Last visit was: 10 Jul 2025, 19:18 It is currently 10 Jul 2025, 19:18
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
afyl128
Joined: 30 Mar 2010
Last visit: 13 Nov 2013
Posts: 74
Own Kudos:
599
 [511]
Given Kudos: 5
GMAT 1: 730 Q48 V42
29
Kudos
Add Kudos
482
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 Jul 2025
Posts: 102,631
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,631
Kudos: 740,202
 [146]
49
Kudos
Add Kudos
96
Bookmarks
Bookmark this Post
avatar
bhoglund
Joined: 03 Nov 2016
Last visit: 27 Apr 2017
Posts: 1
Own Kudos:
84
 [84]
Given Kudos: 12
Posts: 1
Kudos: 84
 [84]
60
Kudos
Add Kudos
24
Bookmarks
Bookmark this Post
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 10 July 2025
Posts: 21,070
Own Kudos:
26,129
 [31]
Given Kudos: 296
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,070
Kudos: 26,129
 [31]
20
Kudos
Add Kudos
11
Bookmarks
Bookmark this Post
Accountant
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

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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 Jul 2025
Posts: 102,631
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,631
Kudos: 740,202
 [28]
4
Kudos
Add Kudos
24
Bookmarks
Bookmark this Post
afyl128

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.
User avatar
GMATGuruNY
Joined: 04 Aug 2010
Last visit: 08 Jul 2025
Posts: 1,345
Own Kudos:
3,660
 [26]
Given Kudos: 9
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,345
Kudos: 3,660
 [26]
15
Kudos
Add Kudos
11
Bookmarks
Bookmark this Post
afyl128
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

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%

General Discussion
User avatar
afyl128
Joined: 30 Mar 2010
Last visit: 13 Nov 2013
Posts: 74
Own Kudos:
599
 [1]
Given Kudos: 5
GMAT 1: 730 Q48 V42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
afyl128
Bunuel
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
User avatar
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 885
Own Kudos:
1,693
 [1]
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 885
Kudos: 1,693
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Accountant
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
\(rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\)

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


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


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


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

Regards,
Fabio.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 Jul 2025
Posts: 102,631
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,631
Kudos: 740,202
 [12]
2
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
avatar
gurudabl
Joined: 09 Jun 2019
Last visit: 20 Nov 2022
Posts: 75
Own Kudos:
Given Kudos: 315
GMAT 1: 570 Q42 V29
Products:
GMAT 1: 570 Q42 V29
Posts: 75
Kudos: 48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 10 July 2025
Posts: 21,070
Own Kudos:
Given Kudos: 296
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,070
Kudos: 26,129
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gurudabl
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.
avatar
pat93
Joined: 04 Nov 2019
Last visit: 21 Mar 2021
Posts: 14
Own Kudos:
Given Kudos: 11
GMAT 1: 660 Q47 V34
GMAT 2: 720 Q49 V40
GMAT 2: 720 Q49 V40
Posts: 14
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ScottTargetTestPrep
Accountant
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

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

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!
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 10 July 2025
Posts: 21,070
Own Kudos:
Given Kudos: 296
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,070
Kudos: 26,129
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pat93
ScottTargetTestPrep
Accountant
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

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

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.
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 10 Jul 2025
Posts: 4,847
Own Kudos:
8,628
 [2]
Given Kudos: 225
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,847
Kudos: 8,628
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
jabhatta2
Joined: 15 Dec 2016
Last visit: 21 Apr 2023
Posts: 1,304
Own Kudos:
Given Kudos: 188
Posts: 1,304
Kudos: 281
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 10 Jul 2025
Posts: 16,101
Own Kudos:
74,247
 [1]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,101
Kudos: 74,247
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jabhatta2
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")
User avatar
avigutman
Joined: 17 Jul 2019
Last visit: 06 Jul 2025
Posts: 1,294
Own Kudos:
Given Kudos: 66
Location: Canada
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Expert
Expert reply
GMAT 3: 770 Q50 V45
Posts: 1,294
Kudos: 1,888
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
User avatar
egmat
User avatar
e-GMAT Representative
Joined: 02 Nov 2011
Last visit: 10 Jul 2025
Posts: 4,601
Own Kudos:
Given Kudos: 686
GMAT Date: 08-19-2020
Expert
Expert reply
Posts: 4,601
Kudos: 32,346
Kudos
Add Kudos
Bookmarks
Bookmark this Post
afyl128
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


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
User avatar
Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 440
Own Kudos:
Given Kudos: 5,899
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
GMAT 1: 450 Q42 V34
Posts: 440
Kudos: 42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

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

\(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 Jul 2025
Posts: 102,631
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,631
Kudos: 740,202
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kimberly77
Bunuel
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.

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.
 1   2   
Moderators:
Math Expert
102631 posts
PS Forum Moderator
686 posts