Question

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?

Answer 1

Answer:

Mary has to put 5 oranges back so that the average price of the pieces of fruit that she keeps is 52 cents.

Step-by-step explanation:

From the given information it is clear that

Price of each apple = 40 cents

Price of each orange = 60 cents

Let the number of apple be x and number of oranges be y.

Mary selects a total of 10 apples and oranges from the fruit stand.

$x+y=10$              .... (1)

The average (arithmetic mean) price of the 10 pieces of fruit is 56 cents.

$\frac{40x+60y}{10}=56$

$4x+6y=56$         .... (2)

On solving (1) and (2), we get

$x=2,y=8$

Let she put z oranges back so that the average price of the pieces of fruit that she keeps is 52 cents.

$\frac{40(2)+60(8-z)}{10-z}=52$

On cross multiplication, we get

$80+480-60z=52(10-z)$

$560-60z=520-52z$

$560-520=60z-52z$

$40=8z$

Divide both sides by 8.

$\frac{40}{8}=z$

$5=z$

Therefore, Mary has to put 5 oranges back so that the average price of the pieces of fruit that she keeps is 52 cents.