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The rate of a chemical reaction is directly proportional to

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The rate of a chemical reaction is directly proportional to [#permalink] New post 23 Mar 2009, 10:03
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Mar 2012, 13:04, edited 1 time in total.
Edited the question and added the OA
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Re: PS Rate of chemical reaction GMAT prep [#permalink] New post 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.
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Re: PS Rate of chemical reaction GMAT prep [#permalink] New post 23 Mar 2009, 15:13
CA: Concentration A
CB: Concentration B
R: Reaction rate

Formula for reaction rate:

R = (CA^2) / CB

Thus if CB is increased by 100% >> means concentration doubles >> 2xCB

Thus, for R to remain the same (CA^2) also has to double.

>> 2 x (CA^2) >> to see what happens to CA, take the 2 into the bracket by taking its root

>> (SQRT2 CA)^2

SQRT 2 is roughly 40% >> answer D
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Re: GMATPrep Percent change of chemical reaction [#permalink] New post 27 Nov 2009, 18:43
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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.
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Re: GMATPrep Percent change of chemical reaction [#permalink] New post 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?
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Re: GMATPrep Percent change of chemical reaction [#permalink] New post 28 Nov 2010, 01:12
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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 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.
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Re: GMATPrep Percent change of chemical reaction [#permalink] New post 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
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Re: GMATPrep Percent change of chemical reaction [#permalink] New post 28 Nov 2010, 09:27
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afyl128 wrote:

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


DS problems about this concept:
ds-question-93667.html
og-proportional-index-63570.html

PS problems about this concept:
easy-proportion-question-88971.html
vic-80941.html

Hope it helps.
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Re: GMATPrep Percent change of chemical reaction [#permalink] New post 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 ? :?: :o

if its a ratio than it should be multiplied and divided by same value in numerator and denominator (i.e 2)
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Re: The rate of a chemical reaction is directly proportional to [#permalink] New post 01 Oct 2013, 11:22
Hi Bunuel...waiting for reply.. have exam this week so a bit nervous :)
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Re: The rate of a chemical reaction is directly proportional to   [#permalink] 01 Oct 2013, 11:22
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