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.
r be 100% of radius.
Radius is decreased by 30%, then 70% of old radius will be new radius.
Area of circle = πr^{2}
= π(70% of r)^{2}
= π(0.70r)^{2}
= π(0.49r^{2})
= 49% of πr^{2}
= (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
radius = 21/2
= 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 = πr^{2}
= 3.14(8.4)^{2}
= 221.55 cm^{2}
So, the required area of the circle is 221.55 cm^{2}.
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
a^{2} = 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 cm^{2}
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
radius = 70 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.
May 21, 24 08:51 PM
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