Enforcement Officer / Accounts Officer (EO/AO) 2020 recruitment test, advertisement 421. Transcribed from the Commission's own booklet (series A, NMJX-U-GTA); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

Suppose x and y are two positive numbers such that when x is reduced by 2 and y is increased by 2, the ratio becomes 2 : 1; and when x is increased by 2 and y is reduced by 2, the ratio becomes 3 : 1. Which one of the following is equal to x – y ?

(a)20
(b)24
(c)18
(d)22

Answer

Why

Correct — A, (a) 20. Turn each ratio into an equation and the pair solves in three lines. The first condition says (x – 2) : (y + 2) = 2 : 1, so x – 2 = 2(y + 2) = 2y + 4, giving x = 2y + 6. The second says (x + 2) : (y – 2) = 3 : 1, so x + 2 = 3(y – 2) = 3y – 6, giving x = 3y – 8. Setting the two expressions for x equal: 2y + 6 = 3y – 8, so y = 14, and then x = 2(14) + 6 = 34. Hence x – y = 34 – 14 = 20. Check it against the stem before trusting it, because a ratio question is easy to set up backwards: x – 2 = 32 and y + 2 = 16, and 32 : 16 is indeed 2 : 1; x + 2 = 36 and y – 2 = 12, and 36 : 12 is indeed 3 : 1. Both numbers are positive, as the stem requires. One thing to notice on the page: the four numbers are printed 20, 24, 18, 22 — not in ascending order. Work out the value first and then hunt for it in the list, rather than assuming the smallest sits at (a).

Why the others are wrong

(b)24 — 24 is the difference of the SECOND pair of adjusted numbers: (x + 2) – (y – 2) = 36 – 12 = 24. It is what you get by solving the system correctly and then subtracting the two quantities that appeared in the second ratio instead of x and y themselves. The stem asks for x – y, the difference of the original numbers, and the adjustments of +2 and –2 were only ever a device for generating the two equations.
(c)18 — 18 is (x – 2) – y = 32 – 14 — one number taken in its adjusted form and the other in its original form. This is the commonest slip on a question of this shape: after the algebra you are holding four numbers, 32, 16, 36 and 12, as well as x = 34 and y = 14, and it is easy to reach for the wrong pair. Write x and y down on their own line the moment you find them, before computing anything the stem asks for.
(d)22 — 22 is x – (y – 2) = 34 – 12, the other mixed pairing — the original x against the reduced y. Note that 18, 20 and 22 sit two apart, which is exactly the size of the adjustment in the stem, so the three wrong-pair answers cluster around the right one and none of them looks obviously absurd. That is the design of the option set, and the only defence is to name what you are subtracting.

Concept

A ratio is a statement about division, and the single move that makes ratio problems mechanical is turning the colon into an equals sign. 'p : q = 2 : 1' means p/q = 2, which means p = 2q — and the trap is that the 2 multiplies the WHOLE of q, brackets and all. Here q is not y but (y + 2), so the first condition is x – 2 = 2(y + 2) and not x – 2 = 2y + 2. Once both conditions are written this way you have two linear equations in two unknowns, and the fastest route is to make each one say 'x = something in y', then equate the two right-hand sides so that x disappears in a single step. That gives 2y + 6 = 3y – 8 and hence y = 14; substituting back into either equation gives x = 34. Every step is reversible, which is why the check at the end costs nothing and catches almost every error: put 34 and 14 back into the words of the stem and see whether the two ratios come out as 2 : 1 and 3 : 1.

Two-condition ratio problems are the workhorse of the EPFO EO/AO quantitative block because they test three things at once — reading a worded condition into an equation, distributing a multiplier across a bracket, and answering the question actually asked. The last of those is where most marks go. The stem does not ask for x, or for y, or for x + y; it asks for x – y, and the option set is built out of the near-misses. The habit worth building is to underline the final ask before starting the algebra and to write it out again at the end. The option ordering here is a second, quieter test: the numbers run 20, 24, 18, 22, so a candidate who solves correctly but picks by position rather than by value can still lose the item.

Key facts

'a : b = m : n' is the same statement as a/b = m/n, and hence as n·a = m·b.
When the ratio is against a compound expression, the multiplier applies to the whole bracket: x – 2 = 2(y + 2), not x – 2 = 2y + 2.
The system here is x = 2y + 6 and x = 3y – 8; equating them removes x in one step and gives y = 14, x = 34.
Verification is free: 32 : 16 = 2 : 1 and 36 : 12 = 3 : 1, and both x and y are positive as the stem requires.
The wrong options are all differences of adjusted quantities — 36 – 12 = 24, 34 – 12 = 22, 32 – 14 = 18 — while the ask is the difference of the originals, 34 – 14 = 20.
Reading the ratios the other way round, as (y + 2) : (x – 2), forces y = 2x – 6 and y = 3x + 8, whose solution is x = –14: negative, and therefore ruled out by the stem's word 'positive'.
The options are printed 20, 24, 18, 22 — the Commission does not order numeric options by size on this paper.

Turn each colon into an equals sign — then name what you are subtracting

First condition→(x – 2) : (y + 2) = 2 : 1, so x – 2 = 2(y + 2) = 2y + 4, giving x = 2y + 6. The multiplier applies to the WHOLE bracket: x – 2 = 2y + 2 is the standing trap on ratio questions.

Second condition→(x + 2) : (y – 2) = 3 : 1, so x + 2 = 3(y – 2) = 3y – 6, giving x = 3y – 8.

Equate the two expressions for x→2y + 6 = 3y – 8, so y = 14; substituting back, x = 2(14) + 6 = 34.

x – y = 34 – 14 = 20→The stem asks for the difference of the ORIGINAL numbers. The adjustments of +2 and –2 were only ever a device for generating the two equations.

Verification costs nothing→x – 2 = 32 and y + 2 = 16, and 32 : 16 is indeed 2 : 1. x + 2 = 36 and y – 2 = 12, and 36 : 12 is indeed 3 : 1. Both numbers are positive, as the stem requires.

Every wrong option is a wrong pair→36 – 12 = 24 subtracts the second adjusted pair. 34 – 12 = 22 mixes the original x with the reduced y. 32 – 14 = 18 mixes the reduced x with the original y. They sit two apart because two is the size of the adjustment, so none of them looks absurd — write x and y on their own line the moment you find them. The options are also printed 20, 24, 18, 22, not in ascending order.

Study next

Ratio and proportion: compounded, duplicate and inverse ratios, and dividing a quantity in a given ratioSimultaneous linear equations by substitution, elimination and cross-multiplicationAges problems, which are the same algebra dressed as 'x years ago' and 'y years hence'Partnership and profit-sharing, where the ratio is of capital multiplied by timeMixtures and alligation, where a ratio changes when a quantity is added or removed

Common traps

Writing x – 2 = 2y + 2 instead of x – 2 = 2(y + 2) — forgetting that the multiplier crosses the bracket.
Inverting a ratio: reading 'the ratio becomes 2 : 1' as (y + 2) : (x – 2). Here that produces a negative solution, which the stem's 'positive numbers' rules out.
Solving correctly for x and y and then answering a different question — x, or x + y, or the difference of the adjusted quantities.
Assuming numeric options ascend from (a) to (d). On this item they are printed 20, 24, 18, 22.
Skipping the substitution check at the end, which on a two-condition problem catches almost every arithmetic slip for the cost of ten seconds.

The EO/AO quantitative block reliably carries one or two worded algebra items of this exact shape: two conditions, each stated as a ratio after some quantity has been added to or taken from the unknowns, and a final ask for a combination of the unknowns rather than for an unknown itself. The stems are short and the options are bare numerals, so all the work is in the setting-up. Expect the numbers to change and the structure not to.

Related PYQs

EPFO_EOAO_2020_Q22Open & attempt →

Which one of the following is the remainder when 74^100 is divided by 9 ?

(a) 2
(b) 5
(c) 3
(d) 7

Answer(d) 7

The other item in this paper's opening quantitative block whose numeric options are printed out of ascending order — there they run 2, 5, 3, 7, so correct working still has to be matched to a position on the page.

Practice

practice — not a real PYQ
Suppose x and y are two positive numbers such that when x is reduced by 3 and y is increased by 3, the ratio becomes 3 : 1; and when x is increased by 3 and y is reduced by 3, the ratio becomes 5 : 1. Which one of the following is equal to x – y ?
- (a)38
- (b)42
- (c)40
- (d)45
Answer(b) 42
practice — not a real PYQ
The ratio of two positive numbers is 5 : 3. If each of them is increased by 4, the ratio becomes 3 : 2. What is the difference between the two numbers ?
- (a)6
- (b)8
- (c)10
- (d)12
Answer(b) 8

Why