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25 May 2011, 02:55
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. 63: Just recall table of 7 and stop where units digit is either 3 or 8. 7,14,21,28(We got it). Four 7's. Rest 5's. Another answer could be 9. 7*9=63 but buyer purchased both. Ignore this. _________________ Kudos [?]: 2051 [0], given: 376 Manager Joined: 08 Jun 2011 Posts: 90 Kudos [?]: 58 [0], given: 65 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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03 Dec 2011, 11:07
Here's a quick way to do this one.

7a + 5b=63

You basically subtract 7 from 63 which leads us to

7 + 56 = 63. Is 56 a multiple of 5? No. Then move on to
2 (7 ) + 49 = 63. Is 49 a multiple of 5? No. Then move on to
3 (7 ) + 42 = 63. Is 42 a multiple of 5? No. Then move on to
4 (7 ) + 35 = 63. Is 35 a multiple of 5? Yes. Save this answer
5 (7) + 28 = 63. Is 28 is a multiple of 5? No. Then move on to
6 (7) + 21 = 63. Is 21 a multiple of 5? No. Then move on to
7 (7) + 14 = 63. Is 14 a multiple of 5? No. Then move on to
8 (7) + 7 = 63. Is 7 a multiple of 5? No. Then move on to
9 (7) + 0. 63. Does not work because he/she bought both Apples and Bananas.

So the answer is 4 + 7 = 11.
The reason you start with 7 in your assessment is that it's the bigger number and 63 is a multiple of 7. This is my opinion, I hope it helps.

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22 Aug 2014, 10:44
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 Given: $$0.7b+0.5a=6.3$$ Question: $$a+b=?$$ $$0.7a+0.5b=6.3$$ --> $$7a+5b=63$$ --> $$5b=63-7a$$ --> $$5b=7(9-a)$$ --> $$5b$$ 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 >13 (as $$5b$$ in this case would be more than$6.30), so $$b=7$$ --> $$a=4$$ --> $$a+b=4+7=11$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-140732.html
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 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/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... 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] 24 Aug 2014, 22:50 Go to page Previous 1 2 [ 25 posts ] Display posts from previous: Sort by # A certain fruit stand sold apples for$0.70 each and bananas

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