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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 [#permalink] New post 30 Sep 2010, 04:08
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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, 22:43, edited 1 time in total.
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Re: help me to solve it..... [#permalink] New post 30 Sep 2010, 04:22
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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


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: help me to solve it..... [#permalink] New post 30 Sep 2010, 18:49
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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.
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Re: help me to solve it..... [#permalink] New post 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...
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Re: help me to solve it..... [#permalink] New post 01 Oct 2010, 04:49
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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] New post 10 Dec 2010, 06:42
Thanks for the explanation
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Re: help me to solve it..... [#permalink] New post 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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Re: help me to solve it..... [#permalink] New post 10 Dec 2010, 23:01
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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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Re: help me to solve it..... [#permalink] New post 11 Dec 2010, 07:20
haha crystal as usual.
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Re: A certain fruit stand sold apples for $0.70 [#permalink] New post 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
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Re: Alternate approaches to OG 12 PS #65? [#permalink] New post 21 Dec 2010, 15:20
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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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Re: help me to solve it..... [#permalink] New post 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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Fruit Stand [#permalink] New post 29 Apr 2012, 01:09
A certain fruit stand sold apples for $ 0.7 each and bananas for $0.5 each. If a customer purchased both apples and bananas from the stand for a total of $6.3, what total number of apples and bananas did the customer purchase?

A) 10

B) 11

C) 12

D) 13

E) 14

Relatively easy question, but I am struggling to solve.

Let say x be the number of apples and y be the number of bananas

0.7x + 0.5y = 6.3 -------------------------(1)

But how can this be solved if we don't have any other data in the question?
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Re: Fruit Stand [#permalink] New post 29 Apr 2012, 02:24
Expert's post
enigma123 wrote:
A certain fruit stand sold apples for $ 0.7 each and bananas for $0.5 each. If a customer purchased both apples and bananas from the stand for a total of $6.3, what total number of apples and bananas did the customer purchase?

A) 10

B) 11

C) 12

D) 13

E) 14

Relatively easy question, but I am struggling to solve.

Let say x be the number of apples and y be the number of bananas

0.7x + 0.5y = 6.3 -------------------------(1)

But how can this be solved if we don't have any other data in the question?


Merging similar topics. Please ask if anything remains unclear.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: help me to solve it..... [#permalink] New post 03 Jun 2012, 06:09
Expert's post
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] New post 12 Sep 2013, 09:08
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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink] New post 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.
Re: A certain fruit stand sold apples for $0.70 each and bananas   [#permalink] 17 Dec 2013, 02:46
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