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A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 30 Mar 2009, 19:16 1 This post received KUDOS 3 This post was BOOKMARKED 00:00 Difficulty: 25% (medium) Question Stats: 75% (02:34) correct 25% (01:45) wrong based on 141 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-140732.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Aug 2014, 09:42, edited 1 time in total.
Edited the question and added the OA.
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 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.
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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.
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19 Aug 2010, 13:06
is there any other esiest or short way to solve the problem?
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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

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 21 Aug 2010, 12:41 1 This post received KUDOS Expert's post 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.
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27 Sep 2010, 14:49
I solved by back tracking:
Main Equation: 7a + 5b = 63
To find: a + b.
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
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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: 7069 Location: Pune, India Followers: 2086 Kudos [?]: 13283 [0], given: 222 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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03 Nov 2010, 19:17
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, 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 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] ### Show Tags 04 Nov 2010, 07:04 1 This post received 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 Get started with Veritas Prep GMAT On Demand for$199

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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!
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24 May 2011, 17:20
70x + 50y = 630

=> 7x + 5y = 63

=> 7*4 + 5*7 = 63

So x+y = 7+4 = 11

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 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: A certain fruit stand sold apples for$0.70 each and bananas   [#permalink] 25 May 2011, 01:38

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