The price of a bottle of cold drink is ₹ 10. One bottle of cold drink can also be bought by returning 10 empty bottles. A person has ₹ 1,000 and 19 empty bottles. Assuming that the person can consume any number of bottles he buys, what will be the number of empty bottles he possesses at the end if he buys maximum number of bottles of cold drink and consumes all?

Correct — C, (c) 2. The quantity to track is not the bottles bought but the empties in hand, and the fact that decides the item is that every exchange consumes ten empties and gives back one — a net cost of nine empties for each extra drink.

Follow the story once. With ₹ 1,000 at ₹ 10 a bottle he buys 100 bottles. He drinks them, so those 100 empties join the 19 he already had, giving 119. Handing over 110 of them brings 11 more bottles and leaves 9 in hand; he drinks the 11, so he now holds 9 + 11 = 20 empties. Handing over all 20 brings 2 more bottles and leaves none; he drinks those 2 and holds 2 empties. Two is fewer than ten, so no further exchange is possible and the process stops. He is left with 2 empty bottles, which is option (c). Along the way he has drunk 100 + 11 + 2 = 113 bottles.

The net-cost view gives the same answer in one line and, more usefully, gives a test that any proposed answer must pass. Since each exchange reduces the stock of empties by exactly nine, the number of empties in hand always leaves the same remainder on division by nine as the 119 he started with. And 119 = 9 × 13 + 2, so he can make thirteen exchanges and must finish with 2 empties. Every candidate answer can therefore be checked without following the chain at all: the final count has to be less than ten and has to leave a remainder of 2 on division by nine, and only one of the four options does.

Why the others are wrong

• (a)0 — Nought cannot be the answer, and the reason is structural rather than arithmetical. Every bottle he obtains is drunk, and drinking it produces an empty; so from the moment he consumes anything at all, he holds at least one empty bottle. An exchange never clears the shelf either, because handing over ten empties brings back a full bottle which becomes an empty in its turn. The stock can therefore fall to one, or to any number below ten, but never to zero. This option is chosen by a candidate who assumes that a problem of this kind is meant to come out exactly, with everything used up — a reasonable instinct on many counting problems and a wrong one here. The invariant confirms it independently: the count of empties always leaves a remainder of 2 on division by nine, and nought does not.
• (b)1 — One is what the chain produces if the 19 empties he already owns are forgotten and only the 100 bottles bought with money are counted. Follow that mistaken version through: 100 empties bring 10 bottles and leave none, those 10 are drunk to give 10 empties, which bring 1 more bottle and leave none, and that last bottle is drunk to leave 1 empty. The stopping state is 1, exactly as offered. The error is a reading error rather than a calculation error — the stem states two resources, the money and the nineteen empties, and the second is easy to lose because it is mentioned only once and never again. The invariant exposes it too, since 100 leaves a remainder of 1 on division by nine while 119 leaves a remainder of 2, and that difference of one is precisely the difference between the two answers.
• (d)3 — Three is not a state this process can end in. The last exchange must convert some multiple of ten empties into that many bottles, and those bottles are then drunk, so the terminal count is whatever was left over plus the number of bottles just obtained — here 0 + 2. To finish with 3 the count would have to leave a remainder of 3 on division by nine at every stage, and it leaves a remainder of 2, because he started with 119. The option is what a candidate produces by abandoning the chain a step early, most often by stopping at the point where 9 empties remain in hand and adding a bottle or two by eye instead of completing the exchange. The discipline that prevents it is to carry on until the stock is genuinely below ten and to write down the stock after each round rather than keeping it in the head.

Exchange problems are counting problems with a feedback loop, and they are answered either by following the loop or by finding an invariant. Following the loop means repeating one step: divide the empties in hand by the number required for an exchange, take the whole-number part as the bottles obtained, keep the remainder, and then add the newly obtained bottles to the remainder because they are drunk in their turn. The process stops when the stock falls below the exchange rate. The invariant is quicker and safer. If k empties buy one bottle, then each exchange removes k empties and returns one, so the stock falls by exactly k − 1 every time; here k = 10 and the stock falls by 9. It follows that the number of free bottles obtainable from an initial stock of N empties is the whole-number part of (N − 1) ÷ (k − 1), and that the final stock is N reduced by (k − 1) times that count. Two consequences are worth remembering: the stock can never reach zero, because the last bottle drunk leaves its own empty behind, and the final stock always leaves the same remainder on division by k − 1 as the starting stock did. That remainder test decides most examination items in a few seconds.

Key facts

• ₹ 1,000 at ₹ 10 a bottle buys 100 bottles, and drinking them adds 100 empties to the 19 already held, giving 119.
• Ten empties buy one bottle, which is itself drunk and returns one empty, so each exchange costs a net nine empties.
• 119 = 9 × 13 + 2, so thirteen exchanges are possible and 2 empties remain at the end.
• Following the chain: 119 empties give 11 bottles with 9 left, then 20 empties give 2 bottles with none left, then 2 empties remain.
• He drinks 100 + 11 + 2 = 113 bottles in all, which is not what the question asks for.
• The stock of empties can never fall to zero, because the last bottle drunk leaves its own empty behind.
• The number of free bottles from N empties at k for one is the whole-number part of (N − 1) ÷ (k − 1).
• The final stock always leaves the same remainder on division by (k − 1) as the starting stock, which here is 2.

Common traps

• Losing one of the two starting resources; the nineteen empties are mentioned once and change the answer.
• Forgetting that a bottle obtained in exchange is drunk in its turn and returns another empty to the stock.
• Answering the number of bottles drunk, which is 113, when the question asks for the empties remaining.
• Assuming the process must end with nothing left, which is impossible once any bottle has been consumed.
• Stopping the chain while ten or more empties are still in hand, which leaves one exchange unmade.

• practice — not a real PYQ

  A person buys 60 bottles of a drink and drinks all of them. Five empty bottles can be exchanged for one full bottle, and every bottle obtained is also drunk. How many bottles in all does the person drink?

  • (a)72
  • (b)74
  • (c)75
  • (d)76

  Answer(b) 74

• practice — not a real PYQ

  In the same arrangement — 60 bottles bought and drunk, with five empty bottles exchangeable for one full bottle which is also drunk — how many empty bottles are left at the end?

  • (a)0
  • (b)1
  • (c)4
  • (d)5

  Answer(c) 4