Indirect Proportionality - Algebra II

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Question

The number of days to construct a house varies inversely with the number of people constructing that house. If it takes 28 days to construct a house with 6 people helping out, how long will it take if 20 people are helping out?

Answer

The statement, 'The number of days to construct a house varies inversely with the number of people constructing that house' has the mathematical relationship $D=\frac{k}{P}$ , where D is the number of days, P is the number of people, and k is the variation constant. Given that the house can be completed in 28 days with 6 people, the k-value is calculated.

$28=\frac{k}{6}$

This k-value can be used to find out how many days it takes to construct a house with 20 people (P = 20).

$D=\frac{168}{P}$

$D=\frac{168}{20}$

So it will take 8.4 days to build a house with 20 people.

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Question

varies directly with , and inversely with the square root of .

If and , then .

Find if and .

Answer

The variation equation can be written as below. Direct variation will put in the numerator, while inverse variation will put $\sqrt{z}$ in the denominator. is the constant that defines the variation.

$x=k* \frac{y}{\sqrt{z}}$

To find constant of variation, , substitute the values from the first scenario given in the question.

$10=k* \frac{6}{\sqrt{64}}=k* \frac{6}{8}=k* \frac{3}{4}$

$k=10\div\frac{3}{4}}= \frac{40}{3}$

We can plug this value into our variation equation.

$x=\frac{40}{3}\cdot \frac{y}{\sqrt{z}}$

Now we can solve for given the values in the second scenario of the question.

$x=\frac{40}{3}* \frac{9}{\sqrt{144}}=\frac{40}{3}* \frac{9}{12}=10$

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Question

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

Answer

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

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

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Question

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

Answer

The variation equation is $A = \frac{K}{x^{2}}$ for some constant of variation .

Substitute the numbers from the first scenario to find :

$45= \frac{K}{15^{2}} = \frac{K}{225}$

$K = 45 \cdot225 = 10,125$

The equation is now $A = \frac{10,125}{x^{2}}$ .

If , then

$A = \frac{10,125}{30^{2}} = \frac{10,125}{900} \approx 11.3$

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Question

The volume of a fixed mass of gas varies inversely as the atmospheric pressure, as measured in millibars, acting on it, and directly as the temperature, as measured in kelvins, acting on it.

A balloon is filled to a capacity of exactly 100 cubic meters at a time at which the temperature is 310 kelvins and the atmospheric pressure is 1,020 millibars. The balloon is released, and an hour later, the balloon is subject to a pressure of 900 millibars and a temperature of 290 kelvins. To the nearest cubic meter, what is the new volume of the balloon?

Answer

If are the volume, pressure, and temperature, then the variation equation will be, for some constant of variation ,

$V = \frac{KT}{P}$

To calculate , substitute :

$100 = \frac{K \cdot 310}{1,020}$

$\frac{K \cdot 310}{1,020} \cdot \frac{1,020}{310} = 100 \cdot \frac{1,020}{310}$

The variation equation is

$V = \frac{329.03 T}{P}$

so substitute and solve for .

$V = \frac{329.03 \cdot 290}{900} \approx 106 \textrm{ m}^{3}$

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Question

The current, in amperes, that a battery provides an electrical object is inversely proportional to the resistance, in ohms, of the object.

A battery provides 1.2 amperes of current to a flashlight whose resistance is measured at 20 ohms. How much current will the same battery supply to a flashlight whose resistance is measured at 16 ohms?

Answer

If is the current and is the resistance, then we can write the variation equation for some constant of variation :

$I = \frac{k}{R}$

or, alternatively,

To find , substitute :

$k=1.2 \cdot 20 = 24$

The equation is $I = \frac{24}{R}$ . Now substitute and solve for :

$I = \frac{24}{16} = 1.5 \textrm{ A}$

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Question

If is inversely proportional to and knowing that when , determine the proportionality constant.

Answer

The general formula for inverse proportionality for this problem is

$c=\frac{k}{d}$

Given that when , we can find by plugging them into the formula.

$3=\frac{k}{5}$

Solve for by multiplying both sides by 5

$3\cdot5=\frac{k}{5}\cdot5$

So .

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Question

The number of days needed to construct a house is inversely proportional to the number of people that help build the house. It took 28 days to build a house with 7 people. A second house is being built and it needs to be finished in 14 days. How many people are needed to make this happen?

Answer

The general formula of inverse proportionality for this problem is

$d=\frac{k}{p}$

where is the number of days, is the proportionality constant, and is number of people.

Before finding the number of people needed to build the house in 14 days, we need to find . Given that the house can be built in 28 days with 7 people, we have

$28=\frac{k}{7}$

Multiply both sides by 7 to find .

$28\cdot7=\frac{k}{7}\cdot7$

So . Thus,

$d=\frac{196}{p}$

Now we can find the how many people are needed to build the house in 14 days.

$14=\frac{196}{p}$

Solve for . First, multiply by on both sides:

$14\cdot p=\frac{196}{p}\cdot p$

Divide both sides by 14

$\frac{14p}{14}=\frac{196}{14}$

So it will take 14 people to complete the house in 14 days.

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Question

varies inversely with three times the square root of . If , then

Find if . Round to the nearest tenth if applicable.

Answer

In order to find the value of when , first determine the variation equation based on the information provided:

$y=\frac{K}{3\sqrt{x}}$ , for some constant of variation .

Insert the and values from the first variance to find the value of :

$2=\frac{K}{3\sqrt{81}}$

$2=\frac{K}{(3\times 9)}$

$2=\frac{K}{27}$

Now that we know , the variation equation becomes:

$y=\frac{54}{3\sqrt{x}}$

or

$y=\frac{18}{\sqrt{x}}$ .

Therefore, when :

$y=\frac{18}{\sqrt{100}}$

$y=\frac{18}{10}$

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Question

varies directly with two times and varies indirectly with three times . When

and .

What is the value of when and Round to the nearest tenth if needed.

Answer

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:

$2 = \frac{(K)(2\times 4)}{(3\times 8)}$

$2 = \frac{8K}{24}$

Now that we have the value of , we can solve for in the second scenario:

$z=\frac{(6)(2y)}{3x}$

$z=\frac{12y}{3x}$

$z=\frac{4y}{x}$

$z=\frac{4(6)}{3}$

$z=\frac{24}{3}$

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Question

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?

Answer

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.

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Question

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 people, how many slices would you get if there are people?

Answer

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.

Plugging in 100 for n and solving for R you get:

$\frac{3000}{100}=R$

The answer R is 30 tickets.

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Question

The budget per committee varies indirectly with the total number of committees created. If each committee is allotted when four committees are established, what would the budget per committee be if there were to be committees?

Answer

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.

$\frac{2000}{2}=B$

The answer of B is $1000.

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Question

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?

Answer

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.

$\frac{48}{16}=H$

Therefore H is 3 hours.

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Question

$\small x$ varies inversely with . If , . What is the value of if ?

Answer

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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Question

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

Answer

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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Question

$\small x$ 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 $\small z = 2$ .

Answer

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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Question

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

Answer

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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Question

varies directly with $z^{2}$ and the square root of . If , and $\small z=12$ then . Find the value of if and .

Answer

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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Question

The speed of a turtle is indirectly proportional to its weight in pounds. At 10 pounds, the turtle's speed was 0.5. What is the speed of the turtle if it grew and weigh 50 pounds?

Answer

Write the formula for the indirect proportional relationship. If one variable increases, the other variable must also decrease.

Using speed and weight as and respectively, the equation becomes:

Use the initial condition of the turtle's speed and weight to solve for the constant.

Substitute this value back into the formula. The formula becomes:

We want to know the speed of the turtle when it is 50 pounds. Divide the variable on both sides to isolate the speed variable.

$s=\frac{5}{w}$

Substitute the new weight of the turtle.

$s=\frac{5}{50} = \frac{1}{10} = 0.1$

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