Rate&Percentage

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Akansha: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Wed Sep 01, 2010 10:04 am; Thanked: 1 times; Followed by:20 members

Rate&Percentage

by Akansha » Mon May 23, 2011 9:14 pm

The rate of a reaction is directly proportional to the square of the concentration of
A and inversely proportional to concentration of B. If B increases by 100%,
which of the following is closest to the % change in concentration of A required
to keep the rate unchanged?
a. 100% decrease
b. 50% decrease
c. 40% decrease
d. 40% increase
e. 50% increase

OA is D

Can anyone help me with this?

Quote

Source: Beat The GMAT — Problem Solving | Mon May 23, 2011 9:14 pm

GMAT/MBA Expert

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

by Anurag@Gurome » Mon May 23, 2011 9:25 pm

Akansha wrote:The rate of a reaction is directly proportional to the square of the concentration of A and inversely proportional to concentration of B. If B increases by 100%,
which of the following is closest to the % change in concentration of A required
to keep the rate unchanged?
a. 100% decrease
b. 50% decrease
c. 40% decrease
d. 40% increase
e. 50% increase

OA is D

Can anyone help me with this?

Let rate of reaction = r, so r Î± AÂ² and r Î± 1/B
The, r Î± AÂ²/B or r = k(AÂ²/B)
Let the new concentration of A = A', and new concentration of B = B'
Now rate remains unchanged means rate is the same, so k(AÂ²/B) = k[(A')Â²/(B')]
B is increased by 100% means B becomes double, which means B' = 2B
So, k(AÂ²/B) = k[(A')Â²/(2B)]
AÂ² = (A')Â²/2 or (A')Â² = 2AÂ²
So, A' = Aâˆš2
Since âˆš2 = 1.414, so A' = 1.4A, which implies there is 40% increase.

The correct answer is D.

Anurag Mairal, Ph.D., MBA
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