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

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At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   26 Aug 2020, 09:32
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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
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   26 Aug 2020, 09:33
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   26 Aug 2020, 09:44
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Carcass 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

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

Cheers,
Brent
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   27 Aug 2020, 01:32
GreenlightTestPrep wrote:
Carcass 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

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

Cheers,
Brent



Hey Brent!

Thanks for the solution.

I was wondering if you were really able to do all that algebra in a little over a minute...

Is there another way where we can do this quickly ?

Ahmad
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   30 Sep 2020, 11:10
GreenlightTestPrep wrote:
Carcass 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

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

Cheers,
Brent


Hi Brent,

I did the following calculation which is logically correct but I am getting answer B,please let me know what seems wrong

Let x be number of apples
10-x be number of oranges

(40x + 60(10-x))/10 = 560
x=2

that means,2 apples and 8 oranges were picked

However,If I want to reduce avg to 52,I will have to choose 4 apples and 6 oranges 4x40 + 6x60 / 10 = 52

so originally,8 oranges were choosen and now for avg 52,6 oranges must be there.

Number of oranges to be removed = 8-6 = 2 hence B

I am not able to figure out how this answer is logically incorrect
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   20 Nov 2020, 01:32
[quote="GreenlightTestPrep"]
How does one recognize and implement the solve for this in under 90 seconds? Is there a generalized process for "change in average" to be aware of?

I don't believe there's a general process for this question type.
It's really a matter of assigning variables and creating the appropriate equations to solve.
Keep practicing!
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   20 Nov 2020, 06:01
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vaishar3 wrote:
Hi Brent,

I did the following calculation which is logically correct but I am getting answer B,please let me know what seems wrong

Let x be number of apples
10-x be number of oranges

(40x + 60(10-x))/10 = 560
x=2

that means,2 apples and 8 oranges were picked

However,If I want to reduce avg to 52,I will have to choose 4 apples and 6 oranges 4x40 + 6x60 / 10 = 52

so originally,8 oranges were choosen and now for avg 52,6 oranges must be there.

Number of oranges to be removed = 8-6 = 2 hence B

I am not able to figure out how this answer is logically incorrect

The issue with your solution is highlighted above.

You're correct to say that, originally, 2 apples and 8 oranges were picked (for a total of 10 pieces of fruit).
When we remove oranges (to get a new average price of 52 cents), we end up with fewer than 10 pieces of fruit.
In your solution, you are assuming that, after we remove some oranges, we still have 10 pieces of fruit remaining.

Does that help?
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink] New post   23 Jul 2024, 21:45
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Re: At a certain fruit stand, the price of each apple is 40 cent [#permalink]
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