The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the precent change in the concentration of chemical A required to keep the reaction rate unchanged?

A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

x is directly proportional to y, translated into math:
x = ky, where k is a constant.
x is inversely proportional to z, translated into math:
x = k/z, where is k is a constant.
x is directly proportional to y and inversely proportional to z, translated into math:
x = k(y/z), where k is a constant.

In the problem above, the rate is directly proportional to the square of A and inversely proportional to B.
Thus:
R = k(A²/B).

Original values:
Let:
R = 100
A = 10
B = 1.
Solving for k, we get:
100 = k(10²/1)
100 = k(100)
k = 1.

New values:
Since the rate doesn't change, R = 100.
B increased by 100% = 2.
k = 1. (Since k is a constant.)
Solving for A, we get:
100 = (1)(A²/2)
A² = 200
A = √200 = √100 * √2 = 10√2 ≈ 10(1.4) = 14.

Percent increase in A:
(increase in A)/(original A) = (14-10)/10 = 4/10 = 40%.

The correct answer is D.

Rate ~ (A)^2
~ 1/(B)
= K (A^2 / B)

New Rate = K ((xA)^2 / (2B))

Equating:

K (A^2 / B) = K (xA^2 / 2B)

=> 2 * A^2 = x^2 * A^2 ==> 2 = X^2 ==> 1.4

Percent Increase = (1.4 - 1)* 100 = 40%
Answer [spoiler]{D}[/spoiler]