Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 21 Aug 2014, 04:45

GMAT Club Daily Prep

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

PS - Certain Fruit stand - OG 12

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Senior Manager
Joined: 30 Nov 2008
Posts: 494
Schools: Fuqua
Followers: 10

Kudos [?]: 121 [1] , given: 15

PS - Certain Fruit stand - OG 12 [#permalink]  30 Mar 2009, 19:16
1
KUDOS
00:00

Difficulty:

(N/A)

Question Stats:

46% (02:13) correct 54% (00:57) wrong based on 12 sessions
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.
Director
Joined: 01 Apr 2008
Posts: 909
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 15

Kudos [?]: 203 [0], given: 18

Re: PS - Certain Fruit stand - OG 12 [#permalink]  30 Mar 2009, 20:34
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?

a) 10
b) 11
c) 12
d) 13
e) 14.
Director
Joined: 25 Oct 2006
Posts: 652
Followers: 7

Kudos [?]: 186 [0], given: 6

Re: PS - Certain Fruit stand - OG 12 [#permalink]  30 Mar 2009, 22:58
7x+5y = 63...I

I did back calculation after that
first x+y = 12 ... II

I find that C is too low so chose B and got the answer
_________________

If You're Not Living On The Edge, You're Taking Up Too Much Space

Current Student
Joined: 03 Aug 2006
Posts: 117
Location: Next to Google
Schools: Haas School of Business
Followers: 4

Kudos [?]: 104 [0], given: 3

Re: PS - Certain Fruit stand - OG 12 [#permalink]  31 Mar 2009, 13:56
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.
Senior Manager
Joined: 30 Nov 2008
Posts: 494
Schools: Fuqua
Followers: 10

Kudos [?]: 121 [0], given: 15

Re: PS - Certain Fruit stand - OG 12 [#permalink]  31 Mar 2009, 15:49
Yes. OA is B.

Here is another approach of mine.

It is given, 0.7A + 0.5B = 6.3 ==> 7A + 5B = 63. The question asked is "what is the value of A + B? "

From the above equation we can derive, 5(A+B) + 2A = 63

==> A+B = (63-2A) / 5. This means A + B is nothing but the a multiple of 5 when some multiple of 2 is deducted from 63.

We also know that A can be at the max 8. (Considering atleast one banana).

Now I guessed what are the possible values of A where 63-2A is divisible by 5. And A results in 4. Once we know A, we can find A+B.
Director
Status: GMAT Learner
Joined: 14 Jul 2010
Posts: 652
Followers: 34

Kudos [?]: 213 [0], given: 32

Re: PS - Certain Fruit stand - OG 12 [#permalink]  19 Aug 2010, 13:06
is there any other esiest or short way to solve the problem?
_________________

I am student of everyone-baten
Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

GRE Forum Moderator
Affiliations: PMP certified, IT professional
Joined: 21 Jun 2010
Posts: 217
Location: USA
Schools: CMU, Kelley
Followers: 7

Kudos [?]: 67 [1] , given: 2

Re: PS - Certain Fruit stand - OG 12 [#permalink]  20 Aug 2010, 09:37
1
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

So, a = 4, b = 7 ; a+b=11
Manager
Joined: 29 Jul 2010
Posts: 127
Followers: 1

Kudos [?]: 3 [0], given: 47

Re: PS - Certain Fruit stand - OG 12 [#permalink]  20 Aug 2010, 10:17
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

B
GMAT Instructor
Joined: 24 Jun 2008
Posts: 967
Location: Toronto
Followers: 253

Kudos [?]: 652 [1] , given: 3

Re: PS - Certain Fruit stand - OG 12 [#permalink]  21 Aug 2010, 12:41
1
KUDOS
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.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Intern
Joined: 20 Apr 2010
Posts: 38
Followers: 0

Kudos [?]: 1 [0], given: 7

Re: PS - Certain Fruit stand - OG 12 [#permalink]  27 Sep 2010, 10:50
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.
Manager
Joined: 16 Jun 2010
Posts: 187
Followers: 2

Kudos [?]: 29 [0], given: 5

Re: PS - Certain Fruit stand - OG 12 [#permalink]  27 Sep 2010, 14:49
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.

Manager
Status: GMAT Preperation
Joined: 04 Feb 2010
Posts: 105
Concentration: Social Entrepreneurship, Social Entrepreneurship
GPA: 3
WE: Consulting (Insurance)
Followers: 1

Kudos [?]: 10 [2] , given: 15

Re: PS - Certain Fruit stand - OG 12 [#permalink]  29 Oct 2010, 22:55
2
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

so 11 is the answer
Senior Manager
Joined: 11 May 2010
Posts: 289
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT Date: 10-22-2011
GPA: 3
WE: Information Technology (Internet and New Media)
Followers: 4

Kudos [?]: 29 [0], given: 12

Re: PS - Certain Fruit stand - OG 12 [#permalink]  03 Nov 2010, 12:08
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 ...

this is really doing my head in!

thanks
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 4668
Location: Pune, India
Followers: 1072

Kudos [?]: 4776 [0], given: 163

Re: PS - Certain Fruit stand - OG 12 [#permalink]  03 Nov 2010, 19:17
Expert's post
n2739178 wrote:
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 ...

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save \$100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 4668
Location: Pune, India
Followers: 1072

Kudos [?]: 4776 [1] , given: 163

Re: PS - Certain Fruit stand - OG 12 [#permalink]  04 Nov 2010, 07:04
1
KUDOS
Expert's post
n2739178 wrote:

this is really doing my head in!

I have put up this theory on this link:
http://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html

See if it makes sense now.
If there are doubts, get back to me on my blog itself or here...
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save \$100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Joined: 11 May 2010
Posts: 289
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT Date: 10-22-2011
GPA: 3
WE: Information Technology (Internet and New Media)
Followers: 4

Kudos [?]: 29 [0], given: 12

Re: PS - Certain Fruit stand - OG 12 [#permalink]  04 Nov 2010, 14:04
great thanks Karishma!
Senior Manager
Joined: 21 Jul 2010
Posts: 272
Followers: 8

Kudos [?]: 27 [0], given: 17

Re: PS - Certain Fruit stand - OG 12 [#permalink]  24 May 2011, 11:09
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.

Thanks for the insight!
Director
Joined: 01 Feb 2011
Posts: 770
Followers: 14

Kudos [?]: 81 [0], given: 42

Re: PS - Certain Fruit stand - OG 12 [#permalink]  24 May 2011, 16:16
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

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.

Thanks for the insight!
SVP
Joined: 16 Nov 2010
Posts: 1692
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 30

Kudos [?]: 285 [0], given: 36

Re: PS - Certain Fruit stand - OG 12 [#permalink]  24 May 2011, 17:20
70x + 50y = 630

=> 7x + 5y = 63

=> 7*4 + 5*7 = 63

So x+y = 7+4 = 11

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1365
Followers: 11

Kudos [?]: 135 [0], given: 10

Re: PS - Certain Fruit stand - OG 12 [#permalink]  25 May 2011, 01:38
apples = 4 and bananas = 7.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Re: PS - Certain Fruit stand - OG 12   [#permalink] 25 May 2011, 01:38
Similar topics Replies Last post
Similar
Topics:
19 A certain fruit stand sold apples for \$0.70 each and bananas 9 15 Oct 2012, 03:33
13 A certain fruit stand sold apples for \$0.70 each and bananas 16 30 Sep 2010, 04:08
6 PS - Certain Fruit stand - OG 12 21 30 Mar 2009, 19:16
PS: Remainders - OG 12 4 30 Mar 2009, 19:28
1 From ETS Paper test A certain fruit stand sold apples for 2 03 Jun 2006, 10:14
Display posts from previous: Sort by

PS - Certain Fruit stand - OG 12

 Question banks Downloads My Bookmarks Reviews Important topics
 Go to page    1   2    Next  [ 22 posts ]

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.