SSC CGL 2024 Tier-I · 13 September 2024, Shift 1 · Quantitative Aptitude · Question 22 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 10 September 2026.

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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 boat can travel 78 km upstream and back in a total of 32 hours. It can travel 15 km upstream and 52 km downstream in a total of 9 hours. How much distance will the boat cover in 12 hours in still water?

(a)92 km
(b)104 km
(c)96 km
(d)100 km

Answer

Why

Correct — C. Work in reciprocals of speed, because the data are times — that makes both conditions linear.

Let p = 1⁄u for upstream and q = 1⁄d for downstream.
78p + 78q = 32 → p + q = 32⁄78 = 16⁄39
15p + 52q = 9

Substitute p = 16⁄39 − q into the second:
240⁄39 + 37q = 9
37q = 351⁄39 − 240⁄39 = 111⁄39
q = 3⁄39 = 1⁄13 → d = 13 km/h

p = 16⁄39 − 3⁄39 = 13⁄39 = 1⁄3 → u = 3 km/h
Still water = (13 + 3) ⁄ 2 = 8 km/h

Distance in 12 hours = 8 × 12 = 96 km, option (c).

Why the others are wrong

(a)
92 km — 92 km in 12 hours is 7⅔ km/h. The two conditions pin upstream at 3 km/h and downstream at 13, so still water is exactly (3 + 13) ⁄ 2 = 8, and the distance is fixed.
(b)
104 km — 104 km needs 8⅔ km/h in still water. Nothing in the two conditions produces that rate — the reciprocals solve to u = 3 and d = 13, whose mean is 8, not 8⅔.
(d)
100 km — 100 km needs 8⅓ km/h, and it is the roundest of the four, which is what makes it the easy guess. The solved pair u = 3, d = 13 gives 8 km/h and 96 km.

Concept

The data here are times, and time is distance divided by speed — so the system is linear in 1⁄speed, never in speed itself.

Writing p = 1⁄u and q = 1⁄d turns '78 km up and back in 32 hours' into 78p + 78q = 32 and '15 km up, 52 km down in 9 hours' into 15p + 52q = 9. Two ordinary linear equations, solved in three lines.

Once u and d are known, the boat's own speed is their mean and the stream's speed is half their difference.

The round trip is deliberately awkward: 156 km in 32 hours is an average of 4.875 km/h, far below the 8 km/h still-water speed, because far more of the time is spent crawling upstream at 3 km/h than running downstream at 13.

Key facts

Downstream speed is boat speed plus stream speed.
Upstream speed is boat speed minus stream speed.
Boat speed in still water = (downstream + upstream) ⁄ 2.
Here the two conditions give upstream 3 km/h and downstream 13 km/h, so the boat is 8 km/h and the stream 5 km/h.

Study next

Two-condition boat-and-stream systems solved through reciprocal speedsRelating still-water speed, stream speed and the round-trip average speedTime-and-work problems, which use the identical reciprocal substitution

Common traps

Treating the round-trip average, 156 ⁄ 32 = 4.875 km/h, as the still-water speed.
Solving correctly for u and d and then answering with the downstream speed, giving 13 × 12 = 156 km.
Reversing the second condition, where 15 km is upstream and 52 km downstream.

This stem gives two mixed distance-and-time conditions and then asks for something a step beyond the speeds. Quant Q.7 of the 13 Sep 2024, 09:00 paper uses the same reciprocal-rate move on two pipes that together fill a tank in 20 minutes.

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