Mathematical Relationships

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SAT Math › Mathematical Relationships

Questions 1 - 10

1

x varies inversely with y. When x=10, y=6. When x=3, what is y?

Explanation

Inverse variation takes the form:

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

Plugging in:

$6=\frac{k}{10}$

Then solve when x=3:

2

varies inversely as the square root of . If , then . Find if (nearest tenth, if applicable).

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 , then

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

3

$\small x$ 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 $\small k = 2$ .

All of these answers are correct

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

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

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

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"

4

Sarah notices her map has a scale of $\frac{1}{4}; in=1; mile$ . She measures between Beaver Falls and Chipmonk Cove. How far apart are the cities?

Explanation

$\frac{1}{4}; in=1; mile$ is the same as

So to find out the distance between the cities

$12.5; in \cdot \frac{4; miles}{1;in }=50; miles$

5

$\small x$ 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 $\small k = 2$ .

All of these answers are correct

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

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

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

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"

6

varies inversely as the square root of . If , then . Find if (nearest tenth, if applicable).

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 , then

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

7

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

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

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

None of these answers are correct

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 :

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

8

Sarah notices her map has a scale of $\frac{1}{4}; in=1; mile$ . She measures between Beaver Falls and Chipmonk Cove. How far apart are the cities?

Explanation

$\frac{1}{4}; in=1; mile$ is the same as

So to find out the distance between the cities

$12.5; in \cdot \frac{4; miles}{1;in }=50; miles$

9

If an object is hung on a spring, the elongation of the spring varies directly as the mass of the object. A 20 kg object increases the length of a spring by exactly 7.2 cm. To the nearest tenth of a centimeter, by how much does a 32 kg object increase the length of the same spring?

$11.5 \textrm{ cm}$

$4.5 \textrm{ cm}$

$18.4 \textrm{ cm}$

$9.1\textrm{ cm}$

$8.4 \textrm{ cm}$

Explanation

Let be the mass of the weight and the elongation of the spring. Then for some constant of variation ,

We can find by setting from the first situation:

$7.2 = k \cdot 20$

$k = 7.2 \div 20 = 0.36$

so

In the second situation, we set and solve for :

$L = 0.36 \cdot32 =11.52$

which rounds to 11.5 centimeters.

10

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

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

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

None of these answers are correct

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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