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

 It is currently 16 Oct 2018, 05:48

### 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

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 19 Oct 2008
Posts: 87
The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

Updated on: 06 Aug 2016, 23:02
4
20
00:00

Difficulty:

65% (hard)

Question Stats:

60% (01:13) correct 40% (01:17) wrong based on 600 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%, 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

Originally posted by Accountant on 23 Mar 2009, 11:03.
Last edited by Bunuel on 06 Aug 2016, 23:02, edited 2 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 49911
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

27 Nov 2009, 19:43
9
7
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. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}$$

_________________
##### General Discussion
Senior Manager
Joined: 06 Mar 2006
Posts: 463
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

23 Mar 2009, 13:19
Accountant wrote:
The rate of a chemical reaction is directly proportional to the square of concentration of chemcial A present and invesrsly proportional to the concentration of chemical B present. If chemical B is increased by 100% which of the following is the change in concentration of chemical A required to keep teh reaction rate unchanged:

A. 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase

The rate of reaction is invesrsly proportional to the concentration of chemical B present. It used to have B=1 . Now that B is increased by 100%. So the new equation would be 2B=(1/2). In order for the rate of reaction to still be 1, we need to change the concentration of A to yield a 2. It used to be A^2=1, now the new equation should be (sqrt(2)*A)^2=2. The change in the concentration of A can be calculated as (sqrt(2) -1)/1 or approximately 40% increase. Answer D.
Intern
Joined: 25 Dec 2008
Posts: 18
Schools: HBS, Stanford
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

23 Mar 2009, 16:13
1
2
CA: Concentration A
CB: Concentration B
R: Reaction rate

Formula for reaction rate:

R = (CA^2) / CB

Thus if CB is increased by 100% >> means concentration doubles >> 2xCB

Thus, for R to remain the same (CA^2) also has to double.

>> 2 x (CA^2) >> to see what happens to CA, take the 2 into the bracket by taking its root

>> (SQRT2 CA)^2

SQRT 2 is roughly 40% >> answer D
Manager
Joined: 30 Mar 2010
Posts: 81
GMAT 1: 730 Q48 V42
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

27 Nov 2010, 18:06
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?
Math Expert
Joined: 02 Sep 2009
Posts: 49911
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

28 Nov 2010, 02:12
3
2
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 $$b$$ 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.
_________________
Manager
Joined: 30 Mar 2010
Posts: 81
GMAT 1: 730 Q48 V42
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

28 Nov 2010, 09:56
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
Joined: 02 Sep 2009
Posts: 49911
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

28 Nov 2010, 10:27
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.
_________________
Manager
Joined: 04 Dec 2011
Posts: 66
Schools: Smith '16 (I)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

29 Sep 2013, 10:12
Hi Bunuel, I am a bit shaky with variation concepts and hence decided to get the basics clear,

I was referring to Karishma's blog here http://www.veritasprep.com/blog/2013/02 ... g-jointly/
Now I understand that if a rate varies directly for Eg X varies directly with Y than we have X/Y = K(some value) because in direct variation the ratio remains same.
and in inverse variation it will be XY = K(some value) because x=1/y
Please correct me if I misunderstood any concept till this point.

Bunuel wrote:
NOTE: Put directly proportional in nominator and inversely proportional in denominator

how can we put a direct variation in numerator? because if I understand the concept correctly it should be in denominator? and an inverse in numerator..how did you arrived at this quick formula?

another question to may be both karishma and you Bunuel ( sorry karishma I am asking questions pertaining to your blog on this forum, but I thought this question can serve as a common reference.)

As per me It should be B/A^2 (infect If I look at karishma's sample question in same page its essentially the same question with just values swapped with N and M)

another point of confusion when B becomes double (i.e 2B) why don't we simply say A^2 also doubles(i.e 2 A^2) why do we say if a^2 has to double it has to be A^2 = 2 ?

if its a ratio than it should be multiplied and divided by same value in numerator and denominator (i.e 2)
_________________

Life is very similar to a boxing ring.
Defeat is not final when you fall down…
It is final when you refuse to get up and fight back!

1 Kudos = 1 thanks
Nikhil

Manager
Joined: 04 Dec 2011
Posts: 66
Schools: Smith '16 (I)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

01 Oct 2013, 12:22
Hi Bunuel...waiting for reply.. have exam this week so a bit nervous
_________________

Life is very similar to a boxing ring.
Defeat is not final when you fall down…
It is final when you refuse to get up and fight back!

1 Kudos = 1 thanks
Nikhil

Intern
Joined: 06 Nov 2015
Posts: 22
GMAT 1: 690 Q49 V35
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

29 Apr 2016, 07:54
Bunuel wrote:

$$a$$ 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.

Hi,

Could someone help to explain this one, how can we come up with the final result, which is $$R=\frac{(A^2x)y}{B}$$?

As far as I understand, according to the case in question, we have:
1. The rate is directly proportional to the square of concentration of chemical A
-> R = $$xA^2$$ (1)
2. The rate is inversely proportional to the concentration of chemical B
-> R = $$\frac{y}{B}$$ (2)

So next, how can we infer that $$R=\frac{(A^2x)y}{B}$$? What steps used to modify/combine (1) and (2) to get this one? Actually, from (1) and (2), I am thinking of $$\frac{(A^2x)*y}{B}$$ as $$R*R$$ = $$R^2$$ instead

_________________

"Chance favors the prepared mind"

Intern
Joined: 06 Jun 2014
Posts: 47
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

06 May 2016, 02:06
1
thuyduong91vnu wrote:
Bunuel wrote:

$$a$$ 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.

Hi,

Could someone help to explain this one, how can we come up with the final result, which is $$R=\frac{(A^2x)y}{B}$$?

As far as I understand, according to the case in question, we have:
1. The rate is directly proportional to the square of concentration of chemical A
-> R = $$xA^2$$ (1)
2. The rate is inversely proportional to the concentration of chemical B
-> R = $$\frac{y}{B}$$ (2)

So next, how can we infer that $$R=\frac{(A^2x)y}{B}$$? What steps used to modify/combine (1) and (2) to get this one? Actually, from (1) and (2), I am thinking of $$\frac{(A^2x)*y}{B}$$ as $$R*R$$ = $$R^2$$ instead

Hello my friend.

This is how I solved this question.
First I didnt assign those variebels x and y as above in the quote. I think those are used just as reference in order to show the proportonality.

Now to the question.
I did assign variable x to be the factor of percent increase or decrease, r to be the rate, and a and b for the concetrations.
So start with first formula $$r=a^2/b$$ where a is directly proportonal ans is nuumerator and b is denominator since it is inversly proportinal. note u have to put $$a^2$$ because it is given that the square root is directly proportonal
now the second equation , the question asks to have same value for the rate but b is doubled or increase for 100%
$$r=(xa)^2/2b$$ or $$r=x^2a^2/2b$$ now from here we can use the short way given by Bunuel or the long way to solve for x, either way it will come out as x^2=2 and$$x=1.41$$. now earlier we said x is a factor of percentage change $$1+0.41$$ or we have an increase of 41%
Intern
Joined: 03 Nov 2016
Posts: 1
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

26 Dec 2016, 14:26
1
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$$\sqrt{2}=1.4$$
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 267
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

29 Aug 2017, 03:57
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. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}$$

hi

your solution to the problem is very clear, no doubt out there .....

can you please, however, speak some more words about the concept "Put directly proportional in nominator and inversely proportional in denominator."

Math Expert
Joined: 02 Sep 2009
Posts: 49911
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

29 Aug 2017, 04:07
gmatcracker2017 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. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}$$

hi

your solution to the problem is very clear, no doubt out there .....

can you please, however, speak some more words about the concept "Put directly proportional in nominator and inversely proportional in denominator."

Check Variations On The GMAT - All In One Topic.

Hope it helps.
_________________
Intern
Joined: 21 Mar 2017
Posts: 40
Location: Zimbabwe
Concentration: General Management, Entrepreneurship
GMAT 1: 680 Q45 V38
GMAT 2: 750 Q49 V42
GPA: 3.3
WE: Accounting (Accounting)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

05 Sep 2017, 04:33
This is a tough question, especially if you aren't well versed with variations.

This is how i approached it:
Since B increased, in order for the rate to stay the same, A also has to increase, therefore we can eliminate A, B, and C.

Now my guess is between D and E. 50% felt wrong so I went with D
_________________

Kudos if you like my response please

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3825
Location: United States (CA)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

07 Sep 2017, 16:32
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

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Joined: 04 May 2014
Posts: 161
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

12 Sep 2017, 00:41
Let Rate be R
Let concentrations
Ca=10
Cb=10

R=K$$\frac{(Ca)²}{Cb}$$ where K is a constant which we can ignore for this question.

R=$$\frac{(10)²}{10}$$=10

Now Cb is doubled =20
We have to keep R=10

10=$$\frac{(Ca)²}{20}$$

or (Ca)²=200
or Ca=√200
or Ca is approx=14 which is 40% more than 10 hence answer is D
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 360
Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

### Show Tags

30 Sep 2018, 09:21
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 :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

Re: The rate of a certain chemical reaction is directly proportional to th &nbs [#permalink] 30 Sep 2018, 09:21
Display posts from previous: Sort by