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A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
30 Mar 2009, 19:16

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Question Stats:

70% (02:36) correct
30% (01:54) wrong based on 93 sessions

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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?

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
30 Mar 2009, 20:34

B. 11? I got the answer somehow:) Someone can show a proper way. 0.7*a + 0.50*b = 6.30.

I listed the multiples of 7 and 5. After adding the last two digits of these two multiples the sum should have 3 as the last digit. so only 7*4 = 28 and some other multiple of 5 ( 8+5 = 13 , last digit 3 => for 6.30 ) is possible. So 4 apples are fixed Hence, we get 4 + 7 = 11.

mrsmarthi 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?

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
31 Mar 2009, 13:56

B

This is how I approached it..The first thing I noticed is that 6.3 is a multiple of 0.7 so if the customer bought all apples he or she would get 9 of them. Then I listed the multiples of 0.7 like below...I just listed the multiple of 7 up to 63. Now I looked at the difference for each multiple and the last multiple where the difference is a multiple of 5

7 14 21 28 35 42 49 56 63

The only place that happens is for 28 where 63 - 28 = 35 which is a multiple of 5, hence 28 contributes for apples and 35 contributes for bananas leading to 4 apples (28/7) and 7 bananas (35/5) to a total of 11.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
20 Aug 2010, 09:37

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0.7a+0.5b = 6.3

7a+5b= 63

Now we know that the unit's place of b has to be 5 or 0 (since it is a multiple of 5). So, 7*a has to have a unit's place of 8 or 3 to get a total sum of 63

One possibility is a=9, b=0 But since the questions states that he bought BOTH, and since 9 is not in answer choice, this is ruled out.

The next possibility is trying to find a multiple of 7 that has unit's place = 8 ; 28 a=4 7*4 = 28 28 + 5*7 = 63

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
20 Aug 2010, 10:17

First you see this problem you should guess that from 50 and 70 you need to get something that ends with 80 and 50 and first numbers that come to hat are 350 and 280 therefore 4*70 and 7*50

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
21 Aug 2010, 12:41

1

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Expert's post

You can use divisibility properties here to get to the answer. The idea is this: if you see an equation like the following, and you know that r and s are positive integers:

5r = 7s

this means that 5r and 7s are exactly the same number, so they must of course have the same divisors. So, since 7 is a divisor of 7s, it must also be a divisor of 5r, and therefore of r. Similarly s must be divisible by 5. That is, when you have equations involving only integers, the primes which divide the left side of the equation must divide the right, and vice versa.

So onto this question. We know:

0.7a + 0.5b = 6.3 5b = 63 - 7a

and since 7 is a factor of the right side of this equation, it must be a factor of the left, so b must be a multiple of 7. Since b is greater than 0, and b cannot be 14 (then the total cost would be greater than $6.30), b must be 7, from which we can find that a is 4. _________________

GMAT Tutor in Toronto

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Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
27 Sep 2010, 10:50

This is a stupid question, but why is this not solvable by algebra? When you do the algebra you get a non integer number. I understand, two variables and one equation means you cant solve, but we are looking for a+b so when you remove the numbers next to a and b, you have a final value a+b=x.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
27 Sep 2010, 14:49

I solved by back tracking: Main Equation: 7a + 5b = 63 To find: a + b. As per the answers: if a+ b >12 ... then the main equation is not satisfied (i.e if a + b = 13 then 5(a + b) = 65 ... which is greater than 63 and hance impossible ) So the answer is 10, 11 or 12. Now similarly using 10 or 11 the answer is in fraction which cannot be the case( as we are dealing with whole fruits ) ... hence answer is 11 or B _________________

Please give me kudos, if you like the above post. Thanks.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
29 Oct 2010, 22:55

2

This post received KUDOS

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

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
03 Nov 2010, 12:08

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 ...

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
03 Nov 2010, 19:17

Expert's post

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. _________________

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
24 May 2011, 11:09

Hm. I like others, was hoping there was a "faster" way of doing this problem. Seems like at a minimum it would take about 2 mins to 2:30 to crunch those numbers.

Re: A certain fruit stand sold apples for $0.70 each and bananas [#permalink]
24 May 2011, 16:16

There are several ways to solve these kind of problems . You can try this approach

5B = 63-7A

=> B = \((7*(9-A))/5\)

=> 9-A must be a multiple of 5 => 9-A can be 0 , 5 ,10 , 15...

=> A-9 will be 0, -5, -10,-15...respectively

=> A will be 9, 4, -1, -6.... respectively

=> As A cannot be -ve , only possible options are 9 or 4

=> If A =9 then B becomes 0 which is not valid here. Hence A=4,B=7

=> A+B =11

Answer is B.

humphy wrote:

Hm. I like others, was hoping there was a "faster" way of doing this problem. Seems like at a minimum it would take about 2 mins to 2:30 to crunch those numbers.

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