GreenlightTestPrep wrote:
Carcass 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
It turns out that the cost per apple is irrelevant. Here's why:
The average (arithmetic mean) price of the 10 pieces of fruit is 56 centsSo, (total value of all 10 pieces of fruit)/10 = 56 cents
This means, total value of all 10 pieces of fruit = 560 cents
How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?Let x = the number of oranges to be removed.
Each orange costs 60 cents, so the value of the x oranges to be removed = 60x
This means
560 - 60x = the value of the REMAINING fruit
Also, if we remove x oranges, then
10 - x = the number of pieces of fruit REMAINING.
We want the REMAINING fruit to have an average value of 52 cents.
We can write: (value of REMAINING fruit)/(number of pieces of fruit REMAINING) = 52
Rewrite as: (
560 - 60x)/(
10 - x) = 52
Multiply both sides by (10-x) to get: 560 - 60x = 52(10 - x)
Expand right side to get: 560 - 60x = 520 - 52x
Add 60x to both sides: 560 = 520 + 8x
Subtract 520 from both sides: 40 = 8x
Solve: x = 5
Answer: E
Cheers,
Brent
Hi Brent,
I did the following calculation which is logically correct but I am getting answer B,please let me know what seems wrong
Let x be number of apples
10-x be number of oranges
(40x + 60(10-x))/10 = 560
x=2
that means,2 apples and 8 oranges were picked
However,If I want to reduce avg to 52,I will have to choose 4 apples and 6 oranges 4x40 + 6x60 / 10 = 52
so originally,8 oranges were choosen and now for avg 52,6 oranges must be there.
Number of oranges to be removed = 8-6 = 2 hence B
I am not able to figure out how this answer is logically incorrect