Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 07 Jul 2015, 08:19

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# sorry if this problem has already been posted, but I do not

Author Message
TAGS:
Senior Manager
Joined: 30 May 2007
Posts: 496
Followers: 7

Kudos [?]: 70 [0], given: 0

sorry if this problem has already been posted, but I do not [#permalink]  19 Jun 2007, 05:52
(sorry if this problem has already been posted, but I do not feel like searching for it)

The rate of a chemical reaction is directly proportional to the square of the concentration of Chemical A and inversely proportional to the concentration of Chemical B. If the concentration of Chemical B is increased by 100%, which is closest to the % change in concentration of Chemical A required to keep the reaction unchanged.

A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

OA is D
Senior Manager
Joined: 30 May 2007
Posts: 496
Followers: 7

Kudos [?]: 70 [0], given: 0

Just in case you cannot see the white font at the bottom of my original post, the OA is D
Manager
Joined: 12 Apr 2007
Posts: 170
Followers: 1

Kudos [?]: 7 [0], given: 0

I dont have a formal method for solving this, but I'll try to explain how I would do this on an exam.

Since chemical B increased by 100%, or doubles, then the rate of the chemical reaction should be halved.

so what I did was assume the rate began at 100, and was halved to 50.

then because the rate is directly proportional to the square of the concentration of chemical A, let's say the concentration started at 10 (because 10^2 = 100). because the rate was halved, we need to double it again to get it from 50 back to 100, so essentially you'd have to get x^2 = 200.

with x^2 = 200, x is about 14. so we must increase the concentration from 10 to 14, about 40%

sorry for the shabby explanation. anyone have a formal method for doing this?
VP
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 143 [0], given: 0

Re: Another GMAT Prep problem [#permalink]  19 Jun 2007, 15:38
djhouse81 wrote:
(sorry if this problem has already been posted, but I do not feel like searching for it)

The rate of a chemical reaction is directly proportional to the square of the concentration of Chemical A and inversely proportional to the concentration of Chemical B. If the concentration of Chemical B is increased by 100%, which is closest to the % change in concentration of Chemical A required to keep the reaction unchanged.

A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

OA is D

You know, if you didn't post OA, I would have done this quesiton incorrectly. So thanks for that, now I have to remember the "inversely" proportional better.

Set r = rate, a = concentration of a, b = concentration of b, c = constant
r = c * a^2
r = (1/c) * b
increase b by 100%...you have
r = (1/c) * 2b
=> r/2 = (1/c) * b
Plug in r/2 for equation for a:
r/2 = c * a^2
=> r = c * (sqrt(2) * a)^2
=> r = c * (1.4 * a)^2
Increase by 40%!!!
VP
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 143 [0], given: 0

Just to let you know, my problem before was thinking that "inversely proportional" means this:

r = 1 / b which is incorrect.

The equation for "inversely proportional" is

r = (1/c) * b
Senior Manager
Joined: 30 May 2007
Posts: 496
Followers: 7

Kudos [?]: 70 [0], given: 0

Thank you for the help. I hope I never see that type of problem on the real thing, but if I do I will know to use a constant in the equation.

It goes to show that without an error log, I would never know where I went wrong, and that could have come back to haunt me on the real thing.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5068
Location: Singapore
Followers: 23

Kudos [?]: 195 [0], given: 0

R = k(A^2/B ) where k is a constant

B increases by 100%, so now R = k(A^2/2B)

To keep the rate constant, we need 2A^2, which can be written as sqrt(2)A * sqrt(2)A. So A needs to increase by 41% ( which is 1-sqrt(2)).
Senior Manager
Joined: 21 Jun 2006
Posts: 287
Followers: 1

Kudos [?]: 34 [0], given: 0

40 % increase

Rate = A^2/B

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

A2^2 = 2A1^2

A2 = 1.41A1

0r 41% increase, approx. 40% increase
Similar topics Replies Last post
Similar
Topics:
Edit - Sorry everyone, the problem is attached to this post 3 18 May 2008, 11:07
sorry if this problem has already been posted, but I do not 5 19 Jun 2007, 05:47
I am so sorry! I do not know why I cannot post the right 5 26 Apr 2007, 20:30
I know this one has been posted before by Kevincan but i 15 10 Aug 2006, 11:10
Sorry for post it again, but I really want to clear this 2 16 Nov 2005, 11:39
Display posts from previous: Sort by

# sorry if this problem has already been posted, but I do not

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.