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

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Re: At a certain fruit stand, the price of each apple is 40 cents and the  [#permalink]

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New post 12 Dec 2018, 10:39
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.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the  [#permalink]

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New post 12 Dec 2018, 10:49
AkhilRao wrote:
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!
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Re: At a certain fruit stand, the price of each apple is 40 cents and the  [#permalink]

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New post 11 Jan 2019, 10:39
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Bunuel wrote:
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


Kudos for a correct solution.


It turns out that the cost per apple is irrelevant. Here's why:

The average (arithmetic mean) price of the 10 pieces of fruit is 56 cents
So, (total value of all 10 pieces of fruit)/10 = 56 cents
This means, total value of all 10 pieces of fruit = 560 cents

How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
Let x = the number of oranges to be removed.
Each orange costs 60 cents, so the value of the x oranges to be removed = 60x
This means 560 - 60x = the value of the REMAINING fruit
Also, if we remove x oranges, then 10 - x = the number of pieces of fruit REMAINING.

We want the REMAINING fruit to have an average value of 52 cents.
We can write: (value of REMAINING fruit)/(number of pieces of fruit REMAINING) = 52
Rewrite as: (560 - 60x)/(10 - x) = 52
Multiply both sides by (10-x) to get: 560 - 60x = 52(10 - x)
Expand right side to get: 560 - 60x = 520 - 52x
Add 60x to both sides: 560 = 520 + 8x
Subtract 520 from both sides: 40 = 8x
Solve: x = 5

Answer: E

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Re: At a certain fruit stand, the price of each apple is 40 cents and the  [#permalink]

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New post 26 Jan 2019, 14:13
This can be done very quickly using weighted average principles.

since the average ends up being 52 cents.

40 60
52

(60-52) (52-40)

8/12=2/3. So the number of apples is 2x and the number of oranges is 3x. The sum has to be a multiple of 5. After some number of oranges were removed there must have been 5 left. Therefore, he must have removed 5 oranges.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the  [#permalink]

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New post 10 May 2019, 17:23
Here's an easy way to solve this:

Let A = apples, B = oranges

(40A + 60B)/10 = 56
(40A +60 (10-A))/10 = 56
40A + 600 - 60A = 560
-20A = -40
A = 2

Therefore, B = 8

So, how many of the 8 oranges do we need to remove to bring down the average?
let x = the number of oranges we need to remove
(40*(2apples) + 60 (8-x) )/10-x = 52
80+480 -60x = 520 - 52x
560 =520+8x
40= 8x
x = 5

(E)
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Re: At a certain fruit stand, the price of each apple is 40 cents and the   [#permalink] 10 May 2019, 17:23

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