Direct and Inverse Variation - SAT Math

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Question

and are the radius and volume, respectively, of a given sphere.

$r ^{6} = Q$

$W = \sqrt[3]{V}$ .

Which of the following is a true statement?

Answer

The volume of a sphere can be calculated from its radius as follows:

$V = \frac{4}{3} \pi r^{3}$

Therefore, squaring both sides, we get

$V ^{2}=\left ( \frac{4}{3} \pi r^{3} \right )^{2}$

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{3\cdot 2}$

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W = \sqrt[3]{V}$

$W ^{6 }= \left (\sqrt[3]{V} \right ) ^{6}$

$W ^{6 }= \left ( V ^{\frac{1}{3}}\right ) ^{6} = V ^{\frac{1}{3} \cdot 6} = V^{2}$

Substituting:

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W ^{6 }=\left ( \frac{4}{3} \pi \right )^{2}Q$

$\frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }= \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}} \cdot \left ( \frac{4}{3} \pi \right )^{2}Q$

$Q = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }$

If we let the constant of variation be $K = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}$ , we see that

$Q = K W ^{6 }$ ,

and varies directly as $W^{6}$ , the sixth power of .

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Question

Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?

Answer

Every minute Phillip completes another _ square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.

There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.

$(60\frac{min}{hr})(2.5hr)=150min$

If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.

$(y\frac{ft^2}{min})(150min)=150y\ ft^2$

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Question

The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?

Answer

Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:

y = kx, where k is a constant.

Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:

y = _kx_2_z_3, where k is a constant.

The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.

24 = k(1)2(2)3 = k(1)(8) = 8_k_

24 = 8_k_

Divide both sides by 8.

3 = k

k = 3

Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.

y = _kx_2_z_3

y = 3(3)2(1)3 = 3(9)(1) = 27

y = 27

The answer is 27.

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Question

In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?

Answer

We know that the initial population is 3, and that every week the population will triple.

The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be .

Alternatively, you can evaluate for each consecutive week.

Week 1:

Week 2:

Week 3:

Week 4:

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Question

and are the diameter and circumference, respectively, of the same circle.

$d = 12L^{2}$

$C = t^{2}$

Which of the following is a true statement? (Assume all quantities are positive)

Answer

If and are the diameter and circumference, respectively, of the same circle, then

$C = \pi d$ .

By substitution,

$t^{2} = \pi \cdot 12 L^{2}$

$t^{2} = 12 \pi L^{2}$

Taking the square root of both sides:

$\sqrt{t^{2} }= \sqrt{12 \pi L^{2}}$

$t = \sqrt{12 \pi} \cdot \sqrt{L^{2}}$

$t = \sqrt{12 \pi} \cdot L$

Taking $K= \sqrt{12 \pi}$ as the constant of variation, we get

,

meaning that varies directly as .

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Question

is the radius of the base of a cone; is its height; is its volume.

$n = 4r = h^{2}$ ; $m = V ^{2}$ .

Which of the following is a true statement?

Answer

The volume of a cone can be calculated from the radius of its base , and the height , using the formula

$V = \frac{1}{3} \pi r^{2}h$

, so $r = \frac{n}{4}$ .

$n = h^{2}$ , so $h = \sqrt{n}$ .

$V = \frac{1}{3} \pi r^{2}h$ , so by substitution,

$V = \frac{1}{3} \pi\left ( \frac{n}{4} \right ) ^{2} \cdot \sqrt{n}$

$V = \frac{1}{3} \pi\left ( \frac{n^{2}}{16} \right ) \cdot \sqrt{n}$

$V = \frac{1}{48} \pi \cdot n^{2}} \sqrt{n}$

Square both sides:

$V ^{2 }=\left ( \frac{1}{48} \pi \cdot n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \right )^{2} \cdot \left ( \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{4}} \cdot n$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{5}}$

If we take $K = \left ( \frac{1}{48} \pi \right )^{2}$ as the constant of variation, then

$m=K n^{5}}$ ,

and varies directly as the fifth power of .

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Question

The temperature at the surface of the ocean is $25^{\circ}C$ . At meters below the surface, the ocean temperature is $5^{\circ}C$ . By how much does the temperature decrease for every meters below the ocean's surface?

Answer

This may seem confusing, but is pretty straightforward.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{5-25}{-2500-0}$

$\text{Slope}=\frac{-20}{-2500}=\frac{125}{1}$

Thus, for every 125 meters below the surface, the temperature decreases by one degree.

To find how much it decreases with every 100 meters, we need to do the following:

$\frac{100}{125}=\frac{4}{5}$

Thus, the temperature decreases by $\frac{4}{5}^{\circ}$ every 100 meters.

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Question

The square of varies inversely with the cube of . If when , then what is the value of when ?

Answer

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

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Question

varies directly as and inversely as .

$z = y ^{2}$ and $B = \sqrt{A}$ .

Which of the following is true about ?

Answer

varies directly as and inversely as , so for some constant of variation ,

$\frac{xy}{A} =K$ .

We can square both sides to obtain:

$\left (\frac{xy}{A} \right ) ^{2}=K^{2}$

$\frac{x^{2}y^{2}}{A^{2}}= K^{2}$

$z = y ^{2}$ .

$B = \sqrt{A}$ , so $B^{4} =\left ( \sqrt{A} \right )^{4} = A^{2}$ .

By substitution,

$\frac{x^{2}z}{B^{4}}= K^{2}$ .

Using $K^{2}$ as the constant of variation, we see that varies inversely as the square of and directly as the fourth power of .

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Question

The radius of the base of a cylinder is ; the height of the same cylinder is ; the cylinder has volume 1,000.

$t = \sqrt{r}$

$v = \sqrt[4]{h}$

Which of the following is a true statement?

Assume all quantities are positive.

Answer

The volume of a cylinder can be calculated from its height and the radius of its base using the formula:

$\pi r^{2} h = V$

$t = \sqrt{r}$ , so $t ^{2}= r$ ;

$v = \sqrt[4]{h}$ , so $h = v^{4}$ .

The volume is 1,000, and by substitution, using the other equations:

$\pi (t^{2})^{2} v^{4} = 1,000$

$\pi t^{4} v^{4} = 1,000$

$\frac{\pi t^{4} v^{4}}{\pi} = \frac{1,000}{\pi}$

$t^{4} v^{4} = \frac{1,000}{\pi}$

$\sqrt[4]{t^{4} v^{4} }= \sqrt[4]{\frac{1,000}{\pi}}$

$tv= \sqrt[4]{\frac{1,000}{\pi}}$

If we take $K= \sqrt[4]{\frac{1,000}{\pi}}$ as the constant of variation, we get

,

meaning that varies inversely as .

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Question

varies directly as the square of and the cube root of , and inversely as the fourth root of . Which of the following is a true statement?

Answer

varies directly as the square of and the cube root of , and inversely as the fourth root of , so, for some constant of variation ,

$\frac{s \sqrt[4]{w}}{y^{2}\sqrt[3]{t}} = K$

We take the reciprocal of both sides, then extract the square root:

$\frac{y^{2}\sqrt[3]{t}}{s \sqrt[4]{w}} =\frac{1} { K}$

$\frac{y^{2}t^{\frac{1}{3}}}{s w ^{\frac{1}{4}}} =\frac{1} { K}$

$\left ( \frac{y^{2}t^{\frac{1}{3}}}{s w ^{\frac{1}{4}}} \right )^{\frac{1}{2}} =\left (\frac{1} { K} \right )^{\frac{1}{2}}$

$\frac{y t^{\frac{1}{6}}}{s ^{\frac{1}{2}} w ^{\frac{1}{8}}}=\left (\frac{1} { K} \right )^{\frac{1}{2}}$

$\frac{y\sqrt[6]{t}}{\sqrt{s} \cdot \sqrt[8]{w}}=\left (\frac{1} { K} \right )^{\frac{1}{2}}$

Taking $\left (\frac{1} { K} \right )^{\frac{1}{2}}$ as the constant of variation, we see that varies directly as the square root of and the eighth root of , and inversely as the sixth root of .

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Question

If varies inversely as , and when , find when .

Answer

The formula for inverse variation is as follows: $y=\frac{k}{x}$

Use the x and y values from the first part of the sentence to find k.

$14=\frac{k}{3}$

Then use that k value and the given x value to find y.

$y=\frac{42}{28}$

$y=1\frac{1}{2}$

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