Question

51. A student starting from his house
walks at a speed of 2 1/2km/h and
reaches his school 6 minutes late.
Next day starting at the same time he
increases his speed by 1 km/h and
reaches 6 minutes early. The distance
between the school and his house is

Answer 1

Answer:

Let the distance between the home & the school of the boy = x km.

In the first case, the speed is 221km/hr = 25km/hr.

∴  Time taken = x÷25hr.

=52xhr.

In the second case his speed= 1 + 23km/hr. =27km/hr.

∴  Time taken=x÷27hr.=72xhr.

∴ The difference in time = (52x−72x)hr.=354xhr.=354x×60min.=748x×min.

Now , in the first case he is 6 min late = −6 min. and

In the second case, he is 6 min  early = +6 min.

∴ The difference in time=[6−(−6)]min.=12 min.

∴ 748xmin.=12 min.

Or 4x=7

Or x=47=1.75.