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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

GmatPrep - Percentages question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Mon Jul 11, 2011 1:13 pmFollowed by: 1 member
by tanyasethi » Sun Jul 31, 2011 5:00 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 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

This question has been answered before by Stuart before but I really did not understand it. Could someone please help me understand? :(

The approach I did not understand is here
https://www.beatthegmat.com/gmat-prep-ch ... 16793.html
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Mon Jul 11, 2011 1:13 pmFollowed by: 1 member

by tanyasethi » Sun Jul 31, 2011 11:59 pm
noone? :(
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Mon Aug 01, 2011 12:13 am
Hi,
Let r - rate of reaction
a - concentration of A
b - concentration of B
k - constant of proportionality
So, it is given that r = (ka^2)/b
the concentration of chemical B is increased by 100 percent. So, it becomes 2b
rate remains unchanged. So, r remains same. Let c be the new concentration of A.
So, r = kc^2/2b
ka^2/b = kc^2/2b =>c^2 = 2a^2
So, c = a(sqrt2) = 1.41a
So, concentration of A is increased by [(1.4a - a)/a]*100 = 40%

Hence, D
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Sat Jul 16, 2011 11:56 pmFollowed by: 1 member
Location: hyderabad
Thanked: 2 times

by naveen451 » Mon Aug 01, 2011 12:43 am
Let's start with A = 10 and B = 100, so the original ratio is 10^2/100 = 1.

If we double B, we have 10^2/200.

Now let's backsolve:

If we increase A by 50% (choice e), the new ratio is 15^2/200 = 225/200.

If we increase A by 40% (choice d), the new ratio is 14^2/200 = 196/200.

The 40% increase is much closer to our 1:1 ratio - choose (D).
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1