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

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Mar 2012, 23:43, edited 1 time in total.
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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.
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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: help me to solve it..... [#permalink]

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New post 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 \(b=7\) --> \(a+b=11\).

Answer: B.


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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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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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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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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New post 11 Dec 2010, 08:20
haha crystal as usual.
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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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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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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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New post 03 Jun 2012, 07:09
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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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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

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New post 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]

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

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New post 21 Dec 2014, 13:38
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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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]

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

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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

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

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Re: A certain fruit stand sold apples for $0.70 each and bananas   [#permalink] 16 Jan 2016, 21:15
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