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Re: The rate of a certain chemical reaction is directly
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01 Jul 2015, 20:30
Sample numbers really help one in these percentage problems. r = A^2/B A = 10, B=5 => r = 20 r = A^2/10=20 A^2=20 => A=14 (1410)/10=4/10=40% increase => D is the best answer



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GMAT Prep Test  Proportion Problem
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25 Jun 2016, 09:48
I encountered the following proportion question in the GMAT Prep test.
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 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
From what I understand A^2 is inversely proportional to B. So if B increases that A^2 decreases and consequently A decreases. However, the answer is 40% increase. I am not quite able to figure this one out. Can anyone help me on this....Thanks!



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Re: The rate of a certain chemical reaction is directly
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25 Jun 2016, 09:51
gourav16183 wrote: I encountered the following proportion question in the GMAT Prep test.
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 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
From what I understand A^2 is inversely proportional to B. So if B increases that A^2 decreases and consequently A decreases. However, the answer is 40% increase. I am not quite able to figure this one out. Can anyone help me on this....Thanks! Your post is merged. Please refer to the above discussion and search before posting. Rules for posting: rulesforpostingpleasereadthisbeforeposting133935.html#p1092822



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Re: The rate of a certain chemical reaction is directly
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25 Jun 2016, 10:22
balboa wrote: 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 percent, 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 R =\(\frac{a^2}{b}\) a = 4 ; b = 2 R = 16/2 =>8 Now B increases by 100% So, a = x ; b = 4 R = 8 So, 8 = \(\frac{a^2}{4}\) or, a^2 = 32 Or, a ~ 5.65 So % increase in a is \(\frac{(5.65  4)}{4}\)*100 => 41.xx % So, Answer will be (D) 40%
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Re: The rate of a certain chemical reaction is directly
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25 Jun 2016, 23:42
Bunuel wrote: 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
NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).
We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means \(x^2=2\) > \(x\approx{1.41}\), which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}\)
Answer: D. This is the credited response. I have nothing to add.
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Re: The rate of a certain chemical reaction is directly
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09 Apr 2017, 10:18
Bunuel wrote: 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
NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).
We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means \(x^2=2\) > \(x\approx{1.41}\), which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}\)
Answer: D. @Bunnuel, can you please help me understand why this set up below is incorrect? Thank you. R=A^2=1/B



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Re: The rate of a certain chemical reaction is directly
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09 Apr 2017, 20:55
Andy24 wrote: Bunuel wrote: 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
NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).
We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means \(x^2=2\) > \(x\approx{1.41}\), which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}\)
Answer: D. @Bunnuel, can you please help me understand why this set up below is incorrect? Thank you. R=A^2=1/B You should put directly proportional in nominator and inversely proportional in denominator.
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Re: The rate of a certain chemical reaction is directly
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15 May 2017, 17:14
balboa wrote: 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 percent, 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 Using "=" as the "directly proportional to" symbol. R=C(A)^2=1/C(B) Subbing in values for C(A) and C(B): R=2^2=1/2 If we increase C(B) by 100%), we must halve C(A)^2. So: C(A)^2 will become 2 and C(A) root2. Now, we need to make C(A) 2 again. root2=1.41. So we need to add 0.6 to 1.4. That's an increase of approx. 40%. Agree? Kudos.



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Re: The rate of a certain chemical reaction is directly
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15 Aug 2017, 04:45
Its a easy question if the though process is right. I could not get it initially but was able to get the logical flow Let rate be R and it is directly proportional to A² ie if A² increases than R also increases we can represent it as follows R∝A²❶ R is also inversely proportion to B ie if B increases than R reduces R∝1/B❷ combining ❶ and❷ R∝A²/B or R=A²₁/B₁ Now B₁ is increased by 100% or is 2B₁ R=A²₁/B₁=A²₂/B₂ B₂=2B₁ A²₁/B₁=A²₂/2B₁ A²₂=A²₁*2B₁/B₁ A²₂=A²₁*2 A₂=√A²₁*2 A₂=A₁√2 A₂=A₁*1.41(√2=1.41) or A has to increase by 40% to maintain the same rate.
The other way is just plug in the values we have the equation A²/B=R Let A=10 and B=5 R=100/5=20 ie the rate is 20 and has to remain the same when B increase by 100% or is 2B 20=A²/2*5 20=A²/10 20*10=A² A²=200 A=√200=14 ie increase of (1410)/100=40% increase



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Re: The rate of a certain chemical reaction is directly
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12 Oct 2017, 13:29
Can someone help me explain why it shouldn't be a 100% increase? See below for my workflow...
R = A^2 / B A = 2 B = 1 R = 2^2 / 1 = 4 / 1 = 4 B is increased by 100% = 1 > 2 R = 2^2 / 2 = 2 If you increased 2^2 by 4 > 8 / 2 = 4 (the same rate)
Thanks for your help.



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Re: The rate of a certain chemical reaction is directly
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The rate of a certain chemical reaction is directly
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09 Aug 2018, 07:22
Let, A=3 and B=2. So, the Rate= A^2/B= (3)^2/2 = 9/2=4.5 (reaction rate) B is increased by 100%. So the new ‘chemical B’ will be 2B. Let, changed A=x. So, New Rate= x^2/2B. We need to keep 'reaction rate' unchanged that is 4.5. Therefore, x^2/2B=4.5. Solve x. => x^2/4=4.5...=> x^2=18...=>x=√ 2 (3)...=> 1.41A [As, A=3]. Hence, A is increased by 41%...closest to 40%. [My first post to GMAT Club. ]



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The rate of a certain chemical reaction is directly
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04 Sep 2018, 22:13
Hi, I am not able to find a post by mikemcgarry which he had written to explain the direct variation and indirect variation. So if anyone can help on that link this question would be the base to understand the application of that post. Conttinuing with our question at hand we are told that Rate lets say R , and Concentration of A is x and Concentration of b is y We are given that R \(\propto\) \(x^2\) This means as the conc of A inc the rate of the Reaction increases. and R \(\propto\) \(\frac{1}{y}\) This means as the conc of B inc the rate of the Reaction decreases and and the conc of B decreases the rate of reaction increases. Combining both we have R \(\propto\) \(\frac{x^2}{y}\) or R = K \(\frac{x^2}{y}\) for some Constant K. Now question asks if the conc B is incresed and to maintain the rate what must increase and by how much. Clearly as conc of B increase the rate must have decreased, hence to maintain the rate as R we must increase the conc of A. Clearly options (A) (B) (C) are out . Now Conc of B is increased by 100% which means the new conc of B now is = y+100%(y) which is = 2y R = K \(\frac{something }{2y}\) Clearly if we can cencel the effect of the 2 in the denominator we can bring the rate back to original which is R . Also numerator is square of concentration of A that means if Conc A becomes \(\sqrt{2}\) x which upon squaring becomes 2\(x^2\) would bring back the rate to original R here is what i mean New conc of B is 2y the for the rate to be R New conc of A must be \(\sqrt{2}\) x because R \(\propto\) \(\frac {(\sqrt 2x)^2}{2y}\) R \(\propto\) \(\frac { 2x^2}{2y}\) which is same as original rate. So we have increased Conc of A from x to 1.414 x ( because \(\sqrt2\) = 1.414) Percentage increase would be \(\frac{1.4141}{1}\) *100 that is nearly equal to 40 % increase I am not able to find the link of the article. But if i do come across the same will update the post. Its nicely explained . In case any of members happen to find the link can they can share, Hello mikemcgarry , would request you if you could share the link of the proportionality concept explained by you. Probus.



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Re: The rate of a certain chemical reaction is directly
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10 Sep 2018, 10:27
balboa wrote: 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 percent, 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.
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Re: The rate of a certain chemical reaction is directly
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14 Sep 2018, 08:29
balboa wrote: 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 percent, 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
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Re: The rate of a certain chemical reaction is directly
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30 Oct 2018, 04:07
I think that one of the biggest challenges in this problem is trying to solve for an answer algebraically. Yes  we can get down to A^2 = 2, but unless you remember that the sqrt(2) = 1.4141 then you're SOL?
No. If we look at the answer choices, they are actually cutting us a ton of slack. We know A HAS to increase, so 3/5 answer choices are eliminated. We know that A^2 = 2 so all we need to do is pick a number for A. (A*1.5)^2=? (A*1.4)^2=?
Obviously, let A=1. 1.5*1.5=2.25 (write it out if you must) 1.4*1.4=1.96
An increase of 40% is the closest we can get to A^2=2.
D.



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Re: The rate of a certain chemical reaction is directly
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02 Nov 2018, 05:48
does the following method seem reasonable? i was a bit confused about how to get the proportionality equations down without errors.
when the question gives us two different relationships of R with A and B, would it not be that R = A^2 and that R = 1/B and that 1/B = A^2 and that 1/B * 1/2 = A^2 * 1/2 hence to reset the relationship back to original, multiply both sides by 2 which would mean (1+40%)*A*(1+40%)*A ~ 1.9^ * A^2 ~ 2 * A^2 for the right side of the equation and thus leading us to choice D



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The rate of a certain chemical reaction is directly
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24 Nov 2018, 17:38
A detailed video explanation can be found here: https://www.youtube.com/watch?v=uZfIsJ078cLet A = concentration of chemical A Let B = concentration of chemical B Since B, the denominator, has been dubbled, the total value of the numerator would have to be doubled, as well. But the value of the numerator isn't A  it's A squared. The question is therefore, "By what percentage would we increase the concentration of chemical A such that, when the new value is squared, it is equal to twice the value of B?" Let p equal the amount by which A is increased: (Ap)^2 = 2B (A^2)(p^2) = 2B In order for the proportion A^2/B to remain unchanged, p squared must equal 2: p^2 = 2 p = sqrt2 The square root of 2 is approximately equal to 1.4 The value of A would INCREASE by approximately .4, i.e 40% Answer D
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Re: The rate of a certain chemical reaction is directly
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01 Dec 2018, 01:36
Also given that the concentration of B is increased by 100 %. Hence B becomes 2B.
New concentration = \(A^2/2B\). In order to keep the new concentration the same as the original concentration, A should be equal to \(\sqrt{2}A\) so that A^2/2B becomes \(A^2/B\)
\(\sqrt{2}A\) == 1.414 * A or approximately 141% of A.
141% of A is 41 % increase of A.

Hi, am using a similar approach by feeding sample values and calculating the same but my answer is different. Please can you/anyone advice me where am I going wrong.
A = 5 B = 10 C = 25/10 = 2.5
New values  > C = A^2/20
=> 2.5 = A^2/20 => 50 = A^2 ; A = 5underroot2.
I understand you are using root 2 as 1.41 and hence 41% increase approx. Good, make sense but shouldn't we use below approach.
New A value  Old A / Old A * 100 for % increase i.e. (5root2  5) / 5 *100 = 2223% aprox.
Please help.




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