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

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A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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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.
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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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New post 30 Sep 2010, 05:22
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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 \(b=7\) --> \(a+b=11\).

Answer: B.
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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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New post 30 Sep 2010, 19:49
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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.
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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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New post 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 \(b=7\) --> \(a+b=11\).

Answer: B.


thank you, but can you explain me how this (9,0) and (4,7) to be solve...
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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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New post 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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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 29 Oct 2010, 23:55
8
hI ALL,

this is one of the first good questions i encountered.. here is what i did:

7a + 5b = 63
2a+5a+5b=63
2a+5(a+b)=63
i started substituting the answer choices
lets start wit 12
2a+5(12)=63 => 2a=3 .. a has to be an integer.. so no
2a+5(11)=63 => 2a=8 => a = 4 good.. but still need to check 13
2a+5(10)=63 => 2a=13 .. a has to be an integer.. so no
2a+5(13)=63 => 2a=-2 ==> a had to be +ve .. so no good

so 11 is the answer
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 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
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 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.
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 04 Nov 2010, 08:04
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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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New post 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?
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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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New post 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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New post 12 Dec 2010, 05:30
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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
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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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New post 21 Dec 2010, 16:20
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17
For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.
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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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New post 21 Dec 2010, 16:58
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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: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 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)
Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 15 Oct 2012, 08:39
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7A + 5B = 63

5B = 63 - 7A

B= (63 - 7A)/5 => must be an integer, as we are dealing with quantities unless we can buy 1/4 of an apple or 1/3 of a banana (That could be funny :D )

so 63 - 7A must yield to a multiple of 5.

Now, let's pick some numbers.

if, A= 1, 63-7= 56 (not divisible by 5)
A= 2, 63-14= 49 (not divisible by 5)
A= 3, 63-21= 42 (not divisible by 5)
A= 4, 63-28= 35 (divisible by 5)

so A= 4, B= 7
A+B= 11 , Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 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

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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 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

Answer B
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 22 Aug 2014, 10:39
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!
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Re: A certain fruit stand sold apples for $0.70 each and bananas  [#permalink]

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New post 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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Re: A certain fruit stand sold apples for $0.70 each and bananas   [#permalink] 24 Aug 2014, 22:50

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