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This question already has an answer here:

In a snack bar there were two baskets of pears of different sorts, 60 pears in each. It was proposed to sell pears from the first basket at 4 for 30¢, and pears from the second basket at 6 for 50¢. Thus the two baskets of pears would be sold for \$9.50 in all.

However, the cashier, in order to simplify her work, mixed together the pears of both kinds and proceeded to sell ten pears for 80¢.

As a result of the sale of pears, there turned out 10¢ too much: not \$9.50 but \$9.60.

Where did this extra dime come from?

(Source: Lapses in Mathematical Reasoning​ by V. M. Bradis, V. L. Minkovskii, A. K. Kharcheva.)

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marked as duplicate by f'', Peregrine Rook, ffao, Deusovi, Mike Earnest Apr 21 '16 at 4:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ −1, unclear what the question is. "Where did this extra dime come from?" From selling 120 pears at ten for 80¢, of course. What's the question? ("As a result of the sale of pears, there turned out 10¢ too much: not \$9.50 but \$9.60." Well, yeah. If you change the price on things, then the total amount taken in will change.) $\endgroup$ – msh210 Apr 20 '16 at 22:44
  • $\begingroup$ What's the word between extortion and bait-and-switch? Fraud; that's where. $\endgroup$ – Mazura Apr 20 '16 at 23:43
  • $\begingroup$ @msh210. I was with you for a while wondering that the puzzling interest was here as the $9.60 answer seemed valid with no issues... I think people are thrown by the fact that if you added the 30¢ and the 50¢ you get 80¢, so the math should 'add up'. They're forgetting that you had more 30¢ units to sell. Suspect we just made the jump through that confusion directly (and with math intact). Maybe people with our er, 'logic units' will be stumped about something else in turn... :) $\endgroup$ – Lamar Latrell Apr 21 '16 at 2:12
  • $\begingroup$ simply from selling cheap pear at a higher price... $\endgroup$ – njzk2 Apr 21 '16 at 2:59
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Explanation:

The pears in the first basket cost 0.30/4 = 0.075 each. The pears in the second basket cost 0.50/6 = 0.0833... each. If we average these prices together, the mixed pears should have been sold at 0.079166... each (selling 120 of them at this price produces 9.50 in revenue). However, the cashier opted instead to sell them at 0.08 a piece, which is 1/12 of a cent more than their price should be. That's where the additional dime comes from.

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It seems rather obvious to me. Am I missing something?

The extra 10 is from selling them at a higher average price.
The plan was to sell one basket at a price of 7 1/2 per pear (albeit in batches of 4), and the other at 8 1/3 per pear (in batches of 6).
This averages just under 8 per pear, which is what the cashier actually sells them for, making an extra 10 cents.

Edit to add this:
I see now where there could be an incorrect way to reason about it that some people (like the cashier maybe) might do. It would have been nice if that were more clearly stated in the question.
The cashier may have thought as follows: Putting together a batch of 4 at \$0.30 and 6 at \$0.50 gives 10 at \$0.80. Therefore I'll sell them as 10 for \$0.80 and get the same price.

This reasoning fails because:

There are more batches of 4 than of 6, and the cashier's method assumes that there are equally many. The batches of 4 are the cheaper ones, so those extra batches of 4 are essentially sold at the incorrect higher average price that the cashier arrives at.

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    $\begingroup$ Both I and the original authors thought that the cashier’s (incorrect) line of reasoning did not need to be explained. $\endgroup$ – MJD Apr 20 '16 at 16:00
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12 of the cheaper pears are sold as the more expensive pears.

She had 60 of each pear, which would be 15 sets of 4 cheap pears and 10 sets of 6 expensive pears. By selling them in groups of 10s, each purchase would yield (on average) 5 cheap pears and 5 expensive pears, but would be the price of 4 cheap pears and 6 expensive pears.

If we assume the first 10 sets sold are 4 cheap pears and 6 expensive pears, that would leave 2 sets of 10 cheap pears. Of those 20 pears, 6 out of 10 would be overpriced. So 12 of them are cheaper pears being sold for the price of the more expensive pears. $$\mathrm{12\ pears \times \left(\frac{50¢}{6\ pears} - \frac{30¢}{4\ pears}\right) = 10¢}$$

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