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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 ## A box of balls originally contained 2 blue balls for every r tagged by: M7MBA ##### This topic has 3 expert replies and 0 member replies ### Top Member ## A box of balls originally contained 2 blue balls for every r A box of balls originally contained 2 blue balls for every red ball. If after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2, how many balls were in the box before the additional 12 balls were added? (A) 9 (B) 12 (C) 15 (D) 18 (E) 24 The OA is the option A. Experts, may you help me here? I don't know how to start to solve this PS question. I'd appreciate your help. ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10197 messages Followed by: 497 members Upvotes: 2867 GMAT Score: 800 Hi MY7MBA, We're told that a box of balls originally contained 2 blue balls for every red ball and that after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2. We're asked for the TOTAL number of balls in the box BEFORE the additional 12 balls were added. This question can be solved by TESTing THE ANSWERS. Let's TEST Answer B: 12 total balls Since we have 2 blue balls for every 1 red ball, that means that 2/3 of the balls are blue. Starting Blue = (2/3)(12) = 8 Starting Red = 12 - 8 = 4 Adding 12 red balls gives us.... Blue = 12 Red = 4 + 12 + 16 Red:Blue = 16:12 = 4:3 This does NOT match what we were told (the ratio is supposed to be 5:2). To increase this ratio, we need the 'impact' of adding 12 red balls to be greater - which means that we need FEWER red balls to start. There's only one answer that does that, but here's the proof that it's correct: Let's TEST Answer A: 9 total balls Since we have 2 blue balls for every 1 red ball, that means that 2/3 of the balls are blue. Starting Blue = (2/3)(9) = 6 Starting Red = 9 - 6 = 3 Adding 12 red balls gives us.... Blue = 6 Red = 3 + 12 + 15 Red:Blue = 15:6 = 5:2 This is an exact match for what we were told, so this MUST be the answer. Final Answer: A GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Quote: A box of balls originally contained 2 blue balls for every red ball. If after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2, how many balls were in the box before the additional 12 balls were added? (A) 9 (B) 12 (C) 15 (D) 18 (E) 24 The OA is the option A. Experts, may you help me here? I don't know how to start to solve this PS question. I'd appreciate your help. Hi M7MBA, Let's take a look at your question. Let b represents blue balls, r represents red balls and x represents the total balls originally in the box. A box of balls originally contained 2 blue balls for every red ball can be represented as: $$r+2b=x\ ...\ \left(i\right)$$ After 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2 $$5r+2b=x+12\ ...\ \left(ii\right)$$ Subtracting eq(i) from (ii): $$5r+2b-\left(r+2b\right)=x+12-x\$$ $$5r+2b-r-2b=12$$ $$5r-r=12$$ $$4r=12$$ $$r=3$$ Therefore, there are 3 red ball in the box originally. Since, the box of balls originally contained 2 blue balls for every red ball. Therefore, there will be 6 blue balls. Total balls in the box originally = 3 + 6 = 9 Therefore, Option A is correct. Hope it helps. I am available if you'd like any follow up. _________________ 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 09 Apr 2015 Posted: 1465 messages Followed by: 19 members Upvotes: 39 M7MBA wrote: A box of balls originally contained 2 blue balls for every red ball. If after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2, how many balls were in the box before the additional 12 balls were added? (A) 9 (B) 12 (C) 15 (D) 18 (E) 24 The ratio of red : blue = x : 2x. We can create the following equation: (x + 12)/2x = 5/2 2(x + 12) = 10x 2x + 24 = 10x 24 = 8x 3 = x Thus, there were originally 3 + 6 = 9 balls in the box. 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