Multiple choice

At a certain fruit stand, the price of each apple is $40$ cents and the price of each orange is $60$ cents. Mary selects a total of $10$ apples and oranges from the fruit stand, and the average (arithmetic mean) price of the $10$ pieces of fruit is $56$ cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $52$ cents?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
  5. $5$
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E Correct answer
Explanation

Initial: 10 fruits, avg 56 cents, total 560 cents. Let a be apples, o be oranges. 40a + 60o = 560 and a + o = 10. Solving gives a=2, o=8. New average 52 cents for k fruits: (40a' + 60o') / k = 52. Since a' + o' = k, 40a' + 60o' = 52a' + 52o' => 8o' = 12a' => 2o' = 3a'. If k=5, a'=2, o'=3. Original had 8 oranges, now 3, so 5 must be put back.

AI explanation

The total initial cost of the 10 pieces of fruit is 10 times 56, which equals 560 cents. Since apples cost 40 cents and oranges cost 60 cents, let a be the number of apples; the equation 40a plus 60 times 10 minus a equals 560 yields 8 apples and 2 oranges. To find how many oranges she must put back to make the average 52 cents, solve 40 times 8 plus 60 times 2 minus x equals 52 times 10 minus x. This simplifies to 440 minus 60x equals 520 minus 52x, which means x equals 5.

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