BPSC 70th CCE Re-exam Prelims General Studies · Question 149 of 142

Correct — A, 3. Number the chairs 1 to 15 from the left and place both people before counting anything. P is at the middle. With an odd number of chairs the middle is exact: for 15 chairs it is the 8th, because (15 + 1) ÷ 2 = 8, and that seat has seven chairs on each side of it, which is what 'middle' means. Q is 12th counting from the right, and to compare the two positions they must be measured from the same end. The rule for converting is that a position from the left and the same object's position from the right always add up to one more than the total, so position from the left = 15 − 12 + 1 = 4. Q therefore sits on chair 4 and P on chair 8. Now count what lies strictly between them: chairs 5, 6 and 7 — three chairs. Written as a formula, the number of items strictly between two positions is the difference minus one, that is 8 − 4 − 1 = 3. The stem is asking how many chairs are 'between' P and Q, and 'between' excludes the two chairs the people are themselves sitting on, which is the single point on which the whole question turns. Every wrong option here comes from one identifiable slip, and each is worth recognising: 4 is the plain difference 8 − 4, which counts one endpoint; 5 is the count from chair 4 to chair 8 inclusive, which counts both; and 2 comes from taking the middle of 15 to be the 7th chair rather than the 8th. A note on the printed text: the booklet sets the ordinal in '12th' with a raised 'th', and our copy stores it on the line. Nothing about the meaning changes.

• (b)2 — Comes from misplacing P rather than from miscounting. If the middle of 15 chairs is taken as the 7th — by halving 15 and dropping the fraction — then only chairs 5 and 6 lie between it and chair 4, giving 2. But 15 is odd, so the middle is exact and is the 8th chair, with seven chairs on either side; the 7th has six on one side and eight on the other.
• (c)5 — The inclusive count. Chairs 4, 5, 6, 7 and 8 make five chairs, but that range includes the seats P and Q are sitting on. 'Between' asks for the chairs strictly in the gap, so both endpoints must be dropped. This is the answer for a candidate who places both people correctly and then counts the wrong span.
• (d)4 — The plain difference between the two positions, 8 − 4, which is a count of one endpoint too many — it includes either chair 4 or chair 8 but not both. Off-by-one errors of exactly this kind are what these questions are built to catch, and the defence is to write out the intervening numbers rather than subtracting.

Linear position questions are governed by three small rules, and almost every item of this type is one of them in disguise. First, the conversion rule: for n objects in a line, an object's position from one end plus its position from the other end equals n + 1, so position from the left = n − position from the right + 1. Second, the middle rule: when n is odd the middle position is (n + 1) ÷ 2 and it is unique, and when n is even there is no single middle — the two central positions are n ÷ 2 and n ÷ 2 + 1, which is why examiners set these puzzles with odd totals. Third, the gap rule: the number of objects strictly between positions a and b is the difference minus one, while the number of objects from a to b inclusive is the difference plus one. Those two differ by two, and the four options of a question like this one are usually built so that all three counts appear among them. A related family uses the same conversion in reverse: given a person's position from both ends, the total is the sum of the two positions minus one.

The habit that makes these free is to convert everything to one reference end immediately, before reading the question again. Here that means turning '12th from the right' into 'the 4th from the left' at once, so that both people are described on the same scale, and only then asking what is being counted. The second habit is to read the counting word with care: 'between' excludes both endpoints, 'from A to B' includes both, and 'after A up to B' includes one. In a four-option question the examiner has usually supplied all three answers, so a candidate who does the arithmetic correctly but reads the counting word carelessly will find their wrong number waiting on the page — which is exactly what 4 and 5 are doing here. The third habit, on any question with fifteen or fewer positions, is simply to write the numbers out. Fifteen chairs can be drawn as fifteen dashes in a few seconds, P and Q marked on them, and the answer read off directly with no formula and no chance of an off-by-one. Formulas are for when the numbers are too large to draw.

• For n objects in a row, position from the left + position from the right = n + 1; here 15 − 12 + 1 puts Q at chair 4
• For an odd n the middle position is (n + 1) ÷ 2, so the middle of 15 chairs is the 8th, with seven chairs on each side
• The number of objects strictly between positions a and b is |a − b| − 1; here 8 − 4 − 1 = 3
• The inclusive count from a to b is |a − b| + 1, which is 5 here — two more than the 'between' count, and offered as option (c)
• For an even number of positions there is no single middle, which is why puzzles of this kind are almost always set with an odd total; for even n the two central places are n ÷ 2 and n ÷ 2 + 1
• P at chair 8 is 8th from the left and 8th from the right, which is the check that confirms it really is the middle of a row of 15
• Run in reverse, the same rule gives the total from two positions: if a person is pth from one end and qth from the other, the row holds p + q − 1 people, because that person is otherwise counted twice
• All three counts a candidate might make are present among the options — 3 strictly between, 4 as the bare difference and 5 inclusive of both endpoints — so a correct calculation read against the wrong counting word still finds a wrong answer on the page

• Reading 'between' as inclusive; the chairs P and Q occupy are not between them
• Taking the middle of 15 as the 7th chair by halving and rounding down; for an odd total the middle is (n + 1) ÷ 2
• Converting a position from the right by subtracting alone — it is n − k + 1, not n − k, and the missing 1 is a classic off-by-one

BPSC sets one or two of these on every paper as short arithmetic-reasoning items with small numbers, so they are the cheapest marks available to anyone who has drilled the three rules. The 71st CCE paper of 2025 asked the same conversion in a class-ranking dress. UPSC has largely moved on from bare position arithmetic to conditional seating puzzles, where several constraints must be satisfied at once and the position is only the last step.

• practice — not a real PYQ

  In a row of 25 students, Ravi is 9th from the left and Sunil is 11th from the right. How many students are there between them ?

  • (a)5
  • (b)6
  • (c)7
  • (d)8

  Answer(a) 5 — Sunil's position from the left is 25 − 11 + 1 = 15, and the students strictly between positions 9 and 15 are those at 10, 11, 12, 13 and 14.

• practice — not a real PYQ

  In a queue, a boy is 12th from the front and 18th from the back. How many people are in the queue ?

  • (a)28
  • (b)29
  • (c)30
  • (d)31

  Answer(b) 29 — the two positions add to one more than the total, so the total is 12 + 18 − 1 = 29; adding them without subtracting one counts the boy twice.