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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Julie opened a lemonade stand and sold lemonade in two ##### This topic has 4 expert replies and 0 member replies ### Top Member ## Julie opened a lemonade stand and sold lemonade in two ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Julie opened a lemonade stand and sold lemonade in two different sizes, a 52-cent (12oz) and a 58-cent (16oz) size. How many 52-cent (12oz) 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 B

Source: Veritas Prep

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This question illustrates a common trap on the GMAT.

For statement 2, we're able to write the equation 52x + 58y = 492 , and in high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables.

Here's another question that exploits this common misconception: https://www.beatthegmat.com/stamps-t288085.html

Cheers,
Brent

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BTGmoderatorDC wrote:
Julie opened a lemonade stand and sold lemonade in two different sizes, a 52-cent (12oz) and a 58-cent (16oz) size. How many 52-cent (12oz) 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 Let X = the number of 52Â¢ lemonades and Y = the number of 58Â¢ lemonades. Since the question stem asks for the number of 52Â¢ lemonades, we get: What is the value of X? Statement 1: X+Y = 9 Since X can be any value between 0 and 9, inclusive, INSUFFICIENT. Statement 2: 52X + 58Y = 492 Since the two statements cannot contradict each other, an integral solution for the equation above must be yielded when X+Y=9. If a total of 9 lemonades are sold, the equation above implies the following: Average cost per lemonade = 492/9 â‰ˆ 54.66. Since the average cost is just a bit closer to 52 than to 58, the number of 52Â¢ lemonades must be a just bit greater than the number of 58Â¢ lemonades, implying that X=5 and Y=4: 5*52 + 4*58 = 260 + 232 = 492. In the case above, X:Y = 5:4. Check whether OTHER revenue ratios are also possible. Since the lemonade values are 52Â¢ and 58Â¢, the revenue ratio can be altered only by adding a multiple of 52 and 58 to X or Y, while subtracting this multiple from the other variable. Since 52 = 2*26 and 58 = 2*29, the LCM of 52 and 58 = 2*26*29. If 2*26*29 is added to either 52Â¢ or 58Â¢, the sum will exceed 492Â¢. Thus, the revenue ratio CANNOT be altered, implying that only ONE revenue ratio will satisfy statement 2: X=5 and Y=4. Thus, X=5. SUFFICIENT. The correct answer is B. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 BTGmoderatorDC wrote: Julie opened a lemonade stand and sold lemonade in two different sizes, a 52-cent (12oz) and a 58-cent (16oz) size. How many 52-cent (12oz) 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
Source: Veritas Prep
$$\left\{ \matrix{ \,m \ge 1\,\,{\mathop{\rm int}} \,\,\,12{\rm{oz - units}}\,\,,\,\,52\,{\rm{cents}}\,{\rm{each}} \hfill \cr \,n \ge 1\,\,{\mathop{\rm int}} \,\,\,\,16{\rm{oz - units}}\,\,,\,\,58\,{\rm{cents}}\,{\rm{each}}\,\,\, \hfill \cr} \right.\,\,\,\,\left( * \right)$$
$$? = m$$
$$\left( 1 \right)\,\,m + n = 9\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {1,8} \right)\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \cr \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2,7} \right)\,\,\,\, \Rightarrow \,\,\,? = 2\,\, \hfill \cr} \right.$$

Money unit will be CENTS. (All amounts in cents are integers!)
$$\left( 2 \right)\,\,52m + 58n = 492\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,2} \,\,\,\,\,26m + 29n = 246\,\,\,$$
$$\left[ {29n\,\,\mathop = \limits^{\left( * \right)} \,} \right]\,\,{\rm{positive}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,29\,\, = \,\,\,246 - 26m = 2\left( {123 - 13m} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{GCF}}\,\left( {2,29} \right)\,\, = \,\,1} \,\,\,123 - 13m\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,29$$
$\left. \begin{gathered} m = 1\,\,\,\, \Rightarrow \,\,\,123 - 13 = 110\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ m = 2\,\,\,\, \Rightarrow \,\,\,123 - 26 = 97\,\,\left[ { = 110 - 13} \right]\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ m = 3\,\,\,\, \Rightarrow \,\,\,123 - 39 = 84\,\,\left[ { = 97 - 13} \right]\,\,\,\,\,\,\left( {{\text{NO}}} \right)\,\,\,\, \hfill \\ m = 4\,\,\,\, \Rightarrow \,\,\,84 - 13 = 71\,\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ \boxed{m = 5}\,\,\,\, \Rightarrow \,\,\,71 - 13 = 58 = 2 \cdot 29\,\,\,\,\,\left( {{\text{YES}}} \right)\,\,\,\,\,\,\,\,\,\,\, \hfill \\ m = 6\,\,\,\, \Rightarrow \,\,\,58 - 13 = 45\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ m = 7\,\,\,\, \Rightarrow \,\,\,45 - 13 = 32\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ m = 8\,\,\,\, \Rightarrow \,\,\,32 - 13 = 19\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ \end{gathered} \right\}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = 5$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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BTGmoderatorDC wrote:
Julie opened a lemonade stand and sold lemonade in two different sizes, a 52-cent (12oz) and a 58-cent (16oz) size. How many 52-cent (12oz) 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 B Source: Veritas Prep Given: Julie sells lemonade in two different sizes, a 52-cent (12oz) and a 58-cent (16oz) size Question: How many 52-cent (12oz) lemonade drinks did Julie sell? Let's take each statement one by one. (1) Julie sold a total of 9 lemonades. Certainly insufficient. (2) The total value of the lemonade drinks Julie sold was$4.92.

Say, the number of 12oz lemonades sold is x and the number of 16oz lemonades sold is y.

Then, 52x + 58y = 492

=> 26x + 29y = 246

x = (246 - 29y)/26

= 9 + (12 - 29y)/26

= 9 + (12 - 26y - 3y)/26

= 9 + (12 - 3y)/26 - 26y/26

= 9 + (12 - 3y)/26 - y

Since x is a positive integer, (12 - 3y) must be a multiple of 26.

Let's try some values for y to make (12 - 3y) must be a multiple of 26.

@y = 4, we have (12 - 3y) => 12 - 3*4 = 12 - 12 = 0 -- a multiple of 26.

Thus, x = 9 + 0 - 4 = 5.

We must try a few more values for y so that (12 - 3y) must be a multiple of 26 and x is a positive integer.

Note that at higher values of y, the term [(12 - 3y)/26 - y] would be negative, making x a negative integer, which is an invalid value. You'll find that there is no another possible valid value of y. Thus, y = 4 and y = 5. Sufficient.

Hope this helps!

-Jay
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