Question

A student walks from his house at 10 km/hr and reaches his school late by 6 minutes. next day, he increases his speed by 15 km/hr and reaches 4 minutes before school time. How far is the school from his house?​

Answer 1

The diff of time (t1-t2)

=late (6mins) - early (4mins)

=2mins

distn is same in both cases=d

d/10-d/15=2

3d-2d=60

d=60

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Answer 2

The school is 7 km from his house.

Step-by-step explanation:

Let the distance be x km.

A student walks from his house at 10 km/hr and reaches his school late by 6 minutes. Next day, he increases his speed by 15 km/hr.

$\frac{x}{10} - \frac{x}{15} = \frac{14}{60}$       [1 minute = 60 minutes]

=> $\frac{3x - 2x}{30} = \frac{7}{30}$

=> 3x - 2x = 7

=> x = 7 km

Hence, the school is 7 km from his house.