Distance covered by wheel in 1 revolution
= 1 × Circumference
Distance covered by wheel in 2 revolution
= 2 × Circumference
Distance covered by wheel in n revolution
= n × Circumference
Distance covered
= No. of revolutions × Circumference
No. of revolutions =
Distance covered
Circumference
If diameter of wheel is increase,
then its circumference will also increase.
Circumference
= π × Diameter
We know that,
Since π is a constant value,
means circumference changes as the diameter changes
If diameter increase by half, then circumference also increases by half
If diameter increase by 1/4, then circumference also increases by 1/4
If a wheel makes 1 revolution
Distance covered by wheel in 1 revolution is
∴ New circumference =
x
+
1
4
(x)
=
4x + x
4
=
5x
4
If circumference is x m and it increases by 1/4,
then new circumference will be
Original circumference + increase in circumference
Q. The fore wheel of a carriage makes 6 revolutions more than the rear wheel in going
120m. If the diameter of the fore wheel be increased by ¼ its present diameter and the
diameter of the rear wheel be increased by one-fifth of its present diameter, then the
fore wheel makes 4 revolutions more than the rear wheel in going the same distance.
Find the circumference of each wheel of the carriage.
Circumference of fore wheel & rear wheel
Sol.
Let the circumference of fore wheel be x m & rear wheel be y m.
120 m
Fore wheel makes 6 more revolutions than rear wheel
How to find no. of revolutions?
No. of revolutions =
Distance covered
Circumference
| | |
| | |
| | |
| | |
Fore wheel
Rear wheel
Circumference
x m
y m
Distance
120 m
120 m
No. of Revolutions =
Distance covered
Circumference
120
x
120
x
As per the first condition,
=
+ 6
120
y
120
y
120
x
–
= 6
120
y
…(i)
Q. The fore wheel of a carriage makes 6 revolutions more than the rear wheel in going
120m. If the diameter of the fore wheel be increased by ¼ its present diameter and the
diameter of the rear wheel be increased by one-fifth of its present diameter, then the
fore wheel makes 4 revolutions more than the rear wheel in going the same distance.
Find the circumference of each wheel of the carriage.
Sol.
Let the circumference of fore wheel be x m & rear wheel be y m.
Distance covered
Circumference
| | |
| | |
| | |
| | |
Fore wheel
Rear wheel
Circumference
Distance
No. of Revolutions =
Distance covered
Circumference
120
x
–
= 6
120
y
…(i)
As per the second condition,
If diameter increases by ¼ , then circumference also increases by ¼
New circumference =
Original circumference + increase in circumference
∴ New circumference =
x
+
1
4
(x)
=
4x + x
4
=
5x
4
5x
4
If diameter increases by 1/5, then circumference also increases by 1/5
New circumference =
Original circumference + increase in circumference
∴ New circumference =
y
+
1
5
(y)
=
5y + y
5
=
6y
5
6y
5
120 m
120 m
No. of revolutions =
Distance covered
Circumference
120
( )
5x
4
120 × 4
5x
24
=
96
x
=
120 × 5
6y
20
=
100
y
=
120
( )
6y
5
96
x
96
x
=
+ 4
100
y
100
y
96
x
–
= 4
100
y
…(ii)
Lets solve these 2 equations
120
x
–
= 6
120
y
…(i)
96
x
–
= 4
100
y
…(ii)
Since variables are in denominator lets remove it by substitution
20a – 20b = 1 ...(iii)
96a – 100b = 4 …(iv)
Lets make coefficient of b equal
Multiplying (i) by 5 we get,
100a – 100b = 5 …(v)
Subtracting (v) from (iv)
96a – 100b = 4
100a – 100b = 5
– 4a = – 1
(–) (+) (–)
20
– 20b = 1
∴ 5
– 20b = 1
∴ – 20b
= 1 – 5
∴ – 20b
= – 4
Resubstituting the values of a and b
∴
∴ x = 4
∴
∴ y = 5
Circumference of fore wheel is 4m and circumference of rear wheel is 5m
∴