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.

The average weight of 100 students in a class is 46 kg. The average weights of boys and girls are 50 kg and 40 kg respectively. What is the difference between the number of boys and girls ?

(a)30
(b)25
(c)20
(d)10

Answer

Why

Correct — C, (c) 20. The step that decides this item is recognising that the class average is a weighted average of the two group averages, so its position between them fixes the ratio of the group sizes. The class average of 46 kg sits 6 kg above the girls' 40 kg and 4 kg below the boys' 50 kg. Those two gaps are the weights, crossed over: the number of boys to the number of girls is 6 : 4, that is 3 : 2. With 100 students in all, that gives 60 boys and 40 girls, and the difference asked for is 60 - 40 = 20. This is the method called alligation, and it is worth learning as a picture — the mean in the middle, the two group values on either side, the distances taken diagonally — because it answers this whole family of questions in a line. The algebra behind it is just as short if it is preferred. Let there be b boys, so there are 100 - b girls, and total weight is 50b + 40(100 - b) = 46 x 100. That is 10b + 4000 = 4600, so b = 60 and the girls number 40. There is a check to make before either computation, and it takes a second: 46 is closer to 50 than to 40, so the boys, whose average is 50, must be the larger group. Any answer that would make the girls more numerous is wrong before it is calculated. One further relation makes every option on this item testable at sight. With 100 students the class average works out as 45 plus one-twentieth of the difference between the numbers of boys and girls, so a difference of 20 gives exactly 46 — and each of the other three differences gives an average that is not 46. Note finally that the four options here are printed in descending order, 30, 25, 20, 10, which is the Commission's own arrangement.

Why the others are wrong

(a)30 — 30 would mean 65 boys and 35 girls, since the two numbers must add to 100 and differ by 30. Put those into the totals and the class weight comes to 65 x 50 + 35 x 40 = 3250 + 1400 = 4650 kg, which is an average of 46.5 kg, not the 46 kg the stem states. The option is a half kilogram out, which is exactly what the relation above predicts: on a class of 100 with these two group averages, every extra unit of difference between the numbers of boys and girls lifts the class average by one-twentieth of a kilogram, so a difference of 30 instead of 20 raises the average by half a kilo. It is the option a candidate reaches for after getting the direction right — more boys than girls — and then guessing at the magnitude instead of using the two gaps.
(b)25 — 25 cannot be the answer to this question whatever the weights are, and seeing why takes no arithmetic at all. The number of boys and the number of girls add up to 100, an even number, so the two are either both even or both odd, and in either case their difference is even. An odd difference of 25 would require 62.5 boys and 37.5 girls. That parity check is worth carrying into every problem of this shape, because it disposes of an option before the real work begins. For completeness, 62.5 boys and 37.5 girls would give a class average of 46.25 kg, which is not 46 either — so the option fails on the arithmetic as well as on the impossibility of half a student.
(d)10 — 10 would mean 55 boys and 45 girls, giving a class total of 55 x 50 + 45 x 40 = 2750 + 1800 = 4550 kg and a class average of 45.5 kg rather than 46 kg. It has the direction right, with boys outnumbering girls, but it puts the class average too close to the midpoint of 45 kg. That midpoint is the thing to hold on to: if the two groups were equal in size the class average would be exactly halfway between 40 and 50, at 45 kg, and the actual average of 46 is a full kilogram above it. A difference of only 10 shifts it half that far. Comparing the given mean with the midpoint of the two group means is the fastest way to judge whether an offered difference is of the right size.

Concept

The idea being tested is the weighted average, and the useful way to hold it is geometrically. When two groups with averages of x and y are pooled, the combined average always lies between x and y, and where it lies is determined entirely by the relative sizes of the groups: it is pulled towards the larger one. If the two groups are equal, the combined average sits at the midpoint. If one group is twice the size of the other, the combined average sits twice as close to that group's value. Turned around, this is the rule of alligation: the ratio of the two group sizes is the ratio of the distances from the combined mean to the two group values, taken crosswise, so the group nearer the mean is the larger. Here the mean of 46 is 6 away from 40 and 4 away from 50, so boys are to girls as 6 is to 4. The same machinery answers questions about mixtures of two grades of rice, two concentrations of a solution, two rates of interest and two speeds on two stretches of a journey; the words change and the picture does not. The algebraic form — one unknown, the other written as the total minus it, and the totals equated — is equally reliable and is worth keeping as the fallback whenever a stem is complicated enough that the picture is hard to draw.

This is the fourth item of the paper's final block and it is the one that most rewards a technique rather than raw calculation. Solved by alligation it takes about fifteen seconds; solved by setting up and grinding through an equation it takes perhaps a minute; and both get the mark, which is precisely the point about a hundred-and-twenty-question paper where time is the binding constraint. The other thing this item rewards is the habit of a direction check before any computation — the class average of 46 kg lies nearer the boys' 50 kg than the girls' 40 kg, so there must be more boys, and half of the reasoning is done. A word of caution about the option list: it is printed in descending order, 30, 25, 20, 10. That is deliberate on this paper, which prints several numeric sets out of ascending order, and a candidate who assumes options increase down the list and picks by position rather than by value can mark the wrong letter after doing everything else correctly.

Key facts

A combined average always lies between the two group averages and is pulled towards the larger group; if the groups are equal in size it sits exactly at the midpoint of the two values.
Alligation states the rule in reverse: the ratio of the group sizes equals the ratio of the distances from the combined mean to the two group values, taken crosswise, so the group nearer the mean is the bigger one.
The working here: 46 is 6 above 40 and 4 below 50, so boys are to girls as 6 : 4, that is 3 : 2; on 100 students that is 60 boys and 40 girls, and the difference is 20.
The algebraic route gives the same thing: 50b + 40(100 - b) = 4600 reduces to 10b = 600, so there are 60 boys and 40 girls.
For this class the relation between the answer and the average is exact — the class average equals 45 kg plus one-twentieth of the difference between the numbers of boys and girls — so 20 gives 46 kg while 30, 25 and 10 give 46.5, 46.25 and 45.5 kg.
A parity check disposes of one option instantly: two numbers adding to 100 must differ by an even number, so an odd difference such as 25 is impossible whatever the weights are.

The class average is a weighted average — where it sits fixes the ratio

The direction check, before any computation→46 is closer to 50 than to 40, so the boys, whose average is 50, must be the LARGER group. Any answer that would make the girls more numerous is wrong before it is calculated.

Alligation — take the two distances crosswise→The class average of 46 kg sits 6 kg above the girls’ 40 kg and 4 kg below the boys’ 50 kg. Those two gaps are the weights, crossed over: the number of boys to the number of girls is 6 : 4, that is 3 : 2. It is worth learning as a picture — the mean in the middle, the two group values on either side, the distances taken diagonally — because it answers this whole family of questions in a line.

60 boys, 40 girls, and a difference of 20→With 100 students in all, a ratio of 3 : 2 gives 60 and 40.

The algebra, if it is preferred→Let there be b boys, so there are 100 − b girls. Total weight is 50b + 40(100 − b) = 46 x 100, that is 10b + 4000 = 4600, so b = 60 and the girls number 40. Equally reliable, and the fallback whenever a stem is complicated enough that the picture is hard to draw.

One relation makes every option testable at sight→On a class of 100 with these two group averages, the class average equals 45 kg plus one-twentieth of the difference between the numbers of boys and girls. A difference of 20 gives exactly 46 kg, while 30 gives 46·5, 25 gives 46·25 and 10 gives 45·5. That midpoint of 45 is the thing to hold on to: if the two groups were equal in size the class average would sit exactly halfway between 40 and 50.

A parity check disposes of one option instantly→The number of boys and the number of girls add up to 100, an even number, so the two are either both even or both odd — and in either case their difference is EVEN. An odd difference of 25 would require 62·5 boys and 37·5 girls, and is impossible whatever the weights are. Note also that the four options on this item are printed in descending order: 30, 25, 20, 10.

Study next

Alligation and mixtures, including problems where two grades of a commodity are blended to reach a stated price or a stated concentrationWeighted averages in other clothes — average speed over two stretches of a journey, and a combined rate of interest on two sumsProblems in which a member joins or leaves a group and the average shifts, solved by comparing the totals before and afterRepeated dilution and replacement problems, where a quantity of a mixture is removed and replaced by one of the components several times overSimple simultaneous equations from worded data, the fallback method whenever the alligation picture is awkward to draw

Common traps

Assuming the answer is the number of boys rather than the difference asked for. The working produces 60 and 40 on the way, and 60 is a satisfying number to arrive at, but the stem asks for the gap between them.
Getting the alligation ratio the wrong way round. The distances are taken crosswise, so the group whose average is nearer the combined mean is the larger one, and reversing that gives 40 boys and 60 girls.
Reading the option list as ascending. These four are printed 30, 25, 20, 10, in descending order, so a candidate who marks by position after computing correctly can still lose the mark.
Ignoring the parity of the total. With an even number of students in the class, the difference between the two groups must also be even, which rules out an odd option without any calculation at all.

Averages of two combined groups are a fixture of EPFO EO/AO quantitative blocks, usually as weights or marks of boys and girls in a class, or as the price of a mixture of two grades. The numbers are chosen so that the alligation ratio comes out clean, which is a signal in itself: if a mixture question is producing ugly fractions, the method being used is probably not the intended one. The ask varies — the number in one group, the ratio of the two groups, the difference between them, or the combined average when the sizes are given — so the last line of the stem has to be read as carefully as the data. Expect the related mixture and average-speed items in the same block, which use identical machinery under different names.

Related PYQs

EPFO_EOAO_2020_Q81Open & attempt →

Which one of the following is the average of all prime numbers between 21 and 55 ?

(a) 35·85
(b) 36·71
(c) 38·00
(d) 39·00

Answer(c) 38·00

The average item from the earlier quantitative block, which asks for the average of all the primes in a range. That one tests whether the set can be listed correctly; this one tests what an average does once two groups are pooled.

EPFO_EOAO_2020_Q116Open & attempt →

The average age of a husband and his wife was 23 years when they were married 5 years ago. The average age of the husband, the wife and their child is 20 years now. How old is the child now ?

(a) 9 months
(b) 1 year
(c) 3 years
(d) 4 years

Answer(d) 4 years

The average-age item three questions earlier. Both convert averages into totals before doing anything else, and both hide their difficulty in a single structural reading rather than in the arithmetic.

Practice

practice — not a real PYQ
The average weight of 30 students in a class is 55 kg. The average weight of the boys is 60 kg and that of the girls is 45 kg. How many girls are there in the class ?
- (a)8
- (b)10
- (c)12
- (d)15
Answer(b) 10
practice — not a real PYQ
Rice costing 40 rupees per kg is mixed with rice costing 60 rupees per kg to obtain 50 kg of a mixture worth 46 rupees per kg. How many kg of the costlier variety does the mixture contain ?
- (a)15 kg
- (b)20 kg
- (c)25 kg
- (d)30 kg
Answer(a) 15 kg

Why