The rate of a certain chemical reaction is directly proportional to the square of the concetration 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 percentage change in the concentration of chemical A required to keep the reaction rate unchanged?

100 % decrease
50 % decrease
40 % decrease
40 % increase
50 % increase

the answer is a 40 % increase, I don't understand why it isn't a decrease if it is inversely proportional..

Soln:
A2 means A square in the solution below.
Earlier the equation looked like C = A2 / B.
Now u have replaced the B with 2B but you still want no change in the equation
so C = (1.4A)square / 2B = 1.96A2 / 2B = A2/B.

So you need to increase A by 40% for no effective change.

Hope it helps.

The rate of a certain chemical reaction is directly proportional to the square of the concetration 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 percentage change in the concentration of chemical A required to keep the reaction rate unchanged?

100 % decrease
50 % decrease
40 % decrease
40 % increase
50 % increase

the answer is a 40 % increase, I don't understand why it isn't a decrease if it is inversely proportional..

A direct relationship between y and x can be written y = kx.
An inverse relationship between y and x can be written y = k/x.

So in this problem, r = kA^2 / B

Now in our new situation, let's call the new concentration of A as A', and the new concentration of B as B'. In our new situation the rate is the same, so
new rate = old rate
k (A')^2 / B' = k A^2 / B

We know that if B is increased by 100%, that means B is doubled, so B' = 2B.
k (A')^2 / 2B = k A^2 / B

We can multiply both sides by B/k, so
(A')^2 = 2 A^2

This means that A' = A sqrt (2). You should have memorized that:
sqrt (2) = 1.4
sqrt (3) = 1.7
so A' = 1.4 A, or a 40% increase.

Hope that helps.

_________________
Ben Ku
Instructor
ManhattanGMAT

Awesome ... Thanks a ton!

no problem.

_________________
Ben Ku
Instructor
ManhattanGMAT

Hi Ben,

Can you explain to me how you came up with r = kA^2 / B I get lost in that part.

Hi,

I'm not Ben but I will try.

If one thing, r in our case, is directly proportional to something else, A^2 (A squared) in our case, then by definition, if nothing else other than A changes,

(1) r = c1 * A^2.

where c1 is a "constant".


If one thing, r in our case, is inversely proportional to something else, B in our case, then by definition, if nothing else other than B changes,

(2) r = c2 / B

where c2 is "constant".

That "if nothing else changes" is important for combining the two because what it leads to is that if something else changes and then stays fixed that "constant" might change. That is that "constant" might have another relationship with r for another variable. Looking at equations (1) and (2) we see that c1 contains B, that is

c1 = k / B

and c2 contains A^2, that is

c2 = k A^2

where k is another "constant". Or

r = kA^2 / B

This could actually continue, that is in this problem k might actually depend on another chemical C. But, thankfully, they stop here.

Once you do several of these kinds of problems, you get familiar enough with the process that you will read the question and just write down the equation.

Great explanation, thank you!

_________________
Jamie Nelson
ManhattanGMAT Instructor

That "if nothing else changes" is important for combining the two

c1 = k / B



c2 = k A^2



***
Even so, I also don't get how you combined them, mathematically, to get r = kA^2 / B. Can someone else explains this? Thanks!

I can't believe I spent 1 hour on solving this problem, after searching all sorts of different explanations provided by MGMAT staffs, Beat the gmat people and Knewton, eventually figured out a way from my own perspective, please let me know if my reasoning is correct:

r= (A^2) C= (1/B) K
Where C=some constant and K=some other constant

Both constants won't change, so when we double B, (1/B) becomes (1/2B), which if you factor it out, becomes (1/2) x (1/B), and the equation becomes:

r= (1/2) (1/B) K, as mentioned, K won't change, nor will C, so to keep the other side of the equation, where (A^2) stands the same, something must happen to (A^2) to counter the "half" effect on the right side of the equation where B stands. So we must multiple by 2 on the the left side of the equation, where (A^2). OR think of it conceptually as: If B is doubled, then r would be halved, to get r back to what it should be, you have to double (A^2), since r is directly proportional to (A^2), by doubling (A^2), you'll be doubling r.

Now A as a number has to increase to get it to become 2A^2 after squaring A. So I let "a" equals (A + change):

2(A^2) = a^2 where a=A + change
divide by 2 on each side,
A^2 = (a^2) / 2
rewrite the right hand side: (a^2) / ((square root of 2)^2)
since they share same power, we can combine the base by division
(a/(square root of 2) ) ^ 2
and this equals to A^2.

Which means the things inside equal to each other:
(a/(square root of 2) = A
(a/1.4) = A
or a = 1.4 A
As mentioned, a = A + change, so the change is .4 A
(CHANGE/ORIGINAL) X 100 = PERCENT OF CHANGE
(.4A/A) x 100 = 40%.

:)

Looks good Rachel! And great job on working on it until you got it--that is awesome!

_________________
Jamie Nelson
ManhattanGMAT Instructor