Bunuel wrote:
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?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Kudos for a correct solution.
Solution:
We can let the number of apples = x and the number of oranges = y. Using these variables we can create the following two equations:
1) x + y = 10
Using the formula average = sum/quantity, we have:
2) (40x + 60y)/10 = 56
Let’s first simplify equation 2:
40x + 60y = 560
4x + 6y = 56
2x + 3y = 28
Isolating for y in equation one gives us: y = 10 – x.
Since y = 10 – x, we can substitute 10 – x for y in the equation 2x + 3y = 28. This gives us:
2x + 3(10 – x) = 28
2x + 30 – 3x = 28
-x = -2
x = 2
Since x + y = 10, then y = 8.
We thus know that Mary originally selected 2 apples and 8 oranges.
We must determine the number of oranges that Mary must put back so that the average price of the pieces of fruit that she keeps is 52¢. We can let n = the number of oranges Mary must put back.
Let’s use a weighted average equation to determine the value of n.
[40(2) + 60(8-n)]/(10 – n) = 52
(80 + 480 - 60n)/(10 – n) = 52
560 – 60n = 520 – 52n
40 = 8n
5 = n
Thus, Mary must put back 5 oranges so that the average cost of the fruit she has kept would be 52 cents.
Answer: E
Same method as I did.
I spent about 6 mins on this question.