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Algebra II : Basic Single-Variable Algebra

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Example Questions

Example Question #31 : Basic Single Variable Algebra

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

Possible Answers:

$\frac{1}{3}$

675

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 #32 : Basic Single Variable Algebra

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

Possible Answers:

31.8

22.5

11.3

63.6

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 #33 : Basic Single Variable Algebra

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:

$2 \textrm{ A}$

$0.96 \textrm{ A}$

$1.8 \textrm{ A}$

$1.6 \textrm{ A}$

$1.5 \textrm{ A}$

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 #34 : Basic Single Variable Algebra

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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Example Question #35 : Basic Single Variable Algebra

The number of days needed to construct a house is inversely proportional to the number of people that help build the house. It took 28 days to build a house with 7 people. A second house is being built and it needs to be finished in 14 days. How many people are needed to make this happen?

Possible Answers:

Correct answer:

Explanation:

The general formula of inverse proportionality for this problem is

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

where is the number of days, is the proportionality constant, and is number of people.

Before finding the number of people needed to build the house in 14 days, we need to find . Given that the house can be built in 28 days with 7 people, we have

$28=\frac{k}{7}$

Multiply both sides by 7 to find .

$28\cdot7=\frac{k}{7}\cdot7$

196=k

So k=196 . Thus,

$d=\frac{196}{p}$

Now we can find the how many people are needed to build the house in 14 days.

$14=\frac{196}{p}$

Solve for . First, multiply by on both sides:

$14\cdot p=\frac{196}{p}\cdot p$

14p=196

Divide both sides by 14

$\frac{14p}{14}=\frac{196}{14}$

p=14

So it will take 14 people to complete the house in 14 days.

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Algebra II : Basic Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Basic Single Variable Algebra

Example Question #1871 : Algebra Ii

Example Question #31 : Proportionalities

Example Question #1872 : Algebra Ii

Example Question #1873 : Algebra Ii

Example Question #32 : Basic Single Variable Algebra

Example Question #33 : Basic Single Variable Algebra

Example Question #1 : How To Find Direct Variation

Example Question #34 : Basic Single Variable Algebra

Example Question #35 : Basic Single Variable Algebra

All Algebra II Resources

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