PSA_LEVEL-UP_QNR36. In a 100 m race, Pintu runs at 1.66 m/s. If Pintu gives Chhotu a start of 4 metres and still beats him by 12 seconds, what can you say about Chhotu’s speed.
a) Chhotu covers 3 metres in 2 seconds.
b) Chhotu covers 4 metres in 2 seconds.
c) Chhotu covers 4 metres in 3 seconds.
d) Chhotu covers 3 metres in 4 seconds.
Speed of Pintu is 1.66 m/s. You can best represent 1.66 as 10/6 [you should remember percentage to fraction equivalent concepts to do faster calculation here].
Thus, speed of Pintu, Sp = 10/6 m/s.
Let, speed of Chhotu = ‘s’ m/s
Since, Pintu gives Chhotu a start of 4 metres, you can imagine that while Pintu runs 100 metres, Chhotu runs only 96 metres. However, the question further adds that Pintu still beats Chhotu by 12 seconds; this means that while Pintu finishes the race, Chhotu still has some distance to run that he can cover in 12 seconds.
Chhotu can run 12*s distance in these 12 seconds. Thus, you can say that while Pintu runs 100 metres, Chhotu effectively runs only [96-12s] metres in the same time.
If time is constant, ratio of distances is equal to the ratio of speeds.
100/(96-12s) = (10/6) /s
=>s=8/6=4/3 m/s. Option c.