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Algebra II : Basic Single-Variable Algebra

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Example Questions

Example Question #21 : Basic Single Variable Algebra

Suppose a runner's distance is directly proportional to her time. If the runner completes 6 miles in 70 minutes, how many minutes did it take the runner to run 4 miles?

Possible Answers:

$\frac{140}{3}$

105

$\frac{12}{35}$

Correct answer:

$\frac{140}{3}$

Explanation:

Write the equation for direct proportionality.

$y=kx \rightarrow D = kt$

Substitute the distance and time to solve for the constant.

6 =k(70)

Divide by 70 on both sides.

$\frac{6}{70} =\frac{k(70)}{70}$

$k=\frac{3}{35}$

Substitute this value back to D = kt .

The equation becomes:

$D = \frac{3}{35}t$

Substitute D=4 to determine how long it took the runner to run 4 miles.

$4 = \frac{3}{35}t$

Multiply by $\frac{35}{3}$ on both sides to isolate the time variable.

$4 \cdot\frac{35}{3}= \frac{3}{35}t\cdot\frac{35}{3}$

Simplify both sides.

The answer is: $\frac{140}{3}$

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Example Question #21 : Basic Single Variable Algebra

The distance of a cyclist is directly proportional to the time he has traveled. Suppose he has traveled 12 miles in 1.5 hours. How far does he travel in a half hour?

Possible Answers:

4.25

4.5

7.4

Correct answer:

Explanation:

Write the equation for direct proportionality.

$y=kx \rightarrow D=kT$

Substitute the distance and time given to solve for the constant of proportionality, .

12 = k(1.5)

Divide by 1.5 on both sides.

$\frac{12}{1.5} = \frac{k(1.5)}{1.5}$

k=8

Write the equation.

D=8T

Substitute half an hour for the time to determine the distance the biker has traveled.

D=8(0.5) = 4

The biker traveled four miles in a half hour.

The answer is:

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Example Question #1861 : Algebra Ii

varies directly with , and inversely with the square root of .

If y=6 and z =64 , then x = 10 .

Find if y=9 and z=144 .

Possible Answers:

$8\tfrac{1}{3}$

$6\tfrac{2}{3}$

$4\tfrac{4}{9}$

$12\tfrac{1}{2}$

Correct answer:

Explanation:

The variation equation can be written as below. Direct variation will put in the numerator, while inverse variation will put $\sqrt{z}$ in the denominator. is the constant that defines the variation.

$x=k* \frac{y}{\sqrt{z}}$

To find constant of variation, , substitute the values from the first scenario given in the question.

$10=k* \frac{6}{\sqrt{64}}=k* \frac{6}{8}=k* \frac{3}{4}$

$k=10\div\frac{3}{4}}= \frac{40}{3}$

We can plug this value into our variation equation.

$x=\frac{40}{3}\cdot \frac{y}{\sqrt{z}}$

Now we can solve for given the values in the second scenario of the question.

$x=\frac{40}{3}* \frac{9}{\sqrt{144}}=\frac{40}{3}* \frac{9}{12}=10$

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Example Question #21 : Basic Single Variable Algebra

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

Possible Answers:

11.3

31.8

63.6

22.5

Correct answer:

31.8

Explanation:

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

Substitute the numbers from the first scenario to find :

$45= \frac{K}{\sqrt{15}}$

$K = 45 \sqrt{15}$

The equation is now $A = \frac{45 \sqrt{15}}{\sqrt{x}}$ .

If x = 30 , then

$A = \frac{45 \sqrt{15}}{\sqrt{30}} \approx 31.8$

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Example Question #1863 : Algebra Ii

varies inversely with three times the square root of . If x=81 , then y=2.

Find if x=100 . Round to the nearest tenth if applicable.

Possible Answers:

2.5

1.8

Correct answer:

1.8

Explanation:

In order to find the value of when x=100 , first determine the variation equation based on the information provided:

$y=\frac{K}{3\sqrt{x}}$ , for some constant of variation .

Insert the and values from the first variance to find the value of :

$2=\frac{K}{3\sqrt{81}}$

$2=\frac{K}{(3\times 9)}$

$2=\frac{K}{27}$

54=K

Now that we know K=54 , the variation equation becomes:

$y=\frac{54}{3\sqrt{x}}$

$y=\frac{18}{\sqrt{x}}$ .

Therefore, when x=100 :

$y=\frac{18}{\sqrt{100}}$

$y=\frac{18}{10}$

y=1.8

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Example Question #26 : Basic Single Variable Algebra

varies directly with two times and varies indirectly with three times . When x=8,

y = 4

and z=2 .

What is the value of when x=3 and y=6? Round to the nearest tenth if needed.

Possible Answers:

12.2

10.4

8.7

Correct answer:

Explanation:

In order to solve for , first set up the variation equation for and :

$z = \frac{(K)(2y)}{3x}$

where is the constant of variation. The term varies indirectly with and is therefore in the denominator.

Next, we solve for based on the initial values of the variables:

$2 = \frac{(K)(2\times 4)}{(3\times 8)}$

$2 = \frac{8K}{24}$

48 = 8K

6=K

Now that we have the value of , we can solve for in the second scenario:

$z=\frac{(6)(2y)}{3x}$

$z=\frac{12y}{3x}$

$z=\frac{4y}{x}$

$z=\frac{4(6)}{3}$

$z=\frac{24}{3}$

z=8

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Example Question #3 : Indirect Proportionality

The number of slices of pizza you get varies indirectly with the total number of people in the restaurant. If you get slices when there are people, how many slices would you get if there are people?

Possible Answers:

$\text{5 slices.}$

$\text{2 slices.}$

$\text{6 slices.}$

$\text{4 slices.}$

Correct answer:

$\text{4 slices.}$

Explanation:

The problem follows the formula

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

where P is the number of slices you get, n is the number of people, and k is the constant of variation.

Setting P=3 and n = 16 yields k=48.

48=P(n)

Now we substitute 12 in for n and solve for P

48=P(12)

$\frac{48}{12}=P$

4=P

Therefore with 12 people, you get 4 slices.

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Example Question #4 : Indirect Proportionality

The number of raffle tickets given for a contest varies indirectly with the total number of people in the building. If you get tickets when there are 150 people, how many slices would you get if there are 100 people?

Possible Answers:

Correct answer:

Explanation:

The problem follows the formula

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

where R is the number of raffle tickets you get, n is the number of people, and k is the constant of variation.

Setting R=20 and n = 150 yields k=3000.

3000=R(n)

Plugging in 100 for n and solving for R you get:

3000=R(100)

$\frac{3000}{100}=R$

30=R

The answer R is 30 tickets.

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Example Question #5 : Indirect Proportionality

The budget per committee varies indirectly with the total number of committees created. If each committee is allotted $\$500$ when four committees are established, what would the budget per committee be if there were to be committees?

Possible Answers:

$\$750$

$\$300$

$\$1500$

$\$1000$

Correct answer:

$\$1000$

Explanation:

The problem follows the formula

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

where B is the budget per committee, n is the number of committees, and k is the constant of variation.

Setting B=500 and n = 4 yields k=2000.

Now using the following equation we can plug in our n of 2 and solve for B.

2000=B(n)

2000=B(2)

$\frac{2000}{2}=B$

1000=B

The answer of B is $1000.

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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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Algebra II : Basic Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #21 : Basic Single Variable Algebra

Example Question #21 : Basic Single Variable Algebra

Example Question #1861 : Algebra Ii

Example Question #21 : Basic Single Variable Algebra

Example Question #1863 : Algebra Ii

Example Question #26 : Basic Single Variable Algebra

Example Question #3 : Indirect Proportionality

Example Question #4 : Indirect Proportionality

Example Question #5 : Indirect Proportionality

Example Question #6 : Indirect Proportionality

All Algebra II Resources

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