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Lemonade Stand

by Osirus@VeritasPrep » Fri May 28, 2010 7:30 am
Julie opened a lemonade stand and sold lemonade in two different sizes, a 52-cent (12oz) size and a 58-cent (16 oz) size. How many 52-cent lemonade drinks did Julie sell?

(1) Julie sold a total of 9 lemonades.

(2) The total value of the lemonade drinks Julie sold was $4.92

OA [spoiler]B I got this answer correct. My question is how can you determine that there is only one value integer value for the 52 cent drinks that would also allow the value for the 58 cent drinks to be an integer, without testing each individual integer cases? The book says that you don't have to calculate each case, you know that there is only one value for the 52 cent drinks that would allow the 58 cent drinks to also be an integer, my question is how do you know this?[/spoiler]
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Source: — Data Sufficiency |
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by Patrick_GMATFix » Fri May 28, 2010 8:08 am
Hello osirus,

Indeed (2) is sufficient. It allows you to write the equation 52x + 58y = 492 --> 26x+29y=246 where x and y are the number of lemonade drinks sold at 52 cents and 58 cents respectively.

In general when you have two variables, you don't have enough info to solve. The key here is that x and y must be integers; this severely limits the range of possible values (actually it limits the range to a unique solution).

Under certain circumstances, you will have sufficiency even though there are two variables. I've discussed these circumstances in two posts in another thread: In short, because 246 in the equation above is smaller than the LCM(26,29) + 26, and because x and y are integers, you can be certain that there is a unique set of values that can work.

Hope that helps,
-Patrick
Last edited by Patrick_GMATFix on Wed Jun 09, 2010 12:10 am, edited 2 times in total.
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by Osirus@VeritasPrep » Fri May 28, 2010 8:21 am
Thanks, I really appreciate your post. That helps a lot.
https://www.beatthegmat.com/the-retake-o ... 51414.html

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by asamaverick » Fri May 28, 2010 8:23 am
Thanks Patrick. You just saved me some precious time with that little concept. I have always tried to plug in numbers for these.
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by Patrick_GMATFix » Fri May 28, 2010 8:51 am
you guys will make me blush if you don't stop.

Seriously, you're welcome.
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by odod » Sat May 29, 2010 8:59 am
Hi Patrick., I'm having some troubles with this sentence - 246 in the equation above is smaller than the LCM(52,58) + 52.

I'm getting the LCM of 52 and 58 to be 2*13*2*29 which is 1508 and is much greater than 246. Where am I going wrong here?
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by odod » Sat May 29, 2010 8:59 am
Hi Patrick., I'm having some troubles with this sentence - 246 in the equation above is smaller than the LCM(52,58) + 52.

I'm getting the LCM of 52 and 58 to be 2*13*2*29 which is 1508 and is much greater than 246. Where am I going wrong here?
ODOD
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by selango » Sat May 29, 2010 9:04 am
odod wrote:Hi Patrick., I'm having some troubles with this sentence - 246 in the equation above is smaller than the LCM(52,58) + 52.

I'm getting the LCM of 52 and 58 to be 2*13*2*29 which is 1508 and is much greater than 246. Where am I going wrong here?
It is LCM(26,29)+52

-->754+52=806 and 246<806.
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by Patrick_GMATFix » Sat May 29, 2010 10:38 am
selango wrote:
odod wrote:Hi Patrick., I'm having some troubles with this sentence - 246 in the equation above is smaller than the LCM(52,58) + 52.

I'm getting the LCM of 52 and 58 to be 2*13*2*29 which is 1508 and is much greater than 246. Where am I going wrong here?
It is LCM(26,29)+52

-->754+52=806 and 246<806.
Selango is right. It was my mistake. I've changed my explanation to LCM(26,29). In a timed environment I wouldn't even find the LCM. 29 is prime, 26 and 29 don't share any factors (other than 1). So the LCM will be 29*26, which is already much greater than 246.
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by GmatGreen » Tue Jun 08, 2010 8:08 pm
Patrick_GMATFix wrote:Hello osirus,

Indeed (2) is sufficient. It allows you to write the equation 52x + 58y = 492 --> 26x+29y=246 where x and y are the number of lemonade drinks sold at 52 cents and 58 cents respectively.

In general when you have two variables, you don't have enough info to solve. The key here is that x and y must be integers; this severely limits the range of possible values (actually it limits the range to a unique solution).

Under certain circumstances, you will have sufficiency even though there are two variables. I've discussed these circumstances in two posts in another thread: In short, because 246 in the equation above is smaller than the LCM(26,29) + 52, and because x and y are integers, you can be certain that there is a unique set of values that can work.

Hope that helps,
-Patrick
This was extremely helpful. Beats trying to plugin and never being sure about whether there is sufficiency :-)
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by imhimanshu » Tue Jun 08, 2010 10:46 pm
Patrick_GMATFix wrote:
selango wrote:
odod wrote:Hi Patrick., I'm having some troubles with this sentence - 246 in the equation above is smaller than the LCM(52,58) + 52.

I'm getting the LCM of 52 and 58 to be 2*13*2*29 which is 1508 and is much greater than 246. Where am I going wrong here?
It is LCM(26,29)+52

-->754+52=806 and 246<806.
Selango is right. It was my mistake. I've changed my explanation to LCM(26,29). In a timed environment I wouldn't even find the LCM. 29 is prime, 26 and 29 don't share any factors (other than 1). So the LCM will be 29*26, which is already much greater than 246.
Will it be (LCM(26,29)+52) or LCM(26,29)+26.

Pls clarify !
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by Patrick_GMATFix » Wed Jun 09, 2010 12:13 am
My apologies. I was being careless. Thank you for pointing it out. In fact, I meant LCM(26,29)+26. If this value is greater than the sum, then we'll have a unique solution. Otherwise, it depends.

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by singhpreet1 » Wed Jun 09, 2010 2:42 am
the only answer i can get out of this is E. would someone care to take this up for me and explain further.

THANKS.
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by Patrick_GMATFix » Thu Jun 10, 2010 11:04 am
singhpreet1 wrote:the only answer i can get out of this is E. would someone care to take this up for me and explain further.

THANKS.
Explain, in detail, your reasoning for E and someone will help point out exactly where you are mistaken.
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by singhpreet1 » Thu Jun 10, 2010 7:48 pm
osirus0830 wrote:Julie opened a lemonade stand and sold lemonade in two different sizes, a 52-cent (12oz) size and a 58-cent (16 oz) size. How many 52-cent lemonade drinks did Julie sell?

(1) Julie sold a total of 9 lemonades.

(2) The total value of the lemonade drinks Julie sold was $4.92

OA [spoiler]B I got this answer correct. My question is how can you determine that there is only one value integer value for the 52 cent drinks that would also allow the value for the 58 cent drinks to be an integer, without testing each individual integer cases? The book says that you don't have to calculate each case, you know that there is only one value for the 52 cent drinks that would allow the 58 cent drinks to also be an integer, my question is how do you know this?[/spoiler]

oops sorry i got it now, i missed out that the price was provided for each of the bottles and the sizes as well. so St.2 is sufficient independently to answer the question. .52x+. 58y= 4.92 i hope i got right this time around!

thanks.
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