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A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 30 Sep 2010, 04:08 1 This post received KUDOS 16 This post was BOOKMARKED 00:00 Difficulty: 25% (medium) Question Stats: 77% (02:41) correct 23% (02:10) wrong based on 880 sessions ### HideShow timer Statistics 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Mar 2012, 22:43, edited 1 time in total.
Edited the question
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30 Sep 2010, 04:22
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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. _________________ Senior Manager Status: Time to step up the tempo Joined: 24 Jun 2010 Posts: 408 Location: Milky way Schools: ISB, Tepper - CMU, Chicago Booth, LSB Followers: 8 Kudos [?]: 198 [8] , given: 50 Re: help me to solve it..... [#permalink] ### Show Tags 30 Sep 2010, 18:49 8 This post received KUDOS 4 This post was BOOKMARKED 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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01 Oct 2010, 04:06
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... Math Expert Joined: 02 Sep 2009 Posts: 36618 Followers: 7100 Kudos [?]: 93564 [4] , given: 10578 Re: help me to solve it..... [#permalink] ### Show Tags 01 Oct 2010, 04:49 4 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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 $$b=7$$ --> $$a+b=11$$.

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. _________________ Intern Joined: 31 Oct 2010 Posts: 32 Followers: 0 Kudos [?]: 61 [0], given: 25 Re: help me to solve it..... [#permalink] ### Show Tags 10 Dec 2010, 19:14 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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10 Dec 2010, 23:01
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mmcooley33 wrote:
ezhilkumarank wrote:
pzazz12 wrote:

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12 Dec 2010, 04:30
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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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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21 Dec 2010, 15:20
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21 Dec 2010, 15:58
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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 9-5 = 4 +7 fruits.
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Re: help me to solve it..... [#permalink]

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03 Jun 2012, 06:09
Mackieman wrote:
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 9-5 = 4 +7 fruits.

This is really fast, even though trial and error is also a good strategy
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17 Dec 2013, 02:46
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: 13 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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24 Nov 2014, 09:29
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: 111 GMAT Date: 12-03-2014 Followers: 0 Kudos [?]: 30 [0], given: 375 Re: A certain fruit stand sold apples for$0.70 each and bananas [#permalink]

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21 Dec 2014, 12:38
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, 07:02
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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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08 Oct 2016, 19:18

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