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At a certain food stand, the price of each apple is ¢ 40 and

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At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 14 Oct 2007, 03:53
2
6
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A
B
C
D
E

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At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is ¢ 56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is ¢ 52?

A. 1
B. 2
C. 3
D. 4
E. 5
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Re: Food Stand  [#permalink]

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New post 09 Dec 2013, 23:10
5
5
praveenism wrote:
Q3:
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. Mary
selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean)
price of the 10 pieces of fruit is ¢ 56. How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?
A. 1
B. 2
C. 3
D. 4
E. 5

Answer:E. I solved the question and got ans as 2 :(


The correct answer is (E). People who are getting (B) need to think about this: The second time when the average is 52, does Mary still have 10 fruits? She ONLY needs to put back oranges. Not replace them with apples. So when you use alligation again and get the ratio as 2:3, you don't get 6 oranges since total number of fruits are not known. Let me solve it step by step.

When avg = 56

wa/wo = (60 - 56)/(56 - 40) = 1/4
So number of apples = 2, number of oranges = 8

When avg - 52
wa/wo = (60 - 52)/(52 - 40) = 2/3
What is the total number of fruit now? We don't know. We know Mary put back some oranges. We don't know how many. What we do know is that then number of apples she had stayed the same. She had 2 apples before so she still has 2 apples. If number of apples now is 2, number of oranges must be 3. So she must have put back 8 - 3 = 5 oranges.
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Re: Food Stand  [#permalink]

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New post 07 Jun 2010, 04:18
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\(A + O = 10\) .....1

\(\frac{40A + 60O}{10} = 56\)

=> \(2A + 3O = 28\)

Substitute equation 1:

=> \(2A + 3*(10-A) = 28\)

=> \(2A + 30 - 3A = 28\)

=> \(A = 2\) From 1: \(O = 8\)

Let's say she put back "r" oranges.

=> \(\frac{40*2 + 60*(8 - r)}{10 - r} = 52\)

=> \(80 + 480 - 60r = 520 - 52r\)

=> \(8r = 40\)

\(r = 5\)

Pick E
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Re: At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 14 Oct 2007, 04:35
singh_amit19 wrote:
Sorry guys for posting such a easy one.........but i suppose there is smthing wrong with the Question!


I guess Apples should be dropped..

As per the stem, we have (0.4a + 0.6o ) / 10 = 0.56
(0.4a + 0.6o ) = 5.6
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Re: At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 14 Oct 2007, 16:41
1
2
Ok this is how I got it...

Since total number of apples and oranges = 10

Let,
x = number of apples
10 - x = number of oranges

So the average is

.4x + .6(10-x)/10 = .56

Solving this equation give us
x = 2

So we have 2 apples
and 8 oranges...................................................A

Now subsitute the number of apples to get our new average .52 and oranges as y

.4(2) + .6(y)/2 + y = .52

after solving y = 3.............................................B


Hence A - B = 5

Ans = E

Hope this helps.
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Re: At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 15 Oct 2007, 01:13
singh_amit19 wrote:
At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?

(A)1
(B)2
(C)3
(D)4
(E)5


I get B...

S/#=A.

Lets just say A and O equal 40 and 60.

So S/10=56. Thus S=560

From this we can just subtract 2 oranges to get 440. Then by adding 2 more apples we get 520. 440+80=520.

520/10 = 52.

I don't see how this is incorrect???
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Re: At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 15 Oct 2007, 04:21
2
Here it is guys:

(0.4 Apples + 0.6 Oranges)/10 = 0.56

knowing that Apples + Oranges = 10

=> 0.4(10-Oranges) + 0.6(Oranges) = 5.6

=> Oranges = 8

=> Apples = 2

Let x be number of Oranges to be reduced

[0.4 Apples + 0.6 (Oranges-x)]/(10-x) = 0.52

Substitute 8 for Oranges and 2 for Apples

=> [0.8 + 0.6(8-x)]/(10-x) = 0.52

=> x = 5

Therefore E
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Re: At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 15 Oct 2007, 08:12
1
Hmm...your reasoning assumes that Mary always has to end up with 10 items of fruit. But that is not what the question is saying. The question goes:

Mary picks 10 items of fruit (each one is an apple or orange), and the average is 5.6

She THEN removes a certain number of oranges (hence she must have now LESS number of oranges and an EQUAL number of apples than before), and the average is now 5.2

Therefore, your second equation 4*0.4+6*0.6=5.2 doesnt hold true anymore.
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Re: At a certain food stand, the price of each apple is $0.4 and  [#permalink]

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New post 16 Oct 2007, 22:45
3
singh_amit19 wrote:
At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?

(A)1
(B)2
(C)3
(D)4
(E)5


got E too. lets set-up an equation:

(5.60 - 0.6n)/(10-n) = 0.52
solve for n, n = 5


good question.......
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At a certain food stand, the price of each apple is ¢ 40 and  [#permalink]

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New post 07 Jun 2010, 02:31
2
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is ¢ 56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is ¢ 52?

A. 1
B. 2
C. 3
D. 4
E. 5
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Re: Food Stand  [#permalink]

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New post 02 Aug 2010, 15:51
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This is a funny question.
I also solved and got the answer as 2, and after checking AbhayPrasanna's method the answer can also be 5 as demonstrated and if u substitue for 6 Oranges and 4 Apples or 3 Oranges and 7 Apples, the average turns out to be 52cents.

So is the question wrong or am I missing something here ?
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Re: Food Stand  [#permalink]

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New post 02 Aug 2010, 19:56
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devashish wrote:
This is a funny question.
I also solved and got the answer as 2, and after checking AbhayPrasanna's method the answer can also be 5 as demonstrated and if u substitue for 6 Oranges and 4 Apples or 3 Oranges and 7 Apples, the average turns out to be 52cents.

So is the question wrong or am I missing something here ?


As per the first two statements in the question(A+0=10 and (.4A+.6O)/10=5.6) no of apples is 2.
The last statement specified that she must put back only oranges to make the average 52cents. The no of apples should still remain 2 and can't change.
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Re: Food Stand  [#permalink]

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New post 02 Aug 2010, 20:28
Here is my work (after I realized I could only put back oranges and not switch an orange for an apple)

a=apples
o=oranges
t=total

\(10 = a + o\)
\(5.60 = .4a-.6o\)
Substitute
\(5.60 = .4a-.6(10-a)\)
solution is a=2
\(10 = 2 + o\)
o=8

So we have 8 oranges and 2 apples

5.20(t-2)=.4(2)-.6(t-2)
solution is t=5

Total - Apples = Oranges
\(5 - 2 = 3\)

In the beginning there were 8 oranges, now there are 3 so we must have given back 5. E!
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Re: Food Stand  [#permalink]

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New post 02 Aug 2010, 21:08
My answer is B = 2

With the help of the alligation Rule we can determine the original ratio as being 2:8

Alligation Rule (Cost of the Highest item less the Mean Cost) : (Mean Cost minus Lowest Cost Item)

(60 - 56) : (56 - 40 )

(4) : (16) = 2 : 8

So, utilizing the same rule for the new Mean 52

We get

(60 - 52) : (52 - 40)
8 : 12
4 : 6
Therefore the answer is 2... B
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Re: Food Stand  [#permalink]

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New post 09 Dec 2013, 22:32
Answer should be E.

Solving by Alligations:

40------------60

56

(60-56) (56-40)
4-------------16

Apples:Oranges = 1:4

There are 10 fruits so that means Apples =2 Oranges = 8

Now the Average Price is 52 so again using alligation:

40----------60

52

(60-52) (52-40)
8 12

They are in 8:12 or 2:3 ratio. Now no of apples are still 2 since we only need to reduce oranges as per question. So assume there are x no of fruits.

so, 2/5*x = 2

x = 5.

So we have 5 fruits. 2 Apples and 3 Oranges. Intitally we had 8 oranges and now 3 oranges that means 5 oranges needs to be kept back in the basket.

Hence ans should be E.
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Re: At a certain food stand, the price of each apple is ¢ 40 and  [#permalink]

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New post 21 Sep 2014, 02:36
VeritasPrepKarishma wrote:
praveenism wrote:
Q3:
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. Mary
selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean)
price of the 10 pieces of fruit is ¢ 56. How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?
A. 1
B. 2
C. 3
D. 4
E. 5

Answer:E. I solved the question and got ans as 2 :(


The correct answer is (E). People who are getting (B) need to think about this: The second time when the average is 52, does Mary still have 10 fruits? She ONLY needs to put back oranges. Not replace them with apples. So when you use alligation again and get the ratio as 2:3, you don't get 6 oranges since total number of fruits are not known. Let me solve it step by step.

When avg = 56

wa/wo = (60 - 56)/(56 - 40) = 1/4
So number of apples = 2, number of oranges = 8

When avg - 52
wa/wo = (60 - 52)/(52 - 40) = 2/3
What is the total number of fruit now? We don't know. We know Mary put back some oranges. We don't know how many. What we do know is that then number of apples she had stayed the same. She had 2 apples before so she still has 2 apples. If number of apples now is 2, number of oranges must be 3. So she must have put back 8 - 3 = 5 oranges.




can we do it like this

let no of apples be a and no of oranges be b

total fruits initially 10

so 56(10) = 40 ( a)+ 60( 10 -a)

second case when avg is 52 ...let no of organes she kept away be r

so
52( 10-r) = 40 ( a) + 60( 10-a-r) -------------2

we need not solves anything up till this point

now we know that the reduction in total is cause only by no of oranges she kept away , let us say she kept away r orhanges

so reduction is ( 56(10) - 52( 10-r) = 4(10) + 52 r ----3
we know that reduction will be equal to 60 r
as in equation 2 .....is same as equation 1 expcept for the amount accouhnted for reduced oranges

so using 3
4(10)=52r = 60 r

r=5
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Re: At a certain food stand, the price of each apple is ¢ 40 and  [#permalink]

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New post 01 May 2017, 06:30
got E as answer
ph 1
A + O = 10; 40A + 60O = 560 ;
A = 2; O =8 ;
ph 2
2*40 + O*60 = 52(2 + O) ; => O = 3 ;

hence, 5 oranges should be removed. Note : we should not remove apples as he didn't mention to remove apples.
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Re: At a certain food stand, the price of each apple is ¢ 40 and  [#permalink]

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Re: At a certain food stand, the price of each apple is ¢ 40 and &nbs [#permalink] 03 Aug 2018, 00:55
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