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SAT II Math I : Mathematical Relationships

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Example Questions

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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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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SAT II Math I : Mathematical Relationships

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

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

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

All SAT II Math I Resources

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