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Algebra 1 : How to find direct variation

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Example Question #1 : How To Find Direct Variation

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?

Possible Answers:

$4.5 \textrm{ cm}$

$18.4 \textrm{ cm}$

$9.1\textrm{ cm}$

$8.4 \textrm{ cm}$

$11.5 \textrm{ cm}$

Correct answer:

$11.5 \textrm{ cm}$

Explanation:

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

L = kM

We can find by setting M = 20,L=7.2 from the first situation:

$7.2 = k \cdot 20$

$k = 7.2 \div 20 = 0.36$

so L = 0.36 M

In the second situation, we set M = 32 and solve for :

$L = 0.36 \cdot32 =11.52$

which rounds to 11.5 centimeters.

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Example Question #2 : How To Find Direct Variation

varies directly with the square root of . If x = 25 , then y = 48 . What is the value of if x = 81 ?

Possible Answers:

86.4

$26\tfrac{2}{3}$

155.52

None of these answers are correct.

491.52

Correct answer:

86.4

Explanation:

If varies directly with the square root of , then for some constant of variation ,

$\small y = K \sqrt{x}$

If x = 25 , then y = 48 ; therefore, the equation becomes

$\small \small 48 = K \sqrt{25}$ ,

5K = 48 .

Divide by 5 to get K = 9.6 , making the equation

$\small \small y = 9.6 \sqrt{x}$ .

If x = 81 , then $\small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4$ .

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Example Question #3 : How To Find Direct Variation

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?

Possible Answers:

$200 \textrm{ kg}$

$50 \textrm{ kg}$

$40 \textrm{ kg}$

$100 \textrm{ kg}$

$80 \textrm{ kg}$

Correct answer:

$50 \textrm{ kg}$

Explanation:

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

L = kM .

We can find by setting M = 33,L=6.6 :

$6.6 = k \cdot 33$

$k = 6.6 \div 33 = 0.2$

Therefore L = 0.2 M .

Set L = 10 and solve for :

10= 0.2 M

$M = 10 \div 0.2 = 50$ kilograms

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Example Question #2 : How To Find Direct Variation

If is directly proportional to and when a=24 at b=12 , what is the value of the constant of proportionality?

Possible Answers:

1.5

2.5

Correct answer:

Explanation:

The general formula for direct proportionality is

a=kb

where is the proportionality constant. To find the value of this , we plug in a=24 and b=12

24=12k

Solve for by dividing both sides by 12

$\frac{24}{12}=\frac{12k}{12}$

2=k

So k=2 .

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Example Question #4 : How To Find Direct Variation

The amount of money you earn is directly proportional to the nunber of hours you worked. On the first day, you earned $32 by working 4 hours. On the second day, how many hours do you need to work to earn $48.

Possible Answers:

Correct answer:

Explanation:

The general formula for direct proportionality is

M=kh

where is how much money you earned, is the proportionality constant, and is the number of hours worked.

Before we can figure out how many hours you need to work to earn $48, we need to find the value of . It is given that you earned $32 by working 4 hours. Plug these values into the formula

32=4k

Solve for by dividing both sides by 4.

$\frac{32}{4}=\frac{4k}{4}$

8=k

So k=8 . We can use this to find out the hours you need to work to earn $48. With k=8 , we have

M=8h

Plug in $48.

48=8h

Divide both sides by 8

$\frac{48}{8}=\frac{8h}{8}$

6=h

So you will need to work 6 hours to earn $48.

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Example Question #3 : How To Find Direct Variation

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?

Possible Answers:

$106 \textrm{ m}^{3}$

$121 \textrm{ m}^{3}$

$83 \textrm{ m}^{3}$

$100 \textrm{ m}^{3}$

$94 \textrm{ m}^{3}$

Correct answer:

$106 \textrm{ m}^{3}$

Explanation:

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

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

To calculate , substitute V = 100, T = 310, P = 1,020 :

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

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

K = 329.03

The variation equation is

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

so substitute T =290, P = 900 and solve for .

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

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Example Question #4 : How To Find Direct Variation

The monthly cost to insure your cars varies directly with the number of cars you own. Right now, you are paying $420 per month to insure 3 cars, but you plan to get 2 more cars, so that you will own 5 cars. How much does it cost to insure 5 cars monthly?

Possible Answers:

$\$633$

$\$748$

$\$810$

$\$390$

$\$700$

Correct answer:

$\$700$

Explanation:

The statement, 'The monthly costly to insure your cars varies directly with the number of cars you own' can be mathematically expressed as M=kC . M is the monthly cost, C is the number of cars owned, and k is the constant of variation.

Given that it costs $420 a month to insure 3 cars, we can find the k-value.

420=k * 3

Divide both sides by 3.

k=140

Now, we have the mathematical relationship.

M=140C

Finding how much it costs to insure 5 cars can be found by substituting 5 for C and solving for M.

M=140(5)

M=700

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Example Question #1 : How To Find Direct Variation

Does the equation below represent a direct variation? If it does, find the constant of variation.

7y = 2x

Possible Answers:

Yes, $y= \frac{7}{2}x, \frac{7}{2}$

No, $\frac{7}{2}y = x$ , 1

Yes, y = 14x, 14

No, $y = \frac{1}{14}x, \frac{1}{14}$

Yes, $y = \frac{2}{7}x$ $, \frac{2}{7}$

Correct answer:

Yes, $y = \frac{2}{7}x$ $, \frac{2}{7}$

Explanation:

Direct Variation is a relationship that can be represented by a function in the form

y = kx , where $k \neq 0$

is the constant of variation for a direct variation. is the coefficient of .

7y = 2x

$y = \frac{2}{7}x$

The equation is in the form y = kx , so the equation is a direct variation.

The constant of variation or is $\frac{2}{7}$

Therefore, the answer is,

Yes it is a direct variation, $y = \frac{2}{7}x,$ with a direct variation of $\frac{2}{7}$

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Example Question #2 : How To Find Direct Variation

Suppose x=24 and y=4 , and that is in direct proportion with . What is the value of proportionality?

Possible Answers:

5.5

$\frac{1}{6}$

Correct answer:

Explanation:

The general formula for direct proportionality is

x=ky

where is our constant of proportionality. From here we can plug in the relevant values for and to get

24=4k

Solving for requires that we divide both sides of the equation by , yielding

k=6

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Example Question #5 : How To Find Direct Variation

The cost of a catering company varies directly with the number of people attending. If the cost is $100 when 20 people attend the party, find the constant of variation.

Possible Answers:

k=5

k=8

k=40

k=2000

Correct answer:

k=5

Explanation:

Because the cost varies directly with the number of people attending, we have the equation

C=kN

Where is the cost and is the number of people attending.

We solve for , the constant of variation, by plugging in C=100 and N=20 .

100=k(20)

And by dividing by 20 on both sides

$\frac{100}{20}=\frac{k(20) }{20}$

Yields

k=5

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Algebra 1 : How to find direct variation

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : How To Find Direct Variation

Example Question #2 : How To Find Direct Variation

Example Question #3 : How To Find Direct Variation

Example Question #2 : How To Find Direct Variation

Example Question #4 : How To Find Direct Variation

Example Question #3 : How To Find Direct Variation

Example Question #4 : How To Find Direct Variation

Example Question #1 : How To Find Direct Variation

Example Question #2 : How To Find Direct Variation

Example Question #5 : How To Find Direct Variation

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