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GMAT Prep - Mixtures - percentages

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Senior | Next Rank: 100 Posts
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by tutonaranjo » Thu Aug 30, 2007 11:37 am
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased 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% decrese
c) 40% decrease
d) 40% increase
e) 50% increase

Answer is D.

Can someone help me out with an equation and a picking numbers scenario?
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by ankanas » Fri Aug 31, 2007 5:57 am
Let
A1=original concentration of A
B1= Original concentration of B

Chemical Reaction (CR1)=k*A1^2/B1

Let the new Chemical Reaction be CR2
Let the new Concentration of Chemical B be B2
A2=New concentration of Chemical A
Acc to the question
B2=2B1 (100% increase)

Therefore to keep the reaction rate unchanged CR1=CR2

K*A1^2/B1=K*A2^2/B2
Since B2=2B1

K*A1^2/B1=K*A2^2/2B1
or A1^2=A2*^2/2
or 2A1^2=A2^2
or 1.414A1=A2
Therefore 40% increase

Answer D
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by jaspreet.sharma » Fri Aug 31, 2007 1:02 pm
ankanas wrote:Let
A1=original concentration of A
B1= Original concentration of B

Chemical Reaction (CR1)=k*A1^2/B1

Let the new Chemical Reaction be CR2
Let the new Concentration of Chemical B be B2
A2=New concentration of Chemical A
Acc to the question
B2=2B1 (100% increase)

Therefore to keep the reaction rate unchanged CR1=CR2

K*A1^2/B1=K*A2^2/B2
Since B2=2B1

K*A1^2/B1=K*A2^2/2B1
or A1^2=A2*^2/2
or 2A1^2=A2^2
or 1.414A1=A2
Therefore 40% increase

Answer D
Nice way to calculate it..... :D
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picking numbers

by resilient » Tue Feb 05, 2008 1:34 am
my gut reaction to this is to keep the whole thing in an easier perspective and pick numbers is there a way to do that!?
Appetite for 700 and I scraped my plate!
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