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The rate of a certain chemical reaction is directly [#permalink]
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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 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
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Re: GMAT Prep  Ratios [#permalink]
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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 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.
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Re: Formula Problem [#permalink]
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CR = (A^2) * ( 1/B) put any values 625 = 25*25 B increases by 100% so it becomes 50 CR remains same at 625 1/B becomes 50 so A^2 should be 625/50 = 12.5 or say decreased by 50% (b)
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Re: GMAT PREP PS [#permalink]
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Given equation = \(A^2/B\) 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. Closest answer choice is D.
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Re: Please help!! Difficult problems from GMATPrep [#permalink]
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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? Given: Rate  \((A^2/B)\) B increased by 100%. New B is 2B. Rate  \((A^2)/2B\). In order for the initial rate of chemical reaction to be the same as the final rate. A should be increased such that it should cancel out the 2 in the denominator. Hence A should be \(A\sqrt{2}\) and hence final rate of reaction is \((A\sqrt{2}))^2/2B\) \(2A^2/2B\) which becomes the same as the initial rate of reaction. Hence A should be \(A\sqrt{2}\). Or final concentration of A is \(1.41A\) or an increase of approximately 41%. Hence answer is D.
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Re: ps question [#permalink]
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satishreddy wrote: the rate of certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of the 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.
100% decrease 50% increase 40% decrease 40% increase 50% increase Let the rate of the reaction is r and concentration of chemical A is A and concentration of chemical B is B. According to the question , r is proportional to A^2 and r is also proportional to 1/B Combining these two condition , r =K (A^2/B) , where K is proportionality constant. Now in the changed condition B has increased by 100 % means is has become 2B. So for keeping the rate constant we have to change the concentration of A, from A to ((Square root of 2)*A) As K (((square root of 2)A)^2)/2B = K (A^2/B) =r Now, let after increase of x% ,A has become ((square root of 2)A) i.e 1.414*A A+(xA/100)=1.414 A Hence x=41.4% Answer is D. Please consider KUDOS if it helped u in some way.Thanks



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Re: % increase , proportions [#permalink]
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tradinggenius wrote: The rate of a 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?
100% decrease 50% decrease 40% decrease 40% increase 50% increase Let r be the rate of chemical reaction and A and B be the concentration of chemical A and chemical B respectively. r is proportional to A^2 r is inversely proportional to B thus, r = k(A^2/B), where k is some constant. now B becomes 2B (100% increase). Let x be the new concentration of A. The new equation becomes r=k(x^2/2B) Since r remains the same, we have k(A^2/B)=k(x^2/2B) so, x=1.4A which is an increase of 40% D is the answer



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Re: Direct and Inverse Proportion Q [#permalink]
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Rate= r = k (A^2) (1/B) [where k is a constant] If B is replaced by 2B, then r = k (A^2) (1/2B) => k (A'^2) (1/2B) = k (A^2) (1/B) [where A' is the new concentration of A] => A'^2 = 2(A^2) => A' = sqrt (2) * A = 1.414 *A Therefore the concentration of A must increase by ~40% [option D]
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Re: Direct and Inverse Proportion Q [#permalink]
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A> (sq rt 2)*A as R=K A^2/B HENCE ((sq rt 2)1)/1 =41% increase hence D
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Re: The rate of a certain chemical reaction [#permalink]
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24 Oct 2012, 07:33
danzig 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 From Q Stem Chemical reaction R =K (A^2/B) Now if B is increased by 100%> it becomes R=K (A^2/2B) Now To bring the relationship to its initial condition we need to cancel out 2 in denominator. So my multiplying root 2 we can do that. So root2 = 1.4 approx. That means we need to increase 40% of A. Hence D.
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Re: The rate of a certain chemical reaction is directly [#permalink]
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04 Nov 2012, 01:38
Hello Bunuel I saw this question coming around couple of times. But doesn't the answer depend on how you express the concentration reaction? If I said CR = A^2  B, (or in general a*A^2 + b*B, where a >0 and b <0) the answer will still be D? Thanks for your time Brother Karamazov



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Re: The rate of a certain chemical reaction is directly [#permalink]
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Re: The rate of a certain chemical reaction is directly [#permalink]
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I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this?



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Re: The rate of a certain chemical reaction is directly [#permalink]
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theGame001 wrote: I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this? Lets make your life more easy. Let us say x = 4, y = 2 then R = 16/2 = 8 Now keep R same and double y to see what is the change in X. R = X^2/Y 8 = X^2/4 X = 4\sqrt{2} X got multiplied by \sqrt{2} which means a 40 % increase OR R = X^2/Y If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase
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Re: The rate of a certain chemical reaction is directly [#permalink]
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04 Jan 2014, 04:19
PerfectScores wrote: theGame001 wrote: I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this? Lets make your life more easy. Let us say x = 4, y = 2 then R = 16/2 = 8 Now keep R same and double y to see what is the change in X. R = X^2/Y 8 = X^2/4 X = 4\sqrt{2} X got multiplied by \sqrt{2} which means a 40 % increase OR R = X^2/Y If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase Thank you but what I am unable to understand is why are we multiplying? x=4 and y=2 Squared 4= 16  I understood this Inversely proportional = 1/2  Understood this 16*1/2 Unable to understand this part



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Re: The rate of a certain chemical reaction is directly [#permalink]
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theGame001 wrote: PerfectScores wrote: theGame001 wrote: I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this? Lets make your life more easy. Let us say x = 4, y = 2 then R = 16/2 = 8 Now keep R same and double y to see what is the change in X. R = X^2/Y 8 = X^2/4 X = 4\sqrt{2} X got multiplied by \sqrt{2} which means a 40 % increase OR R = X^2/Y If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase Thank you but what I am unable to understand is why are we multiplying? x=4 and y=2 Squared 4= 16  I understood this Inversely proportional = 1/2  Understood this 16*1/2 Unable to understand this part R is directly proportional to X^2 and inversely proportional to Y which means that R is directly proportional to X^2/Y So if x = 4 and y = 2 R is directly proportional to (4^2)/2 which is equivalent to X^2/Y
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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 \(Rate = \frac{k*A^2}{B}\) \(Rate = \frac{k*NewA^2}{2B}\) \(\frac{k*NewA^2}{2B} = \frac{k*A^2}{B}\) \(NewA = sqrt{2} * A\) \(%change = \frac{sqrt{2}*A  A}{A}*100\) \(%change= 0.414*100=41.4% increase\) Answer : D
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Ratio A to B : A^2 = B Say B = 100 Then A = 10^2 B Increases 100% ==> 200 x^2 = 200 ? 14^2=196 (almost 200, so x is a slightly bigger) 10 ==> 14 ==> increase of 40%. Hence Answer D.
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