SSC CGL 2024 Tier-I · 18 September 2024, Shift 3 · Quantitative Aptitude · Question 16 of 98

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.

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 man can row a distance of 6 km in 1 hour in still water. He can row the same distance in 45 minutes with the current. Find the total time taken by him to row a distance of 16 km with the current and to return to the starting point.

(a)4 hours
(b)4 hours and 40 minutes
(c)6 hours and 40 minutes
(d)6 hours

Answer

Why

Correct — D. Still water: 6 km in 1 hour, so the boat alone does 6 km/h.

Downstream: 6 km in 45 minutes, and 45 minutes is 3⁄4 hour.
6 ÷ 3⁄4 = 8 km/h

Current = 8 − 6 = 2 km/h
Upstream = 6 − 2 = 4 km/h

Going 16 km: 16 ⁄ 8 = 2 hours
Returning 16 km: 16 ⁄ 4 = 4 hours
Total = 2 + 4 = 6 hours → option (d).

Why the others are wrong

(a)
4 hours — 4 hours is the return leg on its own. Upstream at 4 km/h takes 4 hours for 16 km, and the 2-hour downstream leg still has to be added.
(b)
4 hours and 40 minutes — 4 h 40 min is 16⁄8 + 16⁄6, which rows home in still water. The return is against the current, at 6 − 2 = 4 km/h.
(c)
6 hours and 40 minutes — 6 h 40 min is 16⁄6 + 16⁄4, which sends the outward leg at the still-water 6 km/h instead of the downstream 8 km/h.

Concept

Every boat question hides just two numbers: the boat in still water (b) and the stream (s).

Downstream speed is b + s and upstream speed is b − s. A question hands you two of those four quantities and asks for a third.

This paper gives b directly, from 6 km in 1 hour, and gives b + s through the 45-minute run. So s = 8 − 6 = 2 falls out by subtraction, and b − s = 4 follows.

Then time is distance ÷ speed on each leg separately. A round trip is never distance ÷ average speed, because the slow leg occupies more of the clock.

The distance changes between the data and the question — 6 km sets the speeds, 16 km is what is actually rowed. Convert to speeds first and the change of distance costs nothing.

Key facts

Downstream speed = boat speed + stream speed.
Upstream speed = boat speed − stream speed.
Stream speed = (downstream − upstream) ÷ 2, and boat speed = (downstream + upstream) ÷ 2.
45 minutes is 3⁄4 hour, so 6 km in 45 minutes is 8 km/h.

Study next

Problems giving the two leg times and asking for the ratio of boat speed to stream speedRound-trip average speed as the harmonic mean of the two leg speedsTime, speed and distance with minute-to-hour conversions inside the data

Common traps

Averaging 8 and 4 to 6 km/h and dividing 32 km by it, which gives 5 h 20 min and hides the long upstream leg.
Rowing the return leg at the still-water 6 km/h rather than the upstream 4 km/h.
Answering with one leg when the question asks for the total time.

SSC gives two of the four speeds — still water, stream, downstream, upstream — and asks for a time, a distance or a ratio, so the first move is always to write down b and s.

The same setup is asked on 10 Sep 2024, 16:00, Quant Q.12, where the boat does 18 km/h and the stream 3 km/h, and on 23 Sep 2024, 12:30, Quant Q.9, where 50 km down and back takes 7 hours 30 minutes.

Related PYQs

No directly related past PYQ was found.