Discussion

Explanation:

Let the speeds of river, faster and slower boats be r, f and s km/hr respectively and distance between A and B be 12 kms.

For Faster boat:
⇒ f + r = 12/2 = 6   ...(1)
⇒ f - r = 12/3 = 4   ...(2)

(1) - (2)
⇒ 2r = 6 - 4 = 2
⇒ r = 1

For Slower boat:
⇒ $\frac{12}{\mathrm{s}-1}$ + $\frac{12}{\mathrm{s}+1}$ = 6

⇒ $\frac{12\times 2\mathrm{s}}{{\mathrm{s}}^2-{\left(1\right)}^2}$ = 6

⇒ s² - 1 = 4s

⇒ s² - 4s - 1 = 0

∴ s = $\frac{4\pm \sqrt{16+4}}{2}$ = 2 ± √5 = 2 + √5 [negative value of s is rejected]

Now, time taken by the slower boat to go from A to B

= $\frac{12}{(2+\sqrt{5})+1}$ = $\frac{12}{(3+\sqrt{5})}$ = $\frac{12(3-\sqrt{5})}{9-5}$ = 3(3 - √5).

Hence, option (c).

Concept: Boats & Streams

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