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


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.


To elaborate more: \(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) 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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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
Show Spoiler
D

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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We can let n = the rate of a certain chemical reaction, a = the concentration of chemical A, and b = the concentration of chemical B. We are given that 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, so, for some positive constant k, we have:

n = ka^2/b

When b is increased by 100 percent, b becomes 2b. To keep the reaction rate unchanged, we can let a become c, so we have:

ka^2/b = kc^2/(2b)

2bka^2 = bkc^2

2a^2 = c^2

c = √(2a^2)

c = a√2

Since √2 ≈ 1.4, c ≈ 1.4a or approximately 140% of a, i.e., a 40% increase in the concentration of chemical A.

Answer: D
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\(rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\)

\(\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)\)


\(?\,\, \cong \,\,k - 1\)


\({\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
{k^2} = 2 \hfill \\
\sqrt 2 \cong 1.41 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2 - 1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2 - 1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Theory:
Variations On The GMAT - All In One Topic

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Ultimate GMAT Quantitative Megathread

Hope it helps.
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Given: 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.

Asked: 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?

Rate of chemical reaction = constant * (concentration of chemical A)^2 / (concentration of chemical B)
r = k * Ca^2 / Cb

Cb' = 2Cb

r = k Ca'^2/2Cb

Ca'^2 = 2Ca^2
Ca' = \sqrt 2 * Ca
Ca' = 1.4 * Ca
40% increase in concentration of chemical A

IMO D
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x is directly proportional to y, translated into math:
\(x = ky\), where k is a constant.
x is inversely proportional to z, translated into math:
\(x = \frac{k}{z}\), where is k is a constant.
x is directly proportional to y and inversely proportional to z, translated into math:
\(x = \frac{ky}{z}\), where k is a constant.

In the problem above, the rate is directly proportional to the square of A and inversely proportional to B.
Thus:
\(R = \frac{{kA²}}{B}\)

Original values:
Let:
A = 10
B = 1
k = 1
Then:
\(R = \frac{1*10²}{1}\)
R = 100

New values:
Since the rate doesn't change, R = 100.
B increased by 100% = 2.
k = 1. (Since k is a constant.)
Solving for A, we get:
\(100 = \frac{1*A²}{2}\)
\(200 = A²\)
\(A = √200 = √100 * √2 = 10√2 ≈ 10(1.4) = 14\)

Percent increase in A:
(increase in A)/(original A) \(= \frac{14-10}{10} = \frac{4}{10} =\) 40%


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Hi Bunuel ScottTargetTestPrep Could anyone clear my small doubt?
Why did you guys ignore the constant?
Nonetheless, isn't there a possibility for two different constants for both A and B?

Thank you,
Dablu

My solution did not ignore the constants; I used "k" to denote the constant of proportionality, but it canceled out later. The constant can, on the other hand, be ignored for this question because you are looking for a value which will keep a ratio unchanged. If the value (A^2)/B is unchanged without the constant, it will still be unchanged with the constant (which is (k*A^2)/B).

In regards to why there are not different constants for A and B, there actually is. If you kept B fixed, then R = p*A^2 for some p and if you kept A fixed, then R = s/B for some s. The values of p and s are not necessarily equal and you will obtain different values of p and s depending on which value of B and A you are fixing. When A and B are both varying, then the equation becomes R = k*(A^2)/B where k is yet another constant. Whatever value you choose for A and B, this k will not change.
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Thank you ScottTargetTestPrep!

Or we can simply say, p and s can be written as p*s = k in the equation and this constant k will stay the same even if the value of B were to double.
p*A^2 * s/ B = p*A^2 * s/2B

p and s can multiply can be another constant "k", which will stay the same even when B doubles. I think I got it!
Thank you for giving a great explanation and keeping up with my queries.

Happy to help!
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Hi ScottTargetTestPrep,

I followed the plug in approach, with A=1 and B=1. So the first value was 1^2/1, then the rate changed to x^2/2=1, finally having x√2. At this point, I guessed for 50% increase thinking that 1.5^2 would be closer to 2. Obviously if I just memorized that 1.4^2 is approximately 2, I would have gotten this question correct. My question is, would any other numbers be better for this type of question or the decision would have always come down to knowing the value of 1.4^2? THANK YOU!

You really should know that 1.4^2 is about 2. But more importantly, you should know that 14^2 is 196. If you knew that it would not be hard to see that 1.4^2 = 1.96, which is about 2.
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Solution:

In questions of proportionality, first create the equation and then proceed ahead.

Rate(R1) = k * Concentration A^2/Concentration B

=k* A^2/B where k is the constant of proportionality.

Concentration of chemical B is increased by 100 percent=>New value = 2B

For R1 to be equal to new rate R2 we should have

k * A1^2/B = k* A2^2/2B

=>A1^2 = A2^2/2

=> A2^2 = 2*A1^2

=> A2 = Sqrt(2*A1) =>The new value A2 is sqrt2 times or approximately 1.4 times A1 (old value)

=> 40% increase(approx.) (option d)

Devmitra Sen
GMAT SME
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I did it a bit differently.
As the rate of the chemical reaction goes up, the concentration of A^2 goes directly up and the concentration of B goes down. So if B goes up, then A^2 goes down. So say B is 10, and it doubled to 20. And let us say A^2 = 100, then A^2 will go from 100 to 50 (cut in half since B was doubled).
We had A^2= 100; A = 10
Now we have A^2 = 50; A = 7
Remember that A^2 is directly proportional to the rate of the reaction. So we need to know by what percentage do we need to increase A (which is 7 after the increase of B) to get it back to what it originally was before B was added (which was 10). To figure this out, just do ((10-7)/7) * 100 = increase of 40%.
So if you increase A by 40%, A^2 will increase in such a way that a directly proportional increase of the rate of the reaction (with respect to A) will place you in a position where the rate of the reaction is the same as before.

Hope this helps!

Bunuel VeritasKarishma - Per my method, A is reducing (not increasing)

Could you please let me know where is flaw in my thinking

eq1) Rate = k x \(A^2\)
eq2) Rate = m x \(\frac{1}{ B}\)

Given the Left hand side of the equations is Rate in both cases then
k x \(A^2\) = m x \(\frac{1}{ B}\)

Moving around, we get
\(A^2\) x B = \(\frac{m}{ k }\)

Now m and k are constants
One constant (m) over another constant (k) will be another constant obviously

Hence --> \(A^2\) x B = Constant (\(\frac{m}{ k }\))

Now if we double B as per the question, A needs to change from A to \(\frac{A}{\sqrt{2}}\) to keep the equation as-is [Given chemical A is squared]

Hence in my scenario -- A is reducing (not increasing) as going from A to \(\frac{A}{\sqrt{2}}\) is a reduction of A by 33 %, not an increase of A

When A varies with B (C constant) and A varies with C (B constant), then the relation between B and C (keeping A constant) is obtained by joint variation.

\(\frac{Rate}{A^2} = Constant\) (When B is constant)
\(Rate*B = Constant\) (when A is constant)

means

\(\frac{Rate*B}{A^2} = Constant\)

So when Rate is kept the same, B varies directly with A^2. So if B doubles, A increases by 40%.

Note that you cannot equate the right hand sides of these equations as you have done because they hold under different constraints ("when B is constant" or "when A is constant")
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