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Algebra II : Identifying Variable Relationships

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Example Questions

Example Question #1 : How To Find Direct Variation

varies directly with the square root of . If x = 25 , then y = 48 . What is the value of if x = 81 ?

Possible Answers:

86.4

155.52

$26\tfrac{2}{3}$

None of these answers are correct.

491.52

Correct answer:

86.4

Explanation:

If varies directly with the square root of , then for some constant of variation ,

$\small y = K \sqrt{x}$

If x = 25 , then y = 48 ; therefore, the equation becomes

$\small \small 48 = K \sqrt{25}$ ,

5K = 48 .

Divide by 5 to get K = 9.6 , making the equation

$\small \small y = 9.6 \sqrt{x}$ .

If x = 81 , then $\small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4$ .

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Example Question #1 : Identifying Variable Relationships

If varies directly with $x^{3}$ and when x=5 , y=61.125 due to the effect of a constant, what is the value of when x=11 ?

Possible Answers:

y=650.859

y=134.475

y=1,331

y=43.331

y=672.375

Correct answer:

y=650.859

Explanation:

Since varies directly with $x^{3}$ , $y=Kx^{3}$ where is a constant.

1. Solve for when x=5 and y=61.125 .

$y=Kx^{3}$

$61.125=K(5^{3})$

$K=\frac{61.125}{125}=0.489$

2. Use your equation to solve for when x=11 .

$y=0.489x^{3}$

$y=0.489(11^{3})=650.859$

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Example Question #2 : Identifying Variable Relationships

If varies indirectly with $\sqrt{x}$ and when x=4 , y=12.3 due to the effect of a constant, what is the value of when x=17 ?

Possible Answers:

y=5.97

y=25.36

y=50.71

y=16.49

y=0.24

Correct answer:

y=5.97

Explanation:

Since varies indirectly with $\sqrt{x}$ , $y=\frac{K}{\sqrt{x}}$

1. Solve for when y=12.3 and x=4 .

$y=\frac{K}{\sqrt{x}}$

$12.3=\frac{K}{\sqrt{4}}$

$12.3= \frac{K}{2}$

K=24.6

2. Use the equation you found to solve for when x=17 .

$y=\frac{24.6}{\sqrt{x}}$

$y=\frac{24.6}{\sqrt{17}}=5.97$

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Example Question #1 : Variable Relationships

varies directly with $x^{2}$ . If x=5, y=75 , what is if x=10 ?

Possible Answers:

300

150

250

100

Correct answer:

300

Explanation:

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

75=K(25)

K=3

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

y=300

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Example Question #4 : Identifying Variable Relationships

varies inversely with $x^{3}$ . If x=2, y=2 , then what is equal to when x=5 ?

Possible Answers:

0.654

0.128

0.226

0.197

Correct answer:

0.128

Explanation:

1. Since varies indirectly with $x^{3}$ :

$y=\frac{K}{x^{3}}$

2. Use the given x and y values to determine the value of K:

$2=\frac{K}{2^{3}}$

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

K=16

3. Using the equation along with the value of K, find the value of y when x=5:

$y=\frac{16}{x^{3}}$

$y=\frac{16}{5^{3}}$

y=0.128

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Example Question #5 : Identifying Variable Relationships

varies directly with $\sqrt[3]{x}$ and when x=8, y=3 . What is when x=27 ?

Possible Answers:

4.5

5.5

Correct answer:

4.5

Explanation:

1. Since varies directly with $\sqrt[3]{x}$ :

$y=K\sqrt[3]{x}$

2. Use the values given for x and y to solve for K:

$y=K\sqrt[3]{x}$

$3=K\sqrt[3]{8}$

3=K(2)

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

3. Use your new equation with the K you solved for to solve for y when x=27:

$y=(\frac{3}{2})\sqrt[3]{27}$

$y=(\frac{3}{2})(3)=4.5$

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Example Question #6 : Identifying Variable Relationships

varies inversely with $\sqrt{x}$ . When $x=4,\ y=0.5$ . What is the value of when x=100 ?

Possible Answers:

$\frac{1}{20}$

$\frac{1}{8}$

$\frac{1}{10}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{10}$

Explanation:

1. Since y varies indirectly with $\sqrt{x}$ :

$y=\frac{K}{\sqrt{x}}$

2. Solve for K using the x and y values given:

$0.5=\frac{K}{\sqrt{4}}$

$0.5=\frac{K}{2}$

K=1

3. Using the equation you created by solving for K, find y when x=100:

$y=\frac{1}{\sqrt{x}}$

$y=\frac{1}{\sqrt{100}}$

$y=\frac{1}{10}$

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Algebra II : Identifying Variable Relationships

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : How To Find Direct Variation

Example Question #1 : Identifying Variable Relationships

Example Question #2 : Identifying Variable Relationships

Example Question #1 : Variable Relationships

Example Question #4 : Identifying Variable Relationships

Example Question #5 : Identifying Variable Relationships

Example Question #6 : Identifying Variable Relationships

All Algebra II Resources

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