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Updated on: 03 Mar 2012, 23:43
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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. 15 Originally posted by pzazz12 on 30 Sep 2010, 05:08. Last edited by Bunuel on 03 Mar 2012, 23:43, edited 1 time in total. Edited the question ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 47951 Re: help me to solve it..... [#permalink] ### Show Tags 30 Sep 2010, 05:22 6 15 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 $$b=7$$ --> $$a+b=11$$.

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Re: help me to solve it.....  [#permalink]

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30 Sep 2010, 19:49
16
9
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. _________________ Support GMAT Club by putting a GMAT Club badge on your blog ##### General Discussion Manager Joined: 22 Sep 2010 Posts: 84 Re: help me to solve it..... [#permalink] ### Show Tags 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 $$b=7$$ --> $$a+b=11$$.

thank you, but can you explain me how this (9,0) and (4,7) to be solve...
Math Expert
Joined: 02 Sep 2009
Posts: 47951
Re: help me to solve it.....  [#permalink]

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01 Oct 2010, 05:49
8
5
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: $$7a+5b=63$$ --> $$5b=63-7a$$ --> $$5b=7(9-a)$$ --> $$5b$$ 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 >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: help me to solve it.....  [#permalink]

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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? Math Expert Joined: 02 Sep 2009 Posts: 47951 Re: help me to solve it..... [#permalink] ### Show Tags 11 Dec 2010, 00:01 3 1 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: help me to solve it.....  [#permalink]

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11 Dec 2010, 08:20
haha crystal as usual.
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Re: A certain fruit stand sold apples for $0.70 [#permalink] ### Show Tags 12 Dec 2010, 05:30 1 3 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/integral-solutions-of-ax-by-c.html
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17 Dec 2013, 03:46
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 He buys both, and the only combination of multiples of 5 and 7 that end in the single digit 3 is the single digits of the respective multiples ending in 8 and 5, which means 28 = 4*7 for bananas, and 7*5 = 35 for apples.. From here you can simply just add a decimal inbetween so that the restriction is upheld. 4 bananas and 7 apples gives us 11 fruits. Intern Joined: 06 Nov 2014 Posts: 12 Re: A certain fruit stand sold apples for$0.70 each and bananas  [#permalink]

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24 Nov 2014, 10:29
I just started by thinking ok at least 5*7 and 7*5 is an easy approach to start with to find multiples. If you look at the answer choices that would be 12 fruits, so somewhere in the middle of your options. In total that would be 70 (or $7), so clearly "one seven less" is 63, so that's 4*7 + 7*5 = fruits 4+7=11. Manager Joined: 07 Dec 2009 Posts: 99 GMAT Date: 12-03-2014 Re: A certain fruit stand sold apples for$0.70 each and bananas  [#permalink]

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21 Dec 2014, 13:38
1
from the question we get the following equation :
0.7A+0.5B = 6.3
this can be written as : 7A+5B = 63
Now rather than randomly plugging in values , we can do the following :
5B = 63 - 7A ==> 5B = 7(9-A)

this tells us that B should be a multiple of 7. it cannot be 14 as then the value will be greater than 63. hence B = 7 and A = 4
Ans is 11
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05 Jan 2015, 08:02
1
1
A = Apples
B = Bananas

0.7A + 0.5B = 6.3
7A + 5B = 63
5B = 63 - 7A
5B = 7 (9 - A)

Therefore, B has to be = 7
And, 5 = (9 - A) i.e. A = 4

A + B = 7 + 4 = 11

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Updated on: 09 Jan 2017, 15:33
5
The algebraic explanations in this thread are valid but needlessly complicated. Plus, it is unlikely that you will be able to come up with a similar approach to such a problem on your own. There is a much simpler and more broadly applicable approach that doesn't require trial and error.

Because the numbers of apples and bananas have to be integers, the easiest thing to do here is start with the total of 6.3 and subtract off .7 until you arrive at a multiple of .5.

6.3 - .7 = 5.6
5.6 - .7 = 4.9
4.2 - .7 = 4.2
4.2 - .7 = 3.5

We are now at a multiple of .5, having subtracted off 4 apples. Because the bananas are 50 cents each, this gives us 7 bananas, for a total of 11 pieces of fruit.
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Originally posted by dannythor6911 on 05 Jan 2017, 15:24.
Last edited by dannythor6911 on 09 Jan 2017, 15:33, edited 1 time in total.
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06 Jan 2017, 18:34
0.7a + 0.5 b = 6.30$Assume a = 5, b = 5 3.5 + 2.5 = 6.0$
more 0.3$more 10 11 B _________________ I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you Re: A certain fruit stand sold apples for$0.70 each and bananas &nbs [#permalink] 06 Jan 2017, 18:34

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