never seen q like this before...would appreciate explanation

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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 porportional to the concentration of chemcial B present. if the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentation in chemical A required to keep the reation rate unchanged?

A 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase
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by cramya » Sun Dec 28, 2008 8:36 pm
Has to do with the wording than the actual math

Direct proportional means as one qty increase the other quantitiy increase so A is in the numerator

Inversely proportional means when 1 quantity increases the other decreases or vice vers

R is poroportional to A^2

R is porportional to 1/B


Old Rate = A^2/ B

New Rate = A^2 /2B

Equate 2 rates to find A in new rate:

A^2/B = A^2/2B

2 A^2 (OLD RATE A) = A^2 (NEW RATE A)

Take square root of both sides

sqrt(2) * A(OLD RATE A) = A(NEW RATE A)

1.4 * A(OLD RATE A) = A(NEW RATE A)


1.4 is approx 40% increase.

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by jdotsuka » Sun Dec 28, 2008 8:42 pm
ah, thanks for explaining...very much appreciated.

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by ssy » Sun Jan 11, 2009 8:07 am
Hi Cramya - Can you please explain from this step onwards? I'm not sure how you got 2A^2 and A^2, what happened to B and 2B in the equation? Many thanks.


2 A^2 (OLD RATE A) = A^2 (NEW RATE A)

Take square root of both sides

sqrt(2) * A(OLD RATE A) = A(NEW RATE A)

1.4 * A(OLD RATE A) = A(NEW RATE A)


1.4 is approx 40% increase.

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by GMATkid913 » Mon Mar 02, 2009 12:52 am
Wouldn't this be a 40% decrease?

If one equation is inversely proportionate to another.

A = 1/B, if you increase B, that means the right side gets smaller, so A has to get smaller as well.

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by ithoughtshewas18 » Sun Mar 08, 2009 11:35 am
I got decrease

Say the chemical reaction rate is 100.
That means it's directly proportional to the square of A, so say A is 10.
The reaction rate is inversely related to B. So 1/100

1/100 = 100 = 10^2

If B is increase 100% = 2/100 = 1/50
So reaction time must be 50, and the A compound must be 7.xxx

I got a 30% something drop.

If its in inverse relation, that means the reaction rate will be lower when B increase, when it is lower, A, which is identical in its square root form will also lower.

Say the reaction rate is 4, that means an inversely related number would be 1/4.

If 1/4 increase by 100%, that equals 1/2, making the reaction rate only 2. Now whatever is related to the reaction rate that is directly proportional, square or not will not increase unless it changes its original relationship form. Meaning the square value at the beginning takes on a whole value after the problem.


Whats the answer in the book say ?