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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 ## The rate of a certain chemical reaction is directly tagged by: AAPL ##### This topic has 3 expert replies and 0 member replies ### Top Member ## The rate of a certain chemical reaction is directly ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult GMAT Prep The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase OA D. ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1295 messages Followed by: 29 members Upvotes: 59 Top Reply AAPL wrote: GMAT Prep The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase $rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}$ $\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)$ $?\,\, \cong \,\,k - 1$ ${\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered} {k^2} = 2 \hfill \\ \sqrt 2 \cong 1.41 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2 - 1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2 - 1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)$ This solution follows the notations and rationale taught in the GMATH method. Regards, fskilnik. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15105 messages Followed by: 1859 members Upvotes: 13060 GMAT Score: 790 AAPL wrote: GMAT Prep The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase The trick is translate the chemical relationship into the following equation: R = A²/B Here's why the equation above works: 1)The problem states that R is directly proportional to A². Directly proportional means that as one value increases, the other value also increases by a proportionate amount. In the equation above, if we increase R, we'll have to increase A² by a proportionate amount in order for the equation to remain valid. 2) The problem states that R is inversely proportional to B. Inversely proportional means that as one value increases, the other value decreases by a proportionate amount. In the equation above, if we increase R, we'll have to decrease B by a proportionate amount in order for the equation to remain valid. Now let's plug in values. Let A = 10 and B = 2. R = 10²/2 = 100/2 = 50. If we increase B by 100%, new B = 4. R = 50 must be unchanged. 50 = A²/4 A² = 200 New A = √200 = 10√2 ≈ 14 % change in A = Difference/(Original A) * 100 = (14-10)/10 * 100 = 4/10 * 100 = 40%. The correct answer is D. _________________ 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 25 Apr 2015 Posted: 2012 messages Followed by: 15 members Upvotes: 43 AAPL wrote: GMAT Prep The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase We can let n = the rate of a certain chemical reaction, a = the concentration of chemical A, and b = the concentration of chemical B. We are given that the rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present, so, for some positive constant k, we have: n = ka^2/b When b is increased by 100 percent, b becomes 2b. To keep the reaction rate unchanged, we can let a become c, so we have: ka^2/b = kc^2/(2b) 2bka^2 = bkc^2 2a^2 = c^2 c = √(2a^2) c = a√2 Since √2 ≈ 1.4, c ≈ 1.4a or approximately 140% of a, i.e., a 40% increase in the concentration of chemical A. Answer: D _________________ Scott Woodbury-Stewart Founder and CEO • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE 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 • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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