EPFO APFC 2023 General Studies · Question 37 of 120

Correct — B, (b) 2 only. Convert both means into totals, which is the move that solves almost every question of this family. The mean of the weights of P, Q, R and S is 50 kg over four people, so P + Q + R + S = 200. The mean of the weights of Q, R, S and T is also 50 kg over four people, so Q + R + S + T = 200. Subtract the second equation from the first: the common block Q + R + S cancels and what remains is P − T = 0, that is, P = T. So statement 2 is correct — the weights of P and T are equal — and it is correct without any information about the individual weights at all. Statement 1 claims something stronger, that T weighs 50 kg, and nothing in the data forces it. The two equations give one relation between five unknowns; they pin down the total of the block Q + R + S only in combination with P or T, and they leave the split entirely open. A single counter-example settles it: let P = 80 and Q = R = S = 40. Then P + Q + R + S = 80 + 120 = 200, so the first mean is 50; and Q + R + S + T = 120 + T must also be 200, giving T = 80. Both conditions in the question hold, and T is 80 kg, not 50. Notice that P = T = 80 in this example, exactly as statement 2 requires — the counter-example to statement 1 obeys statement 2, which is the clearest possible demonstration that the second follows from the data and the first does not. The underlying principle generalises and is worth carrying: if two groups of the same size have the same mean and share all but one member, then the two unshared members are equal, because each group's total is the same and the shared block contributes the same amount to both. Everything else about the groups — the individual values, the spread, whether anyone weighs 50 kg — remains undetermined.

• (a)1 only — '1 only' inverts the truth of both statements. It takes the claim the data cannot support, that T weighs 50 kg, and rejects the one the data proves, that P and T are equal. The reasoning behind it is usually a slip from 'the mean is 50' to 'each member is 50', which is the single commonest misunderstanding of an average: a mean is a statement about a total shared out, not about any individual value in the set. Four people averaging 50 kg may weigh 20, 40, 60 and 80; nothing about the average tells you that anyone in the group weighs 50, and adding a second group with the same average does not change that.
• (c)Both 1 and 2 — 'Both 1 and 2' accepts the correct deduction and then adds the unfounded one. It is the natural choice for a candidate who has spotted that P = T and has assumed that the shared mean of 50 must also fix the individual values, or who has quietly assumed the simplest case in which all five friends weigh 50 kg. That case is consistent with the data, but consistency is not entailment: a statement is correct in this format only if it must hold for every arrangement the conditions allow, and the arrangement P = T = 80 with Q = R = S = 40 satisfies both conditions while making statement 1 false.
• (d)Neither 1 nor 2 — 'Neither 1 nor 2' rejects a deduction that follows in one line of algebra. It is the option for a candidate who sees that five unknowns cannot be determined from two equations and concludes that nothing at all can be said. But under-determination is not ignorance: two equations in five unknowns still constrain the unknowns, and here the constraint is exactly the relation P = T, obtained by subtracting one equation from the other so that the three shared terms cancel. Looking for what cancels, rather than for what can be solved, is the technique this item is testing.

The arithmetic mean of n values is their sum divided by n, and the practical consequence for examination work is that a statement about a mean is a statement about a total: mean × count = sum. Almost every mean-based reasoning question is solved by turning each given average into a total and then adding or subtracting the totals. Two structural facts follow and are worth holding as results rather than rederiving each time. First, if two groups of equal size have equal means and overlap in all but one member, the two unshared members are equal, because the shared block contributes identically to two identical totals. Second, if one value in a set of n is changed by an amount d, the mean changes by d/n; equivalently, replacing a member of weight x by one of weight y shifts the mean by (y − x)/n, which is the same result written the other way round. Neither result says anything about individual values inside the shared block, and that is the boundary the question is probing: the data fix a relation between the two people who differ between the groups, and leave the three people common to both groups completely free. Keeping straight what a mean determines and what it leaves open is the whole content of this item.

The quantitative block of this paper uses the two-statement code format for statistics as well as for geometry, and the two require the same discipline: a statement counts as correct only if it must be true under every arrangement the conditions permit, so one statement is proved by algebra and the other is disproved by a single constructed example. Building that example is a skill worth rehearsing, because it is far faster than arguing about whether something is determined. Here the construction takes seconds — choose any value for the common block, let the two ends adjust — and it converts an argument into a decided fact. The format also carries a scoring implication. Because a wrong answer costs one-third of the 2.5 marks on offer, and because the four code options are mutually exclusive, deciding both statements confidently is worth the extra fifteen seconds; a candidate who has settled statement 2 but is unsure about statement 1 is choosing between two live options and should ask whether an example can be built rather than guessing between them.

• The arithmetic mean multiplied by the number of values gives the total, so a mean of 50 kg for four people means their weights sum to 200 kg; converting means into totals is the first step in almost every problem of this kind.
• If two groups of the same size have the same mean and share all but one member, the two unshared members must be equal — subtracting the two totals cancels the shared block and leaves the relation directly.
• A mean tells you nothing about any individual value in the set: four people averaging 50 kg may weigh 20, 40, 60 and 80, and no member of the group need weigh 50 kg at all.
• P = 80 with Q = R = S = 40 satisfies both conditions of this question and gives T = 80, which disproves the claim that T must weigh 50 kg while confirming that P and T are equal.
• Changing one value in a set of n by an amount d changes the mean by d/n, which is the general form of the swap rule used when one member of a group is replaced by another.

• Reading a mean as a statement about individuals: an average of 50 kg does not put anyone at 50 kg, and this item's first statement exists to catch exactly that slip
• Assuming the simplest arrangement — everyone equal — and treating what is consistent with the data as though it were entailed by the data
• Concluding that nothing can be deduced because there are more unknowns than equations; the relation P = T falls straight out of subtracting the two totals
• Deciding one statement and guessing the other in a code-set item; with four mutually exclusive codes and a one-third deduction for a wrong answer, constructing a counter-example is faster and safer than a coin toss

Averages appear in this paper both as direct computation and, more often, as two-statement code items testing whether a candidate knows what an average determines. The recurring devices are a shared block between two groups, a replacement of one member by another, and a claim that some individual value must equal the mean. Prepare by rehearsing the two structural results — equal means with a shared block force the unshared members to be equal, and a change of d in one value moves the mean by d/n — and by practising the construction of counter-examples, since half the marks in this format come from disproving rather than proving.

No directly related past PYQ was found.

• practice — not a real PYQ

  The mean weight of five students A, B, C, D and E is 52 kg. If the mean weight of A, B, C and D is 50 kg, what is the weight of E?

  • (a)52 kg
  • (b)56 kg
  • (c)60 kg
  • (d)62 kg

  Answer(c) 60 kg — the five students together weigh 5 × 52 = 260 kg and the first four weigh 4 × 50 = 200 kg, so E weighs 260 − 200 = 60 kg. The method is the one that solves the whole family: convert every stated mean into a total, and then take the difference of the totals to isolate the member that appears in one group but not the other.

• practice — not a real PYQ

  The mean weight of a group of six friends is 60 kg. One member weighing 54 kg leaves the group and is replaced by a new member, after which the mean is still 60 kg. The weight of the new member is

  • (a)54 kg
  • (b)57 kg
  • (c)60 kg
  • (d)66 kg

  Answer(a) 54 kg — the group size is unchanged at six and the mean is unchanged at 60 kg, so the total is unchanged at 360 kg, which means the member who left and the member who joined must weigh the same. This is the same structural result as in the original question: two groups of equal size with equal means that share all but one member have unshared members of equal value.