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A person travels a certain distance on a bicycle at a certain speed. Had he moved 3 km/hour faster, he would have taken 40 minutes less. Had he moved 2km/hour slower, he would have taken 40 minutes more. Find the distance.
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- 45 km.
- 40 km.
- 50 km.
- 55 km.
Correct Option: B
Let the original speed of the person be x km/hr. and the distance be y km.
Case : I
− | = 40 minutes | ||
x | x + 3 |
or, | hr. | |
60 |
or, | − | = | = | ||||
x | x + 3 | 60 | 3 |
or, y | ![]() | − | ![]() | = | ||||
x | (x + 3) | 3 |
or, y | ![]() | ![]() | = | |||
x (x + 3) | 3 |
or, | = | ||
x (x + 3) | 3 |
or, 2 x (x + 3) = 9y ...(i)
Case : II
= | − | = | |||
x − 2 | x | 60 |
or, y | ![]() | + | ![]() | + | |||
x − 2 | x | 3 |
or, y | ![]() | ![]() | = | |||
x (x − 2) | 3 |
or, | = | ||
x (x − 2) | 3 |
or, x (x – 2) = 3y ... (ii)
On dividing equation (i) by (ii) we have,
= | ||
x (x − 2) | 3y |
or, | = 3 | |
(x − 2) |
or, 2x + 6 = 3x – 6
or, 3x – 2x = 6 + 6 = 12
or, x = 12 km/hr.
∴ Original speed of the person
= 12 km/hr.
Putting the value of x in equation (ii)
12 (12 – 2) = 3y
or, 3y = 12 × 10
or, y = | = 40 | |
3 |
∴ The required distance = 40km.