How to use the direct variation formula - PSAT Math

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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.

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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:

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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 ?

Answer

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

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

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

Setting as the new constant of variation, the variation equation becomes

,

so varies inversely as .

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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:

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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 ?

Answer

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

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

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

Setting as the new constant of variation, the variation equation becomes

,

so varies inversely as .

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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:

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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 ?

Answer

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

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

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

Setting as the new constant of variation, the variation equation becomes

,

so varies inversely as .

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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:

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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 ?

Answer

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

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

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

Setting as the new constant of variation, the variation equation becomes

,

so varies inversely as .

Compare your answer with the correct one above

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