BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

% / word problem GMAT P1

Expert replies
by smallsorrow » Thu Nov 06, 2008 1:23 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 B is increased by 100%, which of the following is closest to the percent change in the concentration of A required to keep the reaction rate unchanged?

1)100% decrease
2)50% decrease
3)40% decrease
4)40% increase
5)50% increase
Join the discussion
Source: — Problem Solving |

by scoobydooby » Thu Nov 06, 2008 2:17 am
option 4

let rate be R, or R=(A^2)/B
B increased by 100% indicates B doubled or B2=2B1, rate has to be kept constant

A1^2/B1=A2^2/B2
A1^2/B1=A2^2/2B1, rearranging
A2^2/A1^2=2B1/B1=2
or A2/A1=sq rt 2/1
or A2=sq rt 2*A1
(A2-A1)/A1*100=(sq rt 2*A1-A1)/A1=(1.414-1)/1*100=41.4% (positive)
hence option 4: 40% increase
Join the discussion

by raunekk » Thu Nov 06, 2008 3:21 am
imo:D

40% increase..

LEt concentration of A (Ca)= 10 and conc. of B(Cb) = 5

acc. to question rate R = Ca^2/Cb = 100/5 = 20

now B conc. doubled = 10

thus to reach a rate R as close as 20,we need A to be around 200...

so, if we take Ca=14 , R = 196 / 10 = 19.6 = 20 (approxi)

Thus % change = (14 - 10)/ 100 = 40 % increase.

i hope this helps..
Join the discussion