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

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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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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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15 Oct 2007, 08:12
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] ### Show Tags 16 Oct 2007, 22:45 1 This post received KUDOS 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....... Manager Joined: 28 Feb 2010 Posts: 174 WE 1: 3 (Mining Operations) Followers: 7 Kudos [?]: 34 [1] , given: 33 At a certain food stand, the price of each apple is ¢ 40 and [#permalink] ### Show Tags 07 Jun 2010, 02:31 1 This post received KUDOS 1 This post was BOOKMARKED 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 _________________ Regards, Invincible... "The way to succeed is to double your error rate." "Most people who succeed in the face of seemingly impossible conditions are people who simply don't know how to quit." Manager Joined: 04 May 2010 Posts: 88 WE 1: 2 yrs - Oilfield Service Followers: 12 Kudos [?]: 116 [2] , given: 7 Re: Food Stand [#permalink] ### Show Tags 07 Jun 2010, 04:18 2 This post received KUDOS 1 This post was BOOKMARKED $$A + O = 10$$ .....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 Manager Joined: 16 Jun 2010 Posts: 188 Followers: 2 Kudos [?]: 88 [1] , given: 5 Re: Food Stand [#permalink] ### Show Tags 02 Aug 2010, 15:51 1 This post received KUDOS 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 ? _________________ Please give me kudos, if you like the above post. Thanks. Manager Joined: 20 Mar 2010 Posts: 84 Followers: 2 Kudos [?]: 92 [1] , given: 1 Re: Food Stand [#permalink] ### Show Tags 02 Aug 2010, 19:56 1 This post received KUDOS 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. _________________ ___________________________________ Please give me kudos if you like my post Manager Status: Waiting to hear from University of Texas at Austin Joined: 24 May 2010 Posts: 76 Location: Changchun, China Schools: University of Texas at Austin, Michigan State Followers: 5 Kudos [?]: 57 [0], given: 4 Re: Food Stand [#permalink] ### Show Tags 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! Manager Joined: 06 Oct 2009 Posts: 105 Location: Mexico Concentration: Entrepreneurship, Finance GPA: 3.85 WE: Sales (Commercial Banking) Followers: 1 Kudos [?]: 139 [0], given: 4 Re: Food Stand [#permalink] ### Show Tags 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 Regards from Mexico City! Intern Joined: 14 Dec 2012 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 4 Re: Food Stand [#permalink] ### Show Tags 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7246 Location: Pune, India Followers: 2201 Kudos [?]: 14317 [2] , given: 222 Re: Food Stand [#permalink] ### Show Tags 09 Dec 2013, 23:10 2 This post received KUDOS Expert's post 3 This post was BOOKMARKED 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 [Reveal] Spoiler: 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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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

[Reveal] Spoiler:
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 $0.4 and [#permalink] ### Show Tags 22 Jul 2015, 01:29 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 14379 Followers: 602 Kudos [?]: 174 [0], given: 0 Re: At a certain food stand, the price of each apple is ¢ 40 and [#permalink] ### Show Tags 25 Jul 2016, 01:39 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: At a certain food stand, the price of each apple is ¢ 40 and [#permalink] 25 Jul 2016, 01:39 Similar topics Replies Last post Similar Topics: 59 At a certain fruit stand, the price of each apple is 40 cents and the 24 20 Oct 2015, 03:19 4 The price of each hair clip is ¢ 40 and the price of each hair band is 10 16 Apr 2015, 04:36 48 A certain fruit stand sold apples for$0.70 each and bananas 17 15 Oct 2012, 04:33
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