Quantitative Reasoning – Practice – Calculate the distance mentally if you can!

Time Speed Distance 1

PSA_LEVEL-UP_QNR14. Walking at 5 kmph I missed my train by 7 min . Walking at 6 kmph I reached the station 5 min early . How far is the station from the house?

A) 7 km
B) 6.5 km
C) 6 km
D) 5km

Ans: Let us solve using the conventional method first:

Let the station is ‘D’ units away from the house and assume I take ‘t’ minutes to reach station on time from my home .

We know, distance = speed * time

In first case,

D=5kmph * (t+7)/60 [Remember speed is in kmph and time in minutes].
Similarly, in second case

D= 6kmph*(t-5)/60

Since D is same in both the cases,

(t+7)/12= (t-5)/10
=> 10t+70=12t-60
=>2t=130=65 minutes

D= (65-5)/10=5 km.

Alternatively, you really DO NOT need any calculation here:

Speed x Time = Distance [constant].

The speed increases from 5kmph to 6 kmph i.e. an increment of 1 kmph on top of 5kmph.
Thus, you can rephrase this as:
Speed increases by 1/5th of the initial speed. [1/5th of 5 is 1 kmph].
Speed becomes 6/5th of the initial speed. [1/5 + 1 = 6/5].

Time will be 5/6th of the initial time. [Reciprocal of 6/5 is 5/6].
Time decreases by 1/6th of the initial time. ———–(A)

If you notice in the question this decrease is nothing but 12 minutes. [ Difference between “7 minutes late” and “5 minutes early” is 12 minutes] ———-(B)

If you observed (A) and (B), you would have a question “whose 1/6th is 12 minutes?” and the answer will be 72 minutes. [6*12=72].
Therefore, distance = initial speed * initial time = 5kmph * 72/60 hrs = 6km.

Once, you understand the above explanation you will be solving questions much faster and that too mentally [where the relation AXB= Constant is involved]. Another similar question to practise:
Time, Speed and Distance question

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