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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 25 May 2011, 01:55 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.
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22 Aug 2014, 09:39
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/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!
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 22 Aug 2014, 09:44 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 $$b=7$$ --> $$a=4$$ --> $$a+b=4+7=11$$. Answer: B. OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-140732.html _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13676 [0], given: 222 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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24 Aug 2014, 21:50
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.
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