The rate of a certain chemical reaction is directly

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

05 Feb 2010, 13:06

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Re: GMAT Prep - Ratios [#permalink]

05 Feb 2010, 13:16

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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.
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Re: Formula Problem [#permalink]

20 Aug 2010, 23:14

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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)
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Re: GMAT PREP PS [#permalink]

23 Aug 2010, 19:06

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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.
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Re: Please help!! Difficult problems from GMATPrep [#permalink]

14 Sep 2010, 19:40

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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.
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Re: ps question [#permalink]

19 Oct 2010, 02:37

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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.

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Re: % increase , proportions [#permalink]

31 Jan 2011, 07:40

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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

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Re: Direct and Inverse Proportion Q [#permalink]

18 Mar 2012, 21:26

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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]
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Re: Direct and Inverse Proportion Q [#permalink]

19 Mar 2012, 01:16

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A-> (sq rt 2)*A

as R=K A^2/B

HENCE
((sq rt 2)-1)/1 =41% increase

hence D
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Re: Direct and Inverse Proportion Q [#permalink]

19 Mar 2012, 01:37

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dddmba2012 wrote:

Also discussed here: the-rate-of-a-chemical-reaction-is-directly-proportional-to-76921.html

DS problems about this concept:
the-amount-of-coal-a-train-burns-each-mile-is-directly-93667.html
in-a-certain-business-production-index-p-is-directly-63570.html

PS problems about this concept:
a-is-directly-proportional-to-b-when-a-8-b-88971.html
in-a-certain-formula-p-is-directly-proportional-to-s-and-80941.html

Hope it helps.
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Re: The rate of a certain chemical reaction [#permalink]

24 Oct 2012, 08:33

danzig wrote:

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.
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Re: The rate of a certain chemical reaction is directly [#permalink]

04 Nov 2012, 02:38

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

Re: The rate of a certain chemical reaction is directly [#permalink] 04 Nov 2012, 02:38

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The rate of a certain chemical reaction is directly

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