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GMATPrep Rates Problem

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GMATPrep Rates Problem

by Schrute Beets » Wed Jul 16, 2008 6:41 pm
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 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

Thanks for your help in advance!

Answer is D.
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by moliver » Wed Jul 16, 2008 7:38 pm
rate of a certain chemical reaction = (concentration of A)^2/concentration B

if concentration of B increased by 100 % we have now the double of B = 2B

so, ( X x concentration of A)^2/2B = (concentration of A)^2/concentration B
X^2/2 = 1
X^2 = 2
X = (2)^(1/2)
now we look at the answers and we have increase of 40%
so if the concentration of A increase 40% = 1.4 A = aprox to (2)^(1/2)

or 1.4^2 = aprox to 2
Ax1.4^2/2B = A/B
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by dferm » Thu Aug 21, 2008 11:53 am
Is there a better explanation for this prob.
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by pre-gmat » Thu Aug 21, 2008 3:04 pm
I just posted this on one other thread...not sure if this is helpful to you:

Another explanation:

R1=KA^2/B
R2=KA1^2/2B Since B increases by 100%



Given: R1=R2

KA^2/B=KA1^2/2B

(A^2 * 2B)/B=A1^2

A^2*2=A1^2

taking the sqrt

A1= Sqrt (A^2*2)

A1= A*sqrt2

Sqrt 2 is approx 1.4

A1= 1.4 A

which is approx 40% increase in A.
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