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GMAT Prep - Direct Proportions

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GMAT Prep - Direct Proportions

by tonebeeze » Fri May 20, 2011 12:06 pm
Can someone walk me through this problem. Thanks.

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

OA = D
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by gmatdriller » Fri May 20, 2011 1:41 pm
Start: A^2/B
Changed: A^2/2B
Question: what do we do to A to maintain A^2/B?
N:B: we need a "2" as numerator to cancel the "2" at the
denominator (B is doubled)
We need multiply A by a root(2) factor because
the square of A returns the "2"

[A*root(2)]^2 / (2B): this is what we need,
root(2) = 1.4 (approx)
So, A will be increased by a factor of 0.4 or increased by 40%

Hope this explanation is helpful.
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by GMATGuruNY » Fri May 20, 2011 1:46 pm
The rate of a chemical reaction is directly proportional to the square of concentration of chemical A and inversely proportional to the concentration of chemical B. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percentage change in concentration of chemical A required to keep the reaction unchanged.

A 100% decrease
B 50% decrease
C 40% decrease
D 40% increase
E 50% increase
This problem describes the following relationship:

R = (A²)/B

R is directly proportional to A².
Directly proportional means that as one value increases, the other increases by a proportionate amount. In the equation above, if R doubles, then A² must double for the equation to remain valid.

R is inversely proportional to B.
Inversely proportional means as one value increases, the other decreases by a proportionate amount. In the equation above, if B doubles, then R must be halved in order for the equation to remain valid.

We can plug values into the equation above.

Let A = 10 and B = 2.
R = (10²)/2 = 100/2 = 50.
If we increase B by 100%, new B = 4.
Since R=50 must stay the same, we get:
50 = (A²)/4
A² = 200
New A = √200 = 10√2 ≈ 14.

The value of A increases from 10 to 14, a 40% increase.

The correct answer is D.
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by clock60 » Fri May 20, 2011 1:51 pm
one of the ways to solve is to plug numbers
say rate of reaction is- x, concentration of chemcial A-a, and concentration of chemical B-b. we are given that x=a^2/b

a^2=10^2, b=2, and a^2/b=10^2/2=50,
if chemical b increased by 100%, new concentration of a, will be a1
(a1)^2/(2*2)=50 (we need rate x=50 unchanged)
(a1)^2=(2*100)
a1=10*1.4=14 (2^1/2=1.4)
and change will be (14-10)/10=0,4*100=40%
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