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A student starting from his house walks at a
school 6 minutes late. Next day starting atspeed of 2 1 km/hour and reaches his 2
the same time he increases his speed by 1 km/hour and reaches 6 minutes early. The distance between the school and his house is
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- 4 km
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3 1 km 2 -
1 3 km 4 - 6 km
Correct Option: C
Distance between school and house = x km (let)
Time = | |
Speed |
According to the question,
− | = | = | ||||
5/2 | 7/2 | 60 | 5 |
(Difference of time = 6 + 6 =12 minutes)
⇒ | − | = | |||
5 | 7 | 5 |
⇒ | = | ⇒ | = | ||||
35 | 5 | 35 | 5 |
⇒ 4x = | = 7 | |
5 |
⇒ x = | = 1 | km. | ||
4 | 4 |
Second Method :
Here, S1 = 2 | , t1 = 6 | |
2 |
S2 = 3 | , t2 = 6 | |
2 |
Distance = | |
S2 − S1 |
= | × | (6 + 6) | ||||
2 | 2 | |||||
- | ||||||
2 | 2 |
= | × | ||
4 | 60 |
= | = 1 | km | ||
4 | 4 |