EPFO EO/AO 2023 General Ability Test · Question 75 of 120

Correct — C, (c) 10. What is being counted is the number of different collections of coins that add to ₹ 10, using ₹ 1, ₹ 2 and ₹ 5 coins with no limit on how many of each. The order in which the coins are handed over does not make a new way of paying, so this is a counting of combinations, not arrangements. The reliable method is to fix the largest coin first, because it has the fewest possible counts, and then let the rest follow. Let c be the number of ₹ 5 coins, b the number of ₹ 2 coins and a the number of ₹ 1 coins, so that a + 2b + 5c = 10. c = 2 uses up the whole ₹ 10, so a = b = 0. One way: 5 + 5. c = 1 leaves ₹ 5 to be made from ones and twos. The number of twos can be 0, 1 or 2, and the ones fill the rest — 5 + 2 + 2 + 1, 5 + 2 + 1 + 1 + 1, and 5 + 1 + 1 + 1 + 1 + 1. Three ways. c = 0 leaves the whole ₹ 10 to be made from ones and twos. The number of twos can be 0, 1, 2, 3, 4 or 5, and again the ones fill the rest. Six ways, from ten ones up to five twos. That is 1 + 3 + 6 = 10 ways in all, and the ten can be written out in full: (5,5); (5,2,2,1); (5,2,1,1,1); (5,1×5); (2×5); (2×4,1×2); (2×3,1×4); (2×2,1×6); (2,1×8); (1×10). The structure worth noticing is that once the ₹ 5 coins are fixed, the remaining amount R can be made in exactly one way for each admissible number of ₹ 2 coins, that is in ⌊R/2⌋ + 1 ways, because the ₹ 1 coins have no freedom left. So the count is (10/2 + 1) + (5 ÷ 2 rounded down, + 1) + 1 = 6 + 3 + 1 = 10. The same reasoning generalises to any target amount with these three denominations. One check on the reading of the question: if the order of the coins did matter, the count would be 128, which is nowhere near any option — confirming that the paper means collections of coins, as the phrase 'in how many ways can one pay' ordinarily does.

• (a)8 — An undercount by two. It is what a candidate gets by enumerating loosely and dropping the two extreme cases — paying with ten ₹ 1 coins and paying with two ₹ 5 coins — both of which are perfectly good ways of paying ₹ 10 and both of which are easy to skip because they use a single denomination. The defence is the systematic sweep: fix the number of ₹ 5 coins at 2, then 1, then 0, and inside each case run the ₹ 2 coins from the maximum down to zero. Nothing can be missed that way.
• (b)9 — An undercount by one, and the single most likely wrong answer, because it is what an otherwise correct enumeration produces when exactly one case slips. The usual casualty is 5 + 5, which uses no small coins at all and so does not look like 'making change', or the all-ones payment at the other extreme. Counting the three cases as 6 + 3 + 1 rather than listing ten collections in a row makes the omission impossible: the 1 is the case that keeps disappearing.
• (d)11 — An overcount by one, which comes from counting the same collection twice — usually by letting order creep in, so that a ₹ 5 and a ₹ 5, or a ₹ 2 followed by a ₹ 1 and a ₹ 1 followed by a ₹ 2, are treated as different payments. Order genuinely does not matter here; if it did, the answer would be 128, not 11. The other route to 11 is including an inadmissible case such as using no coins at all or overshooting ₹ 10 by a rupee.

This is the coin-change problem, one of the standard counting questions. The general statement is: in how many ways can an amount N be written as a sum of given denominations, where order is irrelevant and repetition is allowed. The systematic method is to work from the largest denomination downwards, fixing its count and recursing on what is left, which guarantees that every collection is generated exactly once. With three denominations the recursion collapses into an easy sum, and with two it becomes a single division: an amount R made from ₹ 1 and ₹ 2 coins alone can be paid in ⌊R/2⌋ + 1 ways, because choosing the number of twos determines everything. The distinction between combinations and arrangements matters more here than anywhere else in the topic: the same question asked about ordered sequences of coins is a completely different problem, solved by the recurrence f(n) = f(n−1) + f(n−2) + f(n−5), which for ₹ 10 gives 128. Examiners rely on the ordinary meaning of 'ways to pay' to settle which is intended, and the size of the options confirms it.

Counting questions in the EO/AO quantitative block are deliberately small enough to enumerate — the answer is always in single or low double figures — because what is being tested is the discipline of a systematic sweep, not a formula. A candidate who lists cases in a haphazard order will be within one of the right answer, which is exactly why the option list is 8, 9, 10, 11: being close is worth nothing. The habit rewarded is fixing the largest denomination first and never counting the same collection twice.

• The count is the number of non-negative integer solutions of a + 2b + 5c = 10, where a, b and c are the numbers of ₹ 1, ₹ 2 and ₹ 5 coins.
• c = 2 gives 1 way, c = 1 leaves ₹ 5 and gives 3 ways, c = 0 leaves ₹ 10 and gives 6 ways — a total of 10.
• An amount R made from ₹ 1 and ₹ 2 coins alone can be paid in ⌊R/2⌋ + 1 ways, since the number of twos determines the number of ones.
• Order is irrelevant in a 'ways to pay' question: two ₹ 5 coins is one way, not two.
• If order did matter, the number of ordered payments of ₹ 10 with ₹ 1, ₹ 2 and ₹ 5 coins would be 128, given by f(n) = f(n−1) + f(n−2) + f(n−5).
• The ten collections are 5+5; 5+2+2+1; 5+2+1+1+1; 5 and five ones; five twos; four twos and two ones; three twos and four ones; two twos and six ones; one two and eight ones; and ten ones.

• Letting order in. Two ₹ 5 coins is a single way of paying, however the coins are handed over.
• Dropping the single-denomination cases — two ₹ 5 coins, or ten ₹ 1 coins — which produce the answers 9 and 8.
• Enumerating without a fixed order of cases. Fix the largest coin first, then the next, and nothing repeats or vanishes.
• Assuming a formula. With three denominations, a three-line case split is faster and safer than any formula a candidate is likely to half-remember.

EO/AO counting items keep the numbers small on purpose, so the answer can always be reached by enumeration if the enumeration is disciplined. Expect option lists of consecutive integers around the true answer, which means partial credit for being close does not exist — practise the largest-denomination-first sweep until it is reflexive.

No directly related past PYQ was found.

• practice — not a real PYQ

  In how many ways can one pay ₹ 8 using coins of denomination ₹ 1, ₹ 2 and ₹ 5, assuming enough coins of each are available ?

  • (a)5
  • (b)6
  • (c)7
  • (d)8

  Answer(c) 7

• practice — not a real PYQ

  In how many ways can ₹ 12 be paid using only ₹ 1 and ₹ 2 coins ?

  • (a)6
  • (b)7
  • (c)12
  • (d)13

  Answer(b) 7