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 Sep 2010, 05:08

4

This post received KUDOS

24

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

80% (01:46) correct
20% (02:01) wrong based on 1251 sessions

HideShow timer Statistics

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 ?

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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

Given: \(0.7b+0.5a=6.3\) Question: \(a+b=?\)

\(0.7a+0.5b=6.3\) --> \(7a+5b=63\). After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: \(a=4\) and \(b=7\) --> \(a+b=11\).

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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

Without calculating anything in paper you could approach this problem.

Know -- Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7.

63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7 -- Answer 11 (B). -- 35 seconds straight.
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

Given: \(0.7b+0.5a=6.3\) Question: \(a+b=?\)

\(0.7a+0.5b=6.3\) --> \(7a+5b=63\). After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: \(a=4\) and \(b=7\) --> \(a+b=11\).

Answer: B.

thank you, but can you explain me how this (9,0) and (4,7) to be solve...

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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

Given: \(0.7a+0.5b=6.3\) Question: \(a+b=?\)

\(0.7a+0.5b=6.3\) --> \(7a+5b=63\). After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: \(a=4\) and \(b=7\) --> \(a+b=11\).

Answer: B.

thank you, but can you explain me how this (9,0) and (4,7) to be solve...

Trial and error would be good for it, but here is another way:

\(7a+5b=63\) --> \(5b=63-7a\) --> \(5b=7(9-a)\) --> \(5b\) must be multiple of 7 --> \(b\) must be multiple of 7 --> \(b\) can not be 0 (as "a customer purchased both apples and bananas") or >14 (as \(5b\) in this case would be more than $6.30), so \(b=7\) --> \(a=4\).

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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

Without calculating anything in paper you could approach this problem.

Know -- Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7.

63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7 -- Answer 11 (B). -- 35 seconds straight.

i get some multipule of 5 and 7 make 63... but why the multiple of 7 with a 3 or 8?

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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

Without calculating anything in paper you could approach this problem.

Know -- Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7.

63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7 -- Answer 11 (B). -- 35 seconds straight.

i get some multipule of 5 and 7 make 63... but why the multiple of 7 with a 3 or 8?

ezhilkumarank means that as multiple of 5 ends with 5 or 0 then multiple of 7 must end with 8 or 3 in order their sum to end with 3 (63). There is another approach in my previous post.

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? A. 10 B. 11 C. 12 D. 13 E. 14

Is there a faster way to do these problems other than brute force ?

0.7y + 0.5x = 6.3 is a straight line, we want the 'y' on the left side, rewriting gives: 7y + 5x = 63 y = 9 - (5/7)x

Almost instantly you should see that the right term (5/7) can only be an integer if X is either 7, 14, 28 and so forth. Ruling out fourteen we only have 7 left which gives 9-5 = 4 +7 fruits.
_________________

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

Show Tags

17 Dec 2013, 03:46

pzazz12 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 ?

A. 10 B. 11 C. 12 D. 13 E. 15

He buys both, and the only combination of multiples of 5 and 7 that end in the single digit 3 is the single digits of the respective multiples ending in 8 and 5, which means 28 = 4*7 for bananas, and 7*5 = 35 for apples.. From here you can simply just add a decimal inbetween so that the restriction is upheld.

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

Show Tags

24 Nov 2014, 10:29

I just started by thinking ok at least 5*7 and 7*5 is an easy approach to start with to find multiples. If you look at the answer choices that would be 12 fruits, so somewhere in the middle of your options. In total that would be 70 (or $7), so clearly "one seven less" is 63, so that's 4*7 + 7*5 = fruits 4+7=11.

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

Show Tags

21 Dec 2014, 13:38

from the question we get the following equation : 0.7A+0.5B = 6.3 this can be written as : 7A+5B = 63 Now rather than randomly plugging in values , we can do the following : 5B = 63 - 7A ==> 5B = 7(9-A)

this tells us that B should be a multiple of 7. it cannot be 14 as then the value will be greater than 63. hence B = 7 and A = 4 Ans is 11

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

Show Tags

22 Dec 2014, 00:51

7x+5y=63

multiple of 5 gives numbers only ending 0 or 5, so multiple of 7 should end 3 or 8 respectively. 7*9=63 (eliminate - no option for 5) 7*8=56 7*7=49 7*6=42 7*5=35 7*4=28 7*3=21 7*2=14 7*1=7

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

Show Tags

05 Jan 2017, 15:24

2

This post received KUDOS

The algebraic explanations in this thread are valid but needlessly complicated. Plus, it is unlikely that you will be able to come up with a similar approach to such a problem on your own. There is a much simpler and more broadly applicable approach that doesn't require trial and error.

Because the numbers of apples and bananas have to be integers, the easiest thing to do here is start with the total of 6.3 and subtract off .7 until you arrive at a multiple of .5.

We are now at a multiple of .5, having subtracted off 4 apples. Because the bananas are 50 cents each, this gives us 7 bananas, for a total of 11 pieces of fruit.
_________________

Dan the GMAT Man Offering tutoring and admissions consulting in the NYC area and online danthegmatman.squarespace.com danthegmatman@gmail.com

Last edited by dannythor6911 on 09 Jan 2017, 15:33, edited 1 time in total.

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

Show Tags

05 Jan 2017, 17:48

My approach: A-->Apples:0.70 B-->Bananas: 0.50 so.. Bought Spent 1A+1B= 1.20 2A+2B= 2.40 3A+3B= 3.60 4A+4B= 4.80 5A+5B= 6.00 -> 5+5 =10 but still have 0,30 that were spent so it cant be (A), the next possible value is 11 according to the answers (B) I hope that works!

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...