EPFO EO/AO 2023 General Ability Test · Question 104 of 120

Correct — D, (d) 20. Work in the three disjoint groups the school actually splits into rather than in the two overlapping ones, and the problem solves itself in two lines. Every student plays at least one game, so there are only three kinds of student: those who play only cricket, those who play only football, and those who play both. Let the number who play only football be f and the number who play both be b. The stem says the number who play only cricket is three times the number who play only football, so only-cricket is 3f. Since the three groups exhaust the school, 3f + f + b = 100, that is, 4f + b = 100. The second condition is about the totals, not the exclusive groups. The cricket total is only-cricket plus both, which is 3f + b, and the football total is only-football plus both, which is f + b. Cricket is twice football: 3f + b = 2(f + b), so 3f + b = 2f + 2b, so f = b. That is the whole problem. The number who play only football equals the number who play both. Substituting f = b into 4f + b = 100 gives 5b = 100, so b = 20. The full picture is worth writing out, because it is what a checker needs: 20 play both, 20 play only football, 60 play only cricket. Cricket in total is 60 + 20 = 80 and football in total is 20 + 20 = 40, and 80 is indeed twice 40. Only-cricket 60 is three times only-football 20. The three groups add to 60 + 20 + 20 = 100. Every condition in the stem is satisfied, and the answer is 20. The same result comes from the inclusion-exclusion identity if you prefer it: n(C or F) = n(C) + n(F) - n(C and F), and because nobody is outside both games the left-hand side is 100. With n(C) = 2 n(F) this gives 100 = 3 n(F) - b, and the exclusive-group condition supplies the second equation.

• (a)30 — Thirty fails on a whole-number test before any ratio is checked. If 30 played both, then the remaining 70 students split into only-cricket and only-football in the ratio 3 to 1, which needs only-football to be 70 / 4 = 17.5 students. People do not come in halves, so 30 is impossible. Even ignoring that, the ratios do not hold: with only-football 17.5 and only-cricket 52.5, the cricket total is 82.5 and the football total is 47.5, and 82.5 is not twice 47.5. The option is placed to catch a candidate who has combined the two conditions incorrectly and arrived at a comfortable round number, and it is a reminder that in counting problems an answer producing fractional people has already disproved itself.
• (b)28 — With 28 playing both, the other 72 divide three to one as 54 who play only cricket and 18 who play only football, which at least keeps the numbers whole. But then the cricket total is 54 + 28 = 82 and the football total is 18 + 28 = 46, and twice 46 is 92, not 82. The second condition fails. This option is the trap for a candidate who has used only the exclusive-group ratio and never tested the totals, and it shows why the two conditions in the stem are genuinely different statements: one is about students who play only that game, the other is about everyone who plays it. Confusing the two is the single commonest error in set-overlap questions.
• (c)25 — Twenty-five leaves 75 students to divide three to one, which needs only-football to be 18.75, so this option fails the whole-number test in the same way as (a). It is the most tempting wrong value because 25 is a quarter of 100 and the arrangement feels natural, and because a candidate who has silently assumed that the two conditions say the same thing may reach it by treating the cricket total as three times the football total instead of twice. Testing that reading is instructive: if the totals were in the ratio 3 to 1 and the exclusive groups in the ratio 3 to 1 as well, the overlap would have to be zero, which contradicts the existence of students who play both.

Problems about two overlapping groups are solved by keeping strictly separate two kinds of quantity: the total in a set and the number in that set alone. If C and F are the two sets, then n(C) counts everyone who plays cricket, including those who also play football, while 'only cricket' counts n(C) minus the overlap. The relation that ties them together is the inclusion-exclusion principle, n(C or F) = n(C) + n(F) - n(C and F), and a phrase such as 'every student plays either cricket or football or both' is the examiner telling you that n(C or F) equals the whole population, so nobody sits outside the two circles. Once that is fixed, the cleanest variables are the three disjoint regions of the Venn diagram: only-C, only-F and both. They add to the total with no correction term, every condition in the stem can be written in terms of them, and no double counting is possible. Where a stem also allows people outside both sets, a fourth region has to be carried, and the total becomes only-C plus only-F plus both plus neither. The extension to three sets follows the same logic with seven regions and the longer inclusion-exclusion formula, in which the triple overlap is subtracted three times and added back once.

Set-overlap questions appear regularly in the reasoning and quantitative blocks of the EO/AO General Ability Test because they can be built entirely out of one sentence and still separate careful readers from quick ones. The discrimination is almost always the same: one condition is stated about a whole set and another about the exclusive part of it, and a candidate who reads both as the same kind of statement gets a clean-looking wrong answer. The habit rewarded is to draw two overlapping circles, label the three regions, and translate each sentence of the stem into a relation among those labels before solving anything.

• Inclusion-exclusion for two sets: n(C or F) = n(C) + n(F) - n(C and F).
• 'Every student plays either game or both' means nobody is outside the union, so the union equals the total.
• The three disjoint regions here are only-cricket, only-football and both, and they add to 100 with no correction.
• Only-cricket = 3 x only-football gives 4f + b = 100 when the regions are summed.
• Cricket total = 2 x football total reduces to f = b, so those who play only football equal those who play both.
• The consistent solution is both = 20, only-football = 20, only-cricket = 60, cricket = 80 and football = 40.
• In counting problems, any candidate answer that forces a fractional number of people is disproved by that alone.

• Reading 'the number who play cricket' as 'the number who play only cricket'; the two differ by the overlap.
• Forgetting to subtract the intersection once when adding the two set totals.
• Assuming there is a 'neither' group when the stem has ruled it out, or ignoring one when the stem allows it.
• Accepting an answer that makes some group a fraction of a person.
• Solving for a variable and reporting it without checking every condition in the stem.

The EO/AO paper sets this family either as a pure counting question like this one, or as a percentage version in which the population is not given and the answer is a proportion. A third variant supplies three activities and asks for the number in exactly one or exactly two of them, which needs the three-set formula. In all versions the first move is the same: decide whether the numbers given are set totals or exclusive counts, and label a diagram accordingly. Committing that step to paper costs fifteen seconds and removes the only error the question is designed to produce.

No directly related past PYQ was found.

• practice — not a real PYQ

  In a class of 60 students, every student studies at least one of Hindi and Sanskrit. If 45 study Hindi and 30 study Sanskrit, how many study both ?

  • (a)10
  • (b)15
  • (c)20
  • (d)25

  Answer(b) 15

• practice — not a real PYQ

  In a group of 80 people, everyone drinks tea or coffee or both. The number who drink tea is three times the number who drink coffee, and 20 people drink both. How many people drink coffee ?

  • (a)20
  • (b)25
  • (c)30
  • (d)40

  Answer(b) 25