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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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
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
Master GMAT
Master GMAT
- vineeshp
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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.
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.
Just telling you what I know and think. I am not the expert.
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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
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
Roy
Master GMAT
Master GMAT