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

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23 Mar 2009, 11:03

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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

Re: PS Rate of chemical reaction GMAT prep [#permalink]

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23 Mar 2009, 13:19

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.

Re: GMATPrep Percent change of chemical reaction [#permalink]

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27 Nov 2009, 19:43

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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. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)

Re: GMATPrep Percent change of chemical reaction [#permalink]

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27 Nov 2010, 18:06

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?

Re: GMATPrep Percent change of chemical reaction [#permalink]

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28 Nov 2010, 02:12

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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 \(R=\frac{(A^2x)y}{B}\).

Re: GMATPrep Percent change of chemical reaction [#permalink]

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28 Nov 2010, 09:56

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: GMATPrep Percent change of chemical reaction [#permalink]

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29 Sep 2013, 10:12

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 chemical reaction is directly proportional to [#permalink]

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17 Oct 2014, 00:38

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

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

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29 Apr 2016, 07:54

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 chemical reaction is directly proportional to [#permalink]

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06 May 2016, 02:06

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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%

gmatclubot

Re: The rate of a chemical reaction is directly proportional to
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06 May 2016, 02:06

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