VeritasPrepKarishma wrote:

praveenism wrote:

Q3:

At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. 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 ¢ 56. How many oranges must Mary put back so that the average

price of the pieces of fruit that she keeps is ¢ 52?

A. 1

B. 2

C. 3

D. 4

E. 5

Answer:E. I solved the question and got ans as 2

The correct answer is (E). People who are getting (B) need to think about this: The second time when the average is 52, does Mary still have 10 fruits? She ONLY needs to put back oranges. Not replace them with apples. So when you use alligation again and get the ratio as 2:3, you don't get 6 oranges since total number of fruits are not known. Let me solve it step by step.

When avg = 56

wa/wo = (60 - 56)/(56 - 40) = 1/4

So number of apples = 2, number of oranges = 8

When avg - 52

wa/wo = (60 - 52)/(52 - 40) = 2/3

What is the total number of fruit now? We don't know. We know Mary put back some oranges. We don't know how many. What we do know is that then number of apples she had stayed the same. She had 2 apples before so she still has 2 apples. If number of apples now is 2, number of oranges must be 3. So she must have put back 8 - 3 = 5 oranges.

can we do it like this

let no of apples be a and no of oranges be b

total fruits initially 10

so 56(10) = 40 ( a)+ 60( 10 -a)

second case when avg is 52 ...let no of organes she kept away be r

so

52( 10-r) = 40 ( a) + 60( 10-a-r) -------------2

we need not solves anything up till this point

now we know that the reduction in total is cause only by no of oranges she kept away , let us say she kept away r orhanges

so reduction is ( 56(10) - 52( 10-r) = 4(10) + 52 r ----3

we know that reduction will be equal to 60 r

as in equation 2 .....is same as equation 1 expcept for the amount accouhnted for reduced oranges

so using 3

4(10)=52r = 60 r

r=5