Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
At a certain food stand, the price of each apple is $0.4 and [#permalink]
14 Oct 2007, 02:53
2
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
75% (hard)
Question Stats:
59% (03:51) correct
41% (02:39) wrong based on 208 sessions
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?
Re: At a certain food stand, the price of each apple is $0.4 and [#permalink]
15 Oct 2007, 00:13
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.
Re: At a certain food stand, the price of each apple is $0.4 and [#permalink]
15 Oct 2007, 07:12
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.
Re: At a certain food stand, the price of each apple is $0.4 and [#permalink]
16 Oct 2007, 21:45
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?
At a certain food stand, the price of each apple is ¢ 40 and [#permalink]
07 Jun 2010, 01:31
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."
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.
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
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
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. _________________
Re: At a certain food stand, the price of each apple is ¢ 40 and [#permalink]
21 Sep 2014, 01:36
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
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
Re: At a certain food stand, the price of each apple is $0.4 and [#permalink]
22 Jul 2015, 00:29
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. _________________
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...