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SAT Math : How to use the direct variation formula

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Example Questions

Example Question #1 : How To Use The Direct Variation Formula

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

Possible Answers:

$300y\ ft^2$

$50y\ ft^2$

$2.5y\ ft^2$

$25y\ ft^2$

$150y\ ft^2$

Correct answer:

$150y\ ft^2$

Explanation:

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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Example Question #2 : How To Use The Direct Variation Formula

The value of varies directly with the square of and the cube of . If y=24 when x=1 and z=2 , then what is the value of when x=3 and z=1 ?

Possible Answers:

Correct answer:

Explanation:

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 = kx²z³, 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) = 8k

24 = 8k

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 = kx²z³

y = 3(3)²(1)³ = 3(9)(1) = 27

y = 27

The answer is 27.

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Example Question #2 : How To Use The Direct Variation Formula

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?

Possible Answers:

$729$

$81$

$27$

$2187$

$243$

Correct answer:

$243$

Explanation:

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

The equation to model this growth will be i(r)^n , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be 3(3)^4 .

3(3)^4=3(81)=243

Alternatively, you can evaluate for each consecutive week.

Week 1: 3(3)=9

Week 2: 3(9)=27

Week 3: 3(27)=81

Week 4: 3(81)=243

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Example Question #8 : Variables

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)

Possible Answers:

varies directly as .

varies directly as the fourth root of .

varies inversely as the fourth power of .

varies directly as the fourth power of .

varies inversely as the fourth root of .

Correct answer:

varies directly as .

Explanation:

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

t =K L ,

meaning that varies directly as .

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Example Question #2 : How To Use The Direct Variation Formula

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?

Possible Answers:

varies directly as the cube root of .

varies directly as the third power of .

varies directly as .

varies directly as the fifth root of .

varies directly as the fifth power of .

Correct answer:

varies directly as the fifth power of .

Explanation:

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

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

n = 4r , 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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Example Question #3 : How To Use The Direct Variation Formula

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?

Possible Answers:

varies directly as $W\;$ .

varies inversely as the sixth power of $W\;$ .

varies directly as the sixth power of $W\;$ .

varies directly as the sixth root of $W\;$ .

varies inversely as the sixth root of $W\;$ .

Correct answer:

varies directly as the sixth power of $W\;$ .

Explanation:

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 $W \:$ .

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Example Question #3 : Direct And Inverse Variation

The temperature at the surface of the ocean is $25^{\circ}C$ . At 2500 meters below the surface, the ocean temperature is $5^{\circ}C$ . By how much does the temperature decrease for every 100 meters below the ocean's surface?

Possible Answers:

$1^{\circ}$

$\frac{3}{4}^{\circ}$

$\frac{4}{5}^{\circ}$

$2^{\circ}$

Correct answer:

$\frac{4}{5}^{\circ}$

Explanation:

This may seem confusing, but is pretty straightforward.

x_1=0

x_2=-2500

y_1=25

y_2=5

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

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

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

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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SAT Math : How to use the direct variation formula

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : How To Use The Direct Variation Formula

Example Question #2 : How To Use The Direct Variation Formula

Example Question #2 : How To Use The Direct Variation Formula

Example Question #8 : Variables

Example Question #2 : How To Use The Direct Variation Formula

Example Question #3 : How To Use The Direct Variation Formula

Example Question #3 : Direct And Inverse Variation

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