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
This is a good question that tests your understanding about the weighted average. Also gives an idea about how comfortable you are in applying the alligation method in weighted average-based questions.
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.From the above statement, we have information about the price per apple and the price per orange. Also, the weighted average price of the fruits Mary had selected.
Do you think this information is sufficient to calculate the ratio of number of apples to oranges Mary had selected?
Since the total number of fruits she selected is given, Don't you think you can use the above ratio to calculate the exact number of apples and oranges she had selected. ?
if you are confused about how to answer the above questions, that means you definitely need to revisit the concept of weighted average.
Let me explain.
The average price of the 10 pieces of fruits is 56 cents. Let's say she had selected x apples and y oranges.
INFO we have : x + y = 10 , Price/apple = 40 cents , Price/Orange = 60 cents
Average price /fruit = 56 cents = Total cost of all fruits / number of fruits = (Total apples cost + Total oranges cost)/ Total number of apples and oranges
Weighed Average Formulae = A = \(\frac{( n1*A1 + n2*A2 )}{(n1 + n2) }\). The same can be used here.
56 = \(\frac{(x*40 + y*60)}{(x+y) } \)
56x + 56 y = 40x + 60y
x(56-40) = y(60-56)
\(\frac{x}{y}\) = \(\frac{(60-56)}{(56-40)}\) = \(\frac{4}{16 }\)= \(\frac{1}{4}\) . If you are clear with the alligation method, you can bypass the initial steps and get the final ratio of apples and oranges instantly.
i.e
using alligation method , x/y = \(\frac{(A2- A)}{(A- A1)}\)
Substituting the values here, \(\frac{x}{y} \)= \(\frac{(60-56)}{(56-40) }\)= \(\frac{4}{16 }\)= \(\frac{1}{4}\).
Note : Alligation method is less time consuming compared to conventional weighted average formula based method.
Moving forward to the next part, now we know the ratio of apples to oranges and the total number of apples and oranges i.e x:y = 1:4 and x + y = 10
x = 1/5 *10 = 2
y = 10 -2 = 8
We found that Mary had selected 2 apples and 8 oranges.
But, Mary returned back some oranges out of these 8 and the average price/fruit decreased to 52 cents.
Let's assume that Mary returned 'z ' oranges, the number of oranges Mary has right now is 8-z. There is no change in the number of apples i.e 2 apples will be there. Also the price/apple and price/orange are the same and the new weighted average is 52 cents.
The above information is sufficient to calculate the new ratio of apples to oranges by using the Weighted Average/Alligation method.
No of apples/No of oranges = \(\frac{x}{(8-z)}\) = \(\frac{(A2- A)}{ (A- A1)}\) = \(\frac{(60-52)}{(52-40)}\) = \(\frac{8}{12}\) =\( \frac{2}{3}\)
Since x =2, \(\frac{2}{(8 -z)}\) = \(\frac{2}{3}\)
3 = 8-z
z = 8-3 = 5 .
Therefore, the number of oranges Mary returned is 5.
Option E is the correct answer.Thanks,
Clifin J Francis
GMAT Mentor.
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