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30 Sep 2010, 05:08
16
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25% (medium)

Question Stats:

78% (02:39) correct 22% (02:11) wrong based on 720 sessions

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, 23:43, edited 1 time in total. Edited the question Math Expert Joined: 02 Sep 2009 Posts: 35275 Followers: 6636 Kudos [?]: 85566 [1] , given: 10237 Re: help me to solve it..... [#permalink] ### Show Tags 30 Sep 2010, 05:22 1 This post received KUDOS Expert's post 7 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

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
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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 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 Manager Joined: 22 Sep 2010 Posts: 92 Followers: 2 Kudos [?]: 133 [0], given: 0 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...
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Re: help me to solve it..... [#permalink]

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01 Oct 2010, 05:49
3
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Expert's post
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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: 35275 Followers: 6636 Kudos [?]: 85566 [1] , given: 10237 Re: help me to solve it..... [#permalink] ### Show Tags 11 Dec 2010, 00:01 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 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.
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22 Dec 2014, 00:51
7x+5y=63

multiple of 5 gives numbers only ending 0 or 5, so multiple of 7 should end 3 or 8 respectively.
7*9=63 (eliminate - no option for 5)
7*8=56
7*7=49
7*6=42
7*5=35
7*4=28
7*3=21
7*2=14
7*1=7

it is 4+(63-28)/5=4+7=11

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16 Jan 2016, 21:15
Hello from the GMAT Club BumpBot!

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] ### Show Tags 08 Oct 2016, 20:18 _________________ SPIKE! Re: A certain fruit stand sold apples for$0.70 each and bananas   [#permalink] 08 Oct 2016, 20:18
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