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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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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
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Originally posted by Bunuel on 30 Sep 2010, 05:08.
Last edited by Bunuel on 20 Feb 2019, 04:26, edited 2 times in total.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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30 Sep 2010, 05:22
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 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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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30 Sep 2010, 19:49
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 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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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01 Oct 2010, 05:06
Bunuel wrote: 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 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...



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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01 Oct 2010, 05:49
pzazz12 wrote: Bunuel wrote: 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 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: \(7a+5b=63\) > \(5b=637a\) > \(5b=7(9a)\) > \(5b\) 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 \(b=7\) > \(a=4\). Hope it's clear.
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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29 Oct 2010, 23:55
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



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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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 ... this is really doing my head in! thanks



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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03 Nov 2010, 20:17
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, coefficient of b, from a above and add 7, coefficient 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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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04 Nov 2010, 08:04
n2739178 wrote: this is really doing my head in!
I have put up this theory on this link: http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.htmlSee if it makes sense now. If there are doubts, get back to me on my blog itself or here...
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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10 Dec 2010, 20:14
ezhilkumarank wrote: 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 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?



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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11 Dec 2010, 00:01
mmcooley33 wrote: ezhilkumarank wrote: 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 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. Hope it's clear.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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12 Dec 2010, 05:30
ajit257 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
Is there a faster way to do these problems other than brute force ? The first solution will invariably involve some brute force. But (9, 0) is easy to get since 63 is a multiple of 7. Check out this post for clarification on these type of questions: http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.html
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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21 Dec 2010, 16:20



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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21 Dec 2010, 16:58
Maybe it could be of any help to realise that 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 95 = 4 +7 fruits.
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Oct 2012, 04:51
Let the no of apples sold = a no of bananas sold = b Question is a+b=? Thus 70a + 50b = 630 7a + 5b = 63 Quick Tip In order to find out the value of a & b its better to find the value of 'a' as "63 7a" must leave a number which will end either with 0 or with 5. (Think about it for a second) Thus the only value which satisfies above equation is a=4 & b=7 a+b=11 (Other values of a & b will lie outside the answer choices) Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Oct 2012, 08:39
7A + 5B = 63
5B = 63  7A
B= (63  7A)/5 => must be an integer, as we are dealing with quantities unless we can buy 1/4 of an apple or 1/3 of a banana (That could be funny :D )
so 63  7A must yield to a multiple of 5.
Now, let's pick some numbers.
if, A= 1, 637= 56 (not divisible by 5) A= 2, 6314= 49 (not divisible by 5) A= 3, 6321= 42 (not divisible by 5) A= 4, 6328= 35 (divisible by 5)
so A= 4, B= 7 A+B= 11 , Answer B



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Oct 2012, 10:50
To solve this question  we can take numbers, as price of apple 7, 5 for Banana and 63 total for ease. Now we can determine quickly that total number should range between 63/7 <= N <=63/5, so ans should be between 9 and 12. Now solving the expression 7A+5B =63 first possibility with 9 apples, 0 banana we get 6.30 total amount, but question says customer purchased both, apple and banana. so not correct. So next choice, for 7A+5B =63 would come by decreasing 63 in multiple of 5 and checking divisibility of that number by 7. this way we get 4 Apples *0.70 + 7 banana *050 = 6.30 Hence total number is 7+4 =11 Ans B Bunuel wrote: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectA 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 Practice Questions Question: 64 Page: 161 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a solution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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16 Oct 2012, 03:53
so apples is 0.70 * A
Bananas 0.50 * B
then 0.70A + 0.50B = 6.30
multiply by 10 we get
7A + 5B = 63
5B = 63  7A B = 7(9A)/5
now to satisfy this equation we need 9  A = 5 only then it will be divisible by 5 therefore A is 4 and when solve we get B is 7
7(94)/5 = 7*5/5 then we need the sum of A + B = 7 + 4 = 11
Answer B



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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22 Aug 2014, 10:39
VeritasPrepKarishma wrote: n2739178 wrote: this is really doing my head in!
I have put up this theory on this link: http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.htmlSee if it makes sense now. If there are doubts, get back to me on my blog itself or here... Hi Karishma, Interesting post  makes complete sense. A question though: In your hypothetical question about " And, a trickier thing to think about  how many integral solutions would 3x  5y = 42 have?"  both have to go up, right? So x would have to go up from 14, to 19, to 24 etc. Conversely, y would also go up from 0 to 3, to 6 etc. Neither of those values can be negative since we have the positive integer constraint. Am I correct? Can you recommend other questions similar to this? Thanks!



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Re: A certain fruit stand sold apples for $0.70 each and bananas
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24 Aug 2014, 22:50
russ9 wrote: VeritasPrepKarishma wrote: n2739178 wrote: this is really doing my head in!
I have put up this theory on this link: http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.htmlSee if it makes sense now. If there are doubts, get back to me on my blog itself or here... Hi Karishma, Interesting post  makes complete sense. A question though: In your hypothetical question about " And, a trickier thing to think about  how many integral solutions would 3x  5y = 42 have?"  both have to go up, right? So x would have to go up from 14, to 19, to 24 etc. Conversely, y would also go up from 0 to 3, to 6 etc. Neither of those values can be negative since we have the positive integer constraint. Am I correct? Can you recommend other questions similar to this? Thanks! Yes, the first easy solution would be 14, 0. Both x and y will move in same direction. Since neither can be negative, they must move up only.
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