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Algebra II : Algebra II

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Example Questions

Example Question #6 : Indirect Proportionality

The number of hours needed for a contractor to finish a job varies indirectly with the total number of people the contractor hires. If the job is completed in hours when there are people, how many hours would it take if there were people?

Possible Answers:

$\text{3 hours}$

$\text{4 hours}$

$\text{2 hours}$

$\text{5 hours}$

Correct answer:

$\text{3 hours}$

Explanation:

The problem follows the formula

$H = \frac{k}{n}$

where H is the number of hours to complete the job, n is the number of people hired, and k is the constant of variation.

Setting H=6 and n = 8 yields k=48.

Therefore using the following equation we can plug 16 in for n and solve for H.

48=H(16)

$\frac{48}{16}=H$

3=H

Therefore H is 3 hours.

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Example Question #1 : Indirect Proportionality

$\small x$ varies inversely with . If x= 15 , y= 2 . What is the value of if y= 90 ?

Possible Answers:

675

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

varies inversely with , so the variation equation can be written as:

$x = k\cdot\frac{1}{y}$

can be solved for, using the first scenario:

$15 = k\cdot\frac{1}{2}$

$k = 15\cdot2 = 30$

Using this value for = 30 and = 90, we can solve for :

$x = 30\cdot\frac{1}{90} = \frac{30}{90} = \frac{3}{9} = \frac{1}{3}$

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Example Question #1871 : Algebra Ii

$\small x$ varies directly with $\small A$ and inversely with the square root of $\small B$ . Find values for $\small A$ and $\small B$ that will give $\small x = 3$ , for a constant of variation $\small k = 2$ .

Possible Answers:

$\small A = 7.5$ and $\small B = 25$

All of these answers are correct

$\small A = 3$ and $\small B = 4$

$\small A = 9$ and $\small B = 36$

Correct answer:

All of these answers are correct

Explanation:

From the first sentence, we can write the equation of variation as:

$\small x = k \cdot \frac{A}{\sqrt{B}}$

We can then check each of the possible answer choices by substituting the values into the variation equation with the values given for $\small x$ and $\small k$ .

$\small 3 = 2 \cdot \frac{3}{\sqrt{4}} = 2 \cdot \frac{3}{2} = 3$

Therefore the equation is true if $\small A = 3$ and $\small B = 4$

$\small 3 = 2 \cdot \frac{7.5}{\sqrt{25}} = 2 \cdot \frac{7.5}{5} = 2 \cdot 1.5 = 3$

Therefore the equation is true if $\small A = 7.5$ and $\small B = 25$

$\small 3 = 2 \cdot \frac{9}{\sqrt{36}} = 2 \cdot \frac{9}{6} = 2 \cdot 1.5 = 3$

Therefore the equation is true if $\small A = 9$ and $\small B = 36$

The correct answer choice is then "All of these answers are correct"

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Example Question #31 : Proportionalities

$\small x$ varies directly with $\small y$ and $\small z^{2}$ . If $\small y = \frac{1}{3}$ and $\small z = 3$ , then $\small x = 3$ . Find $\small x$ if $\small y = 12$ and $\small z = 2$ .

Possible Answers:

None of these answers are correct

Correct answer:

Explanation:

From the relationship of $\small x$ , $\small y$ , and $\small z$ ; the equation of variation can be written as:

$\small x = k \cdot y \cdot z^{2}$

Using the values given in the first scenario, we can solve for k:

$\small 3 = k \cdot \frac{1}{3} \cdot 3^{2} = k \cdot \frac{1}{3} \cdot 9 = k \cdot 3$

$\small k = 1$

Using the value of k and the values of y and z, we can solve for x:

$\small x = 1 \cdot 12 \cdot 2^{2} = 12 \cdot 4 = 48$

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Example Question #1872 : Algebra Ii

varies inversely with and the square root of . When x=4 and z=81 , $y=\frac{2}{9}$ . Find when x=8 and z=4 .

Possible Answers:

$\small \frac{1}{4}$

None of these answers are correct

$\small \frac{1}{2}$

Correct answer:

$\small \frac{1}{2}$

Explanation:

First, we can create an equation of variation from the the relationships given:

$y = k \cdot \frac{1}{x \cdot \sqrt{z}}$

Next, we substitute the values given in the first scenario to solve for $\small k$ :

$\frac{2}{9} = k \cdot \frac{1}{4 \cdot \sqrt{81}} = k \cdot \frac{1}{4 \cdot 9} = k \cdot \frac{1}{36}$

$\small \rightarrow k = 36 \cdot \frac{2}{9} = \frac{36}{9} \cdot 2 = 4 \cdot 2 = 8$

Using the value for , we can now use the second values for and to solve for :

$\small Y = 8 \cdot \frac{1}{8 \cdot \sqrt{4}} = 8 \cdot \frac{1}{8 \cdot 2} = 8 \cdot \frac{1}{16} = \frac{1}{2}$

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Example Question #1873 : Algebra Ii

varies directly with $z^{2}$ and the square root of . If y=100 , and $\small z=12$ then x=5 . Find the value of if y=4 and z=6 .

Possible Answers:

$\small \frac{1}{4}$

None of these answers are correct

$\small \frac{1}{2}$

$\small \frac{16 \sqrt6}{288}$

Correct answer:

$\small \frac{1}{4}$

Explanation:

From the relationship given, we can set up the variation equation

$\small x = k \cdot \sqrt{y} \cdot z^{2}$

Using the first relationship, we can then solve for $\small k$

$\small 5 = k \cdot \sqrt{100} \cdot 12^{2}$

$\small 5 = k \cdot 10 \cdot 144$

$\small k = \frac{5}{10 \cdot 144} = \frac{1}{2\cdot144} = \frac{1}{288}$

Now using the values from the second relationship, we can solve for x

$\small x = \frac{1}{288}\cdot\sqrt4\cdot6^2 = \frac{1}{288}\cdot2\cdot36 = \frac{72}{288} = \frac{1}{4}$

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Example Question #1 : How To Find Inverse Variation

varies inversely as the square of . If x = 15 , then A = 45 . Find if x = 30 (nearest tenth, if applicable).

Possible Answers:

63.6

11.3

31.8

22.5

Correct answer:

11.3

Explanation:

The variation equation is $A = \frac{K}{x^{2}}$ for some constant of variation .

Substitute the numbers from the first scenario to find :

$45= \frac{K}{15^{2}} = \frac{K}{225}$

$K = 45 \cdot225 = 10,125$

The equation is now $A = \frac{10,125}{x^{2}}$ .

If x = 30 , then

$A = \frac{10,125}{30^{2}} = \frac{10,125}{900} \approx 11.3$

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Example Question #2 : How To Find Inverse Variation

The current, in amperes, that a battery provides an electrical object is inversely proportional to the resistance, in ohms, of the object.

A battery provides 1.2 amperes of current to a flashlight whose resistance is measured at 20 ohms. How much current will the same battery supply to a flashlight whose resistance is measured at 16 ohms?

Possible Answers:

$0.96 \textrm{ A}$

$1.8 \textrm{ A}$

$1.6 \textrm{ A}$

$1.5 \textrm{ A}$

$2 \textrm{ A}$

Correct answer:

$1.5 \textrm{ A}$

Explanation:

If is the current and is the resistance, then we can write the variation equation for some constant of variation :

$I = \frac{k}{R}$

or, alternatively,

k=IR

To find , substitute I = 1.2,R=20 :

$k=1.2 \cdot 20 = 24$

The equation is $I = \frac{24}{R}$ . Now substitute R = 16 and solve for :

$I = \frac{24}{16} = 1.5 \textrm{ A}$

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Example Question #1 : How To Find Direct Variation

The volume of a fixed mass of gas varies inversely as the atmospheric pressure, as measured in millibars, acting on it, and directly as the temperature, as measured in kelvins, acting on it.

A balloon is filled to a capacity of exactly 100 cubic meters at a time at which the temperature is 310 kelvins and the atmospheric pressure is 1,020 millibars. The balloon is released, and an hour later, the balloon is subject to a pressure of 900 millibars and a temperature of 290 kelvins. To the nearest cubic meter, what is the new volume of the balloon?

Possible Answers:

$83 \textrm{ m}^{3}$

$100 \textrm{ m}^{3}$

$121 \textrm{ m}^{3}$

$94 \textrm{ m}^{3}$

$106 \textrm{ m}^{3}$

Correct answer:

$106 \textrm{ m}^{3}$

Explanation:

If V,P,T are the volume, pressure, and temperature, then the variation equation will be, for some constant of variation ,

$V = \frac{KT}{P}$

To calculate , substitute V = 100, T = 310, P = 1,020 :

$100 = \frac{K \cdot 310}{1,020}$

$\frac{K \cdot 310}{1,020} \cdot \frac{1,020}{310} = 100 \cdot \frac{1,020}{310}$

K = 329.03

The variation equation is

$V = \frac{329.03 T}{P}$

so substitute T =290, P = 900 and solve for .

$V = \frac{329.03 \cdot 290}{900} \approx 106 \textrm{ m}^{3}$

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Example Question #7 : How To Find Inverse Variation

If is inversely proportional to and knowing that c=3 when d=5 , determine the proportionality constant.

Possible Answers:

Correct answer:

Explanation:

The general formula for inverse proportionality for this problem is

$c=\frac{k}{d}$

Given that c=3 when d=5 , we can find by plugging them into the formula.

$3=\frac{k}{5}$

Solve for by multiplying both sides by 5

$3\cdot5=\frac{k}{5}\cdot5$

15=k

So k=15 .

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #6 : Indirect Proportionality

Example Question #1 : Indirect Proportionality

Example Question #1871 : Algebra Ii

Example Question #31 : Proportionalities

Example Question #1872 : Algebra Ii

Example Question #1873 : Algebra Ii

Example Question #1 : How To Find Inverse Variation

Example Question #2 : How To Find Inverse Variation

Example Question #1 : How To Find Direct Variation

Example Question #7 : How To Find Inverse Variation

All Algebra II Resources

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