SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · Quantitative Aptitude · Question 12 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

A motorboat takes 18 hours to go downstream, and it takes 36 hours to return the same distance. Find the ratio of the speed of the boat in still water to the speed of the stream.

(a)3 : 2
(b)3 : 1
(c)2 : 1
(d)2 : 3

Answer

Why

Correct — B.

The distance is the same both ways, so speed is inversely proportional to time. The times are 18 hours down and 36 up:
downstream : upstream = 1⁄18 : 1⁄36 = 2 : 1

Write downstream = 2k and upstream = k.
Still water b = (2k + k) ⁄ 2 = 3k ⁄ 2
Stream s = (2k − k) ⁄ 2 = k ⁄ 2

b : s = 3k ⁄ 2 : k ⁄ 2 = 3 : 1 → option (b).

Why the others are wrong

(a)
3 : 2 — 3 : 2 would make the stream two-thirds of the boat, so the speeds would run downstream : upstream = 5 : 1. The paper's times — 18 hours down against 36 up — fix that speed ratio at 2 : 1 instead.
(c)
2 : 1 — 2 : 1 is the ratio of the two speeds, downstream to upstream — the first line of the working, not the answer. Had b : s been 2 : 1, the times would have run 3 : 1.
(d)
2 : 3 — 2 : 3 makes the stream faster than the boat, so the upstream speed 2 − 3 would be negative and the return trip impossible. A boat that does come back in 36 hours must satisfy b > s.

Concept

Boats-and-streams runs on two sums: downstream speed = b + s and upstream speed = b − s, where b is the boat in still water and s is the stream.

Invert them and you get the boat and the stream back: b = (down + up) ⁄ 2 and s = (down − up) ⁄ 2. Because both are halves of the same pair, the halves cancel in a ratio.

When the same distance is covered each way, times are inverse to speeds, so 18 hours down against 36 up hands you the speed pair 2 : 1 with no distance ever needed.

The distance is never given and never needed. Any distance you invent — 36 km, 360 km — produces the same 3 : 1, because the ratio depends on the times alone.

Key facts

Downstream speed = b + s and upstream speed = b − s.
For a fixed distance, speed is inversely proportional to time, so downstream : upstream = 1⁄18 : 1⁄36 = 2 : 1.
The shortcut b : s = (upstream time + downstream time) : (upstream time − downstream time) gives (36 + 18) : (36 − 18) = 54 : 18 = 3 : 1.
Any valid boat-and-stream setup needs b > s, or the upstream journey never finishes.

Study next

Deriving b : s = (t_up + t_down) : (t_up − t_down) from the two sum equationsBoats-and-streams problems that give distances rather than times, needing two equationsHarmonic mean of speeds, which is why the still-water time is not the average of the two times

Common traps

Answering 2 : 1, the downstream-to-upstream speed ratio, instead of boat-to-stream.
Using 18 : 36 as the speed ratio, when times run inverse to speeds rather than equal to them.
Averaging 18 and 36 to get a still-water time of 27 hours, when the correct figure is 24.

SSC swaps the unknown around this same pair of equations from shift to shift — sometimes the stream's speed, sometimes a time, sometimes the ratio as here.

Also asked on 12 Sep 2024, 16:00, Quant Q.13 (2 km against the current in 1 hour, 1 km with it in 10 minutes), 13 Sep 2024, 09:00, Quant Q.22 (78 km upstream and back in 32 hours), and 13 Sep 2024, 16:00, Quant Q.12 (30 km each way in 14 hours).

Related PYQs

No directly related past PYQ was found.