GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Nov 2019, 23:15 ### 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

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

Author Message
TAGS:

### Hide Tags

Manager  Joined: 30 Mar 2010
Posts: 79
GMAT 1: 730 Q48 V42 The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

15 00:00

Difficulty:   75% (hard)

Question Stats: 55% (01:57) correct 45% (02:17) wrong based on 1465 sessions

### HideShow timer Statistics

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
Math Expert V
Joined: 02 Sep 2009
Posts: 59125
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

7
5
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. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}$$

To elaborate more: $$a$$ is directly proportional to $$b$$ means that as the absolute value of $$b$$ 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 $$b$$ 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.
_________________
Intern  Joined: 03 Nov 2016
Posts: 1
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

4
1
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$$\sqrt{2}=1.4$$
##### General Discussion
Senior Manager  G
Joined: 04 Aug 2010
Posts: 492
Schools: Dartmouth College
The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

2
afyl128 wrote:
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%

_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8412
Location: United States (CA)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

1
1
Accountant wrote:
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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager  Joined: 30 Mar 2010
Posts: 79
GMAT 1: 730 Q48 V42 Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

Bunuel wrote:
afyl128 wrote:
Bunuel wrote:
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. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}$$

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
Math Expert V
Joined: 02 Sep 2009
Posts: 59125
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

afyl128 wrote:

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

ds-question-93667.html
og-proportional-index-63570.html

easy-proportion-question-88971.html
vic-80941.html

Hope it helps.
_________________
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

Accountant wrote:
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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Math Expert V
Joined: 02 Sep 2009
Posts: 59125
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

SVP  P
Joined: 03 Jun 2019
Posts: 1849
Location: India
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

afyl128 wrote:
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 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
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning
All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com
Manager  G
Joined: 12 Aug 2017
Posts: 52
Concentration: General Management, Marketing
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

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
_________________
The key is not the will to win, it's the will to prepare to win that is important !!
Manager  B
Joined: 09 Jun 2019
Posts: 84
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

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
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8412
Location: United States (CA)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

gurudabl wrote:
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.
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager  B
Joined: 09 Jun 2019
Posts: 84
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

Thank you ScottTargetTestPrep!
Quote:
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.

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. Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8412
Location: United States (CA)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

gurudabl wrote:
Thank you ScottTargetTestPrep!
Quote:
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.

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!
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: The rate of a certain chemical reaction is directly proportional to th   [#permalink] 08 Nov 2019, 13:22
Display posts from previous: Sort by

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