Discussion

Explanation:

Let speed of boat in still water = b
Speed of still water = r

Case 1: Speed while going upstream = b - r
Time taken while going upstream = 28 mins (given)

Case 2: Speed while going downstream = b + r
Time taken while going upstream = t mins (assume)

Case 3: Speed while travelling in still water = b
Time taken while going upstream = (t + 3) mins (given)

In all these cases, distance travelled is same, hence ratio of speeds will be reciprocal ratio of time.

⇒ b – r : b + r : b = $\frac{1}{28} : \frac{1}{\mathrm{t}} : \frac{1}{\mathrm{t}+3}$

Simplifying this ratio, we get

⇒ b – r : b + r : b = t(t + 3) : 28(t + 3) : 28t

b – r = t(t + 3)k   …(1)
b + r = 28(t + 3)k   …(2)
b = 28tk   …(3)

Adding (1) and (2), we get
2b = (t² + 3t + 28t + 28 × 3)k

From (3)
⇒ 2 × 28tk = (t² + 31t + 84)k

⇒ 56t = t² + 31t + 84

⇒ 0 = t² - 25t + 84

⇒ (t - 21)(t - 4) = 0

⇒ t = 21 or 4

Hence, option (d).

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