keep the reaction

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Tue Apr 07, 2009 4:14 am

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. 25
B. 32
C. 36
D. 40
E. 44

OA D

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Source: Beat The GMAT — Problem Solving | Tue Apr 07, 2009 4:14 am

moutar: Senior | Next Rank: 100 Posts; Posts: 92; Joined: Wed Mar 25, 2009 12:13 pm; Thanked: 10 times

by moutar » Tue Apr 07, 2009 4:32 am

Take a = conc of A, b = conc of B and R = rate.

Then R = k(a^2)/b

Increasing b by 100% is the same as doubling

We need to find x in the equation (i.e. the inc/dec of A):
k(a^2)/b = k(xa)^2/2b

Divide by ka^2 and multiply by b

1 = x^2/2
x = sqrt (2)
= 1.415

Therefore increse in A is approx 40% = D

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

by Vemuri » Tue Apr 07, 2009 7:29 am

moutar wrote:Take a = conc of A, b = conc of B and R = rate.

Then R = k(a^2)/b

Increasing b by 100% is the same as doubling

We need to find x in the equation (i.e. the inc/dec of A):
k(a^2)/b = k(xa)^2/2b

Divide by ka^2 and multiply by b

1 = x^2/2
x = sqrt (2)
= 1.415

Therefore increse in A is approx 40% = D

Hi moutar,

Whatever you explained just over my head. Can you please explain or make it easier to understand?

Also, if (concentration of A)^2 is inversely proportional to concentration of B, don't you think any increase in B should reduce the concentration of A to maintain the balance? All the answer choices are positive & this makes me think that maybe I did not understand the question.

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moutar: Senior | Next Rank: 100 Posts; Posts: 92; Joined: Wed Mar 25, 2009 12:13 pm; Thanked: 10 times

by moutar » Tue Apr 07, 2009 8:26 am

R is directly proportional to the square of a.
i.e. R = ca^2 (where c is some constant)

R is inversely proportional to b.
i.e. R = d/b (where d is some constant)

Putting these together gives R = ka^2/b (k is a combination of d and c)

So for an increase of b, a would have to increase by the square root of the factor for R to stay the same.

Anything else?

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

by Vemuri » Tue Apr 07, 2009 7:45 pm

moutar wrote:R is directly proportional to the square of a.
i.e. R = ca^2 (where c is some constant)

R is inversely proportional to b.
i.e. R = d/b (where d is some constant)

Putting these together gives R = ka^2/b (k is a combination of d and c)

So for an increase of b, a would have to increase by the square root of the factor for R to stay the same.

Anything else?

Thanks moutar. It makes perfect sense to me now. Maybe a good night's sleep & your patient explanation did the trick

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