At a certain fruit stand, the price of each apple is 40 cents and the

The question tells you that initial quantity is not maintained. It doesn't talk about picking more fruits/replacing fruits etc.
If you want to maintain the number of fruits to 10 and achieve 52 cents as the average price, you know that

w1/w2 = (60 - 52)/(52 - 40) = 8/12 = 2/3

So there should be 2 apples for every 3 oranges. This means there should be 4 apples and 6 oranges.
But actually she has 2 apples and 8 oranges. So she needs to put back 2 oranges and pick 2 more apples.
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Please see the attached snapshot for a weighted average (teeter-totter or see-saw view of the problem)

For those who are thinking about how to approach this question efficiently, here are my GMAT Timing Tips (the links below have growing lists of questions that you can use to practice these timing tips):

Weighted average mapping strategy: The weighted average mapping strategy is the same as the "teeter-totter" approach shown in the great solution by FANewJersey. This really is a time-saving approach. The key thing that you get from this approach is that the ratio of oranges to apples is 4 to 1 before (so 10 total fruits gives us 8 oranges and 2 apples) and 3 to 2 after putting back the apples.

Track how the parts of a whole change: The key thing to notice is that there are the same number of apples (2) before and after putting back the oranges. So, if the ratio is 3 oranges to 2 apples after putting back the oranges, then we end up with 2 apples and 3 oranges. This means that Mary put back 8 – 3 = 5 oranges.

Please let me know if you have any questions, and if you would like me to post a video solution!

Hi experts, i understand the above mentioned approach but could you please tell me why is the following approach wrong

Let 'x' be the number of Apples, 'y' be the number of Oranges so x+y=10
Since the average (i.e., price of fruits/ the the total number of fruits) is 56
(40x + 60y)/10 = 56
On solving x=2 and y =8

Now the question states that the adjusted average is 52
So (40x + 60y)/10 = 52
On solving x=4 and y=6

So if Mary picks up 8 Oranges the average is 56 but if she picks up 6 Oranges the average drops to 52
Clearly she has to drop 2 Oranges to acquire the desired average.

Sure! Your approach would be correct if the question asked about how many oranges she would need to exchange for apples in order to have the average price be 52 cents. In that case, the number of fruits would remain 10, and 6 oranges/4 apples would give an average price of 52 cents.

However, the question asks how many oranges she must put back, so all she can do is reduce the number of oranges; she cannot increase the number of apples. So, the number of apples must remain 2, and we need to know how many oranges she will need to have in order for the average price to be 52 cents. This is 3 oranges, so she needs to put back 5 oranges to go from having 8 oranges/2 apples to having 3 oranges/2 apples.

Please let me know if you have more questions!

Same method as I did.
I spent about 6 mins on this question.
I wonder how to handle in the real exam?
Just guess? (and to work on other easier questions?)

Hi Teitsuya,

Most GMAT questions are written so that they can be approached in more than one way - so on any given question, you have two goals. First, you want to correctly answer the question if possible (in a reasonable amount of time) - and second, you want to use the most efficient approach possible (so that you don't waste time). Most of the Quant questions that you'll see on Test Day can be solved using Tactics and a bit of Arithmetic (and in this question, that type of approach would be a lot faster than the lengthy, step-heavy Algebra that you may have done). This is all meant to say that you have to consider how you "see" the Exam; the Quant section of the GMAT is NOT a "math test", so if you treat it as a math test then you'll likely end up spending extra time working through a number of the questions that you'll face.

GMAT assassins aren't born, they're made,
Rich

Video solution from Quant Reasoning starts at 0:27
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Answer: Option E

Step-by-Step Video solution by GMATinsight



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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

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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Why is it that the original ratio of apples to oranges is not 40:60 (2/3)?
Why do we calculate the ratio with the difference bt price and average?
Thank you

Hi ruis,

The 40:60 ratio is a ratio of the PRICES of 1 apple to 1 orange. That information - when combined with the facts that there are 10 pieces of fruit in total and the average price of those 10 pieces was 56 cents - allows you to determine how many apples and how many oranges there originally were.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com
www.empowergmat.com

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In the required avg side shouldn't we multiply 2:3 by 2 as well to make it 10, only after that can we subtract the quantities we got right? because 2+3 is 5 and we're given 10 as the total quantity. 

The current average is 56.

That's closer to 60 than to 40.

What percent could be oranges ? 75 ?

75% would be 45 for the oranges and 25% of 40 or 10 for the apples, total

55 cents. 1 cent short.

80% oranges would be 48 cents, so 20% apples would be 8 cents.

Total 56 cents. So 8 oranges, 2 apples.

Some oranges are removed to achieve 52 cent average.

How is 52 cents achieved as an average ? Just like before.

How about 60% oranges equals 36 cents, leaving 40% apples or 16 cents. Total:52 cents.

Apples stay the same number 2 and they're now 40% of the total. So the total must be:

2/.4 = 5

Since there were 10 fruits before and now 5, 5 oranges must have been removed.

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Given : 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.

Asked: How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

The ratio of apple to oranges = (60-56) : (56-40) = 4:16 = 1:4
Number of apples = 1/5*10 = 2
Number of oranges = 4/5*10 = 8

After putting back some oranges, the ratio of apple to oranges = (60-52) : (52-40) = 8:12 = 2:3
Number of oranges = 3
She puts back = 8-3 = 5 oranges

IMO E
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