SSC CGL 2024 Tier-I · 23 September 2024, Shift 2 · Quantitative Aptitude · Question 9 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.

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, whose speed in 15 km/h in still water goes 50 km downstream and comes back in a total of 7 hours 30 minutes. The speed of the stream (in km/h) is:

(a)5
(b)11
(c)9
(d)7

Answer

Why

Correct — A. Rule: with x the stream's speed, downstream is 15 + x and upstream is 15 − x, and the two times add to the given total.

7 hours 30 minutes = 7.5 hours
50 ⁄ (15 + x) + 50 ⁄ (15 − x) = 7.5

Over the common denominator 225 − x², the numerator is
50(15 − x) + 50(15 + x) = 50 × 30 = 1500 — the x terms cancel.

1500 = 7.5 × (225 − x²)
225 − x² = 1500 ⁄ 7.5 = 200

x² = 25, so x = 5 km/h → option (a)

Why the others are wrong

(b)
11 — 11 makes the speeds 26 and 4 km/h. The trip would take 50 ⁄ 26 + 50 ⁄ 4 ≈ 14.4 hours, nearly double the 7.5 the stem allows.
(c)
9 — 9 gives 24 and 6 km/h, so the round trip runs 2.08 + 8.33 ≈ 10.4 hours. The upstream leg alone eats more than the whole budget.
(d)
7 — 7 gives 22 and 8 km/h and a total of about 8.5 hours — the closest wrong option, and the reason this is worth solving rather than eyeballing.

Concept

Boat-and-stream is one substitution. With b the still-water speed and x the stream, downstream is b + x, upstream is b − x, and their product is b² − x².

That difference of squares is what makes the algebra collapse. Adding the two times over the common denominator kills the x terms in the numerator, leaving 2bd on top — here 2 × 15 × 50 = 1500.

What is left is linear in x², so there is no quadratic to factorise. Back-substitution is the fastest check under exam pressure: four options, two divisions each.

The stem reads 'whose speed in 15 km/h in still water' — a typo for 'is'. Nothing in the working turns on it.

Key facts

Downstream speed is b + x and upstream speed is b − x, for still-water speed b and stream speed x.
7 hours 30 minutes is 7.5 hours, not 7.3 hours.
For equal distance d each way, d ⁄ (b + x) + d ⁄ (b − x) = 2bd ⁄ (b² − x²).
Here 2 × 15 × 50 ÷ (225 − x²) = 7.5, so 225 − x² = 200 and x = 5 km/h.

Study next

The ratio form, where downstream and upstream times alone fix b : xAverage speed on a round trip, which is the harmonic mean of the two speedsTwo-equation problems mixing upstream and downstream distances in one journey

Common traps

Writing 7 hours 30 minutes as 7.3 hours
Averaging the two speeds instead of adding the two times
Solving cleanly and then reporting the downstream speed as the stream speed

SSC varies which piece is withheld — the times, the distances or the ratio.

Also asked 17 Sep 2024, 12:30, Quant Q.12 (18 hours down, 36 hours back, find the ratio) and 13 Sep 2024, 09:00, Quant Q.22 (two mixed journeys, then the distance covered in still water).

Related PYQs

No directly related past PYQ was found.