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sorry if this problem has already been posted, but I do not

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sorry if this problem has already been posted, but I do not [#permalink] New post 19 Jun 2007, 05:52
(sorry if this problem has already been posted, but I do not feel like searching for it)

The rate of a chemical reaction is directly proportional to the square of the concentration of Chemical A and inversely proportional to the concentration of Chemical B. If the concentration of Chemical B is increased by 100%, which is closest to the % 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

OA is D
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 [#permalink] New post 19 Jun 2007, 12:56
Just in case you cannot see the white font at the bottom of my original post, the OA is D
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 [#permalink] New post 19 Jun 2007, 15:23
I dont have a formal method for solving this, but I'll try to explain how I would do this on an exam.

Since chemical B increased by 100%, or doubles, then the rate of the chemical reaction should be halved.

so what I did was assume the rate began at 100, and was halved to 50.

then because the rate is directly proportional to the square of the concentration of chemical A, let's say the concentration started at 10 (because 10^2 = 100). because the rate was halved, we need to double it again to get it from 50 back to 100, so essentially you'd have to get x^2 = 200.

with x^2 = 200, x is about 14. so we must increase the concentration from 10 to 14, about 40%

sorry for the shabby explanation. anyone have a formal method for doing this?
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Re: Another GMAT Prep problem [#permalink] New post 19 Jun 2007, 15:38
djhouse81 wrote:
(sorry if this problem has already been posted, but I do not feel like searching for it)

The rate of a chemical reaction is directly proportional to the square of the concentration of Chemical A and inversely proportional to the concentration of Chemical B. If the concentration of Chemical B is increased by 100%, which is closest to the % 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

OA is D


You know, if you didn't post OA, I would have done this quesiton incorrectly. So thanks for that, now I have to remember the "inversely" proportional better.

Set r = rate, a = concentration of a, b = concentration of b, c = constant
r = c * a^2
r = (1/c) * b
increase b by 100%...you have
r = (1/c) * 2b
=> r/2 = (1/c) * b
Plug in r/2 for equation for a:
r/2 = c * a^2
=> r = c * (sqrt(2) * a)^2
=> r = c * (1.4 * a)^2
Increase by 40%!!!
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 [#permalink] New post 19 Jun 2007, 15:42
Just to let you know, my problem before was thinking that "inversely proportional" means this:

r = 1 / b which is incorrect.

The equation for "inversely proportional" is

r = (1/c) * b
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 [#permalink] New post 19 Jun 2007, 16:54
Thank you for the help. I hope I never see that type of problem on the real thing, but if I do I will know to use a constant in the equation.

It goes to show that without an error log, I would never know where I went wrong, and that could have come back to haunt me on the real thing.
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 [#permalink] New post 19 Jun 2007, 23:43
R = k(A^2/B ) where k is a constant

B increases by 100%, so now R = k(A^2/2B)

To keep the rate constant, we need 2A^2, which can be written as sqrt(2)A * sqrt(2)A. So A needs to increase by 41% ( which is 1-sqrt(2)).
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 [#permalink] New post 20 Jun 2007, 12:47
40 % increase

Rate = A^2/B

A1^2/B = A2^2/2B

A2^2 = 2A1^2

A2 = 1.41A1

0r 41% increase, approx. 40% increase
  [#permalink] 20 Jun 2007, 12:47
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