January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 23 Jun 2008
Posts: 138

The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
05 Feb 2010, 12:06
Question Stats:
55% (01:05) correct 45% (01:36) wrong based on 2103 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: GMAT Prep  Ratios
[#permalink]
Show Tags
05 Feb 2010, 12:16
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%, 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 NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case). We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means \(x^2=2\) > \(x\approx{1.41}\), which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}\) Answer: D.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 360
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: GMAT PREP PS
[#permalink]
Show Tags
23 Aug 2010, 18:06
Given equation = \(A^2/B\) Also given that the concentration of B is increased by 100 %. Hence B becomes 2B. New concentration = \(A^2/2B\). In order to keep the new concentration the same as the original concentration, A should be equal to \(\sqrt{2}A\) so that A^2/2B becomes \(A^2/B\) \(\sqrt{2}A\) == 1.414 * A or approximately 141% of A. 141% of A is 41 % increase of A. Closest answer choice is D.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog




Manager
Joined: 18 Feb 2010
Posts: 151
Schools: ISB

Re: Formula Problem
[#permalink]
Show Tags
20 Aug 2010, 22:14
CR = (A^2) * ( 1/B) put any values 625 = 25*25 B increases by 100% so it becomes 50 CR remains same at 625 1/B becomes 50 so A^2 should be 625/50 = 12.5 or say decreased by 50% (b)
_________________
CONSIDER AWARDING KUDOS IF MY POST HELPS !!!



Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 360
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: Please help!! Difficult problems from GMATPrep
[#permalink]
Show Tags
14 Sep 2010, 18:40
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%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged? Given: Rate  \((A^2/B)\) B increased by 100%. New B is 2B. Rate  \((A^2)/2B\). In order for the initial rate of chemical reaction to be the same as the final rate. A should be increased such that it should cancel out the 2 in the denominator. Hence A should be \(A\sqrt{2}\) and hence final rate of reaction is \((A\sqrt{2}))^2/2B\) \(2A^2/2B\) which becomes the same as the initial rate of reaction. Hence A should be \(A\sqrt{2}\). Or final concentration of A is \(1.41A\) or an increase of approximately 41%. Hence answer is D.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog



Manager
Joined: 08 Sep 2010
Posts: 170
Location: India
WE 1: 6 Year, Telecom(GSM)

Re: ps question
[#permalink]
Show Tags
19 Oct 2010, 01:37
satishreddy wrote: the rate of certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of the chemical B present. if the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged.
100% decrease 50% increase 40% decrease 40% increase 50% increase Let the rate of the reaction is r and concentration of chemical A is A and concentration of chemical B is B. According to the question , r is proportional to A^2 and r is also proportional to 1/B Combining these two condition , r =K (A^2/B) , where K is proportionality constant. Now in the changed condition B has increased by 100 % means is has become 2B. So for keeping the rate constant we have to change the concentration of A, from A to ((Square root of 2)*A) As K (((square root of 2)A)^2)/2B = K (A^2/B) =r Now, let after increase of x% ,A has become ((square root of 2)A) i.e 1.414*A A+(xA/100)=1.414 A Hence x=41.4% Answer is D. Please consider KUDOS if it helped u in some way.Thanks



Intern
Joined: 15 Jan 2011
Posts: 6

Re: % increase , proportions
[#permalink]
Show Tags
31 Jan 2011, 06:40
tradinggenius wrote: The rate of a 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?
100% decrease 50% decrease 40% decrease 40% increase 50% increase Let r be the rate of chemical reaction and A and B be the concentration of chemical A and chemical B respectively. r is proportional to A^2 r is inversely proportional to B thus, r = k(A^2/B), where k is some constant. now B becomes 2B (100% increase). Let x be the new concentration of A. The new equation becomes r=k(x^2/2B) Since r remains the same, we have k(A^2/B)=k(x^2/2B) so, x=1.4A which is an increase of 40% D is the answer



SVP
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 1525

Re: Direct and Inverse Proportion Q
[#permalink]
Show Tags
18 Mar 2012, 20:26
Rate= r = k (A^2) (1/B) [where k is a constant] If B is replaced by 2B, then r = k (A^2) (1/2B) => k (A'^2) (1/2B) = k (A^2) (1/B) [where A' is the new concentration of A] => A'^2 = 2(A^2) => A' = sqrt (2) * A = 1.414 *A Therefore the concentration of A must increase by ~40% [option D]
_________________
GyanOne  Top MBA Rankings and MBA Admissions Blog
Top MBA Admissions Consulting  Top MiM Admissions Consulting
Premium MBA Essay ReviewBest MBA Interview PreparationExclusive GMAT coaching
Get a FREE Detailed MBA Profile Evaluation  Call us now +91 98998 31738



Manager
Joined: 12 Mar 2012
Posts: 237
Concentration: Operations, Strategy

Re: Direct and Inverse Proportion Q
[#permalink]
Show Tags
19 Mar 2012, 00:16
A> (sq rt 2)*A as R=K A^2/B HENCE ((sq rt 2)1)/1 =41% increase hence D
_________________
Practice Practice and practice...!!
If my reply /analysis is helpful>please press KUDOS If there's a loophole in my analysis> suggest measures to make it airtight.



Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: Direct and Inverse Proportion Q
[#permalink]
Show Tags
19 Mar 2012, 00:37



Senior Manager
Joined: 15 Jun 2010
Posts: 297
Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)

Re: The rate of a certain chemical reaction
[#permalink]
Show Tags
24 Oct 2012, 07:33
danzig wrote: 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%, 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 From Q Stem Chemical reaction R =K (A^2/B) Now if B is increased by 100%> it becomes R=K (A^2/2B) Now To bring the relationship to its initial condition we need to cancel out 2 in denominator. So my multiplying root 2 we can do that. So root2 = 1.4 approx. That means we need to increase 40% of A. Hence D.
_________________
Regards SD  Press Kudos if you like my post. Debrief 610540580710(Long Journey): http://gmatclub.com/forum/from600540580710finallyachievedin4thattempt142456.html



Intern
Joined: 11 Jul 2012
Posts: 40

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
04 Nov 2012, 01:38
Hello Bunuel I saw this question coming around couple of times. But doesn't the answer depend on how you express the concentration reaction? If I said CR = A^2  B, (or in general a*A^2 + b*B, where a >0 and b <0) the answer will still be D? Thanks for your time Brother Karamazov



Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
08 Jul 2013, 00:09



Manager
Joined: 13 Jul 2013
Posts: 65

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
04 Jan 2014, 00:49
I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this?



Manager
Joined: 20 Dec 2013
Posts: 122

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
04 Jan 2014, 04:12
theGame001 wrote: I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this? Lets make your life more easy. Let us say x = 4, y = 2 then R = 16/2 = 8 Now keep R same and double y to see what is the change in X. R = X^2/Y 8 = X^2/4 X = 4\sqrt{2} X got multiplied by \sqrt{2} which means a 40 % increase OR R = X^2/Y If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase
_________________
76000 Subscribers, 7 million minutes of learning delivered and 5.6 million video views
Perfect Scores http://perfectscores.org http://www.youtube.com/perfectscores



Manager
Joined: 13 Jul 2013
Posts: 65

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
04 Jan 2014, 04:19
PerfectScores wrote: theGame001 wrote: I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this? Lets make your life more easy. Let us say x = 4, y = 2 then R = 16/2 = 8 Now keep R same and double y to see what is the change in X. R = X^2/Y 8 = X^2/4 X = 4\sqrt{2} X got multiplied by \sqrt{2} which means a 40 % increase OR R = X^2/Y If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase Thank you but what I am unable to understand is why are we multiplying? x=4 and y=2 Squared 4= 16  I understood this Inversely proportional = 1/2  Understood this 16*1/2 Unable to understand this part



Manager
Joined: 20 Dec 2013
Posts: 122

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
04 Jan 2014, 04:22
theGame001 wrote: PerfectScores wrote: theGame001 wrote: I am unable to understand this question.
Rate= a^2 Rate= 1/b
Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?
Can someone please help with this? Lets make your life more easy. Let us say x = 4, y = 2 then R = 16/2 = 8 Now keep R same and double y to see what is the change in X. R = X^2/Y 8 = X^2/4 X = 4\sqrt{2} X got multiplied by \sqrt{2} which means a 40 % increase OR R = X^2/Y If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase Thank you but what I am unable to understand is why are we multiplying? x=4 and y=2 Squared 4= 16  I understood this Inversely proportional = 1/2  Understood this 16*1/2 Unable to understand this part R is directly proportional to X^2 and inversely proportional to Y which means that R is directly proportional to X^2/Y So if x = 4 and y = 2 R is directly proportional to (4^2)/2 which is equivalent to X^2/Y
_________________
76000 Subscribers, 7 million minutes of learning delivered and 5.6 million video views
Perfect Scores http://perfectscores.org http://www.youtube.com/perfectscores



Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 772
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
09 Apr 2014, 13:26
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%, 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 \(Rate = \frac{k*A^2}{B}\) \(Rate = \frac{k*NewA^2}{2B}\) \(\frac{k*NewA^2}{2B} = \frac{k*A^2}{B}\) \(NewA = sqrt{2} * A\) \(%change = \frac{sqrt{2}*A  A}{A}*100\) \(%change= 0.414*100=41.4% increase\) Answer : D
_________________
Piyush K
 Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press> Kudos My Articles: 1. WOULD: when to use?  2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".



Intern
Status: Going the extra mile
Joined: 08 Feb 2014
Posts: 16
Location: Netherlands
Concentration: Strategy, International Business
GMAT 1: 470 Q37 V18 GMAT 2: 570 Q36 V32 GMAT 3: 560 Q37 V30 GMAT 4: 610 Q41 V34

Re: The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
03 May 2014, 01:50
Ratio A to B : A^2 = B Say B = 100 Then A = 10^2 B Increases 100% ==> 200 x^2 = 200 ? 14^2=196 (almost 200, so x is a slightly bigger) 10 ==> 14 ==> increase of 40%. Hence Answer D.
_________________
Structural persistence is the key to succes . Party hard, study harder.
Still bashing, will continue to do so , although it's important to chill aswell ; ) STUDY+CHILL=VICTORY



Director
Joined: 03 Feb 2013
Posts: 848
Location: India
Concentration: Operations, Strategy
GPA: 3.88
WE: Engineering (Computer Software)

The rate of a certain chemical reaction is directly
[#permalink]
Show Tags
21 Dec 2014, 07:40




The rate of a certain chemical reaction is directly &nbs
[#permalink]
21 Dec 2014, 07:40



Go to page
1 2
Next
[ 39 posts ]



