Test by Bikram | Mock GATE | Test 4 | Question: 62

Question

A student reaches school 15 minutes early by running at a speed of 5 km/h from his house. By walking at 3 km/h, he is late by 9 minutes. The distance between the school and his house is ______ km.

Answer 1

Let the genuine time taken by a student to reach school from his house $=$$x$ $hr$

Given data; speed $S1$ $=$ $5km/h$

time $T1$=$ (x-15/60) hr$

speed $S2$$=$ $3 km/h$

time $T2$  = $(x+9/60) hr$

Distance in both cases is same. So

$S1$*$T1$ $=$ $S2$ *$T2$

$5 * ( x - 15/60 )$ $=$ $3 * ( x + 9/60 )$

on solving $x$ $=$ $51/60  hours$

The distance between his school and his house $=$ $5 * (51/60-15/60)$

                                                                            $=$ $(5 * 36) /60$

                                                                            $=$ $3 km$

Comments

What do you mean by genuine time ? Genuine in what sense ?

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@VS

Genuine in the sense I was talking about actual time!!!

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Actual time , But w.r.t what are you saying actual time ?

We are given 2 speeds 5km/h and 3km/h now, actual time w.r.t what speed ?

Answer 2

Let, the distance between the house and school be D kms.

Case 1 :  Running at the speed of 5 km/h a student reaches his school from his house 15 minutes early

Time taken = t1= D/5 hrs -->(1)

Case 2 : Walking at 3 km/h, he is late by 9 minutes

Time taken = t2= D/3 hrs -->(2)

Now, t2 - t1= 15/60 + 9/60 = 0.4

0.4 = D/3 - D/5 (From (1) and (2) )

Solving, D=3 kms(ANS)

Answer 3

$Speed=\dfrac{Distance}{Time}$

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$\dfrac{5km}{60min}=\dfrac{D}{(x-15)min}...........(1)$

$\dfrac{3km}{60min}=\dfrac{D}{(x+9)min}...........(2)$

Distance same

$\dfrac{5km}{60min}(x-15)min=\dfrac{3km}{60min}(x+9)min$

$x=51$

Substitute in $(1)\ or\ (2)$ the value of $x$

$D=3km$