How to find inverse variation - PSAT Math

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Question

Find the inverse equation of:

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Answer

To solve for an inverse, we switch x and y and solve for y. Doing so yields:

$\frac{3x-20}{2}=y$

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Question

A school's tornado shelter has enough food to last 20 children for 6 days. If 24 children ended up taking shelter together, for how many fewer days will the food last?

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Answer

Because the number of days goes down as the number of children goes up, this problem type is inverse variation. We can solve this problem by the following steps:

206=24x

120=24x

x=120/24

x=5

In this equation, x represents the total number of days that can be weathered by 24 students. This is down from the 6 days that 20 students could take shelter together. So the difference is 1 day less.

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Question

x y

If varies inversely with , what is the value of ?

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Answer

An inverse variation is a function in the form: or $y = \frac{k}{x}$ , where is not equal to 0.

Substitute each $\left ( x,y \right )$ in .

Therefore, the constant of variation, , must equal 24. If varies inversely as , must equal 24. Solve for .

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Question

varies inversely with the square root of .

If when , then evaluate when . (Nearest tenth)

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Answer

If varies inversely with the square root of , then, if are the initial values of the variables and are the final values,

$A'\sqrt{B'} = A\sqrt{B}$ .

Substitute and find :

$A'\sqrt{100} = 8\sqrt{66}$

$A'=\frac{ 8\sqrt{66}}{\sqrt{100} } \approx \frac{ 8 \cdot 8.1240}{10} \approx 6.5$

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Question

varies inversely as the square of .

If when , then evaluate when . (Nearest tenth)

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Answer

If varies inversely as the square of , then, if are the initial values of the variables and are the final values,

$A'B' ^{2} = AB^{2}$ .

Substitute and find :

$A' \cdot 100 ^{2} = 8 \cdot 66^{2}$

$A' =\frac{ 8 \cdot 66^{2}}{100^{2}} =\frac{ 8 \cdot 66^{2}}{100^{2}} = \frac{ 8 \cdot 4,356}{10,000} =3.4848$

This rounds to 3.5.

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Question

varies inversely as the cube of .

If when , then evaluate when . (Nearest tenth)

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Answer

If varies inversely as the cube of , then, if are the initial values of the variables and are the final values,

$A'B' ^{3} = AB^{3}$

Substitute and find :

$A' \cdot 100 ^{3} = 8 \cdot 66^{3}$

$A' =\frac{ 8 \cdot 66^{3}}{100^{3}} = \frac{ 8 \cdot 287,496}{1,000,000} =2.299968$

This rounds to 2.3.

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Question

varies inversely as the cube root of .

If when , then evaluate when . (Nearest tenth)

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Answer

If varies inversely as the cube root of , then, if are the initial values of the variables and are the final values,

$A'\sqrt[3]{B'} = A\sqrt[3]{B}$

Substitute and find :

$A'\sqrt[3]{100} = 8\sqrt[3]{66}$

$A'=\frac{ 8\sqrt[3]{66}}{\sqrt[3]{100} } \approx \frac{ 8 \cdot 4.0412}{4.6416} \approx 7.0$

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Question

Find the inverse of .

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Answer

To find the inverse of a function we need to first switch the and . Therefore, becomes

We now solve for y by subtracting 1 from each side

From here we divide both sides by 2 which results in

$y=\frac{x-1}{2}$

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Question

Find the inverse of .

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Answer

To find the inverse we first switch the variables then solve for y.

Then we subtract from each side

Now we divide by to get our final answer. When we divide by we are left with . When we divide by we are left with . Thus resulting in:

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Question

Find the inverse of

$y=\frac{1}{2}x-4$

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Answer

To find the inverse we first switch the x and y variables

$x=\frac{1}{2}y-4$

Now we add 4 to each side

$x+4=\frac{1}{2}y$

From here to isolate y we need to multiply each side by 2

By distributing the 2 we get our final solution:

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Question

Find the inverse equation of .

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Answer

1. Switch the and variables in the above equation.

2. Solve for :

$\frac{5y}{5}=\frac{4x+18}{5}$

$y=\frac{4x+18}{5}$

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Question

$\begin{matrix} & \x\ & \y\ & \ & \ & \end{matrix}$ When , .

When , .

If varies inversely with , what is the value of when ?

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Answer

If varies inversely with , $y=\frac{K}{x}$ .

1. Using any of the two combinations given, solve for :

Using :

$18=\frac{K}{2}$

2. Use your new equation $y=\frac{36}{x}$ and solve when :

$y=\frac{36}{12}=3$

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Question

A school's tornado shelter has enough food to last 20 children for 6 days. If 24 children ended up taking shelter together, for how many fewer days will the food last?

Tap to reveal answer

Answer

Because the number of days goes down as the number of children goes up, this problem type is inverse variation. We can solve this problem by the following steps:

206=24x

120=24x

x=120/24

x=5

In this equation, x represents the total number of days that can be weathered by 24 students. This is down from the 6 days that 20 students could take shelter together. So the difference is 1 day less.

← Didn't Know|Knew It →

Question

Find the inverse equation of:

Tap to reveal answer

Answer

To solve for an inverse, we switch x and y and solve for y. Doing so yields:

$\frac{3x-20}{2}=y$

← Didn't Know|Knew It →

Question

x y

If varies inversely with , what is the value of ?

Tap to reveal answer

Answer

An inverse variation is a function in the form: or $y = \frac{k}{x}$ , where is not equal to 0.

Substitute each $\left ( x,y \right )$ in .

Therefore, the constant of variation, , must equal 24. If varies inversely as , must equal 24. Solve for .

← Didn't Know|Knew It →

Question

varies inversely with the square root of .

If when , then evaluate when . (Nearest tenth)

Tap to reveal answer

Answer

If varies inversely with the square root of , then, if are the initial values of the variables and are the final values,

$A'\sqrt{B'} = A\sqrt{B}$ .

Substitute and find :

$A'\sqrt{100} = 8\sqrt{66}$

$A'=\frac{ 8\sqrt{66}}{\sqrt{100} } \approx \frac{ 8 \cdot 8.1240}{10} \approx 6.5$

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Question

varies inversely as the square of .

If when , then evaluate when . (Nearest tenth)

Tap to reveal answer

Answer

If varies inversely as the square of , then, if are the initial values of the variables and are the final values,

$A'B' ^{2} = AB^{2}$ .

Substitute and find :

$A' \cdot 100 ^{2} = 8 \cdot 66^{2}$

$A' =\frac{ 8 \cdot 66^{2}}{100^{2}} =\frac{ 8 \cdot 66^{2}}{100^{2}} = \frac{ 8 \cdot 4,356}{10,000} =3.4848$

This rounds to 3.5.

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Question

varies inversely as the cube of .

If when , then evaluate when . (Nearest tenth)

Tap to reveal answer

Answer

If varies inversely as the cube of , then, if are the initial values of the variables and are the final values,

$A'B' ^{3} = AB^{3}$

Substitute and find :

$A' \cdot 100 ^{3} = 8 \cdot 66^{3}$

$A' =\frac{ 8 \cdot 66^{3}}{100^{3}} = \frac{ 8 \cdot 287,496}{1,000,000} =2.299968$

This rounds to 2.3.

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Question

varies inversely as the cube root of .

If when , then evaluate when . (Nearest tenth)

Tap to reveal answer

Answer

If varies inversely as the cube root of , then, if are the initial values of the variables and are the final values,

$A'\sqrt[3]{B'} = A\sqrt[3]{B}$

Substitute and find :

$A'\sqrt[3]{100} = 8\sqrt[3]{66}$

$A'=\frac{ 8\sqrt[3]{66}}{\sqrt[3]{100} } \approx \frac{ 8 \cdot 4.0412}{4.6416} \approx 7.0$

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Question

Find the inverse of .

Tap to reveal answer

Answer

To find the inverse of a function we need to first switch the and . Therefore, becomes

We now solve for y by subtracting 1 from each side

From here we divide both sides by 2 which results in

$y=\frac{x-1}{2}$

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