There are 20 girls and 30 boys in a class, and their respective average marks are found to be 55 and 58. The average marks of the entire class are
- (a)56·5
- (b)56·6
- (c)56·7
- (d)56·8
Answer
Why
Correct — D, (d) 56·8.
This is a WEIGHTED MEAN, and the whole question is whether you notice that the two groups are not the same size.
Work with totals, never with the two averages directly. An average multiplied by its count gives back the total, and totals can be added:
girls 20 × 55 = 1100 boys 30 × 58 = 1740 class 1100 + 1740 = 2840 over 20 + 30 = 50 pupils 2840 ÷ 50 = 56·8
That is option (d).
THE FASTER ROUTE, and the one worth owning, is ALLIGATION. The class average has to sit somewhere on the line between 55 and 58, and it sits nearer whichever group is bigger. Boys are 30 of the 50, so they pull three-fifths of the way:
55 + (30/50) × (58 − 55) = 55 + 0·6 × 3 = 56·8
Equally, the girls pull two-fifths: 58 − 0·4 × 3 = 56·8. The two computations must agree, which makes this its own check.
TWO SANITY TESTS DISPOSE OF HALF THE PAPER'S OFFER BEFORE ANY ARITHMETIC. First, a combined average always lies strictly between the two group averages, so anything outside 55 to 58 is impossible. Second — and this is what the option set is built on — it lies on the SAME SIDE of the midpoint as the LARGER group. The midpoint of 55 and 58 is 56·5. Because there are more boys than girls, and the boys are the higher-scoring group, the answer must be ABOVE 56·5. That alone kills option (a), which is exactly 56·5.
The decimal points in all four options are printed as raised middle dots rather than full stops, which is this booklet's habit throughout its quantitative items.
Why the others are wrong
- (a)56·5 — This is the trap the whole item is built around: it is the SIMPLE average of the two averages, (55 + 58) ÷ 2 = 56·5, taken as though the two groups were the same size. They are not — 20 girls against 30 boys — and averaging averages is only valid when the counts are equal. The error is easy to see in totals: 56·5 across 50 pupils implies a class total of 2825, whereas the two groups actually contribute 1100 + 1740 = 2840. It is also the one option that can be ruled out without arithmetic at all. The boys are both the larger group and the higher-scoring one, so the class figure must be pulled ABOVE the midpoint of 55 and 58; 56·5 IS that midpoint, so it is the one value the data make impossible.
- (b)56·6 — A near miss with no calculation behind it. An average of 56·6 over 50 pupils implies a class total of 56·6 × 50 = 2830, which is 10 short of the 2840 the two groups actually produce. Read through alligation, 56·6 would mean the higher-scoring group carried a weight of (56·6 − 55) ÷ 3 = 8/15 of the class, that is about 26·7 of the 50 pupils rather than the 30 the question gives. Options like this exist to catch a candidate who has understood that the answer must exceed 56·5 and then guesses the first value above it instead of finishing the sum.
- (c)56·7 — The same near miss one step closer. At 56·7 the implied class total is 2835, still 5 short of 2840. In alligation terms it corresponds to a weight of (56·7 − 55) ÷ 3 = 17/30 for the boys — about 28·3 pupils out of 50, not 30. Three of the four values on offer lie in a tight band 0·1 apart precisely so that an approximate method cannot separate them: once you know the answer is above 56·5 you still have to do the division. The honest defence is to compute the two totals, add them and divide, which takes a few seconds and leaves nothing to guess.
Concept
The WEIGHTED MEAN is the arithmetic mean of a pooled set, computed from the group means and their sizes:
combined mean = (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂)
The idea behind the formula is that an average is a total in disguise. Multiply a mean by its count and you recover the group's total; totals are additive in a way that means are not. That single move — mean to total, add, divide by the pooled count — solves every question of this shape, including ones with three or four groups.
WHY AVERAGING THE AVERAGES FAILS. The simple average (x̄₁ + x̄₂) ÷ 2 is the special case of the weighted mean in which n₁ = n₂. Whenever the groups differ in size, it silently gives the smaller group the same influence as the larger, and the error runs in the direction of the smaller group. Here it understates the class figure by 0·3, because it lets 20 girls count for as much as 30 boys.
ALLIGATION is the same formula rearranged, and it is much quicker by hand. The combined mean divides the interval between the two group means in the RATIO OF THE COUNTS, reckoned crosswise: the distance from the combined mean to a group's mean is inversely proportional to that group's size. So with 20 and 30 the answer sits two-fifths of the way from 58 and three-fifths of the way from 55. Alligation is worth learning because it also runs backwards — given the combined mean and the two group means, it hands you the ratio of the counts, which is how the same idea is asked in reverse.
TWO PROPERTIES THAT WORK AS CHECKS. The combined mean always lies strictly between the two group means, so any option outside that interval is impossible. And it lies on the same side of the midpoint as the larger group, so the midpoint itself is only the answer when the groups are equal in size. On this item the midpoint is printed as an option, and it is the one figure the data rule out.
Quantitative aptitude, data and statistics is the single largest strand on this APFC paper — roughly a third of Part B — and averages recur inside it more than any other single topic. The construction here is the standard one: two groups, two averages, unequal counts, and the unweighted average of the two averages placed on the page as an option.
That construction rewards a specific habit. Before computing anything, ask which way the answer must lean. The larger group pulls the combined figure towards its own mean, so the answer must fall on the larger group's side of the midpoint. Doing that first turns a four-way choice into a two- or three-way one and, on an item like this, immediately exposes which option is the designed error.
The habit also protects against the mirror-image question, which this family of papers asks just as often: the combined average is given and the ratio of the two group sizes is wanted. The same alligation line answers it read from right to left, and a candidate who only knows the totals-and-divide routine has to set up an equation instead.
The three plausible options are spaced 0·1 apart, so estimation cannot finish the job. That spacing is a signal in itself: when the offered values are that close, the paper is asking for an exact computation, and the fastest exact route is totals rather than decimals.
Key facts
- The combined mean of two groups is (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂) — convert each mean back to a total, add the totals, then divide by the pooled count.
- Here 20 × 55 = 1100 and 30 × 58 = 1740, so the class total is 2840 over 50 pupils, giving 56·8.
- Averaging two averages is valid ONLY when the two groups are of equal size; otherwise it gives the smaller group the same influence as the larger.
- Alligation: the combined mean divides the gap between the two group means in the ratio of the counts taken crosswise — 55 + (30/50) × 3 = 56·8, and equally 58 − (20/50) × 3 = 56·8.
- A combined mean always lies strictly between the two group means, which rules out any option outside that interval before any arithmetic is done.
- It lies on the same side of the midpoint as the larger group, so the midpoint of the two means is the answer only when the groups are equal in size.
- The same relation run backwards recovers the ratio of the group sizes from the combined mean, which is how this idea is asked in reverse.
Study next
Common traps
- Averaging the two averages. That is only correct when the groups are the same size, and the paper prints the resulting value as an option.
- Forgetting to check which side of the midpoint the answer must fall on. The larger group always pulls the combined mean towards its own mean.
- Dividing by the number of groups instead of by the pooled count — dividing 2840 by 2 rather than by 50.
- Assuming a rough estimate can separate options spaced 0·1 apart. Values that close are a signal that an exact computation is wanted.
- Reading the raised middle dot in the options as anything other than a decimal point; this booklet prints its decimals that way throughout.
Averages appear on every EPFO paper and usually more than once. The recurring shapes are: two or more groups with different sizes pooled into one (this item); a group average changed by adding, dropping or correcting one member; the average of a run of consecutive integers, where the answer is the midpoint of the first and last; and the reverse question, in which the combined average is supplied and the ratio of the group sizes is wanted. The option set almost always contains the value produced by averaging the averages, because that is the single most common error. The defence is uniform — convert every average into a total before doing anything else, and sanity-check the result against the interval between the group means and against the midpoint rule.
Related PYQs
EPFO_EOAO_2020_Q119The 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(c) 20
The same weighted-average relation run BACKWARDS: a class of 100 averages 46 kg while boys average 50 and girls 40, and the difference between the numbers of boys and girls is wanted. Alligation answers it in one line.
EPFO_APFC_2016_Q99Consider the sequential integers 27 to 93, both included in the sequence. The arithmetic average of these numbers will be
- (a) 61·5
- (b) 61
- (c) 60·5
- (d) 60
Answer(d) 60
The other averaging item on this paper — the arithmetic average of the consecutive integers 27 to 93, where the answer is simply the midpoint of the first and last term.
EPFO_EOAO_2020_Q116The 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
An averages item disguised as an ages problem, solved the same way: turn each stated average into a total, then work with the totals.
Practice
- practice — not a real PYQ
In a section of 40 students, 15 students score an average of 60 and the remaining 25 score an average of 68. What is the average for the whole section ?
- (a)64
- (b)64·5
- (c)65
- (d)66
Answer(c) 65 — the totals are 15 × 60 = 900 and 25 × 68 = 1700, so the section total is 2600 over 40 students, giving 65. Averaging the two averages would give 64, which is the value on offer as option (a); it is wrong because the higher-scoring group is the larger one, so the answer must lie above the midpoint of 60 and 68.
- practice — not a real PYQ
The average of m numbers is 20 and the average of another n numbers is 30. If the average of all the m + n numbers taken together is 26, then m : n is
- (a)2 : 3
- (b)3 : 2
- (c)1 : 2
- (d)2 : 1
Answer(a) 2 : 3 — from 20m + 30n = 26(m + n) we get 4n = 6m, so m : n = 2 : 3. By alligation the same answer falls out at sight: the combined value 26 is 6 above 20 and 4 below 30, and the counts go crosswise, giving 4 : 6 = 2 : 3. The larger group is the one with the mean of 30, which is consistent with 26 lying above the midpoint of 25.