Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

GMATPrep Probability

Expert replies
by razorback » Thu Oct 06, 2011 11:11 pm
Quite baffled...

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

OA: D
Join the discussion
Source: — Problem Solving |

by shankar.ashwin » Thu Oct 06, 2011 11:17 pm
Given,

R = k A^2 / B

If B is increased 100%, we have 2B but the A^2 / 2B ratio should be the same for the rate to be equal.

Since there is a extra 2 in denominator, we should look to increase the Num to compensate for it (or cancel out the 2)

A is squared, so Sqrt2 A, when squared will give 2A^2 and the 2 cancels out.
Sqrt2 is 1.414, so thats a 41% increase. Approx D
Join the discussion

by sl750 » Fri Oct 07, 2011 1:10 am
R(original) = k*A^2/B
Let A' be the new value of A required to keep the Rate same
R(New) = k*A'^2/2B

As we require R(new) = R(original), we have A'^2/2B = A^2/B or A'^2 = 2A^2 or A' = sqrt(2)*A
Percent change in concentration of A is (sqrt(2)*A -A)/A = 40%increase
Join the discussion

by GMATGuruNY » Fri Oct 07, 2011 4:36 am
razorback wrote:Quite baffled...

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

OA: D
R is directly proportional to A² means that as R increases, A² increases by a proportional amount:
R/A² = k.
In the equation above, if R doubles, then A² must double, so that the value of R/A² is always equal to the constant k.

R is inversely proportional to B means that as R increases, B decreases by a proportional amount:
RB = k.
In the equation above, if R doubles, then B must be halved, so that the value of RB is always equal to the constant k.

Thus, the problem above describes the following relationship:
R = A²/B.
In the equation above:
R/A² is always equal to B and RB is always equal to A².

Given R=A²/B, we can plug in values.

Plug A=10 and B=1 into R=A²/B:
Then R = 10²/1 = 100.

If B doubles and R is unchanged, we get:
100 = A²/2
200 = A²
A = √200 = 10√2 ≈ 14.

A increases from 10 to 14.
Percent increase in A = Difference/Original = 4/10 = 40%.

The correct answer is D.

It would be wise to remember the following:

x is directly proportional to y and inversely proportional to z means x = y/z.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by Anurag@Gurome » Fri Oct 07, 2011 5:00 am
razorback wrote:Quite baffled...

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

OA: D

Let the rate of a chemical reaction = R
Then R = k(A^2)/B ...Equation 1
Let the new concentration of chemical A and B be a and b respectively. B is increased by 100% means that B is doubled, which means b = 2B.

Then R = k(a^2)/2B ...Equation 2
From equation 1 and 2, k(a^2)/2B = k(A^2)/B
a^2 = 2.A^2
a = A√2
a = 1.4A, which implies there is 40% increase.

The correct answer is D.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by razorback » Fri Oct 07, 2011 7:52 pm
Thanks! Makes sense now.

In all the quant prep materials Ive come across I havent seen a proportionality question this difficult.

Also, not sure why I labeled the post "Probability", if a mod can and wants to fix it to "Proportionality" that would be sweet. Thanks again.
Join the discussion

by abhi0697 » Sat Oct 08, 2011 6:12 pm
Kindly check out the below link for video solution:

https://www.youtube.com/watch?v=KNR3R78SHPA


_____________________________________________________
I2-Infiniti Education
GMAT Classes in Singapore
www.i2-infiniti.com
Join the discussion

by saketk » Mon Oct 10, 2011 3:53 am
The expression will be

R = K (Ca)^2/(Cb)

let Ca= 4
and Cb = 1
This will give R's value= 16K ---(1)

As per the condition, Cb is increased by 100% i.e. the new value of Cb =2

put this value in the equation (1)

We will get the new value of Ca --

16k= k *(Ca)^2/2 or Ca = 4*1.41

i.e. the value is increased by 41% approx

Answer D
Join the discussion