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

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

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27 Nov 2010, 18:06
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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
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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28 Nov 2010, 02:12
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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.
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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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
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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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.
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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26 Dec 2016, 14:26
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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$$\sqrt{2}=1.4$$
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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07 Sep 2017, 16:32
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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.

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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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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.
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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08 Oct 2019, 08:46
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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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08 Oct 2019, 10:16
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
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The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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08 Oct 2019, 10:38
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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%

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Re: The rate of a certain chemical reaction is directly proportional to th  [#permalink]

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08 Oct 2019, 12:43
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
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Re: The rate of a certain chemical reaction is directly proportional to th   [#permalink] 08 Oct 2019, 12:43
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