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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]

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New post 05 Sep 2019, 05:51
v12345 wrote:
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.


here price of apple = 40 cents and price of orange = 60 cents

average price = 56 cents

so, the apples and oranges are in ratio

apple : orange = 60 - 56 : 56 - 40
=> apple : orange = 4 : 16
=> apple : orange = 1 : 4

total no =10, therefore apples = 2 nos. and oranges = 8 nos.


now we have to make average price = 52 cents
so, the apples and oranges will be in ratio

apple : orange = 60 - 52 : 52 - 40
=> apple : orange = 8 : 12
=> apple : orange = 2 : 3

now we have apples = 2 nos. from above, therefore oranges must be 3 in number

oranges to be put back = 8 - 3 = 5 nos

Answer choice E

kudos if you like the explanation


Why did you assume, in the second ratio, 2:3, that the total apples is 2 and oranges is 3? You are not given total, as in the first ratio (where total given is 10). Here, total is less than 10, but you don't know what.
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At a certain fruit stand, the price of each apple is 40 cents and the  [#permalink]

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New post 07 Oct 2019, 05:03
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.


Hello Bunuel,

I have a very basic doubt.. In one of the solutions

it is given

apple : orange = 60 - 56 : 56 - 40
=> apple : orange = 4 : 16
=> apple : orange = 1 : 4

I want to know how the ratio of Apple is 4 And Oranges 16? How did we arrive at this?

For your help i have pasted one of the answers below from where this is taken

" here price of apple = 40 cents and price of orange = 60 cents

average price = 56 cents

so, the apples and oranges are in ratio

apple : orange = 60 - 56 : 56 - 40
=> apple : orange = 4 : 16
=> apple : orange = 1 : 4

total no =10, therefore apples = 2 nos. and oranges = 8 nos.


now we have to make average price = 52 cents
so, the apples and oranges will be in ratio

apple : orange = 60 - 52 : 52 - 40
=> apple : orange = 8 : 12
=> apple : orange = 2 : 3

now we have apples = 2 nos. from above, therefore oranges must be 3 in number

oranges to be put back = 8 - 3 = 5 nos

Answer choice 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 08 Oct 2019, 02:42
As is given in the quwstion
60x + 40y = 56*10
And x+y = 10
Thus x = 8 and y = 2
Now, 60p + 40*2 = 52 (p+2)
8p = 104 -80 thus p = 3
So the number of orange is to be reduced by 5 =(8-3)
OA: 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 17 Oct 2019, 03:09
There is a much simpler approach to this problem. We do not require any use of weighted averages or ratios or any other complex calculations. This sum can be solved within 10-25sec, after reading the question.

40a+60o = 560 (average of 10 fruits is 56)

a + o = 10

40a + 60(o-x) = 52(10-x) (Since the new average is 52)

We can see from the left side of the equation that the total price is a multiple of 10. This is possible only if 10-x = 10 or 5. 10 can be eliminated because we have to return some oranges.
There is only 1 value of x for which 52 will be a multiple of 10, which is 5.

Solved only with common sense.
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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 07 Dec 2019, 07:28
Confusing question. I read it as if Mary would always select a total of 10 apples and oranges; so that if an orange is removed, it must be replaced with an apple (as if the total must always be 10). Doing this would result in answer B (2); she would replace 2 oranges with 2 apples resulting in a total price of 520 cents.

If Mary only removes, and not replaces, oranges (thus the total number of fruit does not have to be 10), than the correct answer would be E.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the   [#permalink] 07 Dec 2019, 07:28

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