Multiple choice

A swimmer can swim with velocity a in still water. He jumps in a river flowing with speed b. The swimmer goes a distance d downstream and then returns to original position. The time consumed in the process will be

  1. $\dfrac{3ad}{a^{2}-b^{2}}$
  2. $\dfrac{2ad}{a^{2}-b^{2}}$
  3. $\dfrac{ad}{a^{2}-b^{2}}$
  4. $\dfrac{4ad}{3a^{2}-b^{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time downstream = d / (a+b). Time upstream = d / (a-b). Total time = d/(a+b) + d/(a-b) = [d(a-b) + d(a+b)] / (a^2 - b^2) = 2ad / (a^2 - b^2).

AI explanation

The swimmer's downstream speed is the sum of his still water speed and the river speed, giving (a + b), while his upstream speed is their difference, giving (a - b). The total time taken for the round trip is the sum of the downstream and upstream travel times, which is d/(a+b) + d/(a-b). Adding these fractions results in a common denominator of a^2 - b^2, yielding 2ad/(a^2 - b^2).

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