092116164200

Solution 14

Marks	Frequency	C.F
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80	1220252312244836	1232578092116164200
	N =

The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.

Thus the median class is 50 – 60

l = 50, h = 10, f = 24, c = C.F. preceding median class = 92,  = 100

Hence, Median = 53.33

Exercise Ex. 18C

Question 1

Find the mode of the following frequency distribution:

Marks	10-20	20-30	30-40	40-50	50-60
Frequency	12	35	45	25	13

Solution 1

Question 2

Compute the mode of the following data:

Class	0-20	20-40	40-60	60-80	80-100
Frequency	25	16	28	20	5

Solution 2

Question 3

Heights of students of Class X are given in the following frequency distribution:

Height (in cm)	150-155	155-160	160-165	165-170	170-175
Number of students	15	8	20	12	5

Find the modal height.

Also, find the mean height. Compare and interpret the two measures of central tendency.Solution 3

Question 4

Find the mode of the following distribution:

Class interval	Frequency
10 – 1414 – 1818 – 2222 – 2626 – 3030 – 3434 – 3838 – 42	8611202522104

Solution 4

As the class 26 – 30 has maximum frequency so it is modal class

Hence, mode = 28.5Question 5

Given below is the distribution of total household expenditure of 200 manual workers in a city:

Expenditure	No. of manual workers
1000 – 15001500 – 20002000 – 25002500 – 30003000 – 35003500 – 40004000 – 45004500 – 5000	244031283223175

Find the average expenditure done by maximum number of manual workers.Solution 5

As the class 1500 – 2000 has maximum frequency, so it os modal class

Hence the average expenditure done by maximum number of workers = Rs. 1820Question 6

Calculate the mode from the following data:

Monthly salary(in Rs)	No. of employees
0 – 50005000- 1000010000 – 1500015000 – 2000020000 – 2500025000 – 30000	90150100807010

Solution 6

As the class 5000 – 10000 has maximum frequency, so it is modal class

Hence, mode = Rs. 7727.27Question 7

Compute the mode from the following data:

Age (in years)	No. of patients
0 – 55 – 1010 – 1515 – 2020 – 2525 – 3030 – 35	611182417135

Solution 7

As the class 15 – 20 has maximum frequency so it is modal class.

Hence mode = 17.3 yearsQuestion 8

Compute the mode from the following series:

Size	Frequency
45 – 5555 – 6565 – 7575 – 8585 – 9595 – 105105 – 115	712173032610

Solution 8

As the class 85 – 95 has the maximum frequency it is modal class

Hence, mode = 85.71Question 9

Compute the mode of the following data:

Class Interval	Frequency
1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3536 – 4041 – 4546 – 50	381318282013864

Solution 9

The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get

Class	Frequency
0.5 – 5.55.5- 10.510.5 – 15. 515.5 – 20.520.5 – 25. 525.5 – 30.530.5 – 35.535.5 – 40.540.5 – 45.545.5 – 50.5	381318282013863

As the class 20.5 – 25.5 has maximum frequency, so it is modal class

Hence, mode = 23.28Question 10

The age wise participation of students in the Annual Function of a school is shown in the following distribution.

Age (in years)	5-7	7-9	9-11	11-13	13-15	15-17	17-19
Number of students	x	15	18	30	50	48	x

Find the missing frequencies when the sum of frequencies is 181. Also, find the mode of the data.Solution 10

Exercise Ex. 18D

Question 1

Find the mean, mode and median of the following data:

Class	Frequency
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 70	510183020125

Solution 1

Let assumed mean be 35, h = 10, now we have

Class	Frequency	Mid-value		C.F
0-1010-2020-3030-4040-5050-6060-70	510183020125	5152535 = A455565	-3-2-10123	51533638395100	-15-20-180202415
	N = 100

(i)Mean 

(ii)N = 100, 

Cumulative frequency just after 50 is 63

Median class is 30 – 40

l = 30, h = 10, N = 100, c = 33, f = 30

(iii)Mode = 3 × median – 2 × mean

= 3 × 35.67 – 2 × 35.6 = 107.01 – 71.2

= 35.81

Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81Question 2

Find the mean, median and mode of the following data:

Class	0-20	20-40	40-60	60-80	80-100	100-120	120-140
Frequency	6	8	10	12	6	5	3

Solution 2

Question 3

Find the mean, median and mode of the following data:

Class	0-50	50-100	100-150	150-200	200-250	250-300	300-350
Frequency	2	3	5	6	5	3	1

Solution 3

Question 4

Find the mode, median and mean for the following data:

Marks obtained	25-35	35-45	45-55	55-65	65-75	75-85
Number of students	7	31	33	17	11	1

Solution 4

Question 5

A survey regarding the heights of 50 girls of a class was conducted and the following data was obtained:

Height in cm	No. of girls
120 – 130130 – 140140 – 150150 – 160160 – 170	2812208
Total	50

Find the mean, Median and mode of the above data.Solution 5

Let the assumed mean A be 145.Class interval h = 10.

Class	Frequency	Mid-Value			C.F.
120-130130-140140-150150-160160-170	2812208	125135145=A155165	-2-1012	-4-802016	210224250
	N = 50

(i)Mean 

(ii)N = 50, 

Cumulative frequency just after 25 is 42

Corresponding median class is 150 – 160

Cumulative frequency before median class, c = 22

Median class frequency f = 20

(iii)Mode = 3 median – 2 mean

= 3 151.5 – 2 149.8 = 454.5 – 299.6

= 154.9

Thus, Mean = 149.8, Median = 151.5, Mode = 154.9Question 6

The following table gives the daily income of 50 workers of a factory:

Daily income(in Rs)	No. of workers
100 – 120120 – 140140 – 160160 – 180180 – 200	12148610

Find the mean, mode and median of the above dataSolution 6

Class	Frequency	Mid-value			C.F.
100-120120-140140-160160-180180-200	12148610	110130150= A170190	-2-1012	-24-140620	1226344050
	N = 50

     Let assumed mean A = 150 and h = 20

(i)Mean 

(ii) 

Cumulative frequency just after 25 is 26

Corresponding frequency median class is 120 – 140

So, l = 120, f = 14, h = 20, c = 12

(iii)Mode = 3 Median – 2 Mode

= 3 138.6 – 2 145.2

= 415.8 – 190.4

= 125.4

Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4Question 7

The table below shows the daily expenditure on food of 30 households in a locality:

Daily expenditure	No. of households
100 – 150150 – 200200 – 250250 – 300300 – 350	671232

Find the mean and median daily expenditure on foodSolution 7

Class	Frequency	Mid-value			C.F.
100-150150-200200-250250-300300-350	671232	125175225275325	-2-1012	-12-7034	613252830
	N = 30

Let assumed mean = 225 and h = 50

(i)Mean = 

(ii) 

Cumulative frequency just after 15 is 25

corresponding class interval is 200 – 250

Median class is 200 – 250

Cumulative frequency c just before this class = 13

Hence, Mean = 205 and Median = 208.33

Exercise Ex. 18E

Question 1

Find the median of the following data by making a ‘less than ogive’.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Number of students	5	3	4	3	3	4	7	9	7	8

Solution 1

We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the ‘less than type’ ogive as follows:

At y = 26.5, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 68 units

Hence, median marks = 68Question 2

The given distribution shows the number of wickets taken by the bowlers in one -day international cricket matches:

Number of wickets	Less than 15	Less than 30	Less than 45	Less than 60	Less than 75	Less than 90	Less than 105	Less than 120
Number of bowlers	2	5	9	17	39	54	70	80

Draw a ‘less type’ ogive from the above data. Find the median.Solution 2

Number of wickets	Less than 15	Less than 30	Less than 45	Less than 60	Less than 75	Less than 90	Less than 105	Less than 120
Number of bowlers	2	5	9	17	39	54	70	80

We plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the ‘less than type’ ogive as follows:

At y = 40, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 78 units

Hence, median number of wickets = 78Question 3

Draw a ‘more than’ ogive for the data given below which gives the marks of 100 students.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Number of students	4	6	10	10	25	22	18	5

Solution 3

We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the ‘more than type’ ogive as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 47 units

Hence, median marks = 47Question 4

The height of 50 girls of Class X of a school are recorded as follows:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165
Number of girls	5	8	9	12	14	2

Draw a ‘more than type’ ogive for the above data.Solution 4

We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the ‘more than type’ ogive as follows:

At y = 25, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 151 units

Hence, median height = 151 cmQuestion 5

The monthly consumption of electricity (in units) of some families of a locality is given in the following frequency distribution:

Monthly consumption (in units)	140-160	160-180	180-200	200-220	220-240	240-260	260-280
Number of families	3	8	15	40	50	30	10

Prepare a ‘more than type’ ogive for the given frequency distribution.Solution 5

We plot the points (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the ‘more than type’ ogive as follows:

At y = 78, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 226 units

Hence, median consumption of electricity = 226 unitsQuestion 6

The following table gives the production yield per hectare of wheat of 100 farms of a village.

Production yield (kg/ha)	50-55	55-60	60-65	65-70	70-75	75-80
Number of farms	2	8	12	24	38	16

Change the distribution to a ‘more than type’ distribution and draw its ogive. Using ogive, find the median of the given data.Solution 6

We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) to get the ‘more than type’ ogive as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 70.5 units

Hence, median production yield = 70.5 kg/haQuestion 7

The table given below shows the weekly expenditures on food of some households in a locality.

Weekly expenditure (in Rs.)	Number of households
100-200	5
200-300	6
300-400	11
400-500	13
500-600	5
600-700	4
700-800	3
800-900	2

Draw a ‘less tha type’ and a ‘more than type’ ogive for this distribution.Solution 7

Less Than Series:

Class interval	Cumulative Frequency
Less than 200	5
Less than 300	11
Less than 400	22
Less than 500	35
Less than 600	40
Less than 700	44
Less than 800	47
Less than 900	49

We plot the points (200, 5), (300, 11), (400, 22), (500, 35), (600, 40), (700, 44), (800, 47) and (900, 49) to get ‘less than type’ ogive.

More Than Series:

Class interval	Frequency
More than 100	49
More than 200	44
More than 300	38
More than 400	27
More than 500	14
More than 600	9
More than 700	5
More than 800	2

We plot the points (100, 49), (200, 44), (300, 38), (400, 27), (500, 14), (600, 9), (700, 5) and (800, 2) to get more than ogive.

Question 8

From the following frequency distribution, prepare the ‘More than Ogive’

Score	No. of candidates
400 – 450450 – 500500 – 550550 – 600600- 650650 – 700700 – 750750 – 800	2035403224271834
Total	230

Also find the medianSolution 8

More than series

We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)

Hence,

Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M

Then,OM = 590

Hence median = 590Question 9

The marks obtained by 100 students of a class in an examination are given below:

Marks	No. of students
0 – 55 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535 – 4040 – 4545 – 50	25681025201842

Draw cumulative frequency curves by using (1), less than series and (2) more than series

Hence, find the medianSolution 9

(i) Less than series:

Marks	No. of students
Less than 5Less than 10Less than 15Less than 20Less than 25Less than 30Less than 35Less than 40Less than 45Less than 50	2713213156769498100

Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)

Join these points free hand to get the curve representing “less than” cumulative curve.

(ii)From the given table we may prepare the ‘more than’ series as shown below

Marks	No. of students
More than 45More than 40More than 35More than 30More than 25More than 20More than 15More than 10More than 5More than 0	2624446979879398100

Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)

Join these points free hand to get required curve

Here 

Two curves intersect at point P(28, 50)

Hence, the median = 28
Question 10

From the following data, draw the two types of cumulative frequency curves and determine the median:

Height (in cm)	Frequency
140 – 144144 – 148148 – 152152 – 156156 – 160160 – 164164 – 168168 – 172172 – 176176 – 180	392431426475828634

Solution 10

We may prepare less than series and more than series

(i)Less than series

Height in (cm)	Frequency
Less than 140Less than 144Less than 148Less than 152Less than 156Less than 160Less than 164Less than 168Less than 172Less than 176Less than 180	03123667109173248330416450

Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)

(ii)More than series

Height in cm	C.F.
More than 140More than 144More than 148More than 152More than 156More than 160More than 164More than 168More than 172More than 176More than 180	450447438414383341277202120340

Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)

The curves intersect at (167, 225).

Hence, 167 is the median.

Exercise Ex. 18F

Question 1

Write the median class of the following distribution:

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	4	4	8	10	12	8	4

Solution 1

Question 2

What is the lower limit of the modal class of the following frequency distribution?

Age (in years)	0-10	10-20	20-30	30-40	40-50	50-60
Number of patients	16	13	6	11	27	18

Solution 2

Class having maximum frequency is the modal class.

Here, maximum frequency = 27

Hence, the modal class is 40 – 50.

Thus, the lower limit of the modal class is 40.Question 3

The monthly pocket money of 50 students of a class are given in the following distribution:

Monthly pocket money (in Rs.)	0-50	50-100	100-150	150-200	200-250	250-300
Number of students	2	7	8	30	12	1

Find the modal class and also give class mark of the modal class.Solution 3

Question 4

A data has 25 observations arranged in a descending order. Which observation represents the median?Solution 4

Question 5

For a cetain distribution, mode and median were found to be 1000 and 1250 respectively. Find mean for this distribution using an empirical relation.Solution 5

Question 6

In a class test, 50 students obtained marks as follows:

Marks obtained	0-20	20-40	40-60	60-80	80-100
Number of students	4	6	25	10	5

Find the modal class and the median class.Solution 6

Question 7

Find the class marks of classes 10-25 and 35-55.Solution 7

Question 8

While calculating the mean of a given data by the assumed-mean method, the following values were obtained:

Find the mean.Solution 8

Question 9

The distributions X and Y with total number of observations 36 and 64, and mean 4 and 3 respectively are combined. What is the mean of the resulting distribution X + Y?Solution 9

Question 10

In a frequency distribution table with 12 classes, the class-width is 2.5 and the lowest class boundary is 8.1, then what is the upper class boundary of the highest class?Solution 10

Question 11

The observations 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 are arranged in ascending order. What is the value of x if the median of the data is 63?Solution 11

Question 12

The median of 19 observations is 30. Two more observations are made and the values of these are 8 and 32. Find the median of the 21 observations taken together.Solution 12

Question 13

Solution 13

Question 14

What is the cumulative frequency of the modal class of the following distribution?

Class	3-6	6-9	9-12	12-15	15-18	18-21	21-24
Frequency	7	13	10	23	4	21	16

Solution 14

Question 15

Find the mode of the given data:

Class interval	0-20	20-40	40-60	60-80
Frequency	15	6	18	10

Solution 15

Question 16

The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:

Age (in year)	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	60	42	55	70	53	20

Form a ‘less than type’ cumulative frequency distribution.Solution 16

Question 17

In the following data, find the value of p and q. Also, find the median class and modal class.

Class	Frequency (f)	Cumulative frequency (cf)
100-200	11	11
200-300	12	p
300-400	10	33
400-500	q	46
500-600	20	66
600-700	14	80

Solution 17

Question 18

The following frequency distribution gives the monthly consumption of electricity of 64 consumers of a locality.

Monthly consumption (in units)	65-85	85-105	105-125	125-145	145-165	165-185
Number of consumers	4	5	13	20	14	8

Form a ‘more than type’ cumulative frequency distribution.Solution 18

Question 19

The following table gives the life-time (in days) of 100 electric bulbs of a certain brand.

Life-time (in days)	Less than 50	Less than 100	Less than 150	Less than 200	Less than 250	Less than 300
Number of bulbs	7	21	52	79	91	100

From this table, construct the frequency distribution table.Solution 19

Question 20

The following table gives the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.

Marks obtained (in percent)	11-20	21-30	31-40	41-50	51-60	61-70	71-80
Number of students	141	221	439	529	495	322	153

(a) Convert the given frequency distribution into the continuous form.

(b) Find the median class and write its class mark

(c) Find the modal class and write its cumulative frequency.Solution 20

Question 21

If the mean of the following distribution is 27, find the value of p.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	8	p	12	13	10

Solution 21

Question 22

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age (in years)	0-10	10-20	20-30	30-40	40-50
Number of persons	5	25	?	18	7

Solution 22

Exercise MCQ

Question 1

Which of the following is not a measure of central tendency?

(a) Mean

(b) Mode

(c) Median

(d) RangeSolution 1

Correct option: (d)

Range is not a measure of central tendency.Question 2

Which of the following cannot be determined graphically? (a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 2

Correct option: (a)

Mean cannot be determined graphically.Question 3

Which of the following measures of central tendency is influenced by extreme values?

(a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 3

Correct option: (a)

Since mean is the average of all observations, it is influenced by extreme values.Question 4

The mode of a frequency distribution is obtained graphically from

(a) a frequency curve

(b) a frequency polygon

(c) a histogram

(d) an ogiveSolution 4

Correct option: (c)

Mode can be obtained graphically from a histogram.Question 5

The median of a frequency distribution is found graphically with the help of

(a) a histogram

(b) a frequency curve

(c) a frequency polygon

(d) ogivesSolution 5

Correct option: (d)

Ogives are used to determine the median of a frequency distribution.Question 6

The cumulative frequency table is useful in determining the

(a) mean

(b) median

(c) mode

(d) all of theseSolution 6

Correct option: (b)

The cumulative frequency table is useful in determining the median.Question 7

The abscissa of the point of intersection of the Less Than Type and of the More Than Type cumulative frequency curves of a grouped data gives its

(a) mean

(b) median

(c) mode

(d) none of theseSolution 7

Correct option: (b)

Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.Question 8

(a) 1

(b) 0

(c) -1

(d) 2Solution 8

Question 9

Solution 9

Question 10

(a) Lower limits of the classes

(b) upper limits of the classes

(c) midpoints of the classes

(d) none of theseSolution 10

Correct option: (c)

d_i‘s are the deviations from A of midpoints of the classes.Question 11

While computing the mean of the grouped data, we assume that the frequencies are

(a) evenly distributed over the classes

(b) centred at the class marks of the classes

(c) centred at the lower limits of the classes

(d) centred at the upper limits of the classesSolution 11

Correct option: (b)

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.Question 12

The relation between mean, mode and median is

(a) mode = (3 ⨯ mean) – (2 ⨯ median)

(b) mode = (3 ⨯ median) – (2 ⨯ mean)

(c) median = (3 ⨯ mean) – (2 ⨯ mode)

(d) mean = (3 ⨯ median) – (2 ⨯ mode)Solution 12

Correct option: (b)

Mode = (3 x median) – (2 x mean)Question 13

If the ‘less than type’ ogive and ‘more than type’ ogive intersect each other at (20.5, 15.5) then the median of the given data is

(a) 5.5

(b) 15.5

(c) 20.5

(d) 36.0Solution 13

Correct option: (c)

Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5Question 14

Consider the frequency distribution of the height of 60 students of a class:

Height (in cm)	No. of Students	Cumulative Frequency
150-155	16	16
155-160	12	28
160-165	9	37
165-170	7	44
170-175	10	54
175-180	6	60

The sum of the lower limit of the modal class and the upper limit of the median class is

(a) 310

(b) 315

(c) 320

(d) 330Solution 14

Question 15

Consider the following frequency distribution :

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	3	9	15	30	18	5

The modal class is

(a) 10-20

(b) 20-30

(c) 30-40

(d) 50-60Solution 15

Correct option: (c)

Class having maximum frequency is the modal class.

Here, maximum frequency = 30

Hence, the modal class is 30 – 40.Question 16

Mode =?

Solution 16

Question 17

Median = ?

Solution 17

Question 18

If the mean and median of a set of numbers are 8.9 and 9 respectively then the mode will be

(a) 7.2

(b) 8.2

(c) 9.2

(d) 10.2Solution 18

Correct option: (c)

Mean = 8.9

Median = 9

Mode = 3Median – 2Mean

= 3 x 9 – 2 x 8.9

= 27 – 17.8

= 9.2Question 19

Look at the frequency distribution table given below:

Class interval	35-45	45-55	55-65	65-75
Frequency	8	12	20	10

The median of the above distribution is

(a) 56.5

(b) 57.5

(c) 58.5

(d) 59Solution 19

Question 20

Consider the following table :

Class interval	10-14	14-18	18-22	22-26	26-30
Frequency	5	11	16	25	19

The mode of the above data is

(a) 23.5

(b) 24

(c) 24.4

(d) 25Solution 20

Question 21

The mean and mode of a frequency distribution are 28 and 16 respectively. The median is

(a) 22

(b) 23.5

(c) 24

(d) 24.5Solution 21

Question 22

The median and mode of a frequency distribution are 26 and 29 respectively. The mean is

(a) 27.5

(b) 24.5

(c) 28.4

(d) 25.8Solution 22

Question 23

For a symmetrical frequency distribution, we have

(a) mean < mode < median

(b) mean > mode > median

(c) mean = mode = median

(d) Solution 23

Correct option: (c)

For a symmetrical distribution, we have

Mean = mode = medianQuestion 24

Look at the cumulative frequency distribution table given below :

Monthly income	Number of families
More than Rs.10000	100
More than Rs.14000	85
More than Rs.18000	69
More than Rs.20000	50
More than Rs.25000	37
More than Rs.30000	15

Number of families having income range 20000 to 25000 is

(a) 19

(b) 16

(c) 13

(d) 22Solution 24

Correct option: (c)

Number of families having income more than Rs. 20000 = 50

Number of families having income more than Rs. 25000 = 37

Hence, number of families having income range 20000 to 25000 = 50 – 37 = 13Question 25

The median of first 8 prime numbers is

(a) 7

(b) 9

(c) 11

(d) 13Solution 25

Question 26

The mean of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

(a) 0

(b) 1

(c) 10

(d) 19Solution 26

Correct option: (d)

Mean of 20 numbers = 0

Hence, sum of 20 numbers = 0 x 20 = 0

Now, the mean can be zero if

sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),

sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),

…….

sum of 19 numbers is (S) and the 20^th number is (-S), then their sum is zero.

So, at the most, 19 numbers can be greater than zero.Question 27

If the median of the data 4, 7, x – 1, x – 3, 16, 25, written in ascending order, is 13 then x is equal to

(a) 13

(b) 14

(c) 15

(d) 16Solution 27

Question 28

The mean of 2, 7, 6 and x is 5 and the mean of 18, 1, 6, x and y is 10.

What is the value of y?

(a) 5

(b) 10

(c) 20

(d) 30

Note: Question modifiedSolution 28

Question 29

Match the following columns :

	Column I		Column II
(a)	The most frequent value in a data is known as …….	(p)	Standard deviation
(b)	Which of the following cannot be determined graphically out of mean, mode and median?	(q)	Median
(c)	An ogive is used to determine…	(r)	Mean
(d)	Out of mean, mode, median and standard deviation, which is not a measure of central tendency?	(s)	Mode

Solution 29

(a) – (s)

The most frequent value in a data is known as mode.

(b) – (r)

Mean cannot be determined graphically.

(c) – (q)

An ogive is used to determine median.

(d) – (p)

Standard deviation is not a measure of central tendency.Question 30

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R.) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d)Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If the median and mode of a frequency distribution are 150 and 154 respectively, then its mean is 148.	Mean, median and mode of a frequency distribution are related as:mode = 3 median – 2 mean.

The correct answer is: (a)/(b)/(c)/(d).Solution 30

Question 31

Assertion (A)	Reason (R)
Consider the following frequency distribution :Class interval3-66-99-1212-1515-1818-21Frequency2521231012 The mode of the above data is 12.4	The value of the variable which occurs most often is the mode.

The correct answer is: (a)/(b)/(c)/(d).Solution 31

Exercise FA

Question 1

Which one of the following measures is determined only after the construction of cumulative frequency distribution?

(a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 1

Correct option: (b)

The cumulative frequency table is useful in determining the median.Question 2

If the mean of a data is 27 and its median is 33 then the mode is

(a) 30

(b) 43

(c) 45

(d) 47Solution 2

Correct option: (c)

Mean = 27

Median = 33

Mode = 3Median – 2Mean

= 3 x 33 – 2 x 27

= 99 – 54

= 45Question 3

Consider the following distribution :

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of the lower limits of the median class and the modal class is

(a) 15

(b) 25

(c) 30

(d) 35Solution 3

Question 4

Consider the following frequency distribution :

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is

(a) 16.5

(b) 18.5

(c) 18

(d) 17.5Solution 4

Question 5

If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, find the median.Solution 5

Question 6

In the table given below, the times taken by 120 athletes to run a 100-m-hurdle race are given.

Class	13.8-14	14-14.2	14.2-14.4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	15	54	25	20

Find the number of athletes who completed the race in less than 14.6 seconds.Solution 6

Number of athletes who completed the race in less than 14.6 seconds

= 2 + 4 + 15 + 54

= 75Question 7

Consider the following frequency distribution :

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

Find the upper limit of the median class.Solution 7

Question 8

The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below :

Profit (in lakhs Rs.)	Number of shops
More than or equal to 5	30
More than or equal to 10	28
More than or equal to 15	16
More than or equal to 20	14
More than or equal to 25	10
More than or equal to 30	7
More than or equal to 35	3

If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25Solution 8

The frequency table is as follows:

ClassesProfit (in lakhs Rs.)	FrequencyNumber of shops
5 – 10	2
10 – 15	12
15 – 20	2
20 – 25	4
25 – 30	3
30 – 35	4
35 – 40	3

The frequency corresponding to the class 20 – 25 is 4.Question 9

Find the mean of the following frequency distribution :

Class	1-3	3-5	5-7	7-9
Frequency	9	22	27	18

Solution 9

Question 10

The maximum bowling speeds (in km/hr) of 33 players at a cricket coaching centre are given below :

Speed in km/hr	85-100	100-115	115-130	130-145
No. of players	10	4	7	9

Calculate the median bowling speed.Solution 10

Question 11

The arithmetic mean of the following frequency distribution is 25.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	16	p	30	32	14

Find the value of p.

Note: Question modifiedSolution 11

Question 12

Find the median of the following frequency distribution :

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	6	16	30	9	4

Solution 12

Question 13

Following is the distribution of marks of 70 students in a periodical test :

Marks	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
Number of students	3	11	28	48	70

Draw a cumulative frequency curve for the above data.Solution 13

Marks	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
Number of students	3	11	28	48	70

We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:

Question 14

Find the median of the following data.

Class interval	0-10	10-20	20-30	30-40	40-50	Total
Frequency	8	16	36	34	6	100

Solution 14

Question 15

For the following distribution draw a ‘less than type’ ogive and from the curve find the median.

Marks obtained	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60	Less than 70	Less than 80	Less than 90	Less than 100
Number of students	2	7	17	40	60	82	85	90	100

Solution 15

Marks obtained	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60	Less than 70	Less than 80	Less than 90	Less than 100
Number of students	2	7	17	40	60	82	85	90	100

We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 56.

Hence, median = 56  Question 16

The median value for the following frequency distribution is 35 and the sum of all the frequencies is 170. Using the formula for median, find the missing frequencies.

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	10	20	?	40	?	25	15

Solution 16

Question 17

Find the missing frequencies f₁ and f₂ in the table given below, it being given that the mean of the given frequency distribution is 50.

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	f1	32	f2	19	120

Solution 17

Question 18

Find the mean of the following frequency distribution using step-deviation method :

Class	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	15	22	20	18	20	25

Solution 18

Question 19

Find the mean, median and mode of the following data :

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	6	8	10	15	5	4	2

Solution 19

Question 20

Draw ‘less than ogive’ and more than ogive’ on a single graph paper and hence find the median of the following data :

Class interval	5-10	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	2	12	2	4	3	4	3

Solution 20

Less Than Series:

Class interval	Frequency
Less than 10	2
Less than 15	14
Less than 20	16
Less than 25	20
Less than 30	23
Less than 35	27
Less than 40	30

We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.

More Than Series:

Class interval	Frequency
More than 5	30
More than 10	28
More than 15	16
More than 20	14
More than 25	10
More than 30	7
More than 35	3

We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.

The two curves intersect at L. Draw LM ⊥ OX.

Thus, median = OM = 16 Question 21

The production yield per hectare of wheat of some farms of a village are given in the following table :

Production yield (in kg/ha)	40-45	45-50	50-55	55-60	60-65	65-70	70-75	75-80	80-85
Number of farms	1	9	15	18	40	26	16	14	10

Draw a less than type ogive and a more than type ogive for this dataSolution 21

Less Than Series:

Class interval	Frequency
Less than 45	1
Less than 50	10
Less than 55	25
Less than 60	43
Less than 65	83
Less than 70	109
Less than 75	125
Less than 80	139
Less than 85	149

We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.

More Than Series:

Class interval	Frequency
More than 40	149
More than 45	148
More than 50	139
More than 55	124
More than 60	106
More than 65	66
More than 70	40
More than 75	24
More than 80	10

We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.

Question 22

The following table gives the marks obtained by 50 students in a class test :

Marks	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
Number of students	2	3	6	7	14	12	4	2

Calculate the mean, median and mode for the above data.Solution 22

Exercise Ex. 17A

Question 1

Two cubes each of volume 27 cm’ are joined end to end to form a solid. Find the surface area of the resulting cuboid.Solution 1

Question 2

Solution 2

Question 3

If the total surface area of a solid hemisphere is 462 cm², find its volume.Solution 3

Question 4

A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs. 25 per metre.Solution 4

Question 5

If the volumes of two cones are in the ratio of 1: 4 and their diameters are in the ratio of 4 : 5, find the ratio of their heights.Solution 5

Question 6

The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km². Find the height of the mountain.Solution 6

Question 7

The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, find its volume.Solution 7

Question 8

The surface area of a sphere is 2464 cm². If its radius be doubled, what will be the surface area of the new sphere?Solution 8

Question 9

A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 in surmounted by a right circular cone of same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m.Solution 9

s 

Question 10

A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs.80 per square meter. Take Solution 10

Radius of the cylinder = 14 m

And its height = 3 m

Radius of cone = 14 m

And its height = 10.5 m

Let l be the slant height

Curved surface area of tent

                    = (curved area of cylinder + curved surface area of cone)

Hence, the curved surface area of the tent = 1034 

Cost of canvas = Rs.(1034 × 80) = Rs. 82720Question 11

A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent. Take .Solution 11

For the cylindrical portion, we have radius = 52.5 m and height = 3 m

For the conical portion, we have radius = 52.5 m

And slant height = 53 m

Area of canvas = 2rh + rl = r(2h + l)

Question 12

A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5m, its height of 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.Solution 12

Radius o f cylinder = 2.5 m

Height of cylinder = 21 m

Slant height of cone = 8 m

Radius of cone = 2.5 m

Total surface area of the rocket = (curved surface area of cone

                                                              + curved surface area of cylinder + area of base)

Question 13

A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.Solution 13

Question 14

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.Solution 14

Height of cone = h = 24 cm

Its radius = 7 cm

Total surface area of toy

Question 15

A toy is in the shape of a cone mounted on a hemisphere of same base radius. If the volume of the toy is 231 cm³ and its diameter is 7 cm, find the height of the toy.Solution 15

Question 16

A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, find the radius of the ice-cream cone.Solution 16

Question 17

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder. The diameter of the hemisphere is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity.Solution 17

Radius of hemisphere = 10.5 cm

Height of cylinder = (14.5 10.5) cm = 4 cm

Radius of cylinder = 10.5 cm

Capacity = Volume of cylinder + Volume of hemisphere

Question 18

A toy is in the form of a cylinder with hemispherical ends. If the whole length of the toy is 90 cm and its diameter is 42 cm, find the cost of painting the toy at the rate of 70 paise per sq cm.Solution 18

Question 19

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.Solution 19

Question 20

A wooden article was made by scooping out a hemisphere from each end of a cylinder, as shown in the figure. If the height of the cylinder is 20 cm and its base is of diameter 7 cm, find the total surface area of the article when it is ready.

Solution 20

Height of cylinder = 20 cm

And diameter = 7 cm and then radius = 3.5 cm

Total surface area of article

               = (lateral surface of cylinder with r = 3.5 cm and h = 20 cm)

Question 21

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 21. cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is 5 cm and its height is 9.8 cm, find the volume of the water left in the tub.Solution 21

Radius of cylinder 

And height of cylinder 

Radius of cone r = 2.1 cm

And height of cone 

Volume of water left in tub

              = (volume of cylindrical tub – volume of solid)

Question 22

From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. Take = 3.14Solution 22

(i)Radius of cylinder = 6 cm

Height of cylinder = 8 cm

Volume of cylinder

Volume of cone removed

(ii)Surface area of cylinder = 2 = 2× 6 × 8 

Question 23

From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.Solution 23

Question 24

From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.Solution 24

Question 25

Solution 25

Question 26

A spherical glass vessel has a cylindrical neck 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water it can hold. Use = .Solution 26

Diameter of spherical part of vessel = 21 cm

Question 27

The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid.

Solution 27

Height of cylinder = 6.5 cm

Height of cone = 

Radius of cylinder = radius of cone

                          = radius of hemisphere

                         = 

Volume of solid = Volume of cylinder + Volume of cone

+ Volume of hemisphere

Question 28

From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece.Solution 28

Question 29

A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs.5 per 100 sq cm. [Use π = 3.14.]Solution 29

Question 30

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is 30 cm.Solution 30

Question 31

The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.

Solution 31

Question 32

A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. The conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of these colours.

Solution 32

Exercise Ex. 17B

Question 1

The dimensions of a metallic cuboid are 100 cm x 80 cm x 64 cm. It is melted and recast into a cube. Find the surface area of the cube.Solution 1

Question 2

A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.Solution 2

Question 3

Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.Solution 3

Question 4

A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls thus formed.Solution 4

Radius of the cone = 12 cm and its height = 24 cm

Volume of cone = 

Question 5

The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder.Solution 5

Question 6

The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed.Solution 6

Internal radius = 3 cm and external radius = 5 cm

Hence, height of the cone = 4 cmQuestion 7

A copper rod of diameter 2 cm and length 10 cm is drawn into a wire of uniform thickness and length 10 m. Find the Thickness of the wire.Solution 7

Question 8

A hemispherical bowl of internal diameter 30cm contains some liquid. This liquid is to be filled into cylindrical shaped bottles each of diameter 5 cm and height 6 cm. Find the number of bottles necessary to empty the bowl.Solution 8

Inner radius of the bowl = 15 cm

Volume of liquid in it = 

Radius of each cylindrical bottle = 2.5 cm and its height = 6 cm

Volume of each cylindrical bottle

Required number of bottles = 

   Hence, bottles required = 60Question 9

A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.Solution 9

Radius of the sphere= 

Let the number of cones formed be n, then

            Hence, number of cones formed = 504Question 10

A spherical cannon ball 28 cm in diameter is melted and recast into right circular conical mould, base of which is 35 cm in diameter. Find the height of the cone.Solution 10

Radius of the cannon ball = 14 cm

Volume of cannon ball = 

Radius of the cone = 

Let the height of cone be h cm

Volume of cone = 

            Hence, height of the cone = 35.84 cmQuestion 11

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5cm and 2 cm. Find the radius of third ball.Solution 11

Let the radius of the third ball be r cm, then,

           Volume of third ball = Volume of spherical ball volume of 2 small balls

Question 12

A spherical shell of lead whose external and internal diameters are respectively 24 cm and 18 cm, is melted ad recast into a right circular cylinder 37 cm high. Find the diameter of the base of the cylinder.Solution 12

External radius of shell = 12 cm and internal radius = 9 cm

Volume of lead in the shell = 

Let the radius of the cylinder be r cm

Its height = 37 cm

Volume of cylinder = 

Hence diameter of the base of the cylinder = 12 cmQuestion 13

A hemisphere of lead of radius 9 cm is cast intoa right circular cone of height 72 cm. Find the radius of the base of the cone.Solution 13

Volume of hemisphere of radius 9 cm

Volume of circular cone (height = 72 cm)

Volume of cone = Volume of hemisphere

Hence radius of the base of the cone = 4.5 cmQuestion 14

A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm. Find the number of cubes so formed.Solution 14

Diameter of sphere = 21 cm

Hence, radius of sphere = 

Volume of sphere =  = 

Volume of cube = a³ = (1 1 1) 

Let number of cubes formed be n

Volume of sphere = n Volume of cube

Hence, number of cubes is 4851.Question 15

How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?Solution 15

Volume of sphere (when r = 1 cm) =  = 

Volume of sphere (when r = 8 cm) =  = 

Let the number of balls = n

Question 16

A solid sphere of radius 3cm is melted and then cast into small spherical balls, each of diameter 0.6 cm. Find the number of small balls so obtained.Solution 16

Radius of sphere = 3 cm

Volume of sphere = 

Radius of small sphere = 

Volume of small sphere = 

Let number of small balls be n

Hence, the number of small balls = 1000.Question 17

The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.Solution 17

Diameter of sphere = 42 cm

Radius of sphere = 

Volume of sphere = 

Diameter of cylindrical wire = 2.8 cm

Radius of cylindrical wire = 

Volume of cylindrical wire = 

Volume of cylindrical wire = volume of sphere

Hence length of the wire 63 m.Question 18

The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.Solution 18

Diameter of sphere = 18 cm

Radius of copper sphere = 

Length of wire = 108 m = 10800 cm

Let the radius of wire be r cm

But the volume of wire = Volume of sphere

Hence the diameter = 2r = (0.3 2) cm = 0.6 cmQuestion 19

A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.Solution 19

Question 20

Solution 20

Question 21

The rain water from a roof of 44 m x 20 m drains into a cylindrical tank having diameter of base 4 m and height 3.5 m. If the tank is just full, find the rainfall in cm.Solution 21

Question 22

Solution 22

Question 23

A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder, in cubic metres.Solution 23

Question 24

Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour.Solution 24

Question 25

Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 22 m wide. Determine the time in which the level of water in the tank will rise by 7 cm.Solution 25

Question 26

Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 4 km/hr. How much area will it irrigate in 10 minutes if 8 cm of standing water is needed for irrigation?Solution 26

Question 27

A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field which is 12 m in diameter and 2.5 m deep. If water flows through the pipe at the rate of 3.6 km/h, in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate ofRs. 0.07. use Solution 27

Height of cylindrical tank = 2.5 m

Its diameter = 12 m, Radius = 6 m

Volume of tank = 

Water is flowing at the rate of 3.6 km/ hr = 3600 m/hr

Diameter of pipe = 25 cm, radius = 0.125 m

Volume of water flowing per hour

Question 28

Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute. Find the rate of flow of water in the pipe in km/hr.Solution 28

Question 29

150 spherical marbles, each of diameter 14 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.Solution 29

Question 30

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.Solution 30

Let the number of marbles be n

n volume of marble = volume of rising water in beaker

Question 31

In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment. What value of the villagers is reflected here?Solution 31

Question 32

In a corner of a rectangular field with dimensions 35 m x 22 m, a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.Solution 32

Question 33

A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 18 cm and diameter 49 cm to cover its whole surface. Find the length and the volume of the wire. If the density of copper be 8.8 g per cu-cm, find the weight of the wire.Solution 33

Question 34

A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of it as found appropriate)Solution 34

Exercise Ex. 17C

Question 1

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 16 cm and 12 cm. Find the capacity of the glass.Solution 1

Question 2

The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area. [Use π = 3.14.]Solution 2

Question 3

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find

(i) the volume of water which can completely fill the bucket;

(ii) the area of the metal sheet used to make the bucket.Solution 3

Question 4

A container, open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs. 21 per litre.Solution 4

Question 5

A container, open at the top and made up of metal sheet, is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the cost of metal sheet used to make the container, if it costs Rs. 10 per 100 cm².Solution 5

Question 6

The radii of the circular ends of a solid frustum of a cone are 33cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area. Take .Solution 6

Here R = 33 cm, r = 27 cm and l = 10 cm

Capacity of the frustum 

Total surface area = 

Question 7

A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm, respectively. Find how many litres of water can the bucket hold. Take Solution 7

Height = 15 cm, R =  and  

Capacity of the bucket = 

Quantity of water in bucket = 28.49 litresQuestion 8

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the bucket if the cost of metal sheet used is Rs. 15 per . Use Solution 8

R = 20 cm, r = 8 cm and h = 16 cm

Total surface area of container = 

Cost of metal sheet used = Question 9

A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10cm respectively. Find the cost of milk which can completely fill the bucket at the rate of Rs. 20 per litre and the cost of metal sheet used if it costs Rs. 10 per 100Solution 9

R = 15 cm, r = 5 cm and h = 24 cm

(i)Volume of bucket = 

Cost of milk = Rs. (8.164 20) = Rs. 163.28

(ii)Total surface area of the bucket

Cost of sheet = Question 10

A container in the shape of a frustum of a cone having diameters of its two circular faces as 35 cm and 30 cm and vertical height 14 cm, is completely filled with oil. If each cm’ of oil has mass 1.2 g, then find the cost of oil in the container if it costs Rs.40 per kg.Solution 10

Question 11

A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket.Solution 11

Question 12

The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm < 14). If the volume of bucket is 5390 cm³, find the value of r.Solution 12

Question 13

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area. [Use π = 3.14.]Solution 13

Question 14

A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required. Take Solution 14

R = 10cm, r = 3 m and h = 24 m

Let l be the slant height of the frustum, then

Quantity of canvas = (Lateral surface area of the frustum)

                                + (lateral surface area of the cone)

Question 15

A tent consists of a frustum of a cone, surmounted by a cone. If the diameters of the upper and lower circular ends of the frustum be 14 m and 26 m respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12mm, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular ends of the frustum and the base of the surmounted conical portion are equal.)Solution 15

ABCD is the frustum in which upper and lower radii are EB = 7 m and FD = 13 m

Height of frustum= 8 m

Slant height of frustum

Radius of the cone = EB = 7 m

Slant height of cone = 12 m

Surface area of canvas required

Question 16

The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, find its volume and curved surface area.Solution 16

Question 17

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution 20

Question 21

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet requited to make the funnel.

Solution 21

Exercise Ex. 17D

Question 1

A river 1.5 m deep and 36 m wide is flowing at the rate of 3.5 km/hr. Find the amount of water (in cubic metres) that runs into the sea per minute.Solution 1

Question 2

The volume of a cube is 729 cm³. Find its surface area.Solution 2

Question 3

How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?Solution 3

Question 4

Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. Find the edge of the new cube formed.Solution 4

Question 5

Five identical cubes, each of edge 5 cm, are placed adjacent to each other. Find the volume of the resulting cuboid.Solution 5

Question 6

The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their surface areas.Solution 6

Question 7

Solution 7

Question 8

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If the volume of the cylinder is 12936 cm³, find the radius of the base of the cylinder.Solution 8

Question 9

The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. Find the ratio of their volumes.Solution 9

Question 10

66 cubic cm of silver is drawn into a wire 1 mm in diameter. Calculate the length of the wire in metres.Solution 10

Question 11

If the area of the base of a right circular cone is 3850 cm² and its height is 84 cm, find the slant height of the cone.Solution 11

Question 12

A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. Calculate the radius of the base of the cone.Solution 12

Question 13

A right cylindrical vessel is full of water. How many right cones having the same radius and height as those of the right cylinder will be needed to store that water?Solution 13

Question 14

The volume of a sphere is 4851 cm³. Find its curved surface area.Solution 14

Question 15

The curved surface area of a sphere is 5544 cm³. Find its volume.Solution 15

Question 16

The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio of their volumes.Solution 16

Question 17

A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. Find the number of spherical balls obtained.Solution 17

Question 18

How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm x 11 cm x 12 cm?Solution 18

Question 19

A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each. How many spheres are formed?Solution 19

Question 20

A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. Find the radius of the base of the cone.Solution 20

Question 21

A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire.Solution 21

Question 22

The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the slant height of the frustum.Solution 22

Question 23

Find the ratio of the volume of a cube to that of a sphere which will fit inside it.Solution 23

Question 24

Find the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?Solution 24

Question 25

Two cubes each of volume 125 cm³ are joined end to end to form a solid. Find the surface area of the resulting cuboid.Solution 25

Question 26

Three metallic cubes whose edges are 3 cm, 4 cm and 5 cm, are melted and recast into a single large cube. Find the edge of the new cube formed.Solution 26

Question 27

A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, find its width.Solution 27

Question 28

A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used, at the rate of Rs. 25 per metre.Solution 28

Question 29

A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the volume of wood in the toy.Solution 29

Question 30

Solution 30

Question 31

A hollow sphere of external and internal diameters 8 cm and 4 cm respectively is melted into a solid cone of base diameter 8 cm. Find the height of the cone.Solution 31

Question 32

A bucket of height 24 cm is in the form of frustum of a cone whose circular ends are of diameter 28 cm and 42 cm. Find the cost of milk at the rate of Rs. 30 per litre, which the bucket can hold.Solution 32

Question 33

The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of same diameter. The height of the cone is 2.8 m. Find the outer surface area of the building.Solution 33

Question 34

A metallic solid right circular cone is of height 84 cm and the radius of its base is 21 cm. It is melted and recast into a solid sphere. Find the diameter of the sphere.Solution 34

Question 35

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.Solution 35

Question 36

If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm, find its capacity and total surface area.Solution 36

Question 37

A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm³ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket. (Useπ = 3.14.)Solution 37

Question 38

The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs.1.40 per cm².Solution 38

Question 39

Solution 39

Question 40

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water

(i) displaced out of the cylinder

(ii) left in the cylinder.Solution 40

Exercise MCQ

Question 1

Choose the correct answer in each of the following:

A cylindrical pencil sharpened at one edge is the

combination of

(a) a cylinder and a cone

(b) a cylinder and frustum of a cone

(c) a cylinder and a hemisphere

(d) two cylindersSolution 1

Correct option: (a)

A cylindrical pencil sharpened at one edge is the combination of a cylinder and a cone. Observe the figure, the lower portion is a cylinder and the upper tapering portion is a cone.Question 2

A shuttlecock used for playing badminton is the combination of

(a) cylinder and a hemisphere

(b) frustum of a cone and a hemisphere

(c) a cone and a hemisphere

(d) a cylinder and a sphere

Solution 2

Correct option: (b)

A shuttlecock used for playing badminton is the combination of a frustum of a cone and a hemisphere, the lower portion being the hemisphere and the portion above that being the frustum of the cone.Question 3

A funnel is the combination of

(a) a cylinder and a cone

(b) a cylinder and a hemisphere

(c) a cylinder and frustum of a cone

(d) a cone and a hemisphere

Solution 3

Correct option: (c)

A funnel is the combination of a cylinder and frustum of a cone. The lower portion is cylindrical and the upper portion is a frustum of a cone.Question 4

A surahi is a combination of

(a) a sphere and a cylinder

(b) a hemisphere and a cylinder

(c) a cylinder and a cone

(d) two hemispheres Surahi

Solution 4

Correct option: (a)

A surahi is a combination of a sphere and a cylinder, the lower portion is the sphere and the upper portion is the cylinder.Question 5

The shape of a glass (tumbler) is usually in the form of

(a) a cylinder

(b) frustum of a cone

(c) a cone

(d) a sphere Glass

Solution 5

Correct option: (b)

The shape of a glass (tumbler) is usually in the form of a frustum of a cone.Question 6

The shape of a gill in the gilli-danda game is a combination of

(a) a cone and a cylinder

(b) two cylinders Gilli

(c) two cones and a cylinder

(d) two cylinders and a cone

Solution 6

Correct option: (c)

The shape of a gill in the gilli-danda game is a combination of two cones and a cylinder. The cones at either ends with the cylinder in the middle.Question 7

A plumbline (sahul) is the combination of

(a) a hemisphere and a cone

(b) a cylinder and a cone

(c) a cylinder and frustum of a cone

(d) a cylinder and a sphere Plumbline

Solution 7

Correct option: (a)

A plumbline (sahul) is the combination of a hemisphere and a cone, the hemisphere being on top and the lower portion being the cone.Question 8

A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left over is called

(a) a cone

(b) a sphere

(c) a cylinder

(d) frustum of a cone

Solution 8

Correct option: (d)

A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left over is called the frustum of a cone.Question 9

During conversion of a solid from one shape to another, the volume of the new shape will

(a) decrease

(b) increase

(c) remain unaltered

(d) be doubledSolution 9

Correct option: (c)

During conversion of a solid from one shape to another, the volume of the new shape will remain altered.Question 10

In a right circular cone, the cross section made by a plane parallel to the base is a

(a) sphere

(b) hemisphere

(c) circle

(d) a semicircleSolution 10

Correct option: (c)

In a right circular cone, the cross section made by a plane parallel to the base is a circle.Question 11

A solid piece of iron in the form of a cuboid of dimensions (cccm) is moulded to form a solid sphere. The radius of the sphere is

(a) 19 cm

(b) 21 cm

(c) 23 cm

(d) 25 cmSolution 11

Question 12

The radius (in cm) of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is

(a) 2.1

(b) 4.2

(c) 8.4

(d) 1.05Solution 12

Question 13

A metallic solid sphere of radius 9 cm is melted to form a solid cylinder of radius 9 cm. The height of the cylinder is

(a) 12 cm

(b) 18 cm

(c) 36 cm

(d) 96 cmSolution 13

Question 14

A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is,

Solution 14

Question 15

The number of solid spheres, each of diameter 6 cm, that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is

(a) 2

(b) 4

(c) 5

(d) 6Solution 15

Question 16

The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is

(a) 64 : 27

(b) 16:9

(c) 4 :3

(d) 16³ : 9³Solution 16

Question 17

If the surface area of a sphere is 616 cm², its diameter (in cm) is

(a) 7

(b) 14

(c) 28

(d) 56Solution 17

Question 18

If the radius of a sphere becomes 3 times then its volume will become

(a) 3 times

(b) 6 times

(c) 9 times

(d) 27 timesSolution 18

Question 19

If the height of a bucket in the shape of frustum of a cone is 16 cm and the diameters of its two circular ends are 40 cm and 16 cm then its slant height is

Solution 19

Question 20

A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged then the water level rises by

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cmSolution 20

Question 21

A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is  

(a) 1 : 2

(b) 1 : 4

(c) 1 : 6

(d) 1 : 8Solution 21

Question 22

The radii of the circular ends of a bucket of height 40 cm are 24 cm and 15 cm. The slant height (in cm) of the bucket is

(a) 41

(b) 43

(c) 49

(d) 51Solution 22

Question 23

A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two parts are equal then the ratio of its radius and the slant height of the conical part is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1Solution 23

Question 24

If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4: 1Solution 24

Question 25

A cubical ice-cream brick of edge 22 cm is to be distributed among some children by filling ice-cream cones of radius 2 cm and height 7 cm up to its brim. How many children will get the ice-cream cones?

(a) 163

(b) 263

(c) 363

(d) 463Solution 25

Question 26

(a) 11000

(b) 11100

(c) 11200

(d) 11300Solution 26

Question 27

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 6 cmSolution 27

Question 28

The diameters of two circular ends of a bucket are 44 cm and 24 cm, and the height of the bucket is 35 cm. The capacity of the bucket is

(a) 31.7 litres

(b) 32.7 litres

(c) 33.7 litres

(d) 34.7 litresSolution 28

Question 29

The slant height of a bucket is 45 cm and the radii of its top and bottom are 28 cm and 7 cm respectively. The curved surface area of the bucket is

(a) 4953 cm²

(b) 4952 cm²

(c) 4951 cm²

(d) 4950 cm²Solution 29

Question 30

The volumes of two spheres are in the ratio 64:27. The ratio of their surface area is

(a) 9:16

(b) 16:9

(c) 3:4

(d) 4:3Solution 30

Question 31

(a) 142296

(b) 142396

(c) 142496

(d) 142596Solution 31

Question 32

A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively, is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 8 cmSolution 32

Question 33

A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. The capacity of the capsule is 

(a) 0.33 cm³

(b) 0.34 cm³

(c) 0.35 cm³

(d) 0.36 cm³Solution 33

Question 34

The length of the longest pole that can be kept in a room (12 m × 9 m × 8 m) is

(a) 29 m

(b) 21 m

(c) 19 m

(d) 17 mSolution 34

Question 35

Solution 35

Question 36

The volume of a cube is 2744 cm³. Its surface area is

(a) 196 cm²

(b) 1176 cm²

(c) 784 cm²

(d) 588 cm²Solution 36

Question 37

The total surface area of a cube is 864 cm². Its volume is

(a) 3456 cm³

(b) 432 cm³

(c) 1728 cm³

(d) 3456 cm³Solution 37

Question 38

How many bricks each measuring (25 cm × 11.25 cm × 6 cm) will be required to construct a wall (8 m × 6 m × 22.5 cm)?

(a) 8000

(b) 6400

(c) 4800

(d) 7200Solution 38

Question 39

The area of the base of a rectangular tank is 6500 cm² and the volume of water contained in it is 2.6 m³. The depth of water in the tank is.

(a) 3.5 m

(b) 4 m

(c) 5 m

(d) 8 mSolution 39

Question 40

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 m³. The breadth of the wall is

(a) 30 cm

(b) 40 cm

(c) 22.5 cm

(d) 25 cmSolution 40

Question 41

If the areas of three adjacent faces of a cuboid are x, y, z respectively then the volume of the cuboid is

Solution 41

Question 42

(a) 361 cm²

(b) 125 cm²

(c) 236 cm²

(d) 486 cm²Solution 42

Question 43

If each edge of a cube is increased by 50%, the percentage increase in the surface area is

(a) 50%

(b) 75%

(c) 100%

(d) 125%Solution 43

Question 44

How many bags of grain can be stored in a cuboidal granary (8 m× 6m × 3 m), if each bag occupies a space of 0.64 m³?

(a) 8256

(b) 90

(c) 212

(d) 225Solution 44

Question 45

A cube of side 6 cm is cut into a number of cubes each of side 2 cm. The number of cubes formed is

(a) 6

(b) 9

(c) 12

(d) 27Solution 45

Question 46

In a shower, 5 cm of rain falls. The volume of the water that falls on 2 hectares of ground, is

(a) 100 m³

(b) 10 m³

(c) 1000 m³

(d) 10000 m³Solution 46

Question 47

Two cubes have their volumes in the ratio 1: 27. The ratio of their surface areas is

(a) 1 : 3

(b) 1 : 8

(c) 1 : 9

(d) 1 : 18Solution 47

Question 48

The diameter of the base of a cylinder is 4 cm and its height is 14 cm. The volume of the cylinder is

(a) 176 cm³

(b) 196 cm³

(c) 276 cm³

(d) 352 cm³Solution 48

Question 49

The diameter of a cylinder is 28 cm and its height is 20 cm. The total surface area of the cylinder is

(a) 2993 cm²

(b) 2992 cm²

(c) 2292 cm²

(d) 2229 cm²Solution 49

Question 50

The height of a cylinder is 14 cm and its curved surface area is 264 cm². The volume of the cylinder is

(a) 308 cm³

(b) 396 cm³

(c) 1232 cm³

(d) 1848 cm³Solution 50

Question 51

The curved surface area of a cylinder is 1760 cm² and its base radius is 14 cm. The height of the cylinder is

(a) 10 cm

(b) 15 cm

(c) 20 cm

(d) 40 cmSolution 51

Question 52

The ratio of the total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm is

(a) 2 :1

(b) 3:1

(c) 4:1

(d) 5:1Solution 52

Question 53

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. The height of the pillar is

(a) 4 m

(b) 5 m

(c) 6 m

(d) 7 mSolution 53

Question 54

The ratio between the radius of the base and the height of the cylinder is 2 : 3. If its volume is 1617 cm³, the total surface area of the cylinder is

(a) 308 cm²

(b) 462 cm²

(c) 540 cm²

(d) 770 cm²Solution 54

Question 55

The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5 : 3. The ratio of their volumes is

(a) 27 : 20

(b) 20 : 27

(c) 4 :9

(d) 9 : 4Solution 55

Question 56

Two circular cylinders of equal volume have their heights in the ratio 1:2. The ratio of their radii is

Solution 56

Question 57

The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved surface area is

Solution 57

Question 58

The diameter of the base of a cone is 42 cm and its volume is 12936 cm³. Its height is

(a) 28 cm

(b) 21 cm

(c) 35 cm

(d) 14 cmSolution 58

Question 59

The area of the base of a right circular cone is 154 cm² and its height is 14 cm. Its curved surface area is

Solution 59

Question 60

On increasing each of the radius of the base and the height of a cone by 20% its volume will be increased by

(a) 20%

(b) 40%

(c) 60%

(d) 72.8%Solution 60

Question 61

The radii of the base of a cylinder and a cone are in the ratio 3:4. If they have their heights in the ratio 2 : 3, the ratio between their volumes is

(a) 9 :8

(b) 3:4

(c) 8 :9

(d) 4 : 3Solution 61

Question 62

A metallic cylinder of radius 8 cm and height 2 cm is melted and converted into a right circular cone of height 6 cm. The radius of the base of this cone is

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) 8 cmSolution 62

Question 63

The height of a conical tent is 14 m and its floor area is 346.5 m². How much canvas, 1.1 m wide, will be required for it?

(a) 490 m

(b) 525 m

(c) 665 m

(d) 860 mSolution 63

Question 64

The diameter of a sphere is 14 cm. Its volume is

Solution 64

Question 65

The ratio between the volumes of two spheres is 8: 27. What is the ratio between their surface areas?

(a) 2:3

(b) 4:5

(c) 5:6

(d) 4: 9Solution 65

Question 66

A hollow metallic sphere with external diameter 8 cm and internal diameter 4 cm is melted and moulded into a cone having base radius 8 cm. The height of the cone is

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 18 cmSolution 66

Question 67

A metallic cone having base radius 2.1 cm and height 8.4 cm is melted and moulded into a sphere. The radius of the sphere is

(a) 2.1 cm

(b) 1.05 cm

(c) 1.5 cm

(d) 2 cmSolution 67Question 68

The volume of a hemisphere is 19404 cm³. The total surface area of the hemisphere is

(a) 4158 cm²

(b) 16632 cm²

(c) 8316 cm²

(d) 3696 cm²Solution 68

Correct option: (a)

Question 69

The surface area of a sphere is 154 cm². The volume of the sphere is all

Solution 69

Correct option: (a)

Question 70

The total surface area of a hemisphere of radius 7 cm is

(588 𝜋) cm²

(392 𝜋) cm²

(147 𝜋) cm²

(598 𝜋) cm²Solution 70

Question 71

The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm. Its volume is

(a) 60060 cm³

(b) 80080 cm³

(c) 70040 cm³

(d) 80160 cm³Solution 71

Question 72

If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high then its surface area is

(a) 1815.3 cm²

(b) 1711.3 cm²

(c) 2025.3 cm²

(d) 2360 cm²Solution 72

Question 73

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, the total area of canvas required is

(a) 1760 m²

(b) 2640 m²

(c) 3960 m²

(d) 7920 m²Solution 73

Question 74

Match the following columns:

Column I	Column II
A solid metallic sphere of radius 8 cm is melted and the material is used to make solid right cones with height 4 cm and radius of the base 8 cm. How many cones are formed?	(p) 18
A 20-in-deep well with diameter 14 m is dug up and the earth from digging is evenly spread out to form a platform 44 m by 14 in. The height of the platform is …… m.	(q) 8
A sphere of radius 6 cm is melted and recast into the shape of a cylinder of radius 4 cm. Then, the height of the cylinder is…… cm.	(r) 16 : 9
The volumes of two spheres are in the ratio 64: 27. The ratio of their surface areas is …..	(s) 5

The correct answer is

(a)-….., (b)- ….. , (c)- ….., (d)- ……Solution 74

Question 75

Match the following columns:

Column I	Column II
The radii of the circular ends of a bucket in the form of frustum of a cone of height 30 cm are 20 cm and 10 cm respectively. The capacity of the bucket is …….cm3.	(p) 2418 π
The radii of the circular ends of a conical bucket of height 15 cm are 20 cm and 12 cm respectively. The slant height of the bucket is… cm.	(q) 22000
The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is …cm2.	(r) 12
Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ….cm.	(s) 17

The correct answer is

(a)-….., (b)- ….. , (c)- ….., (d)- ……Solution 75

Question 76

Assertional- and-Resons type

Each question consists of two statements, namely,

Assertion (A) and Reason (R). For selecting the correct

answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If the radii of the circular ends of a bucket 24 cm high are 15 cm and 5 cm respectively, then the surface area of the bucket is 545π cm2.	If the radii of the circular ends of the frustum of a cone are R and r respectively and its height is h, then its surface area is π (R2 + r2 + l(R-r), Where l2 = h2+(R-r)2

The correct answer is (a)/(b)/(c) /(d) .Solution 76

Question 77

Assertional- and-Resons type

Each question consists of two statements, namely,

Assertion (A) and Reason (R). For selecting the correct

answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
A hemisphere of radius 7 cm is to be painted outside on the surface. The total cost of painting at it Rs. 5 per cm2 is Rs. 2300.	The total surface area hemisphere is 3πr2.

The correct answer is (a)/(b)/(c) /(d) .Solution 77

Question 78

Assertional- and-Resons type

Each question consists of two statements, namely,

Assertion (A) and Reason (R). For selecting the correct

answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The number of coins 1.75 cm in diameter and 2 mm thick from a melted cuboid (10 cm× 5.5 cm × 3.5 cm) is 400.	Volume of a cylinder of base radius r and height h is given byV = (πr2h) cubic units.And, area of a cuboid= (l × b × h) cubic units.

The correct answer is (a)/(b)/(c) /(d) .Solution 78

Question 79

Assertional- and-Resons type

Each question consists of two statements, namely,

Assertion (A) and Reason (R). For selecting the correct

answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If the volumes of two spheres are in the ratio 27:8 then their surface areas are in the ratio 3:2.

The correct answer is (a)/(b)/(c) /(d) .Solution 79

Question 80

Assertional- and-Resons type

Each question consists of two statements, namely,

Assertion (A) and Reason (R). For selecting the correct

answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The curved surface area of a cone of base radius 3 cm and height 4cm is (15π) cm2.	Volume of a cone = πr2h.

The correct answer is (a)/(b)/(c) /(d) .Solution 80

Exercise FA

Question 1

Find the number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metallic cylinder of height 45 cm and diameter 4 cm.Solution 1

Question 2

Two right circular cylinders of equal volumes have their heights in the ratio 1: 2. What is the ratio of their radii?Solution 2

Question 3

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, find the total area of the canvas required.Solution 3

Question 4

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm respectively. Find the curved surface area of the bucket.Solution 4

Question 5

A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls formed.Solution 5

Question 6

A hemispherical bowl of internal diameter 30 cm is full of a liquid. This liquid is filled into cylindrical-shaped bottles each of diameter 5 cm and height 6 cm. How many bottles are required?Solution 6

Question 7

A solid metallic sphere of diameter 21 cm is melted and recast into Milan cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.Solution 7

Question 8

The diameter of a sphere is 42 cm. it is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.Solution 8

Question 9

A drinking glass is in the shape of frustum of a cone of height 21 cm with 6 cm and 4 cm as the diameters of its two circular ends. Find the capacity of the glass.Solution 9

Question 10

Two cubes, each of volume 64 cm³, are joined end to end. Find the total surface area of the resulting cuboid.Solution 10

Question 11

Solution 11

Question 12

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.Solution 12

Question 13

A hemispherical bowl of internal radius 9 cm is full of water. This water is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm. Find the number of bottles needed to fill the whole water of the bowl.Solution 13

Question 14

Solution 14

Question 15

The slant height of the frustum of a cone is 4 cm and the perimeters (i.e., circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.Solution 15

Question 16

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the surface area of the solid.Solution 16

Question 17

From a solid cylinder whose height is 15 cm and diameter 16 cm, a conical cavity of the same height and same “diameter is hollowed out. Find the total surface area of the remaining solid. (Use 𝜋 = 3.14.)Solution 17

Question 18

A solid rectangular block of dimensions 4.4 m, 2.6 in and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.Solution 18

Question 19

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.Solution 19

Question 20

A fanner connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, in how much time will the tank be filled completely?Solution 20

Exercise Ex. 16A

Question 1

Solution 1

Question 2

The circumference of a circle is 22 cm. Find the area of its quadrant.Solution 2

Question 3

What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm?Solution 3

Question 4

If the area of a circle is numerically equal to twice its circumference, then what is the diameter of the circle?Solution 4

Question 5

What is the perimeter of a square which circumscribes a circle of radius a cm?Solution 5

Since square circumscribes a circle of radius a cm, we have

Side of the square = 2 ⨯ radius of circle = 2a cm

Then, Perimeter of the square = (4 ⨯ 2a) = 8a cmQuestion 6

Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.Solution 6

Question 7

Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm.Solution 7

Question 8

Find the area of a circle whose circumference is 8π.Solution 8

Question 9

Find the perimeter of a semicircular protractor whose diameter is 14 cm.Solution 9

Question 10

Find the radius of a circle whose perimeter and area are numerically equal.Solution 10

Question 11

The radii of two circles are 19 cm and 9 cm. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.Solution 11

Question 12

The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles.Solution 12

Question 13

Find the area of the sector of a circle having radius 6 cm and of angle 30°. [Take π = 3.14.]Solution 13

Question 14

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.Solution 14

Question 15

The circumferences of two circles are in the ratio 2 : 3. What is the ratio between their areas?Solution 15

Question 16

The areas of two circles are in the ratio 4 : 9. What is the ratio between their circumferences?Solution 16

Question 17

A square is inscribed in a circle. Find the ratio of the areas of the circle and the square.Solution 17

Question 18

The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.Solution 18

Question 19

A pendulum swings through an angle of 30 and describes an arc 8.8cm in length. Find the length of the pendulumSolution 19

Length of the pendulum = radius of sector = r cm

Question 20

The minute hand of a clock is 15cm long. Calculate the area swept by it in 20minutes. Take  = 3.14Solution 20

Angle described by the minute hand in 60 minutes = 360°

Angle described by minute hand in 20 minutes

Required area swept by the minute hand in 20 minutes

                                          =Area of the sector(with r = 15 cm and  = 120°)

Question 21

A sector of 56^o, cut out from a circle, contains . Find the radius of the circle.Solution 21

= 56^o and let radius is r cm

Area of sector = 

Hence radius= 6cmQuestion 22

The area of the sector of a circle of radius 10.5cm is . Find the central angle of the sector.Solution 22

Area of the sector of circle = 

Radius = 10.5 cm

Question 23

The perimeter of a certain sector of circle of radius 6.5cm is 31 cm. Find the area of the sector.Solution 23

Let sector of circle is OAB

Perimeter of a sector of circle =31 cm

OA + OB + length of arc AB = 31 cm

 6.5 + 6.5 + arc AB = 31 cm

  arc AB = 31 – 13

            = 18 cm

Question 24

The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44cm in length.Solution 24

Length of arc of circle = 44 cm

Radius of circle = 17.5 cm

Area of sector = 

Question 25

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimensions 14 cm x 7 cm. Find the area of the remaining cardboard.Solution 25

Question 26

In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. [Use π = 3.14.]

Solution 26

Question 27

From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semicircular portion with BC as diameter is cut off.  Find the area of the remaining paper.Solution 27

Question 28

In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region.

Solution 28

Question 29

In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region.

Solution 29

Question 30

In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm with centre O. If ∠POQ = 30°, find the area of the shaded region.

Solution 30

Question 31

In the given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution 31

Question 32

In the given figure, the shape of the top of a table is that of a sector of a circle with centre 0 and ∠A0B =90°. If AO =0B = 42 cm, then find the perimeter of the top of the table.

Solution 32

Question 33

In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm. Find the area of shaded region.

Solution 33

Question 34

In the given figure, AOBCA represents a quadrant of a circle of radius 3.5 cm with centre O. Calculate the area of the shaded portion.

Solution 34

Shaded area = (area of quadrant) – (area of DAOD)

Question 35

Find the perimeter of the shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles.

Solution 35

Question 36

In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square.Solution 36

Question 37

In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the

(i) perimeter,

(ii) area of the shaded region.

Solution 37

Question 38

In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [Use π = 3.14.]

Solution 38

Question 39

In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. [Use π = 3.14.]

Solution 39

Question 40

In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.

Solution 40

Question 41

Find the area of a quadrant of a circle whose circumference is 44 cm.Solution 41

Question 42

In the given figure, find area of the shaded region, where ABCD is a square of side 14cm and all circles are of the same diameter.

Solution 42

Side of the square ABCD = 14 cm

Area of square ABCD = 14 14 = 196 

Radius of each circle = 

Area of the circles = 4 area of one circle

Area of shaded region = Area of square – area of 4 circles

= 196 – 154 = 42Question 43

Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle.

Solution 43

Question 44

A wire is bent to form a square enclosing an area of 484 m². Using the same wire, a circle is formed. Find the area of the circle.Solution 44

Question 45

A square ABCD is inscribed in a circle of radius r. Find the area of the square.Solution 45

Question 46

The cost of fencing a circular field at the rate of Rs.25 per metre is Rs.5500. The field is to be ploughed at the rate of 50 paise per m². Find the cost of ploughing the field.

Solution 46

Question 47

A park is in the form of a rectangle 120m by 90m. At the centre of the park there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950. Find the radius of the circular lawn. (given: =3.14)

Solution 47

Area of rectangle = (120 × 90) 

                          = 10800 

Area of circular lawn = [Area of rectangle – Area of park excluding circular lawn]

                              = [10800 – 2950] = 7850 

Area of circular lawn = 7850 

Hence, radius of the circular lawn = 50 mQuestion 48

In the given figure, PQSR represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed.

Solution 48

Area of flower bed = (area of quadrant OPQ)

-(area of the quadrant ORS)

Question 49

In the given figure O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is draw. If AC = 54cm and BC = 10 cm, find the area of the shaded region.

Solution 49

Diameter of bigger circle = AC = 54 cm

Radius of bigger circle = 

Diameter AB of smaller circle

Radius of smaller circle = 

Area of bigger circle = 

                             = 2291. 14 

Area of smaller circle = 

                              = 1521. 11 

Area of shaded region = area of bigger circle – area of smaller circle

Question 50

From a thin metallic piece in the shape of a trapezium ABCD in which AB ‖⃦ CD and ∠BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.

Solution 50

Question 51

Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.

Solution 51

Exercise Ex. 16B

Question 1

The circumference of a circle is 39.6 cm. find its area.Solution 1

Circumference of circle = 2r = 39.6 cm

Question 2

The area of a circle is 98.56 cm². Find its circumference.Solution 2

Question 3

The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle.Solution 3

Question 4

A copper wire when bent in the form of a square encloses an area of 484 . The same wire is now bent in the form of a circle. Find the area enclosed by the circle.Solution 4

Area of square = 

Perimeter of square = 4 side = 4 22 = 88 cm

Circumference of circle = Perimeter of square

Question 5

A wire when bent in the form of an equilateral triangle encloses an area of 121. The same wire is bent to form a circle. Find the area enclosed by the circle.Solution 5

Area of equilateral = 

Perimeter of equilateral triangle = 3a = (3 22) cm

                                              = 66 cm

Circumference of circle = Perimeter of circle

2r = 66 r = 

Area of circle = 

                   = Question 6

The length of a chain used as the boundary of a semicircular park is 108 m. Find the area of the park.Solution 6

Question 7

The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumferences of the circles.Solution 7

Let the radii of circles be x cm and (7 – x) cm

Circumference of the circles are 26 cm and 18 cmQuestion 8

Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.Solution 8

Area of outer circle = 

                            = 1662.5 

Area of ring = Outer area – inner area

                 = (1662.5 – 452.5) Question 9

A path of 8m width runs around the outside of a circular park whose radius is 17m. Find the area of the path.Solution 9

Inner radius of the circular park = 17 m

Width of the path = 8 m

Outer radius of the circular park = (17 + 8)m = 25 m

Area of path = 

Question 10

A racetrack is in the form of a ring whose inner circumference is 352m and outer circumference is 396m. Find the width and the area of the track.Solution 10

Let r m and R m be the radii of inner circle and outer boundaries respectively.

Then, 2r = 352 and 2R = 396

Width of the track = (R – r) m

Area the track = 

Question 11

A sector is cut from a circle of radius 21cm. The angle of the sector is 150^o. Find the length of the arc and the area of the sector.Solution 11

Length of the arc 

Length of arc = 

Area of the sector = 

Question 12

The area of the sector of a circle of radius 10.5 cm is 69.3 cm². Find the central angle of the sector.Solution 12

Question 13

The length of an arc of a circle, subtending an angle of 54° at the centre, is 16.5 cm. Calculate the radius, circumference and area of the circle.Solution 13

Length of arc = 

Circumference of circle = 2  r

Area of circle = 

Question 14

The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major segments.Solution 14

Question 15

Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment. Take = 3.14, Solution 15

OAB is equilateral.

So, AOB = 60

Length of arc BDA = (2 12 – arc ACB) cm

                         = (24 – 4) cm = (20) cm

                         = (20 3.14) cm = 62.8 cm

Area of the minor segment ACBA

Question 16

A chord 10cm long is drawn in a circle whose radius is . Find the areas of both the segments. Take  = 3.13Solution 16

Let OA = , OB = 

And AB = 10 cm

Area of AOB = 

Area of minor segment = (area of sector OACBO) – (area of OAB)

                                 = 

Question 17

Find the areas of both the segments of a circle of radius 42 cm with central angle 120^o. Solution 17

Area of sector OACBO

Area of minor segment ACBA

Area of major segment BADB

Question 18

A chord of a circle of radius 30cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. Take  = 3.14, Solution 18

Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°

Area of the sector OACBO

Area of OAB = 

Area of the minor segment ACBA

= (area of the sector OACBO) – (area of the OAB)

=(471 – 389.25) = 81.75 

Area of the major segment BADB

= (area of circle) – (area of the minor segment)

= [(3.14 × 30 × 30) – 81.75)] = 2744.25Question 19

In a circle of radius 10.5cm, the minor arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc.Solution 19

Let the major arc be x cm long

Then, length of the minor arc = 

Circumference = 

Question 20

The short and long hands of a clock are 4cm and 6cm long respectively. Find the sum of distances travelled by their tips in 2days. Take = 3.14Solution 20

In 2 days, the short hand will complete 4 rounds

Distance travelled by its tip in 2 days

=4(circumference of the circle with r = 4 cm)

= (4 × 2 × 4) cm = 32 cm

In 2 days, the long hand will complete 48 rounds

length moved by its tip

= 48(circumference of the circle with r = 6cm)

= (48 × 2 × 6) cm = 576 cm

Sum of the lengths moved

= (32 + 576) = 608 cm

= (608 × 3.14) cm = 1909.12 cmQuestion 21

Find the area of a quadrant of a circle whose circumference is 88 cm.Solution 21

Question 22

A rope by which a cow is tethered is increased from 16m to 23m. How much additional ground does it have now to graze?Solution 22

Area of plot which cow can graze  when r = 16 m is 

                                                        = 804.5 m²

Area of plot which cow can graze when radius is increased to 23 m

Additional ground = Area covered by increased rope – old area

                        = (1662.57 – 804.5) = 858 Question 23

A horse is placed for grazing inside a rectangular field 70m by 52m. It is tethered to one corner by a rope 21 m long. On how much area can it graze? How much area is left ungrazed?Solution 23

Area which the horse can graze = Area of the quadrant of radius 21 m

Area ungrazed = 

Question 24

A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12m. If the length of the rope is 7m, find the area of the field which the horse cannot graze. Take . Write the answer correct to 2 places of decimal.Solution 24

Each angle of equilateral triangle is 60

Area that the horse cannot graze is 36.68 m²Question 25

Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed? Take = 3.14Solution 25

Ungrazed area

Question 26

In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is , find the radius of the circle.

Solution 26

OP = OR = OQ = r

Let OQ and PR intersect at S

We know the diagonals of a rhombus bisect each other at right angle.

Therefore we have

Question 27

The side of a square is 10 cm. Find (i) the area of the inscribed circle, and (ii) area of the circumscribed circle. Take  = 3.14Solution 27

Diameter of the inscribed circle = Side of the square = 10 cm

Radius of the inscribed circle = 5 cm

Diameter of the circumscribed circle

= Diagonal of the square

Radius of circumscribed circle = 

(i)Area of inscribed circle = 

(ii)Area of the circumscribed circle Question 28

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.Solution 28

Let the radius of circle be r cm

Then diagonal of square = diameter of circle = 2r cm

Area of the circle = 

Question 29

The area of a circle inscribed in an equilateral triangle is 154 . Find the perimeter of the triangle. Take Solution 29

Let the radius of circle be r cm

Let each side of the triangle be a cm

And height be h cm

Question 30

The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in 19.8 km long journey?Solution 30

Radius of the wheel = 42 cm

Circumference of wheel =2r = 

Distance travelled = 19.8 km = 1980000 cm

Number of revolutions = Question 31

The wheels of the locomotive of a train are 2.1m in radius. They make 75 revolutions in one minute. Find the speed of the train in km per hour.Solution 31

Radius of wheel = 2.1 m

Circumference of wheel = 

Distance covered in one revolution = 13.2 m

Distance covered in 75 revolutions = (13.2 75) m = 990 m

= 

Distance a covered in 1 minute = 

Distance covered in 1 hour = Question 32

The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.Solution 32

Distance covered by the wheel in 1 revolution

The circumference of the wheel = 198 cm

Let the diameter of the wheel be d cm

Hence diameter of the wheel is 63 cmQuestion 33

A boy is cycling in such a way that the wheels of his bicycle are making 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed(in km/h) at which the boy is cycling.Solution 33

Radius of the wheel 

Circumference of the wheel = 2r = 

Distance covered in 140 revolution

Distance covered in one hour = Question 34

The diameter of the wheels of a bus is 140cm, How many revolutions per minute do the wheels make when the bus is moving at a speed of 72.6 km per hour?Solution 34

Distance covered by a wheel in 1minute

Circumference of a wheel = 

Number of revolution in 1 min = Question 35

The diameter of the front and rear wheels of a tractor are 80cm and 2 m respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers in 800 revolutions.Solution 35

Radius of the front wheel = 40 cm = 

Circumference of the front wheel= 

Distance moved by it in 800 revolution

Circumference of rear wheel = (2 1)m = (2) m

Required number of revolutions = Question 36

Four equal circles are described about the four corners of a square so that each touches two of the others, as shown in the figure. Find the area of the shaded region, if each of the square measures 14cm.

Solution 36

Each side of the square is 14 cm

Then, area of square = (14 × 14) 

                               = 196 

Thus, radius of each circle 7 cm

Required area = area of square ABCD

                             -4 (area of sector with r = 7 cm,  = 90°)

Area of the shaded region = 42 Question 37

Four equal circles, each of radius 5 cm, touch each other, as shown in the figure. Find the area included between them. Take  = 3.14.

Solution 37

Let A, B, C, D be the centres of these circles

Join AB, BC, CD and DA

Side of square = 10 cm

Area of square ABCD

Area of each sector = 

                            = 19.625 

Required area = [area of sq. ABCD – 4(area of each sector)]

                     = (100 – 4 19.625) 

                     = (100 – 78.5) = 21.5 Question 38

Four equal circles, each of radius a units, touch each other. Show that the area between them is sq. units.Solution 38

Required area = [area of square – areas of quadrants of circles]

Let the side = 2a unit and radius = a units

Area of square = (side side) = (2a 2a) sq. units

Question 39

Three equal circles, each of radius of 6cm, touchone another as shown in the figure. Find the area enclosed between them. Take  = 3.14 and 

Solution 39

Let A, B, C be the centres of these circles. Joint AB, BC, CA

Required area=(area of ABC with each side a = 12 cm)

                            -3(area of sector with r = 6, = 60°)

The area enclosed = 5.76 cm²Question 40

If three circles of radius a each, are drawn such that each touches the other two, prove thatthe area included between them is equal to . Take .

Solution 40

Let A, B, C be the centers of these circles. Join AB, BC, CA

Required area= (area of ABC with each side 2)

                               -3[area of sector with r = a cm, = 60°]

Question 41

In the given figure, ABCD is a trapezium of area 24.5 cm². If AD ∥ BC, ∠DAB = 90°, AD =10 cm, BC = 4 cm and ABE is quadrant of a circle then find the area of the shaded region.

Solution 41

Question 42

ABCD is a field in the shape of a trapezium, AD ∥ BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m. Find the following:

 i. total area of the four sectors,

 ii. area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.

Solution 42

Question 43

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made. 

Solution 43

Question 44

In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED = 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region.

Solution 44

Question 45

In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE =12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region: Find the area of the shaded region. [Use π = 3.14.]

Solution 45

Question 46

In the given figure, 0 is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of shaded region. [Use π = 3.14.]

Solution 46

Question 47

In the given figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region.

Solution 47

Question 48

On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design.

Solution 48

Question 49

The perimeter of the quadrant of a circle is 25 cm. Find its area.Solution 49

Question 50

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.]Solution 50

Question 51

The radius of a circular garden is 100 m. There is a road 10 m wide, running all around it. Find the area of the road and the cost of levelling it at Rs.20 per m². [Use π = 3.14.]Solution 51

Question 52

The area of an equilateral triangle is .Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. Take.Solution 52

Area of equilateral triangle ABC = 49

Let a be its side

Area of sector BDF = 

Area of sector BDF = Area of sector CDE = Area of sector AEF

Sum of area of all the sectors

Shaded area = Area of ABC – sum of area of all sectors

Question 53

A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP CD, HQ FI and EL DF. IF CD = 8cm, BP = HQ = 4 cm and DE = EF = 5cm, find the area of the whole figure. Take  = 3.14. Solution 53

Question 54

A circular disc of radius 6cm is divided into three sectors with central angles 90^o, 120^o and 150^o. What part of the whole circle is the sector with central angle 150^o? Also, calculate the ratio of the areas of the three sectors.Solution 54

Question 55

A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35cm, then find the total area of the design. Use 

Solution 55

ABCDEF is a hexagon

AOB = 60, Radius = 35 cm

Area of sector AOB

Area of AOB = 

Area of segment APB = (641.083 = 530.425)= 110.658

Area of design (shaded area) = 6 110.658= 663.948

                                                            = 663.95 Question 56

In the given figure, PQ = 24cm, PR = 7cm and O is the centre of the circle. Find the area of the shaded region. Take  = 3.14

Solution 56

In PQR, P = 90, PQ = 24 cm, PR = 7 cm

Area of semicircle

Area of PQR = 

Shaded area = 245.31 – 84 = 161.31 Question 57

In the given figure, ABC is right angled at A. find the area of the shaded region if AB = 6cm, BC = 10 cm and O is the centre of thein circle of ABC. Take  = 3.14.

Solution 57

In ABC, A = 90°, AB = 6cm, BC = 10 cm

Area of  ABC = 

Let r be the radius of circle of centre O

Question 58

In the given figure, ABC is right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3cm and AC = 4cm. Find the area of the shaded region.

Solution 58

Area of shaded region = Area of ABC + Area of semi-circle APB

+ Area of semi circle AQC – Area of semicircle BAC

Further in ABC,A = 90

Adding (1), (2), (3) and subtracting (4)

Question 59

PQRS is a diameter of a circle of radius 6cm. The lengths PQ, QR, and RS are equal. Semicircles are drawn with PQ andQS as diameters, as shown in the given figure. If PS = 12cm, find the perimeter and area of the shaded region. Take  = 3.14

Solution 59

PS = 12 cm

PQ = QR = RS = 4 cm, QS = 8 cm

Perimeter = arc PTS + arc PBQ + arc QES

Area of shaded region = (area of the semicircle PBQ)

+ (area of semicircle PTS)-(Area of semicircle QES)

Question 60

The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track.

Solution 60

Length of the inner curved portion

= (400 – 2 90) m

= 220 m

Let the radius of each inner curved part be r

Inner radius = 35 m, outer radius = (35 + 14) = 49 m

Area of the track = (area of 2 rectangles each 90 m 14 m)

+ (area of circular ring with R = 49 m, r = 35 m

Length of outer boundary of the track

Exercise MCQ

Question 1

The area of a circle is 38.5 cm². The circumference of the circle is

1. 6.2 cm
2. 12.2 cm
3. 11 cm
4. 22 cm

Solution 1

Question 2

The area of a circle is 49πcm². Its circumference is

1. 7π cm
2. 14π cm
3. 21π cm
4. 28π cm

Solution 2

Question 3

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is

a. 111 cm²

b. 184 cm²

c. 154 cm²

d. 259 cm²Solution 3

Question 4

The perimeter of a circular field is 242 m. The area of the field is

1. 9317 m²
2. 18634 m²
3. 4658.5  m²
4. none of these

Solution 4

Question 5

On increasing the diameter of a circle by 40%, its area will be increased by

1. 40%
2. 80%
3. 96%
4. 82%

Solution 5

Question 6

On decreasing the radius of circle by 30%, its area is decreased by

1. 30%
2. 60%
3. 45%
4. none of these

Solution 6

Question 7

The area of a square is the same as the area of a circle. Their perimeters are in the ratio

Solution 7

Question 8

The circumference of a circle is equal to the circumferences of two circle having diameters 36 cm and 20 cm. The radius of the new circle is

1. 16 cm
2. 28 cm
3. 42 cm
4. 56 cm

Solution 8

Question 9

The area of a circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is

1. 25 cm
2. 31 cm
3. 50 cm
4. 62 cm

Solution 9

Question 10

If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is

1. 4:π
2. π:4
3. π:7
4. 7: π

Solution 10

Question 11

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R then

1. R₁ + R₂ = R
2. R₁ + R₂< R
3. R₁² + R₂²< R²
4. R₁² + R₂² = R²

Solution 11

Question 12

If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a radius R then

1. R₁ + R₂ = R
2. R₁ + R₂> R
3. R₁ + R₂< R²
4. none of these

Solution 12

Question 13

If the circumference of a circle and the perimeter of a square are equal then

1. area of the circle = area of the square
2. (area of the circle) > (area of the square)
3. (area of the circle) < (area of the square)
4. none of these

Solution 13

Question 14

The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circle is

1. 320 cm²
2. 330 cm²
3. 332 cm²
4. 340 cm²

Solution 14

Question 15

The areas of two concentric circles are 1386 cm² and 962.5 cm². The width of the ring is

1. 2.8 cm
2. 3.5 cm
3. 4.2 cm
4. 3.8 cm

Solution 15

Question 16

The circumferences of two circles are in the ratio 3:4. The ratio of their areas is

1. 3:4
2. 4:3
3. 9:16
4. 16:9

Solution 16

Question 17

The area of two circles is in the ratio 9:4. The ratio of their circumference is

1. 3:2
2. 4:9
3. 2:3
4. 81:16

Solution 17

Question 18

The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?

1. 2800
2. 4000
3. 5500
4. 7000

Solution 18

Question 19

The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?

1. 140
2. 150
3. 160
4. 166

Solution 19

Question 20

In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is

1. 14 m
2. 24 m
3. 28 m
4. 40 m

Solution 20

Question 21

The area of a sector of angle θ° of a circle with radius R is

Solution 21

Question 22

The length of an arc of a sector of angle θ° of a circle with radius R is

Solution 22

Question 23

The length of the minute hand of a clocks is 21 cm. The area swept by the minute hand in 10 minutes is

1. 231 cm²
2. 210 cm²
3. 126 cm²
4. 252 cm²

Solution 23

Question 24

A chord of a circle of radius 10 cm subtends right angles at the centre. The area of the minor segment

( given, π=3.14) is

1. 32.5 cm²
2. 34.5 cm²
3. 28.5 cm²
4. 30.5 cm²

Solution 24

Question 25

In a circle of radius 21 cm, an arc subtends an angles of 60° at the centre. The length of the arc is

1. 21 cm
2. 22 cm
3. 18.16 cm
4. 23.5 cm

Solution 25

Question 26

1. 120.56 cm²
2. 124.63 cm²
3. 118.24 cm²
4. 130.57 cm²

Solution 26

Exercise FA

Question 1

In the given figure, a square OABC has been inscribed in the quadrant OPBQ. If OA = 20cm then the area of the shaded region is [take π=3.14]

1. 214 cm²
2. 228 cm²
3. 242 cm²
4. 248 cm²

Solution 1

Question 2

The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?

1. 200
2. 250
3. 300
4. 350

Solution 2

Question 3

The area of a sector of a circle with radius r, making an angle of x° at the centre is

Solution 3

Question 4

In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If π = 3.14 then the area of the shaded region is

1. 264 cm²
2. 266 cm²
3. 272 cm²
4. 254 cm²

Solution 4

Question 5

The circumference of a circle is 22 cm. Find its area. Solution 5

Question 6

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.Solution 6

Question 7

The minute hand of a clock is 12 cm long. Find the area swept by it in 35 minutes.Solution 7

Question 8

The perimeter of a sector of a circle radius 5.6 cm is 27.2 cm. Find the area of the sector.Solution 8

Question 9

A chord of a circle of radius 14 cm makes right angle at the centre. Find the area of the sector.Solution 9

Question 10

In the given figure, the sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.

Solution 10

Question 11

Solution 11

Question 12

The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm. find its speed in km per hour. Solution 12

Question 13

OACB is a quadrant of a circle with centre O and its radius is 3.5 cm. If OD = 2 cm, find the area of (i) the quadrant OACB (ii) the shaded region.  

Solution 13

Question 14

In the given, ABCD is a square each of whose side measures 28 cm. Find the area of the shaded region. 

Solution 14

Question 15

In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region.

Solution 15

Question 16

The minute hand of a clock is 7.5 cm long. Find the areaSolution 16

Question 17

A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.Solution 17

Question 18

A chord of circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segment. Solution 18

Question 19

Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed? [Take π = 3.14]Solution 19

Question 20

A square tank has an area of 1600 m². There are four semicircular plots around it. Find the cost of turfing the plots at Rs. 12.50 per m².[Take π = 3.14]Solution 20