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by jscpba » Wed Jan 12, 2011 11:06 am
Another question for you guys. I need a clear explanation for this one. I don't follow the OG's steps.

A certain fruit stand sells apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stanf 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


Thanks guys
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Source: — Problem Solving |

by anshumishra » Wed Jan 12, 2011 11:19 am
jscpba wrote:Another question for you guys. I need a clear explanation for this one. I don't follow the OG's steps.

A certain fruit stand sells apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stanf 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


Thanks guys
x -> apples -> each costs 0.7$
y-> bananas -> each costs 0.5$

0.7x+0.5y = 6.3 (x and y has to be integer)
There are two integral solutions : x=9,y=0 OR x=4,y=7

{Note : If you have problem how to reach to these solutions -- Start with x=1 to x~=6.3/0.7=9 and y~=6.3/0.5=11 to y=1. You have to check only for integral solution. That way you would get a single solution x=4,y=7 only.}

Since it is given that both bananas and apples have been purchased, first solution is not valid.
So, the solution is : x+y = 4+7 = 11 B
Thanks
Anshu

(Every mistake is a lesson learned )
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by GMATGuruNY » Wed Jan 12, 2011 1:07 pm
jscpba wrote:Another question for you guys. I need a clear explanation for this one. I don't follow the OG's steps.

A certain fruit stand sells apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stanf 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


Thanks guys
An efficient approach would be to list the multiples of 70 and 50:

70, 140, 210, 280, 350, 420, 490, 560, 630.
50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600.

Now look for a combination whose sum is 630.

280 + 350 = 630.

280/70 = 4.
350/50 = 7.

Thus, the number purchased was 4+7 = 11.

The correct answer is B.
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by jscpba » Wed Jan 12, 2011 1:35 pm
Thank you to both of you. These are both clear explanations. The second seems to be the one that would be most efficient time wise.
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