The rate of a certain chemical reaction is directly proportional to th

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

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.

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%

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

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

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


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


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


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

Regards,
Fabio.

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Hope it helps.

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.

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.

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

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")

Video solution from Quant Reasoning:
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GIVEN:

• 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.
• The concentration of chemical B is increased by 100 percent.

TO FIND:

• The percent change in the concentration of chemical A required to keep the reaction rate unchanged.

SOLUTION:
Let’s first understand all that is given to us. We will translate all the English into Math and from there, find our path forward.
Let’s read the question part-by-part and translate each part to Math before we move to the next.

TRANSLATION - Understanding the Given information:
Part (I): “The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present.”

Let’s fix some symbols first. Denote:

- The rate of the chemical reaction by R
- Concentration of chemical A by A

Now, per (I) above, we have that R is directly proportional to \(A^2\). Mathematically, this can be written as: R ∝ \(A^2\) ---- (i)


Part (II): “The rate of a certain chemical reaction is inversely proportional to the concentration of chemical B present.”

Again, let’s denote the concentration of chemical B by B.

So, per (II) above, we have that R is inversely proportional to B. Mathematically, this can be written as: R ∝ 1/B ---- (ii)

From (i) and (ii), R ∝ \(A^2\) and R ∝ 1/B. Thus, R ∝ \(A^2\)/B.
Thus, R = k\(\frac{A^2}{B}\), for a constant of proportionality ‘k’.


Part (III): “The concentration of chemical B is increased by 100 percent.”
Since we denote the concentration of B by B, per (III), B is increased by 100%.
So, new concentration of B is 200% of B, that is, 2B --------- (iii)


TRANSLATION - Understanding the asked question:
We need to find “the percent change in the concentration of chemical A required to keep the reaction rate unchanged.”
This means that we’re looking for the change that A must undergo so that reaction rate remains unchanged despite B changing to 2B.


FINDING REQUIRED PERCENTAGE – Putting everything together

- Original rate equation (before B changed): R = k\(\frac{A^2}{B}\). ----- (iv)
- New rate equation (after B changed to 2B): R = k\(\frac{(New A)^2}{2B}\). ----- (v)

From (iv) and (v), k\(\frac{A^2}{B}\) = k\(\frac{(New A)^2}{2B}\).
Simplifying the equation above, we get \(A^2 = \frac{(New A)^2}{2}\).
Thus, \((New A)^2 = 2A^2\), OR New A = √2 A. ----------(vi)


So, the change in concentration of A is an increase from A to √2A. As % change, this is:

• = \(\frac{(√2A − A)}{A}\) × 100
• = \(\frac{(√2 − 1)}{1}\) × 100
• ≈ (1.41 – 1) × 100
• ≈ 40% increase

Correct Answer: Choice D


TAKEAWAYS:

1. Always translate English to Math as you go!
2. Proportionality:
   
   1. If x is directly proportional to y, we write x ∝ y.
   2. If x is inversely proportional to z, we write x ∝ 1/z.
   3. If x is directly proportional to y and inversely proportional to z, we combine (a) and (b) and write x ∝ y/z.
3. Percentage change from A to B = \(\frac{(B−A)}{A}\) × 100

Hope this helps!


Shweta Koshija
Quant Product Creator, e-GMAT

Great explanation Bunuel
One question not sure why new and old rate are being equal to each other in below? Could you kindly help clarify? Thanks

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

Check the highlighted part in the question above.