Answer: Option B
:
B
Approach 1: Conventional Approach:
(b) Let the distance of race be x metres which is covered by A in t seconds. Then, in the same time B covers (x-12)metres and C covers (x-18)metres.
$\therefore SpeedofA=\frac{x}{t}m/sec,$
$SpeedofB=\frac{x-12}{t}m/sec,$
$SpeedofC=\frac{x-18}{t}m/sec.$
Time taken by B to finish the race $=\frac{x}{\frac{x-12}{t}}=\frac{xt}{x-12}$ sec.
Now distance travelled by C in this time,
$=x\times t\times \frac{x-18}{(x-12)t}=x-8$
$\to \frac{x(x-18)}{x-12}=x-8\to x=48m.$
Approach 2: Reverse Gear Approach
Going from answer options:
Note the ratio of distance covered should be the same:
Assume option (c) to be correct: then we get the distances travelled to be in the ratio of60:48:40 and for B and C we get the ratio to be 60:52, here we observe for B and C that the ratio is not the same.
For option B we get A:B:C = 48:36:30 or 8:6:5 for B and C we get 48:40 or 6:5 we see that for both the options we have the same ratio of distance travelled hence this the correct answer option!