Julie opened a lemonade stand and sold lemonade in two diff

Thanks! Hmm.. I think I'm missing something pretty obvious here, how do you conclude that there is only one value for M and N and that it has to be M=4, N=5? When I look at the equation I see 1 equation and 2 unknowns so my knee jerk reaction is "not solvable!!"

Well, when question or context gives you certain constraints to help you. (Or actually to trick you )

If we dont know that number of drinks got to be a non-negative integer then surely we can not solve 1 equation with 2 variables. But in this case that is one underlying constraint. So we can simply check if there is anything that satisifies the equation.

Remember this trick for any such context (Number of drinks/ animals/ trees/ votes/ persons etc)

Hope it helps

Can you please explain the statement again ? I got the answer choice wrong. "If we dont know that number of drinks got to be a non-negative integer then surely we can not solve 1 equation with 2 variables. But in this case that is one underlying constraint. So we can simply check if there is anything that satisifies the equation."

Check these posts:
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
common-gmat-trap-31x-25y-128578.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html

Hope it helps.

This is Value type of DS questions in which we should answer if there is only one value or more than one value.
If only one possible value - sufficient
If more than one value - insufficient

We do not need to count this value as we do in PS

In this case we should answer if it is only one possible value of 52-cent lemonade drinks' number

S1. x+y=9, can be 1+8, 2+7, 3+6, 4+5..., so INSUFFICIENT
S2. 52x+58y=492, we have two different prices per drink, so there is always unique number of X (non-negative integer), so SUFFICIENT

B

It took a lot of time for me to arrive at the values.
I agree this is a DS question and solving till the last line is not required as in case of PS.
but still the equation looks quite complex that I felt it may not have a solution at all.
Is there is any way to solve the equation in less time.

Hi Mechmeera,

In situations such as this (when you THINK that you need two variables and two unique equations to answer the given question), it helps to be on the lookout for "weird" numbers and/or low "totals." You also have to be ready to do some 'brute force' work to get the solution.

In this prompt, we're told that the two sizes cost 52 cents and 58 cents. Fact 2 tells us that the TOTAL value of glasses sold was $4.92. Since each size of lemonade sells for OVER 50 cents, and the total is LESS than $5, there must be FEWER than 10 lemonades sold - this leads to a relatively small number of possibilities.

While the work might seem a little tedious, you CAN list out the various 'multiples' of each size and look for an option that totals $4.92

For the 12-ounce lemonade:
.52
1.04
1.56
2.08
2.60
3.12
3.64
4.16
4.68

For the 16-ounce lemonade:
.58
1.16
1.74
2.32
2.90
3.48
4.06
4.64

How many ways are there to add a number from the first group to the number from the second group and get a TOTAL of $4.92 (hint: the units digit is a '2', so look for a pair of values that SUM to that units digit). You'll find that there's just one pairing. Thus, Fact 2 is SUFFICIENT.

GMAT assassins aren't born, they're made,
Rich

VeritasPrepKarishma

how to conclude that there is only one value for M and N for the equation M*29 = 246-N*26

Please help.

Hi techiesam,

If you read my explanation (the post that immediately appears before your post), you'll see that with a little logic - and some 'brute force' arithmetic - you can prove that there's only one solution when you include the information in Fact 2.

GMAT assassins aren't born, they're made,
Rich

Thanks..But the brute force method is time consuming,specially when you are taking the test.Is there any other way!

Check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -of-thumb/

It discusses in detail how to solve equations with integer solutions and also how you can find out the exact number of solutions an equation such as this will have.

Thank you very much Ma'am.This is what I've been looking for.

\(\left\{ \begin{gathered}\\
\,m \geqslant 1\,\,\operatorname{int} \,\,\,12{\text{oz - units}}\,\,,\,\,52\,{\text{cents}}\,{\text{each}} \hfill \\\\
\,n \geqslant 1\,\,\operatorname{int} \,\,\,\,16{\text{oz - units}}\,\,,\,\,58\,{\text{cents}}\,{\text{each}}\,\,\, \hfill \\ \\
\end{gathered} \right.\,\,\,\,\left( * \right)\)

\(? = m\)

\(\left( 1 \right)\,\,m + n = 9\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {1,8} \right)\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {2,7} \right)\,\,\,\, \Rightarrow \,\,\,? = 2\,\, \hfill \\ \\
\end{gathered} \right.\)


Money unit will be CENTS. (All amounts in cents are integers!)

\(\left( 2 \right)\,\,52m + 58n = 492\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,2} \,\,\,\,\,26m + 29n = 246\,\,\,\)

\(\left[ {29n\,\,\mathop = \limits^{\left( * \right)} \,} \right]\,\,{\text{positive}}\,\,{\text{multiple}}\,\,{\text{of}}\,\,29\,\, = \,\,\,246 - 26m = 2\left( {123 - 13m} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{GCF}}\,\left( {2,29} \right)\,\, = \,\,1} \,\,\,123 - 13m\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{positive}}\,\,{\text{multiple}}\,\,{\text{of}}\,\,29\)

\(\left. \begin{gathered}\\
m = 1\,\,\,\, \Rightarrow \,\,\,123 - 13 = 110\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\\\
m = 2\,\,\,\, \Rightarrow \,\,\,123 - 26 = 97\,\,\left[ { = 110 - 13} \right]\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\\\
m = 3\,\,\,\, \Rightarrow \,\,\,123 - 39 = 84\,\,\left[ { = 97 - 13} \right]\,\,\,\,\,\,\left( {{\text{NO}}} \right)\,\,\,\, \hfill \\\\
m = 4\,\,\,\, \Rightarrow \,\,\,84 - 13 = 71\,\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\\\
\boxed{m = 5}\,\,\,\, \Rightarrow \,\,\,71 - 13 = 58 = 2 \cdot 29\,\,\,\,\,\left( {{\text{YES}}} \right)\,\,\,\,\,\,\,\,\,\,\, \hfill \\\\
m = 6\,\,\,\, \Rightarrow \,\,\,58 - 13 = 45\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\\\
m = 7\,\,\,\, \Rightarrow \,\,\,45 - 13 = 32\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\\\
m = 8\,\,\,\, \Rightarrow \,\,\,32 - 13 = 19\,\,\,\,\,\left( {{\text{NO}}} \right) \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = 5\)


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

Regards,
Fabio.

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