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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
30 Sep 2010, 04:08

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79% (02:23) correct
21% (01:52) wrong based on 241 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 ?

Re: help me to solve it..... [#permalink]
30 Sep 2010, 04:22

1

This post received KUDOS

Expert's post

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.

Re: help me to solve it..... [#permalink]
30 Sep 2010, 18:49

4

This post received KUDOS

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]
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...

Re: help me to solve it..... [#permalink]
01 Oct 2010, 04:49

1

This post received KUDOS

Expert's post

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.

Re: help me to solve it..... [#permalink]
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?

Re: help me to solve it..... [#permalink]
10 Dec 2010, 23:01

1

This post received KUDOS

Expert's post

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.

Re: A certain fruit stand sold apples for $0.70 [#permalink]
12 Dec 2010, 04:30

1

This post received KUDOS

Expert's post

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 ?

Re: help me to solve it..... [#permalink]
21 Dec 2010, 15:58

3

This post received KUDOS

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. _________________

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? _________________

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. _________________

Re: help me to solve it..... [#permalink]
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 _________________

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
12 Sep 2013, 09:08

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

gmatclubot

Re: A certain fruit stand sold apples for $0.70 each and bananas
[#permalink]
17 Dec 2013, 02:46