Assistant Provident Fund Commissioner (APFC) 2016 recruitment test, advertisement 170. Transcribed from the Commission's own booklet (series A, B-CRHE-O-PAPA); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

Walking at 3/4th of his usual speed, a man reaches his office 20 minutes late. What is the time taken by him to reach the office at his usual speed ?

(a)80 minutes
(b)70 minutes
(c)60 minutes
(d)50 minutes

Answer

Why

Correct — C, (c) 60 minutes.

THE STEP THAT DECIDES THIS QUESTION is that over a FIXED DISTANCE, time is inversely proportional to speed. The office does not move, so the distance is the same on both journeys, and whatever multiple is applied to the speed, its reciprocal is applied to the time.

He walks at three-fourths of his usual speed. Invert the fraction: he takes four-thirds of his usual time. So

usual time : slow time = 3 : 4

Read that as three parts against four parts. The difference between them is ONE part, and the stem tells us what that one part is worth — he is twenty minutes late, so one part is 20 minutes. Therefore

usual time = 3 parts = 3 × 20 = 60 minutes slow time = 4 parts = 4 × 20 = 80 minutes

The usual time is 60 minutes, which is option (c).

THE ALGEBRA, for a reader who prefers it written out. Let the usual time be T minutes and the usual speed be s. The distance is sT. Walking at 3s/4, the time taken is

distance / speed = sT ÷ (3s/4) = 4T/3

He is late by the difference between the two:

4T/3 − T = T/3 = 20 → T = 60

THE GENERAL FORMULA that this problem is an instance of, and which is worth memorising because the shape recurs constantly. If a man walks at a/b of his usual speed and is consequently late by t, then his usual time is

T = t × a / (b − a)

Here a = 3, b = 4 and t = 20, so T = 20 × 3 / 1 = 60. The formula is not magic; it is the ratio argument compressed. The slow time is b parts, the usual time is a parts, the lateness is the (b − a) parts between them, and dividing t by (b − a) gives the value of one part.

TWO CAUTIONS ABOUT THE READING. First, "reaches his office 20 minutes late" means twenty minutes later than he usually arrives — the comparison is with his own normal journey, not with some fixed appointed hour. Second, the question asks for the time taken AT HIS USUAL SPEED, not the time taken on the slow walk. Those are 60 and 80 minutes respectively, and both are printed on the option list. Deciding which quantity is wanted before computing is worth more here than any amount of arithmetic skill.

A SANITY CHECK that takes a moment. If the usual time is 60 minutes, then at three-quarters of the speed the journey takes 60 × 4/3 = 80 minutes, which is exactly 20 minutes more. The answer reproduces the condition it was derived from. The fraction is printed inline as three-fourths with the ordinal ending attached, rather than as a stacked fraction, which is how this booklet sets such expressions.

Why the others are wrong

(a)80 minutes — Eighty minutes is the time the slow walk actually takes, and it is by far the most likely wrong answer on this item. The working that produces it is entirely correct as far as it goes — usual time 60, slow time 80, difference 20 — and the error is only in which of the two numbers is handed in. The stem asks for "the time taken by him to reach the office AT HIS USUAL SPEED", and that is the smaller of the pair. There is also a wholly wrong route to the same figure, which is worth knowing so that arriving at 80 is not mistaken for confirmation: a candidate who sets the lateness equal to one QUARTER of the usual time, reasoning loosely that a quarter of the speed has been lost, writes T/4 = 20 and gets T = 80. That reasoning is incorrect, because losing a quarter of the speed adds a THIRD to the time, not a quarter — the reciprocal of 3/4 is 4/3, and 4/3 exceeds 1 by 1/3. The two errors converge on the same number, which is why the option is so effective.
(b)70 minutes — Seventy minutes is not reachable by any consistent treatment of this problem, and it repays a moment's thought as to why it is on the list at all. The four options descend in steps of ten — 80, 70, 60, 50 — and that even spacing is deliberate. A candidate who has half-worked the problem and knows only that the answer lies somewhere between the slow journey and something shorter cannot narrow it down by the shape of the list, because no option is isolated or oddly placed. Seventy also sits between the two figures that a correct solution produces, 60 and 80, and so catches anyone who averages, hedges or guesses towards the middle. The remedy is to finish the calculation rather than to estimate: the ratio of times is exactly 4 to 3, the gap between them is exactly one part, and one part is exactly twenty minutes, so the two admissible answers are 60 and 80 and nothing lies between them.
(d)50 minutes — Fifty minutes is the smallest option and would require the usual journey to be shorter than the answer by ten minutes, which no reading of the stem supports. One way to reach it is to subtract the lateness from a figure that has itself been mis-derived — starting from 70 and taking off twenty, for instance. Another is to invert the relationship between speed and time: a candidate who thinks that walking at three-fourths of the usual speed means taking three-fourths of the usual time will conclude that the slow journey is the shorter one, which contradicts the stem's own statement that he arrives late, and any figures produced afterwards are unmoored. That inversion is the fundamental error to guard against on this whole topic. Over a fixed distance, LESS speed always means MORE time, and the multiplier on the time is the reciprocal of the multiplier on the speed — three-fourths of the speed gives four-thirds of the time.

Concept

SPEED, TIME AND DISTANCE ARE HELD TOGETHER BY ONE RELATION, distance = speed × time, and every problem in the topic is that relation with one of the three quantities held constant.

WHEN DISTANCE IS FIXED, speed and time are INVERSELY proportional: multiply the speed by k and the time is multiplied by 1/k. This is the case in front of us and it is the commonest of the three, because journeys between two fixed points are the natural setting for a word problem. The practical form to remember is that a fractional change in speed inverts into a fractional change in time, so three-fourths of the speed gives four-thirds of the time, five-sixths gives six-fifths, and so on.

WHEN TIME IS FIXED, distance and speed are directly proportional. WHEN SPEED IS FIXED, distance and time are directly proportional. Identifying which quantity the problem holds constant is the first move in every question of this family, and it is usually settled by a single phrase in the stem — "the same journey", "his office", "in the same time".

THE RATIO-PARTS METHOD is the most reliable way to handle these problems by hand, because it avoids algebra and keeps the two journeys visibly distinct. Express the two speeds as a ratio, invert it to get the ratio of times, write the two times as that many parts, and use the stated difference to find the value of one part. Here the speeds are 4 : 3 (usual to slow, since the slow speed is three-quarters), so the times are 3 : 4, the gap is one part, and one part is twenty minutes. The method scales without difficulty to harder versions — three journeys, or a change of speed part-way through — where algebra becomes cumbersome.

THE STANDARD FORMULAE that grow out of it, all derived from the same ratio argument:

At a/b of the usual speed and late by t: usual time = t × a / (b − a) Covering a distance at u and being late by t1, at v and early by t2: the distance is uv(t1 + t2) / (v − u)

The second is the other stock question in this area and it is solved by exactly the same reasoning, with the total swing t1 + t2 playing the part that t plays here.

A NOTE ON AVERAGE SPEED, since it is the classic trap adjacent to this topic. When equal DISTANCES are covered at two different speeds, the average speed is the HARMONIC mean, 2uv/(u + v), not the arithmetic mean. When equal TIMES are spent at two speeds, the average is the arithmetic mean. Confusing the two is one of the most reliable ways to lose a mark on this topic.

Time-speed-distance is a standing topic of the quantitative strand on EPFO papers, and this item is a clean specimen of its most common form: one journey, one fractional change of speed, one stated consequence, one quantity wanted. Nothing needs to be looked up and the arithmetic is trivial; the question is testing whether the candidate knows that speed and time invert, and whether he reads which of the two times is being asked for.

The design of the option list deserves attention. It contains BOTH numbers that a correct solution produces — 60 minutes for the usual journey and 80 for the slow one — which means that a candidate can do every step of the working faultlessly and still hand in the wrong answer. That is the setter's real target here, and it is a target he will aim at again: on quantitative items generally, the wrong option nearest the top of the list is very often the intermediate quantity rather than the final one. The habit that defends against it is to underline the actual ask in the stem before starting, and to check the answer against that underlining before moving on.

The four options descend evenly in steps of ten, which is another deliberate choice. An option list with an obvious outlier gives information away; an evenly spaced ladder gives none, and forces the candidate to finish the calculation.

The fraction in the stem is printed inline, as three-fourths with the ordinal ending attached to it rather than as a stacked fraction, which is this booklet's practice with such expressions throughout.

Key facts

Over a fixed distance, time is inversely proportional to speed, so walking at three-fourths of the usual speed means taking four-thirds of the usual time.
The ratio of usual time to slow time is therefore 3 : 4, the gap between them is one part, and that one part is the twenty minutes of lateness.
The usual time is three parts, 60 minutes, and the slow journey takes four parts, 80 minutes — both figures appear on the option list.
In algebra: 4T/3 − T = T/3 = 20, hence T = 60 minutes.
The general result is that at a/b of the usual speed with lateness t, the usual time is t × a/(b − a); here 20 × 3/1 = 60.
The answer reproduces the condition: 60 minutes at three-quarters speed takes 60 × 4/3 = 80 minutes, which is exactly 20 minutes more.
Losing a quarter of the speed adds a THIRD to the time, not a quarter, because the multiplier on time is the reciprocal of the multiplier on speed.
For equal distances covered at two different speeds the average speed is the harmonic mean 2uv/(u + v), not the arithmetic mean of the two speeds.

Study next

Inverse proportion between speed and time over a fixed distance, and the ratio-parts method of handling two journeys at once.The standard late-and-early problem: a distance covered at one speed arriving late and at another arriving early, and the formula that follows.Average speed for a journey in stages, and why equal distances call for the harmonic mean while equal times call for the arithmetic mean.Relative speed problems — two bodies moving in the same direction and in opposite directions, and trains crossing poles, platforms and each other.Boats and streams, where upstream and downstream speeds are the still-water speed minus and plus the current.Races and head starts, where a distance advantage has to be converted into a time advantage or the reverse.

Common traps

Handing in the slow journey's time of 80 minutes when the stem asks for the usual time of 60. Both numbers are on the option list.
Treating three-fourths of the speed as three-fourths of the time. Less speed means more time, and the time multiplier is the reciprocal, four-thirds.
Setting the lateness equal to a quarter of the usual time. A quarter lost from the speed adds a third to the time, so the lateness is one third of the usual time.
Reading "20 minutes late" as measured against a fixed appointed hour rather than against his own normal arrival.
Guessing towards the middle of an evenly spaced option list. The only two figures the working produces are 60 and 80.

Time-speed-distance appears on every EPFO paper in the quantitative strand, and the forms are limited enough to be prepared exhaustively. The most frequent is the one here — a fractional change of speed on a fixed journey, with the consequence stated as a lateness or a saving. Close behind it are the late-and-early pair, the average speed of a two-stage journey, relative speed with trains or with two people walking, and boats in a stream.

Two setter habits are worth anticipating. The first is to place the intermediate quantity on the option list, as here, so that correct working attached to careless reading still fails. The second is to state the change in a form that must be inverted — three-fourths of the speed, five-sixths of the speed, twenty-five per cent slower — because the inversion is where candidates who have memorised a formula rather than understood the proportion come unstuck.

The preparation that pays is small: know that fixed distance means inverse proportion, work in ratio parts rather than algebra, and underline what the question actually asks for. Those three habits handle the whole topic and take almost no time in the examination hall, which matters on a paper where the quantitative strand is the largest of all and every item has to be finished quickly to leave room for the rest.

Related PYQs

EPFO_APFC_2016_Q103

A and B run a 1 km race. A gives B a start of 50 m and still beats him by 15 seconds. If A runs at 8 km/h, what is the speed of B ?

(a) 4·4 km/h
(b) 5·4 km/h
(c) 6·4 km/h
(d) 7·4 km/h

Answer(d) 7·4 km/h

A race between two runners on this same paper, where a head start and a time margin have to be converted into a speed — the same inverse relation applied to a competitive setting.

EPFO_APFC_2016_Q104

In a race of 1 km, A can beat B by 40 m and B can beat C by 50 m. With how much distance can A beat C in a 0·5 km race ?

(a) 42 m
(b) 43 m
(c) 44 m
(d) 45 m

Answer(c) 44 m

A chained-race item on this paper, where A beats B and B beats C, and the two margins have to be combined and then scaled to a shorter race.

Practice

practice — not a real PYQ
Walking at five-sixths of his usual speed, a man reaches his office 8 minutes later than he usually does. What is the time he takes to reach the office at his usual speed ?
- (a)32 minutes
- (b)40 minutes
- (c)48 minutes
- (d)50 minutes
Answer(b) 40 minutes — at five-sixths of the speed he takes six-fifths of the time, so the two times are in the ratio 5 : 6, the gap of one part is 8 minutes, and the usual time is five parts. The slow journey takes 48 minutes, which is the intermediate figure and not what was asked.
practice — not a real PYQ
A man covering a certain distance at 4 km/h reaches his destination 10 minutes late, and covering the same distance at 5 km/h he reaches it 5 minutes early. What is that distance ?
- (a)4 km
- (b)5 km
- (c)6 km
- (d)8 km
Answer(b) 5 km — the swing between the two journeys is 15 minutes, that is a quarter of an hour, so d/4 − d/5 = 1/4. Since d/4 − d/5 is d/20, the distance is 5 km, and the two journeys take 75 and 60 minutes respectively.

Why