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A certain fruit stand sold apples for $0.70 each and bananas for$0.50

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Joined: 02 Sep 2009
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A certain fruit stand sold apples for $0.70 each and bananas for$0.50  [#permalink]

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Updated on: 20 Feb 2019, 04:26
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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) 14 _________________ Originally posted by Bunuel on 30 Sep 2010, 05:08. Last edited by Bunuel on 20 Feb 2019, 04:26, edited 2 times in total. Updated. Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 59134 Re: A certain fruit stand sold apples for$0.70 each and bananas for $0.50 [#permalink] Show Tags 30 Sep 2010, 05:22 9 18 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: A certain fruit stand sold apples for $0.70 each and bananas for$0.50  [#permalink]

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30 Sep 2010, 19:49
26
1
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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 General Discussion Manager Joined: 22 Sep 2010 Posts: 67 Re: A certain fruit stand sold apples for$0.70 each and bananas for $0.50 [#permalink] Show Tags 01 Oct 2010, 05:06 1 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: 59134
Re: A certain fruit stand sold apples for $0.70 each and bananas for$0.50  [#permalink]

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01 Oct 2010, 05:49
10
9
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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03 Nov 2010, 13:08
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: 9789 Location: Pune, India Re: A certain fruit stand sold apples for$0.70 each and bananas  [#permalink]

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03 Nov 2010, 20:17
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 Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9789 Location: Pune, India Re: A certain fruit stand sold apples for$0.70 each and bananas  [#permalink]

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04 Nov 2010, 08:04
1
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...
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Re: A certain fruit stand sold apples for $0.70 each and bananas for$0.50  [#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: 59134 Re: A certain fruit stand sold apples for$0.70 each and bananas for $0.50 [#permalink] Show Tags 11 Dec 2010, 00:01 4 2 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: A certain fruit stand sold apples for $0.70 each and bananas for$0.50  [#permalink]

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12 Dec 2010, 05:30
1
5
ajit257 wrote:
A certain fruit stand sold apples for $0.70 each and bananas for$0.50 each. If a customer

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15 Oct 2012, 04:51
7
2
Let the no of apples sold = a
no of bananas sold = b
Question is a+b=?
Thus 70a + 50b = 630
7a + 5b = 63
Quick Tip- In order to find out the value of a & b its better to find the value of 'a' as "63- 7a" must leave a number which will end either with 0 or
with 5. (Think about it for a second)
Thus the only value which satisfies above equation is a=4 & b=7
a+b=11 (Other values of a & b will lie outside the answer choices)
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15 Oct 2012, 10:50
4
To solve this question -
we can take numbers, as price of apple 7, 5 for Banana and 63 total for ease.
Now we can determine quickly that total number should range between 63/7 <= N <=63/5, so ans should be between 9 and 12.

Now solving the expression
7A+5B =63

first possibility with 9 apples, 0 banana we get 6.30 total amount, but question says customer purchased both, apple and banana. so not correct.

So next choice, for 7A+5B =63 would come by decreasing 63 in multiple of 5 and checking divisibility of that number by 7. this way we get
4 Apples *0.70 + 7 banana *050 = 6.30

Hence total number is 7+4 =11

Ans B

Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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 Practice Questions Question: 64 Page: 161 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project Each week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a solution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! _________________ Lets Kudos!!! Black Friday Debrief Manager Joined: 21 Sep 2012 Posts: 188 Re: A certain fruit stand sold apples for$0.70 each and bananas  [#permalink]

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16 Oct 2012, 03:53
5
1
so apples is 0.70 * A

Bananas 0.50 * B

then 0.70A + 0.50B = 6.30

multiply by 10 we get

7A + 5B = 63

5B = 63 - 7A
B = 7(9-A)/5

now to satisfy this equation we need 9 - A = 5 only then it will be divisible by 5
therefore A is 4 and when solve we get B is 7

7(9-4)/5 = 7*5/5 then we need the sum of A + B = 7 + 4 = 11

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24 Aug 2014, 22:50
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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