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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 Arithmetic tagged by: BTGmoderatorRO This topic has 2 expert replies and 0 member replies Arithmetic Each of four girls, A,B,C and D, had a few chocolates with her. A first gave 1/3rd of the chocolates with her to B, B gave 1/4th of what she then had to C and C gave 1/5th of what she then had to D. Finally, all the four girls had an equal number of chocolates. If A had 80 more chocolates then B initially, find the difference between the number of chocolates that C and D initially had A) 20 B) 30 C) 15 D) 25 E) 50 OA is A This question could be confusing, do I need a formula to solve this. An Expert is needed on this pls. GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Hello Roland2rule. Let's take a look at your question. Let a, b, c and d the number of chocolates that have A, B, C and D respectively. First, we know that A had 80 more chocolates then B initially, that is to say, $$\left(1\right)\ \ \ \ \ \ a=80+b\ .$$ Now, A gave 1/3rd of the chocolates with her to B, so A at the end had $$\left(2\right)\ \ \ a-\frac{a}{3}=\frac{2a}{3}\ chocolates$$ and B has $$\left(3\right)\ \ b+\frac{a}{3}=\frac{3b+a}{3}\ chocolates.$$ But B gave 1/4th of what she then had to C, this implies that B at the end has $$\left(4\right)\ \ \frac{3b+a}{3}\ -\frac{\frac{3b+a}{3}}{4}=\frac{3b+a}{3}-\frac{3b+a}{12}=\frac{9b+3a}{12}=\frac{3b+a}{4}$$ and C now has $$\left(5\right)\ c+\frac{\frac{3b+a}{3}}{4}=c+\frac{3b+a}{12}.$$ Let's stop here for a moment. We know that all the four girls had an equal number of chocolates at the end. Hence, we have that (2) and (4) are equal, and using (1) we can get $$\frac{2a}{3}=\frac{3b+a}{4}\ \Leftrightarrow\ \ 8a=9b+3a\ \Leftrightarrow\ 5a=9b\ \Leftrightarrow\ 400+5b=9b$$ $$400=4b\ \Leftrightarrow\ \ b=100\ and\ then\ a=180.$$ Now, we can rewrite some of the previous equations as follows: B at the beggining had 100 chocolates, then she had $$100+\frac{a}{3}=100+\frac{180}{3}=160$$ and she gave 1/4 to C, so she gave 40 chocolates to C. Now, C has c+40 and C gave 1/5 to D. Finally, C had $$c+40-\frac{c+40}{5}=\frac{5c+200-c-40}{5}=\frac{4c+160}{5}$$ and it has to be equal to 120. Hence, $$\frac{4c+160}{5}=120\ \Leftrightarrow\ 4c+160=600\ \ \Leftrightarrow\ c=110.$$ Therefore, c+40=150 and C gave 1/5 to D, then she gave 30 chocolates to D. Now, D has d+30 chocolates, and it has to be equal to 120, therefore d=90. In conclusion, the difference between the number of chocolates that C and D initially had is c-d=110-90=20. So, the correct answer is the option A=20. I hope this can help you. I'm available if you'd like a follow-up. Regards. _________________ GMAT Prep From The Economist We offer 70+ point score improvement money back guarantee. Our average student improves 98 points. Free 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. Get started now. GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Hello Roland2rule. Let's take a look at your question. Let a, b, c and d the number of chocolates that have A, B, C and D respectively. First, we know that A had 80 more chocolates then B initially, that is to say, $$\left(1\right)\ \ \ \ \ \ a=80+b\ .$$ Now, A gave 1/3rd of the chocolates with her to B, so A at the end had $$\left(2\right)\ \ \ a-\frac{a}{3}=\frac{2a}{3}\ chocolates$$ and B has $$\left(3\right)\ \ b+\frac{a}{3}=\frac{3b+a}{3}\ chocolates.$$ But B gave 1/4th of what she then had to C, this implies that B at the end has $$\left(4\right)\ \ \frac{3b+a}{3}\ -\frac{\frac{3b+a}{3}}{4}=\frac{3b+a}{3}-\frac{3b+a}{12}=\frac{9b+3a}{12}=\frac{3b+a}{4}$$ and C now has $$\left(5\right)\ c+\frac{\frac{3b+a}{3}}{4}=c+\frac{3b+a}{12}.$$ Let's stop here for a moment. We know that all the four girls had an equal number of chocolates at the end. Hence, we have that (2) and (4) are equal, and using (1) we can get $$\frac{2a}{3}=\frac{3b+a}{4}\ \Leftrightarrow\ \ 8a=9b+3a\ \Leftrightarrow\ 5a=9b\ \Leftrightarrow\ 400+5b=9b$$ $$400=4b\ \Leftrightarrow\ \ b=100\ and\ then\ a=180.$$ Now, we can rewrite some of the previous equations as follows: B at the beggining had 100 chocolates, then she had $$100+\frac{a}{3}=100+\frac{180}{3}=160$$ and she gave 1/4 to C, so she gave 40 chocolates to C. Now, C has c+40 and C gave 1/5 to D. Finally, C had $$c+40-\frac{c+40}{5}=\frac{5c+200-c-40}{5}=\frac{4c+160}{5}$$ and it has to be equal to 120. Hence, $$\frac{4c+160}{5}=120\ \Leftrightarrow\ 4c+160=600\ \ \Leftrightarrow\ c=110.$$ Therefore, c+40=150 and C gave 1/5 to D, then she gave 30 chocolates to D. Now, D has d+30 chocolates, and it has to be equal to 120, therefore d=90. In conclusion, the difference between the number of chocolates that C and D initially had is c-d=110-90=20. So, the correct answer is the option A=20. I hope this can help you. I'm available if you'd like a follow-up. Regards. _________________ GMAT Prep From The Economist We offer 70+ point score improvement money back guarantee. Our average student improves 98 points. Free 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. Get started now. • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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