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.
_________________

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.
_________________

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
_________________

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

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


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


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


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

Regards,
Fabio.
_________________

Theory:
Variations On The GMAT - All In One Topic

Questions:
http://gmatclub.com/forum/a-is-directly ... 88971.html
http://gmatclub.com/forum/the-rate-of-a ... 90119.html
http://gmatclub.com/forum/if-the-price- ... 50508.html
http://gmatclub.com/forum/in-a-certain- ... 46815.html
http://gmatclub.com/forum/a-spirit-and- ... 68909.html
https://gmatclub.com/forum/in-a-certain ... 80941.html
https://gmatclub.com/forum/a-certain-qu ... 47469.html
https://gmatclub.com/forum/recently-fue ... 44188.html
https://gmatclub.com/forum/the-cost-of- ... 44190.html
https://gmatclub.com/forum/the-price-of ... 44191.html
https://gmatclub.com/forum/if-the-ratio ... 44184.html
https://gmatclub.com/forum/20-workmen-c ... 44185.html
https://gmatclub.com/forum/the-variable ... 05761.html
https://gmatclub.com/forum/in-a-certain ... 63570.html
http://gmatclub.com/forum/the-amount-of ... 93667.html

Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________

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
_________________

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%


_________________

Given:
r = (k.(Ca)^2)/Cb (where Ca is conc. of chemical A and Cb is conc. of chemical B)
Cb increases by 100% => Cb' = 2Cb (Cb'is the new conc. of chemical B)
r is same =>
(Ca)^2 / Cb = (Ca')^2/Cb' (Ca'is the new conc. of chemical A)
Cb' = 2Cb
(Ca')^2 = 2(Ca)^2 => Ca' = 1.41Ca (Implying 40% increase in existing conc. of chemical A)

Ans: D
_________________

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.
_________________

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!
_________________

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!

Rate is Directly Proportional to the SQUARE of Concentration A

R = k * (A)^2


Rate is Inversely Proportional to Concentration B


R = (k) / B



If B Increases by 100% ------> This is the Same as 2x (TWICE) B ----- i.e., Doubling B

since the Rate is Inversely Proportional to B -------> the Rate will Decrease by 1/2 (50%) and we will have 1/2 the Rate


In order to keep the Rate the Same, we need to DOUBLE the Rate


R = k * (A)^2 ---------- if we MULTIPLY the Concentration of A by sqrt(2) -----> this will lead to DOUBLING the Rate

sqrt(2) is approximately = 1.414


1.414 * (A) = 1(A) + .414(A) = 100%(A) + 41.4%(A)


Approximately a 40% Increase in A will keep the Rate UNCHANGED


-D-

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.
_________________

@VeritasKarishma/ @Bunuel/@Chetan2u (Any other experts also),

I did not get the highlighted equation (how it is derived), can anyone please let me know how joint variation is derived?

Thanks!

Rate/M^2 = k

Rate*N = k

Rate*N/M^2 = k

If Rate has to remain constant, N/M^2 must remain the same too.

If N is doubled, M^2 must be doubled too i.e. M must become ?2 times. Since ?2 = 1.4 (approximately),

M must increase by 40%.

Why should the increase of A^2 be equal to 2??
_________________