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:

A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

30 Mar 2009, 20:16

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

74% (02:38) correct
26% (01:46) wrong based on 132 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

30 Mar 2009, 21:34

B. 11? I got the answer somehow:) Someone can show a proper way. 0.7*a + 0.50*b = 6.30.

I listed the multiples of 7 and 5. After adding the last two digits of these two multiples the sum should have 3 as the last digit. so only 7*4 = 28 and some other multiple of 5 ( 8+5 = 13 , last digit 3 => for 6.30 ) is possible. So 4 apples are fixed Hence, we get 4 + 7 = 11.

mrsmarthi wrote:

A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

31 Mar 2009, 14:56

B

This is how I approached it..The first thing I noticed is that 6.3 is a multiple of 0.7 so if the customer bought all apples he or she would get 9 of them. Then I listed the multiples of 0.7 like below...I just listed the multiple of 7 up to 63. Now I looked at the difference for each multiple and the last multiple where the difference is a multiple of 5

7 14 21 28 35 42 49 56 63

The only place that happens is for 28 where 63 - 28 = 35 which is a multiple of 5, hence 28 contributes for apples and 35 contributes for bananas leading to 4 apples (28/7) and 7 bananas (35/5) to a total of 11.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

20 Aug 2010, 10:37

2

This post received KUDOS

0.7a+0.5b = 6.3

7a+5b= 63

Now we know that the unit's place of b has to be 5 or 0 (since it is a multiple of 5). So, 7*a has to have a unit's place of 8 or 3 to get a total sum of 63

One possibility is a=9, b=0 But since the questions states that he bought BOTH, and since 9 is not in answer choice, this is ruled out.

The next possibility is trying to find a multiple of 7 that has unit's place = 8 ; 28 a=4 7*4 = 28 28 + 5*7 = 63

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

20 Aug 2010, 11:17

First you see this problem you should guess that from 50 and 70 you need to get something that ends with 80 and 50 and first numbers that come to hat are 350 and 280 therefore 4*70 and 7*50

You can use divisibility properties here to get to the answer. The idea is this: if you see an equation like the following, and you know that r and s are positive integers:

5r = 7s

this means that 5r and 7s are exactly the same number, so they must of course have the same divisors. So, since 7 is a divisor of 7s, it must also be a divisor of 5r, and therefore of r. Similarly s must be divisible by 5. That is, when you have equations involving only integers, the primes which divide the left side of the equation must divide the right, and vice versa.

So onto this question. We know:

0.7a + 0.5b = 6.3 5b = 63 - 7a

and since 7 is a factor of the right side of this equation, it must be a factor of the left, so b must be a multiple of 7. Since b is greater than 0, and b cannot be 14 (then the total cost would be greater than $6.30), b must be 7, from which we can find that a is 4. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

27 Sep 2010, 11:50

This is a stupid question, but why is this not solvable by algebra? When you do the algebra you get a non integer number. I understand, two variables and one equation means you cant solve, but we are looking for a+b so when you remove the numbers next to a and b, you have a final value a+b=x.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

27 Sep 2010, 15:49

I solved by back tracking: Main Equation: 7a + 5b = 63 To find: a + b. As per the answers: if a+ b >12 ... then the main equation is not satisfied (i.e if a + b = 13 then 5(a + b) = 65 ... which is greater than 63 and hance impossible ) So the answer is 10, 11 or 12. Now similarly using 10 or 11 the answer is in fraction which cannot be the case( as we are dealing with whole fruits ) ... hence answer is 11 or B _________________

Please give me kudos, if you like the above post. Thanks.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

29 Oct 2010, 23:55

3

This post received KUDOS

hI ALL,

this is one of the first good questions i encountered.. here is what i did:

7a + 5b = 63 2a+5a+5b=63 2a+5(a+b)=63 i started substituting the answer choices lets start wit 12 2a+5(12)=63 => 2a=3 .. a has to be an integer.. so no 2a+5(11)=63 => 2a=8 => a = 4 good.. but still need to check 13 2a+5(10)=63 => 2a=13 .. a has to be an integer.. so no 2a+5(13)=63 => 2a=-2 ==> a had to be +ve .. so no good

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

03 Nov 2010, 13:08

Hi all

I can't understand MGMAT's OG guide answer for this one...

1st question: why can't you exchange apples with bananas 1 for 1 until you get the right answer?

2nd question: They're banging on about 5 apples having the same value as 7 bananas, then taking 5 apples away from the 'partial solution ~(i.e. when A = 9 and B = 0) and adding 7 bananas... Why would you do that? It makes no sense to me... If you set the vlaue of the apples = bananas you get them both equalling $3.50 in value which equals $7 between both of them... which is clearly over $63 ...

I can't understand MGMAT's OG guide answer for this one...

1st question: why can't you exchange apples with bananas 1 for 1 until you get the right answer?

2nd question: They're banging on about 5 apples having the same value as 7 bananas, then taking 5 apples away from the 'partial solution ~(i.e. when A = 9 and B = 0) and adding 7 bananas... Why would you do that? It makes no sense to me... If you set the vlaue of the apples = bananas you get them both equalling $3.50 in value which equals $7 between both of them... which is clearly over $63 ...

this is really doing my head in!

thanks

I do not know what exactly your book says but I am guessing this is how they have solved it:

7a + 5b = 63 Such equations have infinite solutions. We can get a single solution under particular constraints. (Will explain this later) One thing we notice right away is that one solution to this problem is a = 9 and b = 0 because 63 is divisible by 7. 7a + 5b = 63 a = 9, b = 0 a = 4, b = 7 (To get this solution, subtract 5, co-efficient of b, from a above and add 7, co-efficient of a, to b above) a = -1, b = 14 (Again, do the same to the solution above) a = 13, b = -7 (You will also get solutions when you add 5 to a of any other solution and subtract 7 from b of the same solution) Hence there are infinite solutions. Here the constraints are that a and b should not be negative. Also, they should not be 0 since he buys at least 1 apple and at least 1 banana. Only 1 solution satisfies these constraints so answer is a = 4 and b =7. Why this works is because when you reduce a by 5, the reduction in 7a is offset by the increase in 5b when you increase b by 7. Let this suffice for now. This is the theory of Integral solutions to equations in two variables. I will explain you the complete theory soon. _________________

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

24 May 2011, 12:09

Hm. I like others, was hoping there was a "faster" way of doing this problem. Seems like at a minimum it would take about 2 mins to 2:30 to crunch those numbers.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

Show Tags

24 May 2011, 17:16

There are several ways to solve these kind of problems . You can try this approach

5B = 63-7A

=> B = \((7*(9-A))/5\)

=> 9-A must be a multiple of 5 => 9-A can be 0 , 5 ,10 , 15...

=> A-9 will be 0, -5, -10,-15...respectively

=> A will be 9, 4, -1, -6.... respectively

=> As A cannot be -ve , only possible options are 9 or 4

=> If A =9 then B becomes 0 which is not valid here. Hence A=4,B=7

=> A+B =11

Answer is B.

humphy wrote:

Hm. I like others, was hoping there was a "faster" way of doing this problem. Seems like at a minimum it would take about 2 mins to 2:30 to crunch those numbers.

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Time is a weird concept. It can stretch for seemingly forever (like when you are watching the “Time to destination” clock mid-flight) and it can compress and...