GMAT PREP PS question

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Tue Apr 17, 2012 6:28 pm

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 percent 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

Thanks

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Source: Beat The GMAT — Problem Solving | Tue Apr 17, 2012 6:28 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Apr 17, 2012 7:42 pm

alex.gellatly wrote: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 percent 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

Thanks

Rate of a certain chemical reaction, R, is directly proportional AÂ²/B
When the concentration of chemical B is increased by 100 percent, then concentration of chemical A also increases by, say, a%.
Then, R = AÂ²/B = (aA)Â²/2B
AÂ²/B = aÂ²AÂ²/2B
1 = aÂ²/2
aÂ² = 2
a = 1.4, which means there is 40% increase in the concentration of chemical A.

The correct answer is D.

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nisagl750: Master | Next Rank: 500 Posts; Posts: 145; Joined: Tue Jan 31, 2012 6:50 amFollowed by: 2 members; Location: New Delhi; Thanked: 16 times; GMAT Score:760

by nisagl750 » Thu Apr 19, 2012 6:01 pm

As the concentration of chemical B is increased by 100% i.e doubled, The rate of chemical reaction is halved (inversely prop.)
That means we need to add chemical A such that it doubles the reduced rate (to bring it back to the normal).
Now, Since the rate is prop to square of the chemical A. We need to get
A^2 = 2
so A=1.414

i.e. 40% increase in chemical A is required.

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