A man buys apples at a certain price per dozen and sells them at 8 times that price per hundred. What percentage does he gain or lose ?

Correct — C, (c) 4% loss.

THE STEP THAT DECIDES THIS QUESTION is noticing that the buying price and the selling price are quoted PER DIFFERENT QUANTITIES. He buys by the DOZEN and sells by the HUNDRED, and the multiplier 8 is attached to the second unit, not to the first. "Eight times that price per hundred" does not mean he sells at eight times what he paid; it means the money he receives for a hundred apples equals eight times the money he paid for twelve.

Once that is seen, the arithmetic is a single line. Eight dozen is 96 apples. So the price he receives for 100 apples is the price he paid for 96 apples. He is handing over a hundred and being paid for ninety-six, and he is short by four apples in every hundred:

loss = 4 out of 100 = 4%.

The formal working, for a reader who would rather see it set out. Let the buying price be P per dozen.

Cost of one apple = P / 12 Selling price of one apple = 8P / 100 = 2P / 25

Ratio of selling price to cost price = (2P/25) ÷ (P/12) = (2/25) × 12 = 24/25 = 0·96

A ratio of 0·96 means he recovers 96 paise for every rupee of cost, so the loss is 4 per cent of the cost price. Since a loss percentage is always reckoned ON THE COST PRICE unless a question says otherwise, that 4 per cent is the answer.

A third route avoids fractions entirely and is the fastest of the three under time pressure. Take a convenient number of apples that is a whole number of dozens AND a whole number of hundreds — 300, the lowest common multiple of 12 and 100.

300 apples = 25 dozen, so the cost is 25P. 300 apples = 3 hundreds, so the sale realises 3 × 8P = 24P. Loss = 25P − 24P = P, on a cost of 25P, which is 1/25 = 4%.

Notice what is reassuring about having three routes: all of them give 4 per cent, and none of them ever produces 6. The magnitude in this problem is robust — every consistent reading of the stem lands on four — and the only thing genuinely at risk is the DIRECTION. Because 8 dozen (96) is less than 100, the seller receives less than he paid, so it is a loss, not a profit. That makes option (c) the answer.

Why the others are wrong

• (a)4% profit — This has the magnitude right and the direction wrong, and it is the most likely wrong answer on the paper for exactly that reason. The whole question turns on comparing 8 dozen with 1 hundred: 8 × 12 = 96, which is FEWER apples than the 100 he is selling, so the money he takes in for a hundred apples is the money he paid for ninety-six. He gives away four apples in every hundred and is paid for none of them. A candidate who computes the ratio 24/25 correctly and then reads it in the wrong direction — treating the 4 per cent gap as something he has gained rather than something he has given up — arrives here. The safeguard is to fix the comparison in words before touching the arithmetic: is the number of apples he is paid FOR bigger or smaller than the number he hands over ? If eight dozen had been more than a hundred, this option would have been correct; had the stem said nine times the price per hundred, nine dozen is 108 and the answer would have been an 8 per cent profit.
• (b)6% profit — Six per cent corresponds to no consistent reading of this stem, in either direction, and it is worth being explicit about that because it saves time in the examination hall. The three natural routes through the problem — per apple, per rupee of cost, and per 300 apples — all give four. Even the classic error of dividing the shortfall by the SELLING price rather than the cost price does not produce six; it produces about 4·17 per cent, since the gap of one unit is measured against 24 rather than 25. The four options are really a two-by-two grid: two magnitudes, 4 and 6, crossed with two directions, profit and loss. Only one of the magnitudes is reachable at all, so the grid exists to stop a candidate guessing the direction and walking away with half the question. Compute the ratio, decide the sign, and the grid collapses to a single cell.
• (d)6% loss — This option has the direction right and the magnitude wrong, which makes it the mirror image of the first wrong option and the one that catches a candidate who has reasoned correctly about the sign and then mis-multiplied. It is a loss — 8 dozen is 96 and 96 is less than 100 — but the size of the loss is fixed by the ratio 96 to 100, which is four parts in a hundred and cannot be six. If the working has produced six, the likely cause is that the dozen has been mishandled: using 8 × 12·5 = 100 gives no loss at all, using 8 × 11 = 88 gives 12 per cent, and neither is on the list. The reliable check is the 300-apple version, where the numbers are all whole: 25 dozen costs 25P, three hundreds realise 24P, and the shortfall of one unit on a cost of twenty-five is unambiguously one twenty-fifth.

THE IDEA BEING TESTED IS UNIT CONSISTENCY IN A PROFIT-AND-LOSS PROBLEM. Every profit-and-loss calculation is a comparison between two money amounts, and a comparison is only meaningful if both amounts refer to the SAME quantity of goods. When a question quotes the cost in one unit and the sale price in another — per dozen and per hundred, per kilogram and per quintal, per piece and per score — the first move is always to bring the two onto a common footing. There are three standard ways to do it, and all of them work here:

1. REDUCE TO ONE ITEM. Cost per apple P/12, sale per apple 8P/100. Simple, but it brings fractions. 2. TAKE THE LCM OF THE TWO UNITS. 300 apples is 25 dozen and 3 hundreds, so every figure is a whole number. This is usually the quickest by hand. 3. COMPARE THE COUNTS DIRECTLY. Eight dozen is 96; he is paid for 96 and delivers 100. This is the shortest of all and it turns the problem into mental arithmetic.

THE SECOND IDEA IS THE BASE. Profit and loss percentages in Indian competitive papers are reckoned on the COST PRICE unless the question expressly says otherwise. That convention matters because the same absolute gap gives a different percentage against a different base: a shortfall of one unit against a cost of 25 is 4 per cent, against a sale of 24 it is about 4·17 per cent, and mixing the two is one of the commonest sources of an answer that is close to a listed option without matching it.

THE THIRD IDEA IS THE MULTIPLIER TRAP, which is what makes this a good question rather than a routine sum. The word "times" invites a candidate to multiply the price by eight and conclude that he has made an enormous profit. The multiplier is doing something quite different: it is converting between two quantity units. Whenever a stem says "at k times that price per <different unit>", the real comparison is between k times the FIRST unit's count and the SECOND unit's count. Here that is 96 against 100. The same structure appears with scores and hundreds, with dozens and gross, and with kilograms and quintals, and recognising it is worth far more than remembering this particular answer.

Key facts

• The comparison is between eight dozen and one hundred: 8 × 12 = 96, so the seller is paid for 96 apples for every 100 he actually hands over.
• That gap of four apples in every hundred is a loss of 4 per cent, because a profit or loss percentage is reckoned on the cost price unless a question says otherwise.
• Reducing to a single apple: the cost is P/12 and the sale price 8P/100 = 2P/25, so the ratio of sale to cost is 24/25 = 0·96, again a 4 per cent shortfall.
• Taking 300 apples — the lowest common multiple of 12 and 100 — makes every figure whole: 25 dozen cost 25P, three hundreds realise 24P, and the loss of P on 25P is 4 per cent.
• The phrase "k times that price per <a different unit>" is a unit conversion, not a price multiplication; the real comparison is between k times the first unit's count and the second unit's count.
• Dividing the shortfall by the selling price instead of the cost price gives about 4·17 per cent here, which is why the base has to be settled before the division.
• Had the stem said nine times the price per hundred, nine dozen would be 108 against 100 and the same method would give an 8 per cent profit — the direction follows from the comparison of counts.

Common traps

• Reading "8 times that price per hundred" as eight times the price. The multiplier converts between quantity units; it does not scale the margin.
• Getting the magnitude right and the direction wrong. Eight dozen is 96, which is fewer than 100, so it is a loss.
• Dividing the shortfall by the selling price rather than the cost price, which gives about 4·17 per cent and matches no option.
• Guessing between profit and loss once the number 4 has appeared. The option set offers both directions for both magnitudes precisely to punish that.
• Working in fractions of P when 300 apples would make every figure a whole number and remove the chance of an arithmetic slip.

• practice — not a real PYQ

  A trader buys pencils at a certain price per score and sells them at 4 times that price per hundred. What percentage does he gain or lose on the transaction ?

  • (a)20% loss
  • (b)20% profit
  • (c)25% loss
  • (d)25% profit

  Answer(a) 20% loss — a score is 20, so four scores are 80 pencils. He is paid for 80 pencils for every 100 he delivers, a shortfall of 20 in every 100, which is a 20 per cent loss on the cost price.

• practice — not a real PYQ

  A man buys apples at a certain price per dozen and sells them at 9 times that price per hundred. What percentage does he gain or lose on the transaction ?

  • (a)8% profit
  • (b)8% loss
  • (c)12% profit
  • (d)12% loss

  Answer(a) 8% profit — nine dozen is 108 apples, so he is paid for 108 apples for every 100 he hands over. The surplus of 8 in every 100 is an 8 per cent gain on the cost price, and the direction reverses because 108 exceeds 100.