Enforcement Officer / Accounts Officer (EO/AO) 2017 recruitment test. Transcribed from the Commission's own booklet (series A, GRS-A-TNS); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

A container is filled with 300 litres of hydrogen gas. The first day it loses 100 litres of hydrogen gas and everyday it loses one-third of the volume it lost in the previous day. Then the container

(a)loses entire hydrogen gas in 3 days
(b)loses entire hydrogen gas in 10 days
(c)loses 150 litres of hydrogen gas in 10 days
(d)possesses at least 150 litres of hydrogen gas on 100th day

Answer

Why

Correct — D, (d) possesses at least 150 litres of hydrogen gas on 100th day. The daily losses form a geometric progression — 100 litres, then a third of that, then a third of that again — and the decisive fact is that such a progression has a finite total however long it runs. With first term 100 and common ratio one third, the sum of the whole endless series is 100 ÷ (1 − 1/3) = 100 × 3/2 = 150 litres. The container can therefore never lose more than 150 litres in all, no matter how many days pass.

After n days the loss is exactly 150 × (1 − 1/3ⁿ), which is always a little under 150, and the gas remaining is 300 minus that, namely 150 + 150 ÷ 3ⁿ. That expression is greater than 150 for every finite n and creeps towards 150 without ever arriving. On the hundredth day the container holds 150 litres plus 150 ÷ 3¹⁰⁰, a quantity indistinguishable from 150 in any measurement but certainly not less than it. So the container possesses at least 150 litres of hydrogen on the hundredth day, which is option (d) — and, as the same expression shows, on every other day too.

The first few days make the pattern concrete. Day one loses 100 and leaves 200. Day two loses 33⅓ and leaves 166⅔. Day three loses 11⅑ and leaves 155·56. Day four loses 3·70 and leaves 151·85. The remaining quantity is falling, but each fall is a third of the one before, so the sequence of remainders closes in on 150 and stops nowhere short of it and nowhere below it. Recognising that the ratio is less than one — and therefore that the total loss converges — is the whole of the question; everything else is a substitution into the sum formula.

Why the others are wrong

(a)loses entire hydrogen gas in 3 days — Three days does not empty the container, and the arithmetic is quick. The losses on the first three days are 100, 33⅓ and 11⅑ litres, which come to 144·44 litres in all, leaving 155·56 litres still in the container. Nor could any number of days empty it, since the total possible loss is 150 litres and the container started with 300. This option is chosen by a candidate who sees the first loss of 100 out of 300 and assumes, by a rough proportional reasoning, that three days of losses will account for the whole of it. That reasoning would be right if the loss were 100 litres every day, which is exactly what the stem denies: each day's loss is a third of the previous day's, so the second day loses 33⅓ and not 100, and the daily amounts shrink away to nothing.
(b)loses entire hydrogen gas in 10 days — Ten days brings the total loss to 150 × (1 − 1/3¹⁰) litres, which is 149·9975, so 150·0025 litres of hydrogen are still in the container. It is not empty, and it will not be empty on any day whatever, because the losses of every day from the first to the last cannot together exceed 150 litres while the container began with twice that. The option tests whether a candidate has understood that a converging series has a ceiling. Ten days is chosen as the figure because it feels long enough for anything to run its course, and the intuition it appeals to — a leak that continues must eventually empty the vessel — is true of a constant leak and false of one that shrinks geometrically. When the daily amount falls by a fixed fraction, the process has a finite total and an unreachable limit.
(c)loses 150 litres of hydrogen gas in 10 days — This is the sharpest of the three, because it is very nearly right. The total lost in ten days is 150 × (1 − 1/3¹⁰) = 149·9975 litres, which is short of 150 by 150 ÷ 3¹⁰, about two and a half thousandths of a litre. No instrument would separate the two figures, and yet mathematically the difference is the entire point of the question: 150 is the limit of the total loss and the loss equals it on no day at all. A partial sum of a converging series approaches its limit and never attains it, so a statement that exactly 150 litres have gone after some stated number of days is false for every such number. The keyed option is worded to respect that distinction, since it claims only that the container holds at least 150 litres, which the exact expression 150 + 150 ÷ 3ⁿ establishes for every day.

Concept

A geometric progression is a sequence in which each term is a fixed multiple of the one before, that multiple being the common ratio. Where the ratio lies strictly between −1 and 1, the terms shrink towards zero and the series of all of them has a finite sum, equal to the first term divided by one minus the ratio. Here the first term is 100 litres and the ratio is one third, so the sum of every loss the container will ever suffer is 100 ÷ (2/3) = 150 litres. The partial sum after n terms is a(1 − rⁿ) ÷ (1 − r), which here is 150(1 − 1/3ⁿ), and the essential feature of that expression is that the bracket is always less than 1 and always more than 0 — so the partial sum is always less than the limit and always increasing. It follows that the container, having begun with 300 litres, retains 150 + 150 ÷ 3ⁿ litres after n days: falling, but bounded below by 150 and reaching it only in the limit. The distinction between a limit and a value actually attained is what separates the two most attractive options in this item, and it is the same distinction that underlies compound depreciation, radioactive decay and any process in which a fixed proportion of what remains is removed at each step.

The quantitative half of an EPFO paper usually contains one item that cannot be brute-forced, and on this paper it is this one. There is no arithmetic path to the answer: a candidate can compute the first few days easily enough, but the hundredth day is out of reach unless the structure has been recognised. That is precisely why the Commission sets it. The recognition required is small — a fixed fraction removed each period means a geometric progression, and a ratio under one means a finite total — and it converts an impossible computation into a two-line substitution. The four options are also arranged to reward reading rather than guessing, since three of them assert something exact about a stated day and only one makes a claim of the safer form 'at least'. On items where a limit is involved, an option hedged in that way deserves a second look, because a limit is approached rather than achieved.

Key facts

The daily losses form a geometric progression with first term 100 litres and common ratio one third.
The sum of an infinite geometric series with ratio between −1 and 1 is the first term divided by one minus the ratio.
Here that total is 100 ÷ (1 − 1/3) = 150 litres, so the container can never lose more than 150 litres altogether.
The loss after n days is 150(1 − 1/3ⁿ), which is always less than 150 and increases towards it.
The gas remaining after n days is therefore 150 + 150 ÷ 3ⁿ litres, which exceeds 150 on every finite day.
On the hundredth day the container holds 150 litres plus 150 ÷ 3¹⁰⁰, so it certainly holds at least 150 litres.
Day by day the container holds 200, 166⅔, 155·56 and 151·85 litres at the ends of the first four days.
A partial sum of a converging series approaches its limit but never attains it, which is why no day sees a loss of exactly 150 litres.

A ratio less than one means the total loss has a ceiling

Sum of the whole series = 100 ÷ (1 − 1/3) = 150 litres→The daily losses form a geometric progression with first term 100 litres and common ratio one third, and such a progression has a finite total however long it runs. The container can therefore never lose more than 150 litres in all, no matter how many days pass.

Loss after n days = 150 × (1 − 1/3ⁿ)→The bracket is always less than 1 and always more than 0, so the partial sum is always below the limit and always increasing.

Gas remaining = 300 − that = 150 + 150 ÷ 3ⁿ→Greater than 150 for every finite n, creeping towards 150 without ever arriving. On the hundredth day the container holds 150 litres plus 150 ÷ 3¹⁰⁰ — a quantity indistinguishable from 150 in any measurement, but certainly not less than it.

The first four days, concretely→Day one loses 100 and leaves 200. Day two loses 33⅓ and leaves 166⅔. Day three loses 11⅑ and leaves 155·56. Day four loses 3·70 and leaves 151·85. Each fall is a third of the one before, so the remainders close in on 150 and stop nowhere short of it and nowhere below it.

(a) three days and (b) ten days→In three days the losses come to 144·44 litres, leaving 155·56. In ten days they come to 149·9975, leaving 150·0025. No number of days empties the container, because the total possible loss is 150 litres and it began with twice that. The intuition that a continuing leak must eventually empty a vessel is true of a constant leak and false of one that shrinks geometrically.

(c) exactly 150 litres in ten days — the sharpest miss→Short of 150 by 150 ÷ 3¹⁰, about two and a half thousandths of a litre, and no instrument would separate the two figures. But a partial sum of a converging series approaches its limit and never attains it, so a claim that exactly 150 litres have gone is false for every stated number of days. The keyed option respects that, claiming only that at least 150 litres remain.

Study next

Geometric progressions: the nth term, the sum of n terms and the sum to infinityThe condition for convergence, and what happens when the common ratio is 1 or greaterCompound depreciation and decay problems, which are geometric progressions in disguiseArithmetic progressions and how to tell one from a geometric progression in a word problemLimits in elementary terms — quantities that approach a value without reaching it

Common traps

Assuming that a leak which never stops must eventually empty the vessel, which is true of a constant leak and false of a shrinking one.
Treating the first day's loss as the daily loss, which would empty the container in three days.
Accepting a statement that the total loss equals its limiting value after a stated number of days.
Computing a few terms and extrapolating by eye instead of using the sum formula.
Overlooking the difference between an exact claim about one day and a claim of the form 'at least', which is far easier to satisfy.

Progressions reach EPFO papers most often through a story rather than through a formula, and the story is nearly always a quantity that changes by a fixed proportion each period — a leak, a bouncing ball, a depreciating asset, a population growing at a steady rate, a debt on compound interest. The examiner's question then takes one of three forms: what is the amount after a stated number of periods, what is the total over all periods, or what can be said about the position in the long run. The first two are direct substitutions into the nth-term and sum formulas; the third, as here, tests whether the candidate knows that a ratio below one produces a finite total. A candidate who can recognise the words 'a fixed fraction of the previous' as the signature of a geometric progression has done nine tenths of the work on every item in the family.

Related PYQs

EPFO_APFC_2016_Q33

In a medium-size township, the trend of annual immigration is an addition of 20% of the population as it was at the beginning; also 15% of the population as it was at the beginning is estimated to relocate elsewhere every year. If the current population is 80000, what is the likely population three years hence ?

(a) 90000
(b) 91200
(c) 92000
(d) 92610

Answer(d) 92610

The same structure of a fixed proportion applied period after period, set as a population problem in which each year's change is a percentage of the year's opening figure.

Practice

practice — not a real PYQ
A tank contains 400 litres of water. On the first day 120 litres leak away, and on each following day the quantity that leaks is half of what leaked on the previous day. What is the greatest total quantity that can ever leak away from the tank?
- (a)180 litres
- (b)240 litres
- (c)300 litres
- (d)400 litres
Answer(b) 240 litres
practice — not a real PYQ
What is the sum of the infinite series 9 + 3 + 1 + 1/3 + … ?
- (a)12
- (b)13·5
- (c)18
- (d)27
Answer(b) 13·5

Why