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 present.

This means that the rate couldbe written as R =K * CA^2/CB
Where K is a constant
CA is the concentration ofchemical A
CB is the concentration ofchemical B

* If conentration of B is increased by 100%, it means that the

Let R1 = CA1^2/ CB1, be the rate befre CB changed
and R2 = CA2^2/ CB2, the rate after CB doubled ( increased by 100%)
To keep the rate unchanged, we would have to fit the following equation :
R1=R2, this means that CA2 ^2/CB2 = CA1^2/CB1

AS CB2 = 2 CB1
THEN, CA2^2 = 2 * CA1^2

CA2 = SQUARE Root of 2* CA1 = 1,4 CA1

tHIS MEANS THAT Ans = +40%


GOOD LUCK

hey
Fei,
How have you been?
Why do you still want to solve GMAT Math problems? you are already done with it
Are you thinking of retaking it ?

As Stacey says we should provide the complete problem alongwith source of it. It's for good. 8)

Anyway it seems that bonana has already replied to you.

I got the same question in my prep test.

R= [K(A^2)]/; where K is a constant.

In this question 3 answer choices can be eliminated without solving:

To keep the fraction constant, if the denominator is increased the numerator has to increase.
3 answer choices mentioned a decrease and hence can be eliminated

To calculate the percentage increase of A

B-->2B i.e. B changes from B to 2B
A^2--> 2.A^2 i.e. A^2 has to change from A^2 to 2.A^2

A^2 = 2. A^2
A = sqrt2. A

Thus, A has to increase to sqrt2.A

Percent increase from A to sqrt2.A is approx 40%