SAT Math - Other Mathematical Relationships

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

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

$\small x$ 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"

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:

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:

$7.2 = k \cdot 20$

$k = 7.2 \div 20 = 0.36$

In the second situation, we set and solve for :

$L = 0.36 \cdot32 =11.52$

which rounds to 11.5 centimeters.

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

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

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?

$\text{4 slices.}$

$\text{2 slices.}$

$\text{6 slices.}$

$\text{5 slices.}$

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.

Now we substitute 12 in for n and solve for P

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

Therefore with 12 people, you get 4 slices.

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 :

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

If an object is hung on a spring, the elongation of the spring varies directly with the mass of the object. A 33 kilogram object increases the length of a spring by exactly 6.6 centimeters. To the nearest tenth of a kilogram, how much mass must an object posess to increase the length of that same spring by exactly 10 centimeters?

$50 \textrm{ kg}$