A father picks up his child in a car from school at sharp 1500h everyd

Question

A father picks up his child in a car from school at sharp 1500h everyday following the same route and routine. One day, the school closes at 1400h, and the child instead of waiting at the school starts walking towards the home at a speed of 2mph. The father, not knowing the change, starts from home as per the fixed schedule and at a constant speed, meets the child on the way and picks up the kid, and returns 4 minutes early as compared to the other days.
If the time spent in picking up kid, turning and traffic is nil, what is the speed in mph of the car?

(A) 20
(B) 30
(C) 56
(D) 58
(E) 60

chetan2u · Accepted Answer

*Adding the solution*

Now, saving of 4 minutes means saving 2 minutes both ways.  
This further means that the father picked up kid 2 minutes early, that is at 1458h.

At 1458, the child has been walking for 58 minutes, a distance that is covered by car in 2 minutes.

Two ways hereafter.

(I) Ratio ...A quick and more elegant method
As the time taken for the same distance by the child and the car is in the ratio 58:2, the speed will be in opposite ratio => 2:58
The speed of child is 2mph, so speed of car will be 58mph.

(II) Speed and distance 
In 58 minutes, the child travels  2*\frac{58}{60} 
This distance is covered by the car in 2 minutes, so speed =  \frac{\frac{2*58}{60}} {\frac{2}{60}}=58

D

LaveenaPanchal · Answer

The car returns 4 minutes early as compared to other days. It means it travelled for 2 minutes lesser on each side. So the father must have picked the child at 1458 hours.

Now child walked for 58 minutes. So the distance travelled by child in 58 minutes is travelled by car in two minutes.

The distance travelled by child = 2 x 58 /60 mph

The car travels 58 / 30 mph in 2 minutes so in 60 minutes it travels at the speed of : 58/30 * 60/2 = 58 mph

PrajayM · Answer

Sir I did not understand the part where you mention "At 1458, the child has been walking for 58 minutes, a distance that is covered by car in 2 minutes." Please can you elaborate as I did not understand why the distance travelled by both will be same

hadimadi · Answer

Hi,

let x be the speed of the car, t the time traveled by the father, D the distance traveled by the father.

(1) D=x*t
(2) D-2y=(t-4)(x)

In (2), 2y denotes that the total distance driven is diminished by the distance walked by the kid, and since that has a two way effect, we multiply with 2. (t-4) because total time saved is 4 minutes by the father. We know that y=(58/60)*2, so it follows:

(2) D=tx-4x+4*(58/60)

With (1)=(2) we get:

4x=4*(58/60)
x=(58/60)

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A father picks up his child in a car from school at sharp 1500h everyd

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