Hussain15 wrote:

At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Kindly share your working & also mention the time in which you have solved the question.

I got answer E

I know there is a fast way but I try to solve all these problems the same way

I use the letter "O" for oranges so it's not a zero after the .6

(.4A + .6O)/10 = .56 and A + O = 10

.4A + .6O = 5.6

Now subtract the above equation from A + O = 10

.4A + .6O = 5.6

-.4A - .4O = -4

.2O = 1.6

O = 8 so therefore A = 2

Now we try to find how many to take away to get the average of .52

.4(2) + .6(8-x) / 10-x = .52

.8 + 4.8 - .6x = 5.2 - .52x

5.6 - 5.2 = .6x - .52x

.04 = .08x

x = 5