SSC CGL 2025 Tier-I · 17 September 2025, Shift 3 · General Intelligence and Reasoning · Question 20 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

If 6 machines complete a task in 10 days, how many days for 15 machines at same rate?

(a)4
(b)6
(c)5
(d)8

Answer

Why

Correct — A.
Rule: for the same task at the same rate, machines × days stays fixed.

Work = 6 × 10 = 60 machine-days
Days for 15 machines = 60 ÷ 15 = 4
Check: 15 is 2.5 times 6, and 10 ÷ 2.5 = 4 → option (a).

Why the others are wrong

(b)
6 — 6 days × 15 machines = 90 machine-days, half as much again as the 60 the task needs.
(c)
5 — 5 days × 15 machines = 75 machine-days. The task needs only 60, so 15 machines finish a day sooner.
(d)
8 — 8 days × 15 machines = 120 machine-days, double the task. Eight days would suit 60 ÷ 8 = 7.5 machines.

Concept

At a fixed rate per machine, the task is a fixed amount of work, measured in machine-days. Put more machines on it and the days fall in the same ratio: the two are in inverse proportion.

So M₁ × D₁ = M₂ × D₂: 6 × 10 = 15 × D₂, giving D₂ = 4.

The stem's words 'at same rate' carry the assumption: every machine works as fast as the others, and adding machines costs no time.

Key facts

Same task, same rate: machines × days is constant.
M₁ × D₁ = M₂ × D₂, and with hours per day, M₁ × D₁ × H₁ = M₂ × D₂ × H₂.
Inverse proportion: 2.5 times the machines takes 10 ÷ 2.5 = 4 days.

Study next

Man-days with hours per day added (M₁D₁H₁ = M₂D₂H₂)Work left after some machines join or leave partwayTime and work with different individual rates (A and B together)

Common traps

Using direct proportion, 10 × 15 ⁄ 6 = 25 days, as if more machines needed longer
Stopping at 15 ÷ 6 = 2.5, which is the ratio of machines, not the number of days

14 Sep 2025, 09:00, Quant Q.15 uses the same fixed total of work, in rates instead of machine-days: A takes 8 days and B 12, so 3 days together finish 5⁄8 and leave 3⁄8, the keyed answer.

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