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The rate of a certain chemical reaction is directly [#permalink]
05 Feb 2010, 12:06

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51% (02:13) correct
49% (01:24) wrong based on 359 sessions

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

Re: GMAT Prep - Ratios [#permalink]
05 Feb 2010, 12:16

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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}

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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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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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.

Re: % increase , proportions [#permalink]
31 Jan 2011, 06:40

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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?

Re: Direct and Inverse Proportion Q [#permalink]
18 Mar 2012, 20:26

1

This post received KUDOS

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] _________________

Re: Direct and Inverse Proportion Q [#permalink]
19 Mar 2012, 00:37

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Expert's post

dddmba2012 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 present is increased by 100 percent, which of the following is closest to the percent change in the 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

Re: The rate of a certain chemical reaction [#permalink]
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.

Re: The rate of a certain chemical reaction is directly [#permalink]
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

Re: The rate of a certain chemical reaction is directly [#permalink]
09 Apr 2014, 13:26

Rate = \frac{k*A^2}{B}

Rate = \frac{k*NewA^2}{B}

\frac{k*NewA^2}{B} = \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 _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".