# CIRCUMFERENCE OF CIRCLE WORD PROBLEMS

Problem 1 :

On decreasing the radius of the circle by 30%, its area decreased by

a)  30%      b)  60%     c)  75%     d)  None of these

Solution :

Let r be the radius of the original circle.

Area of circle = πr2

π(70% of r)2

= π(0.70r)2

= π(0.49r2)

= 49% of πr2

= (100 - 49)%

= 51% is decreased.

So, the answer is none of these.

Problem 2 :

In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is

a)  14 m     b)  24 m    c)  28 m     d)  40 m

Solution :

Number of revolutions made = 1000

Distance covered by these revolutions = 88 km

Converting km to m, we get

1 km = 1000 m

88 km = 88 (1000)

= 88000 m

Circle of the wheel x 1000 = 88000 m

2 πr x 1000 = 88000 m

2 πr = 88000/1000

2 πr = 88

πr = 44

r = 44/(22/7)

r = 44 (7/22)

r = 14 m

So, the required radius is 14 m.

Problem 3 :

The diameter of the wheel is 40 cm. How many revolutions will it make an covering 176 m ?

a)  140     b)  150    c) 160     d)  166

Solution :

Diameter of the wheel = 40 cm

Radius of the wheel = 20 cm

Distance covered by this revolutions = 176 m

100 cm = 1 m

= 176 (100)

= 17600 m

Circumference of wheel = 2 πr

= 2 π (20)

= 40π

Number of revolutions make = Distance covered / distance covered by the wheel in one revolution.

= 17600/40π

= 176000 / 40(3.14)

= 140.12

So, the required number of revolutions is 140.

Problem 4 :

The radius of the wheel is 0.25 m. How many revolutions will it take in covering 11 km ?

a)  2800     b)  4000    c) 5500     d)  7000

Solution :

The radius of the wheel = 0.25 m

Distance covered by revolutions = 11 km

1 km = 1000 m

11 km = 11000 m

Circumference of wheel (or) distance covered by the wheel in one revolution

2 πr

= 2(3.14) (0.25)

= 1.57 m

Number of revolutions = 11000/1.57

= 7006

So, the required number of revolutions is 7000.

Problem 5 :

Find the circumference of a circle of diameter 21 cm

a)  62 cm     b)  64 cm    c) 66 cm     d)  68 cm

Solution :

Diameter = 21 cm

= 10.5 cm

Circumference of the circle = 2 πr

= 2(3.14)(10.5)

= 65.94 cm

So, the required circumference of the circle is 66 cm.

Problem 6 :

Find the area of the circle whose circumference is 52.8 cm

Solution :

Circumference = 52.8 cm

2 πr = 52.8

2(3.14) r = 52.8

r = 52.8 / 2(3.14)

r = 8.40

Area of the circle = πr2

= 3.14(8.4)2

= 221.55 cm2

So, the required area of the circle is 221.55 cm2.

Problem 7 :

A steel wire when bent in the form of a square, encloses an area 121 sq.cm. The same wire is bent in the form of circle. Find the area of the circle ?

Solution :

Area of circle = 121 sq.cm

a2 = 121

a = 11

The length of the wire whose side length = 11 cm

The same wire is bent in the form of circle = 4a = 2πr

4(11) = 2(3.14)r

r = 44/2(3.14)

r = 7

Area of circle whose radius is 7 cm

π(7)2

= 49 π cm2

Problem 8 :

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Solution :

Number of revolutions made = 5000

Distance covered = 11 km

11 km = 11000 m

5000 x 2πr = 11000

2πr = 11000 / 5000

2πr = 2.2

r = 2.2/2(3.14)

r = 1.1/3.14

r = 0.35

diameter = 2(r)

=2(0.35)

= 0.7 m

So, the required diameter is 0.7 m.

Problem 9 :

The diameter of the wheels of a bus is 140 cm. How many revolutions per minute must a wheel make in order to move at the speed of 66 km/hr ?

Solution :

Diameter of the wheel = 140 cm

speed = 66 km/hr

Distance to be covered in one hour = 66 km

Number of revolutions = 66/2πr

= 66/2(3.14) (70)

= 0.150

1 hour = 60 minutes

Converting into minutes, 0.15 x 60

= 9 revolutions

So, the required number of revolutions is 9.

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