PSA_PRACTICE_QNR32. John drives from point A to point B at a constant speed. On his way back, he drives 20% faster and spends 12 minutes less. How much time does he spend on the round trip?

Ans: Let, John’s usual speed = ‘s’ units/min and the distance between A & B = ‘d’

d= s*t, where ‘t’ is the usual time taken to travel AB. ——(Y)

In second case,

d= (s+ 20% of s)(t-12)

=> d= 1.2s (t-12) —–(Z)

(Z) divided by (Y),

1= 1.2(t-12)/t

=>t= 1.2t – 14.4

=>0.2t=14.4

=>t= 72

From, A to B, he takes 72 minutes; thus from B to A he takes 72-12 or 60 minutes. Total time spent on round trip= 60+72=132 minutes.

**Faster method:**

While the above method involves some equations and steps; the below is a much faster method that involves only mental calculation. Let me explain it in detail so that we nail the point:

S*T = D [In the form of product of two variable such that R.H.S or result is a constant].

S increases by 20%. The equivalent ratio of 20% =1/5. Speed increases by 1/5. Thus, speed becomes 6/5 of the original values. Time will become 5/6 of the original value (to make sure 5,6 get cancelled to render the same result ‘D’).

Time becomes 5/6th of the original time; time decreases by (1-5/6) or 1/6. This 1/6 is given as 12 minutes. Now, the question is 1/6th of ‘WHAT’ is 12 minutes. ‘WHAT’ = 6*12=72 minutes.

Thus usual time =72 minutes (A to B), From B to A, the time will be 60 minutes. Total time =132 minutes.

The above method does not involve any equation and only mental calculation.