SSC CGL 2024 Tier-I · 13 September 2024, Shift 3 · Quantitative Aptitude · Question 12 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 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 boatman rows to a place 30 km distant and back in 14 hours. He finds that he can row 10 km with the stream at the same time as 4 km against the stream. Find the speed (in km/h) of the stream.

Answer

Why

Correct — D. The second sentence fixes the ratio of the two speeds, not their values.

Equal times over 10 km down and 4 km up → downstream : upstream = 10 : 4 = 5 : 2

Let downstream = 5k and upstream = 2k.
30⁄5k + 30⁄2k = 14
6⁄k + 15⁄k = 14
21⁄k = 14 → k = 1.5

Downstream = 7.5 km/h, upstream = 3 km/h
Stream = (7.5 − 3) ÷ 2 = 9⁄4 km/h → option (d), the picture showing 9⁄4.

Why the others are wrong

(a)
Option (a) shows 4⁄9, the reciprocal of the answer. It comes from inverting the 10 : 4 ratio; the stream here runs at 2.25 km/h, not under half a km/h.
(b)
Option (b) shows 9⁄2, which is 7.5 − 3, the whole difference between the two speeds. Stream speed is half that difference.
(c)
Option (c) shows 2⁄9, the reciprocal of 9⁄2. No step in the working produces it, and a fifth of a km/h could not turn a 30 km round trip into 14 hours.

Concept

Downstream speed is boat + stream and upstream speed is boat − stream, so the two unknowns are recoverable from the pair:

stream = (down − up) ÷ 2 and boat = (down + up) ÷ 2.

The trick in this stem is the phrase "at the same time". Equal times over unequal distances mean the distances stand in the same ratio as the speeds, which hands you 5 : 2 with no algebra.

The 14-hour round trip then fixes the scale factor k, and the two speeds fall out.

All four options are images of fractions rather than text, so the card names them by shape: the keyed option (d) is the fraction 9⁄4, that is 2.25 km/h.

Key facts

Downstream speed = boat speed + stream speed, and upstream speed = boat speed − stream speed.
Stream speed = (downstream − upstream) ÷ 2.
Equal times over unequal distances put the distances in the same ratio as the speeds.
Here downstream = 7.5 km/h and upstream = 3 km/h, giving a stream of 2.25 km/h.

Study next

Two-equation boat problems that mix upstream and downstream legs in a single journeyDistance covered in still water once both speeds are knownRound-trip average speed, which is the harmonic mean 2DU⁄(D + U), not (D + U)⁄2

Common traps

Reading 10 km and 4 km as speeds instead of as a 5 : 2 ratio of speeds
Reporting the difference of the two speeds as the stream speed rather than half of it
Splitting the 14 hours as 7 hours each way, which the unequal speeds rule out

SSC changes which fact is withheld — sometimes the stream, sometimes the boat, sometimes a distance.

Also asked 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, key 96 km) and 18 Sep 2024, 09:00, Quant Q.19 (two mixed journeys, stream 3 km/h).

Related PYQs

No directly related past PYQ was found.