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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

23 Mar 2009, 10:03

1

This post received KUDOS

19

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

56% (02:16) correct
44% (01:16) wrong based on 427 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%, 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

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

23 Mar 2009, 12:19

Accountant wrote:

The rate of a chemical reaction is directly proportional to the square of concentration of chemcial A present and invesrsly proportional to the concentration of chemical B present. If chemical B is increased by 100% which of the following is the change in concentration of chemical A required to keep teh reaction rate unchanged:

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

Please explain your answer.

The rate of reaction is invesrsly proportional to the concentration of chemical B present. It used to have B=1 . Now that B is increased by 100%. So the new equation would be 2B=(1/2). In order for the rate of reaction to still be 1, we need to change the concentration of A to yield a 2. It used to be A^2=1, now the new equation should be (sqrt(2)*A)^2=2. The change in the concentration of A can be calculated as (sqrt(2) -1)/1 or approximately 40% increase. Answer D.

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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

27 Nov 2010, 17:06

Bunuel wrote:

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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)

Answer: D.

Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator?

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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)

Answer: D.

Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator?

\(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some non-zero constant \(x\) such that \(a=xb\);

\(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some non-zero constant constant \(y\) such that \(a=\frac{y}{b}\).

So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

28 Nov 2010, 08:56

Bunuel wrote:

afyl128 wrote:

Bunuel wrote:

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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)

Answer: D.

Brunel, why do you put the directly proportional in the nominator and inversely proportional in the denoimnator?

\(a\) is directly proportional to \(b\) means that as the absolute value of \(a\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some non-zero constant \(x\) such that \(a=xb\);

\(a\) is inversely proportional to \(b\) means that as the absolute value of \(a\) gets bigger, the absolute value of \(b\) gets smaller, so there is some non-zero constant constant \(y\) such that \(a=\frac{y}{b}\).

So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).

Hope it's clear.

Thanks =) do you have any links to similar questions? i'm very shaky on these

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

29 Sep 2013, 09:12

Hi Bunuel, I am a bit shaky with variation concepts and hence decided to get the basics clear,

I was referring to Karishma's blog here http://www.veritasprep.com/blog/2013/02 ... g-jointly/ Now I understand that if a rate varies directly for Eg X varies directly with Y than we have X/Y = K(some value) because in direct variation the ratio remains same. and in inverse variation it will be XY = K(some value) because x=1/y Please correct me if I misunderstood any concept till this point.

now to your explanation

Bunuel wrote:

NOTE: Put directly proportional in nominator and inversely proportional in denominator

how can we put a direct variation in numerator? because if I understand the concept correctly it should be in denominator? and an inverse in numerator..how did you arrived at this quick formula?

another question to may be both karishma and you Bunuel ( sorry karishma I am asking questions pertaining to your blog on this forum, but I thought this question can serve as a common reference.)

As per me It should be B/A^2 (infect If I look at karishma's sample question in same page its essentially the same question with just values swapped with N and M)

another point of confusion when B becomes double (i.e 2B) why don't we simply say A^2 also doubles(i.e 2 A^2) why do we say if a^2 has to double it has to be A^2 = 2 ?

if its a ratio than it should be multiplied and divided by same value in numerator and denominator (i.e 2)
_________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

16 Oct 2014, 23:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

19 Nov 2015, 23:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

29 Apr 2016, 06:54

Bunuel wrote:

\(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some non-zero constant \(x\) such that \(a=xb\);

\(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some non-zero constant constant \(y\) such that \(a=\frac{y}{b}\).

So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).

Hope it's clear.

Hi,

Could someone help to explain this one, how can we come up with the final result, which is \(R=\frac{(A^2x)y}{B}\)?

As far as I understand, according to the case in question, we have: 1. The rate is directly proportional to the square of concentration of chemical A -> R = \(xA^2\) (1) 2. The rate is inversely proportional to the concentration of chemical B -> R = \(\frac{y}{B}\) (2)

So next, how can we infer that \(R=\frac{(A^2x)y}{B}\)? What steps used to modify/combine (1) and (2) to get this one? Actually, from (1) and (2), I am thinking of \(\frac{(A^2x)*y}{B}\) as \(R*R\) = \(R^2\) instead

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

06 May 2016, 01:06

1

This post received KUDOS

thuyduong91vnu wrote:

Bunuel wrote:

\(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(b\) gets bigger too, so there is some non-zero constant \(x\) such that \(a=xb\);

\(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some non-zero constant constant \(y\) such that \(a=\frac{y}{b}\).

So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).

Hope it's clear.

Hi,

Could someone help to explain this one, how can we come up with the final result, which is \(R=\frac{(A^2x)y}{B}\)?

As far as I understand, according to the case in question, we have: 1. The rate is directly proportional to the square of concentration of chemical A -> R = \(xA^2\) (1) 2. The rate is inversely proportional to the concentration of chemical B -> R = \(\frac{y}{B}\) (2)

So next, how can we infer that \(R=\frac{(A^2x)y}{B}\)? What steps used to modify/combine (1) and (2) to get this one? Actually, from (1) and (2), I am thinking of \(\frac{(A^2x)*y}{B}\) as \(R*R\) = \(R^2\) instead

Please help to clarify. Thanks

Hello my friend.

This is how I solved this question. First I didnt assign those variebels x and y as above in the quote. I think those are used just as reference in order to show the proportonality.

Now to the question. I did assign variable x to be the factor of percent increase or decrease, r to be the rate, and a and b for the concetrations. So start with first formula \(r=a^2/b\) where a is directly proportonal ans is nuumerator and b is denominator since it is inversly proportinal. note u have to put \(a^2\) because it is given that the square root is directly proportonal now the second equation , the question asks to have same value for the rate but b is doubled or increase for 100% \(r=(xa)^2/2b\) or \(r=x^2a^2/2b\) now from here we can use the short way given by Bunuel or the long way to solve for x, either way it will come out as x^2=2 and\(x=1.41\). now earlier we said x is a factor of percentage change \(1+0.41\) or we have an increase of 41%

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

06 Aug 2016, 14:04

Accountant wrote:

The rate of a chemical reaction is directly proportional to the square of concentration of chemcial A present and invesrsly proportional to the concentration of chemical B present. If chemical B is increased by 100% which of the following is the change in concentration of chemical A required to keep teh reaction rate unchanged:

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

Please edit the question. It is " which of the following is the closest to the change". Makes a lot of difference.
_________________

It is not who I am underneath but what I do that defines me.

The rate of a chemical reaction is directly proportional to the square of concentration of chemcial A present and invesrsly proportional to the concentration of chemical B present. If chemical B is increased by 100% which of the following is the change in concentration of chemical A required to keep teh reaction rate unchanged:

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

Please edit the question. It is " which of the following is the closest to the change". Makes a lot of difference.

Re: The rate of a certain chemical reaction is directly proportional to th [#permalink]

Show Tags

26 Dec 2016, 13:26

1

This post received KUDOS

This question can also be solved by plugging in values for a and b and comparing the results:

- The rate of this reaction at first is \(r=\frac{A^2}{B}\) - If \(A=1\) and \(B=2\) the result of the equation is \(r=\frac{1}{2}\) - When \(B\) is increased by 100% the equation becomes \(r=\frac{A^2}{2B}\) - Since the question asks by how much \(A\) would need to increase in order for the reaction rate to remain the same, the second equation can be rewritten as \(\frac{1}{2}=\frac{A^2}{4}\) where \(r=\frac{1}{2}\), the same as in the first equation. - The second equation can then be rewritten as \(2=A^2\) and then as \(\sqrt{2}=A\) - The percent increase from \(A=1\) in the first equation to \(A=\sqrt{2}\) in the second equation is approximately 40% since\(\sqrt{2}=1.4\) - The answer is

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...