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PSAT 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:

$2.5y\ ft^2$

$25y\ ft^2$

$300y\ ft^2$

$150y\ ft^2$

$50y\ ft^2$

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 #3 : Variables

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:

$2187$

$729$

$27$

$81$

$243$

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 #2 : How To Use The Direct Variation Formula

varies directly as and inversely as ; $A = 6s^{2}$ and $V= s^{3}$ . Assuming that no other variables affect , which statement is true of concerning its relationship to ?

Possible Answers:

varies directly as .

varies inversely as the fifth power of .

varies directly as the fifth power of .

varies inversely as .

None of the other statements are correct.

Correct answer:

varies inversely as .

Explanation:

varies directly as and inversely as , meaning that for some constant of variation ,

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

Setting $A = 6s^{2}$ and $V= s^{3}$ , the formula becomes

$\frac{Ls^{3}}{6s^{2}}= K$

$\frac{Ls }{6 }= K$

Ls= 6K

Setting K' = 6K as the new constant of variation, the variation equation becomes

Ls= K' ,

so varies inversely as .

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

varies directly as both and the cube of . Which statement is true of concerning its relationship to ?

Possible Answers:

varies inversely as the cube root of .

None of the other statements is true.

varies directly as the cube of .

varies directly as the cube root of .

varies inversely as the cube of .

Correct answer:

varies inversely as the cube root of .

Explanation:

varies directly as both and the cube of , meaning that for some constant of variation ,

$\frac{x}{yz^{3}} = K$ .

Take the reciprocal of both sides, and the equation becomes

$\frac{yz^{3}}{x} =\frac{1}{ K}$

Take the cube root of both sides, and the equation becomes

$\frac{z\sqrt[3]{y}}{\sqrt[3]{x}} = \sqrt[3]{\frac{1}{ K}}$

$K' = \sqrt[3]{\frac{1}{ K}}$ takes the role of the new constant of variation here, and we now have

$\frac{z\sqrt[3]{y}}{\sqrt[3]{x}} =K'$

so varies inversely as the cube root of .

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PSAT Math : How to use the direct variation formula

Study concepts, example questions & explanations for PSAT Math

All PSAT 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 #3 : Variables

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

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

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