SSC CGL 2024 Tier-I · 25 September 2024, Shift 1 · Quantitative Aptitude · Question 20 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.

Suppose P starts walking at a speed of 8 km/h. After 8 hours B started travelling on a bicycle at a speed of 24 km/h. The distance from the starting B can catch A is:

(a)88 km
(b)96 km
(c)72 km
(d)64 km

Answer

Why

Correct — B. The eight-hour delay is a head start measured in distance, and the gap then closes at the difference of the speeds.

Head start = 8 km/h × 8 h = 64 km
Relative speed = 24 − 8 = 16 km/h
Time to close the gap = 64 ⁄ 16 = 4 h
Distance from the start = 24 km/h × 4 h = 96 km

That is option (b). Check it from the walker's side: 8 km/h for 8 + 4 = 12 hours is also 96 km, so the two are at the same point.

Why the others are wrong

(a)
88 km — At 88 km the cyclist has ridden 88 ⁄ 24 = 11⁄3 hours, and by then the walker is at 64 + 8 × 11⁄3 = 93⅓ km — still ahead, so no catch has happened.
(c)
72 km — 72 km is three hours of cycling. The walker is then at 64 + 24 = 88 km, a full 16 km ahead — exactly one hour of closing still to go.
(d)
64 km — 64 km is the head start itself, the point the walker had reached when the cyclist set off. It answers a question the stem did not ask.

Concept

Same direction, so the gap closes at the difference of the speeds, 24 − 8 = 16 km/h.

A delay stated in time has to become a distance before it is any use: eight hours of walking at 8 km/h is a 64 km lead already open when the cyclist starts.

Catch-up time = gap ⁄ relative speed = 4 hours, counted from the cyclist's start, not from the walker's.

For the meeting point, multiply that time by the cyclist's own speed — or by the walker's speed and add his 64 km lead. Both give 96 km.

The stem calls the walker P in its first sentence and A in its last ("the distance from the starting B can catch A"), and never introduces B before using the letter.

It is one person walking and one on a bicycle, and the mismatch does not touch the arithmetic — but read this question by role, not by letter.

Key facts

In same-direction motion the gap closes at the difference of the speeds, here 16 km/h.
A head start given in time becomes a distance: 8 hours at 8 km/h is a 64 km lead.
Catch-up time is gap ÷ relative speed, so 64 ⁄ 16 = 4 hours after the cyclist starts.
The meeting point is 96 km from the common starting point, which both travellers reach.

Study next

Relative speed in the same and opposite directions, including trains crossing each otherHead-start problems stated in metres and seconds, where the units must be matched firstCircular-track meetings, where a lap length replaces the head start

Common traps

Subtracting the 8-hour delay from the answer's time instead of converting it into a 64 km lead
Answering with the head start, 64 km, or with the catch-up time, 4 hours
Computing the distance from the walker's speed but forgetting to add his lead

The lead reaches you in one of two forms — a distance already open, or, as here, a delay in hours you must convert before anything else.

Worded as a policeman chasing a thief with the gap given as a distance, the same relative-speed reading is asked at 19 Sep 2024, 12:30, Quant Q.18 and at 24 Sep 2024, 16:00, Quant Q.10.

Related PYQs

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