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by divya23 » Wed Jun 15, 2011 5:53 am
the rate of a certain chemical reaction is directly prop to square of conc of chemical A and inversely prop to conc of chemical B present.if conc of chemical B is increased by 100% which of the follow is the closest to he percentage change in the conc of chemical a required to keep the reaction rate unchanged.
ans = 40%increase
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by Roy@MasterGmat » Wed Jun 15, 2011 6:08 am
Let x = chemical reaction,

A = Chemical A

B = Chemical B


Since x is directly proportional to the square of A and inversely proportional to B:

x=A^2/B


B increased by 100% becomes 2B. Denote the percentage change in A as n:

x = (nA)^2/2B



Finally, solve for n by equating (nA)^2/2B with the original A^2/B:

(nA)^2/2B = A^2/B

n^2A^2 / 2B = A^2 / B \:A^2

n^2 / 2B = 1/B \*2B

n^2 = 2

n = √2


√2 = approx' 1.4, so the percent increase is approx' 40%.


This is a *very* straightforward algebraic approach to this question. For a more qualitative explanation, see Geva's recent post on the matter:

https://www.beatthegmat.com/mba/2011/05/ ... n#comments
Roy
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by vineeshp » Wed Jun 15, 2011 6:12 am
R = k A^2 / B

R1 = k A1^2 / B
R2 = k A2^2 / 2B

Given R1 = R2.

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

Cancelling common factors.
A1^2 = A2^2 / 2
A2^2 = 2 * A1^2
A2/A1 = root(2) =1.414
In %,
(A2/A1) * 100 = 141.4 ~ 140

i.e. A2 is 140 % of A1.
which is 40% increase over A1.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by divya23 » Wed Jun 15, 2011 6:12 am
i got till root 2 and i cnaged it to 1.414 also but could not figure it out after that :(
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by Roy@MasterGmat » Wed Jun 15, 2011 6:38 am
Let's go over that part:

To keep the reaction unchanged, A needs to increase into 1.4*A. That means A needs to increase by 0.4A (to increase A into 1.4A, add 0.4A, or in math: A+0.4A = 1.4*A).

Since 0.4 is 40%, 0.4A is 40% of A. Therefore, A needs to increase by 40%.

As another example, 1.5x is 50% greater than x, since 0.5 is 50%. 1.7y is 70% greater than y, since 0.7 is 70%.

Hope it's clear now ;)
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