Question

23. Aboy walks from his house at 3 km/hr and reaches his school 10 minute late. If he increases his speed by 1 km/hr, he reaches the school 6minute earlier then the scheduled time. How far is the school from his house ​

Answer 1

Answer:

Correct answer is

1.75 km

Step-by-step explanation:

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

In the first case, the speed is 2

2

1

km/hr =

2

5

km/hr.

∴ Time taken = x÷

2

5

hr.

=

5

2x

hr.

In the second case his speed= 1 +

2

3

km/hr. =

2

7

km/hr.

∴ Time taken=x÷

2

7

hr.=

7

2x

hr.

∴ The difference in time = (

5

2x

−

7

2x

)hr.=

35

4x

hr.=

35

4x

×60min.=

7

48x

×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.

∴

7

48x

min.=12 min.

Or 4x=7

Or x=

4

7

=1.75.

∴ The distance between the home & the school of the boy = 1.75 km.

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