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.

Which one of the following is the arithmetic mean of √3 – √2 and its reciprocal ?

(a)√3
(b)√2
(c)2
(d)1

Answer

Why

Correct — A, (a) √3. The reciprocal has to be rationalised before anything can be averaged. Multiply 1/(√3 – √2) above and below by the conjugate √3 + √2. The denominator becomes (√3 – √2)(√3 + √2) = 3 – 2 = 1, so the reciprocal of √3 – √2 is simply √3 + √2. That is the whole trick of the item: these two surds are exact reciprocals of one another, because their product is 1. Now take the arithmetic mean of the pair — add and halve. The √2 terms are equal and opposite and cancel, the √3 terms double: [(√3 – √2) + (√3 + √2)] / 2 = 2√3 / 2 = √3. A decimal check settles it in seconds: √3 – √2 is about 0.318, its reciprocal is about 3.146, and half their sum is about 1.732, which is √3.

Why the others are wrong

(b)√2 — √2 is what comes out if you subtract where you should add. Half the DIFFERENCE of the pair, [(√3 + √2) – (√3 – √2)] / 2, is 2√2 / 2 = √2 — the √3 terms cancel instead of the √2 terms. An arithmetic mean is always a sum divided by two, and on a conjugate pair the sum keeps the common term and kills the differing one. The decimal check kills this option too: √2 is about 1.414, while the true mean is about 1.732.
(c)2 — The mean of two numbers must lie between them, and here the pair is about 0.318 and about 3.146, so 2 is not impossible on that ground alone — but it is wrong, and a ten-second estimate puts the answer at about 1.73 rather than 2. Mechanically, 2 is what survives if you reach the correct sum 2√3, drop the radical and report the coefficient. The safeguard is to carry the surd all the way through: nothing in this working ever removes the √3, because the two quantities being averaged both contain it with the same sign.
(d)1 — 1 is the GEOMETRIC mean of the pair, not the arithmetic mean. The product of any number and its reciprocal is 1, and the square root of 1 is 1, so the geometric mean of a number and its reciprocal is always exactly 1. The arithmetic mean is a different quantity, and by the AM–GM inequality it is at least as large as the geometric mean, with equality only when the two numbers are equal. Here they plainly are not, so the answer must be strictly greater than 1 — as √3, about 1.732, is.

Concept

Two ideas meet in one line of working. The first is rationalisation: an expression of the form a√m + b√n is cleared out of a denominator by multiplying above and below by its conjugate, a√m – b√n, because the product of a conjugate pair is the difference of two squares and therefore rational. Here (√3 – √2)(√3 + √2) = 3 – 2 = 1, which is why √3 – √2 and √3 + √2 are exact reciprocals — a pair worth memorising, along with √5 – 2 and √5 + 2, and √2 – 1 and √2 + 1, which behave the same way for the same reason. The second idea is that the arithmetic mean of a conjugate pair a – b and a + b is a, since the b terms cancel on addition. Put the two together and the question collapses: the mean of √3 – √2 and its reciprocal is the mean of √3 – √2 and √3 + √2, which is √3.

Surds and their reciprocals are a recurring shape in the quantitative block of EPFO EO/AO papers because they reward one memorised move — multiply by the conjugate — and punish computation by decimals alone. The habit this item rewards is recognising the pair before doing any arithmetic: the moment you see a difference of two square roots and the word 'reciprocal', the conjugate is the next thing to write down. The estimate afterwards is a check, not the method; the four options here are far enough apart that a rough decimal comparison confirms the algebra.

Key facts

The conjugate of √3 – √2 is √3 + √2, and their product is 3 – 2 = 1.
Because that product is 1, √3 – √2 and √3 + √2 are reciprocals of each other, so no division survives into the final step.
Arithmetic mean of two quantities = (sum) / 2; geometric mean = square root of the product.
The geometric mean of any positive number and its reciprocal is exactly 1, since their product is 1.
AM–GM: for positive numbers the arithmetic mean is never less than the geometric mean, and the two are equal only when the numbers are equal.
Useful decimals for checking: √2 is about 1.414, √3 about 1.732, so √3 – √2 is about 0.318 and √3 + √2 about 3.146.
Other reciprocal surd pairs built the same way: √2 – 1 with √2 + 1, √5 – 2 with √5 + 2.

The two surds are exact reciprocals, so the mean collapses in one line

Rationalise the reciprocal→Multiply 1/(√3 – √2) above and below by the conjugate √3 + √2. The denominator becomes (√3 – √2)(√3 + √2) = 3 – 2 = 1, so the reciprocal of √3 – √2 is simply √3 + √2.

Arithmetic mean = (sum) / 2→[(√3 – √2) + (√3 + √2)] / 2 = 2√3 / 2 = √3. The √2 terms are equal and opposite and cancel; the √3 terms double.

Decimal check→√2 is about 1.414 and √3 about 1.732, so √3 – √2 is about 0.318 and its reciprocal about 3.146. Half their sum is about 1.732 — which is √3.

Why 1 is offered→1 is the GEOMETRIC mean of the pair, not the arithmetic one. The product of any number and its reciprocal is 1, and the square root of 1 is 1. By AM–GM the arithmetic mean is never less than the geometric mean, and the two are equal only when the numbers are equal — which here they plainly are not.

Why √2 is offered→It is half the DIFFERENCE rather than half the sum: [(√3 + √2) – (√3 – √2)] / 2 = 2√2 / 2 = √2. On a conjugate pair, addition keeps the common term and kills the differing one, and subtraction does the reverse.

Other reciprocal surd pairs, built the same way→√2 – 1 with √2 + 1, and √5 – 2 with √5 + 2. In each case the product of the pair is 1, which is exactly what makes them reciprocals.

Study next

Rationalising denominators with conjugates, including three-term surdsArithmetic, geometric and harmonic means and the inequality chain that orders themExpressions of the form x + 1/x and x – 1/x, and how squaring one gives the otherComparing and ordering surds without a calculatorSimplifying nested radicals into a difference of two square roots

Common traps

Averaging by subtracting. On a conjugate pair, the sum keeps the common term and the difference keeps the other one, so the two operations return different options that are both on the page.
Confusing the geometric mean with the arithmetic mean. A number and its reciprocal always have geometric mean 1, and 1 is offered here.
Forgetting to halve, which leaves 2√3 rather than √3.
Reading (√3 – √2)(√3 + √2) as 3 + 2 rather than 3 – 2. The difference of squares is the whole point of the conjugate.
Losing the radical in the last step and reporting a bare coefficient.

Surd items in this paper's quantitative block are one-line stems with bare options — often a single radical, a small integer and 1 — so the distractors are the outputs of specific wrong moves rather than random values. Expect either 'the mean of a surd and its reciprocal', 'rationalise this expression', or 'x + 1/x given x = a surd'; all three are solved by the same conjugate step.

Related PYQs

No directly related past PYQ was found.

Practice

practice — not a real PYQ
Which one of the following is the arithmetic mean of √5 – 2 and its reciprocal ?
- (a)2
- (b)√5
- (c)2√5
- (d)1
Answer(b) √5
practice — not a real PYQ
If x = √3 – √2, which one of the following is equal to x + 1/x ?
- (a)2√2
- (b)2√3
- (c)√3
- (d)2
Answer(b) 2√3

Why