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The rate of a certain chemical reaction is directly proportional to th
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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
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Originally posted by Accountant on 23 Mar 2009, 10:03.
Last edited by Bunuel on 06 Aug 2016, 22:02, edited 2 times in total.
Edited the question and added the OA




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27 Nov 2009, 18:43




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23 Mar 2009, 12:19
Accountant wrote: The rate of a chemical reaction is directly proportional to the square of concentration of chemcial A present and invesrsly proportional to the concentration of chemical B present. If chemical B is increased by 100% which of the following is the change in concentration of chemical A required to keep teh reaction rate unchanged:
A. 100% decrease B. 50% decrease C. 40% decrease D. 40% increase E. 50% increase
Please explain your answer. The rate of reaction is invesrsly proportional to the concentration of chemical B present. It used to have B=1 . Now that B is increased by 100%. So the new equation would be 2B=(1/2). In order for the rate of reaction to still be 1, we need to change the concentration of A to yield a 2. It used to be A^2=1, now the new equation should be (sqrt(2)*A)^2=2. The change in the concentration of A can be calculated as (sqrt(2) 1)/1 or approximately 40% increase. Answer D.



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Re: The rate of a certain chemical reaction is directly proportional to th
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23 Mar 2009, 15:13
CA: Concentration A CB: Concentration B R: Reaction rate
Formula for reaction rate:
R = (CA^2) / CB
Thus if CB is increased by 100% >> means concentration doubles >> 2xCB
Thus, for R to remain the same (CA^2) also has to double.
>> 2 x (CA^2) >> to see what happens to CA, take the 2 into the bracket by taking its root
>> (SQRT2 CA)^2
SQRT 2 is roughly 40% >> answer D



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27 Nov 2010, 17:06
Bunuel wrote: 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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D. Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator?



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28 Nov 2010, 01:12
afyl128 wrote: Bunuel wrote: 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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D. Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator? \(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some nonzero constant \(x\) such that \(a=xb\); \(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some nonzero constant constant \(y\) such that \(a=\frac{y}{b}\). So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\). Hope it's clear.
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28 Nov 2010, 08:56
Bunuel wrote: afyl128 wrote: Bunuel wrote: 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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D. Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator? \(a\) is directly proportional to \(b\) means that as the absolute value of \(a\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some nonzero constant \(x\) such that \(a=xb\); \(a\) is inversely proportional to \(b\) means that as the absolute value of \(a\) gets bigger, the absolute value of \(b\) gets smaller, so there is some nonzero constant constant \(y\) such that \(a=\frac{y}{b}\). So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\). Hope it's clear. Thanks =) do you have any links to similar questions? i'm very shaky on these



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29 Sep 2013, 09:12
Hi Bunuel, I am a bit shaky with variation concepts and hence decided to get the basics clear, I was referring to Karishma's blog here http://www.veritasprep.com/blog/2013/02 ... gjointly/Now I understand that if a rate varies directly for Eg X varies directly with Y than we have X/Y = K(some value) because in direct variation the ratio remains same. and in inverse variation it will be XY = K(some value) because x=1/y Please correct me if I misunderstood any concept till this point. now to your explanation Bunuel wrote: NOTE: Put directly proportional in nominator and inversely proportional in denominator how can we put a direct variation in numerator? because if I understand the concept correctly it should be in denominator? and an inverse in numerator..how did you arrived at this quick formula? another question to may be both karishma and you Bunuel ( sorry karishma I am asking questions pertaining to your blog on this forum, but I thought this question can serve as a common reference.) As per me It should be B/A^2 (infect If I look at karishma's sample question in same page its essentially the same question with just values swapped with N and M) another point of confusion when B becomes double (i.e 2B) why don't we simply say A^2 also doubles(i.e 2 A^2) why do we say if a^2 has to double it has to be A^2 = 2 ? if its a ratio than it should be multiplied and divided by same value in numerator and denominator (i.e 2)
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01 Oct 2013, 11:22
Hi Bunuel...waiting for reply.. have exam this week so a bit nervous
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Re: The rate of a certain chemical reaction is directly proportional to th
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29 Apr 2016, 06:54
Bunuel wrote: \(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some nonzero constant \(x\) such that \(a=xb\);
\(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some nonzero constant constant \(y\) such that \(a=\frac{y}{b}\).
So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).
Hope it's clear.
Hi, Could someone help to explain this one, how can we come up with the final result, which is \(R=\frac{(A^2x)y}{B}\)? As far as I understand, according to the case in question, we have: 1. The rate is directly proportional to the square of concentration of chemical A > R = \(xA^2\) (1) 2. The rate is inversely proportional to the concentration of chemical B > R = \(\frac{y}{B}\) (2) So next, how can we infer that \(R=\frac{(A^2x)y}{B}\)? What steps used to modify/combine (1) and (2) to get this one? Actually, from (1) and (2), I am thinking of \(\frac{(A^2x)*y}{B}\) as \(R*R\) = \(R^2\) instead Please help to clarify. Thanks
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Re: The rate of a certain chemical reaction is directly proportional to th
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06 May 2016, 01:06
thuyduong91vnu wrote: Bunuel wrote: \(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some nonzero constant \(x\) such that \(a=xb\);
\(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some nonzero constant constant \(y\) such that \(a=\frac{y}{b}\).
So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).
Hope it's clear.
Hi, Could someone help to explain this one, how can we come up with the final result, which is \(R=\frac{(A^2x)y}{B}\)? As far as I understand, according to the case in question, we have: 1. The rate is directly proportional to the square of concentration of chemical A > R = \(xA^2\) (1) 2. The rate is inversely proportional to the concentration of chemical B > R = \(\frac{y}{B}\) (2) So next, how can we infer that \(R=\frac{(A^2x)y}{B}\)? What steps used to modify/combine (1) and (2) to get this one? Actually, from (1) and (2), I am thinking of \(\frac{(A^2x)*y}{B}\) as \(R*R\) = \(R^2\) instead Please help to clarify. Thanks Hello my friend. This is how I solved this question. First I didnt assign those variebels x and y as above in the quote. I think those are used just as reference in order to show the proportonality. Now to the question. I did assign variable x to be the factor of percent increase or decrease, r to be the rate, and a and b for the concetrations. So start with first formula \(r=a^2/b\) where a is directly proportonal ans is nuumerator and b is denominator since it is inversly proportinal. note u have to put \(a^2\) because it is given that the square root is directly proportonal now the second equation , the question asks to have same value for the rate but b is doubled or increase for 100% \(r=(xa)^2/2b\) or \(r=x^2a^2/2b\) now from here we can use the short way given by Bunuel or the long way to solve for x, either way it will come out as x^2=2 and\(x=1.41\). now earlier we said x is a factor of percentage change \(1+0.41\) or we have an increase of 41%



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Re: The rate of a certain chemical reaction is directly proportional to th
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26 Dec 2016, 13:26
This question can also be solved by plugging in values for a and b and comparing the results:  The rate of this reaction at first is \(r=\frac{A^2}{B}\)  If \(A=1\) and \(B=2\) the result of the equation is \(r=\frac{1}{2}\)  When \(B\) is increased by 100% the equation becomes \(r=\frac{A^2}{2B}\)  Since the question asks by how much \(A\) would need to increase in order for the reaction rate to remain the same, the second equation can be rewritten as \(\frac{1}{2}=\frac{A^2}{4}\) where \(r=\frac{1}{2}\), the same as in the first equation.  The second equation can then be rewritten as \(2=A^2\) and then as \(\sqrt{2}=A\)  The percent increase from \(A=1\) in the first equation to \(A=\sqrt{2}\) in the second equation is approximately 40% since\(\sqrt{2}=1.4\)  The answer is



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29 Aug 2017, 02:57
Bunuel wrote: 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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D. hi your solution to the problem is very clear, no doubt out there ..... can you please, however, speak some more words about the concept "Put directly proportional in nominator and inversely proportional in denominator." thanks in advance ...



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29 Aug 2017, 03:07
gmatcracker2017 wrote: Bunuel wrote: 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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D. hi your solution to the problem is very clear, no doubt out there ..... can you please, however, speak some more words about the concept "Put directly proportional in nominator and inversely proportional in denominator." thanks in advance ... Check Variations On The GMAT  All In One Topic. Hope it helps.
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05 Sep 2017, 03:33
This is a tough question, especially if you aren't well versed with variations. This is how i approached it: Since B increased, in order for the rate to stay the same, A also has to increase, therefore we can eliminate A, B, and C. Now my guess is between D and E. 50% felt wrong so I went with D
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07 Sep 2017, 15:32
Accountant 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 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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11 Sep 2017, 23:41
Let Rate be R Let concentrations Ca=10 Cb=10
R=K\(\frac{(Ca)²}{Cb}\) where K is a constant which we can ignore for this question.
R=\(\frac{(10)²}{10}\)=10
Now Cb is doubled =20 We have to keep R=10
10=\(\frac{(Ca)²}{20}\)
or (Ca)²=200 or Ca=√200 or Ca is approx=14 which is 40% more than 10 hence answer is D



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Re: The rate of a certain chemical reaction is directly proportional to th
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30 Sep 2018, 08:21
Accountant 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
\(rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\) \(\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)\) \(?\,\, \cong \,\,k  1\) \({\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered} {k^2} = 2 \hfill \\ \sqrt 2 \cong 1.41 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2  1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2  1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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