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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 ## If every boy in a kindergarten class buys a soda and every g tagged by: alanforde800Maximus ##### This topic has 3 expert replies and 0 member replies ## If every boy in a kindergarten class buys a soda and every g ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class? A. 1 B. 3 C. 4 D. 12 E. Cannot be uniquely determined ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12655 messages Followed by: 1245 members Upvotes: 5254 GMAT Score: 770 alanforde800Maximus wrote: If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class? A. 1 B. 3 C. 4 D. 12 E. Cannot be uniquely determined Let B = # of boys in kindergarten Let G = # of girls in kindergarten Let J = price (in CENTS) of ONE juice box Let S = price (in CENTS) of ONE soda If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. Let scenario 1 = boys buy sodas and girls buy juice boxes Let scenario 2 = boys buy juice boxes and girls buy sodas We can start with the following "word equation": (total cost of scenario 1) = (total cost of scenario 2) - 1 So, we get: BS + GJ = BJ + GS - 1 It would be nice if we could factor either side of the equation. So, let's move some pieces around. Subtract GS from both sides: BS - GS + GJ = BJ - 1 Subtract GJ from both sides: BS - GS = BJ - GJ - 1 Factor each side: S(B - G) = J(B - G) - 1 Subtract J(B - G) from both sides: S(B - G) - J(B - G) = -1 Simplify: (S - J)(B - G) = -1 IMPORTANT: Since B and G are both POSITIVE INTEGERS, we know that (B - G) is an INTEGER Likewise, we know that S and J are both POSITIVE INTEGERS, since the prices cannot be less than one cent. This means (S - J) is an INTEGER Since (S - J)(B - G) = -1, we know that one value, (S - J) or (B - G), must be 1 and the other value must be -1 There are more boys than girls in the class So, B - G must be positive. This means B - G must equal 1 What is the difference between the number of boys and the number of girls in the class? B - G = 1 Answer: A Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10071 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Hi alanforde800Maximus, This is a thick, layered question, and would likely take most Test Takers more time than average to solve correctly. The key to solving it is to realize that we don't know the prices of each soda and each juice box - they MIGHT be integers, but they MIGHT NOT. Also, we don't know the relative prices (so one might be more expensive than the other, or vice-versa). From the given prompt, we have 4 variables: B = The number of boys G = The number of girls S = The price of 1 soda J = The price of 1 juice box From the prompt, we can create just 1 equation: (B)(S) + (G)(J) = (B)(J) + (G)(S) - 1 Here's how we can TEST VALUES to prove that there's more than one answer. Since this IS such a thick question, the key to doing the work quickly is to keep the values SMALL. We do have a couple of 'restrictions' that we have to work with: 1) B and G are both INTEGERS (since you cannot have a 'fraction' of a boy or girl) 2) We're told that there are MORE boys than girls, so B > G IF.... B=2 G=1 S=1 J=2 (2)(1) + (1)(2) = (2)(2) + (1)(1) - 1 2 + 2 = 4 + 1 - 1 4 = 4 Here, we have 2 boys and 1 girl, so the difference is 1. In the above example, both S and J are INTEGERS and S < J. What happens if we make one of those variables a fraction...... IF.... B=3 G=1 S=1/2 J=1 (3)(1/2) + (1)(1) = (3)(1) + (1)(1/2) - 1 1.5 + 1 = 3 + 0.5 - 1 2.5 = 2.5 Here, we have 3 boys and 1 girl, so the difference is 2. Thus, there's no exact answer.... Final Answer: E GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1282 messages Followed by: 29 members Upvotes: 59 alanforde800Maximus wrote: If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class? A. 1 B. 3 C. 4 D. 12 E. Cannot be uniquely determined $$\left. \matrix{ g\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{girls}} \hfill \cr b\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{boys}}\,\,{\rm{ = }}\,\,\,g + k\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,k\,\,\,\,,\,\,\,k \ge 1\,\,\,{\mathop{\rm int}} \,\,\,\,\,\left( {b > g} \right)$$ $$\left. \matrix{ s\,\, = \,\,\,{\rm{one}}\,\,{\rm{soda}} \hfill \cr j\,\, = \,\,\,{\rm{one}}\,\,{\rm{juce}}\,\, \hfill \cr} \right\}\,\,\,\,\,{\rm{cost}}\,\,\,\left( {{\rm{in}}\,\,{\rm{cents}}} \right)$$ $$g,j,k,s\,\,\,\, \ge \,\,\,1\,\,\,{\rm{ints}}\,\,\,\,\left( * \right)$$ $$\left[ {\left( {g + k} \right)\,j\,\, + \,g\,s} \right]\,\, - \,\,\,\left[ {\left( {g + k} \right)\,s\,\, + \,g\,j} \right]\,\,\, = 1\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{cents}}\,} \right]$$ $$k\left( {j - s} \right) = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,k\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,1\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = k = 1$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br • Award-winning private GMAT tutoring Register now and save up to$200

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