Inverse Variation Worksheet

Inverse Variation Worksheet
• Page 1
1.
If the point (7, 15) is from an inverse variation, then find the constant of variation.
 a. $\frac{7}{8}$ b. 105 c. 107 d. $\frac{1}{105}$

Solution:

An equation in two variables of the form xy = k, where k ≠ 0, is an inverse variation. k is constant of variation.

So, the value of k = (7)(15) = 105

2.
Find the constant of variation if the relationship among the data in the table is an inverse variation.
 $x$ 0.15 0.25 0.35 $y$ 70 42 30
 a. 1.05 b. 5.01 c. 10.5 d. 466.67

Solution:

An inverse variation is a linear function that can be written in the form xy = k, when k ≠ 0.

(0.15) (70) = 10.5
(0.25) (42) = 10.5
(0.35) (30) = 10.5.
[Find the product xy .]

The product xy is constant for every relation. So, x varies inversely with y.

So, constant of variation is 10.5.

3.
Determine whether the relationship among the data in the table is an inverse variation, if so find the constant of variation.
 $x$ 2 - 3 4 $y$ 0.25 - 0.375 0.5
 a. No b. Yes, 2 c. Yes, 0.5 d. Yes, 1.125

Solution:

An inverse variation is a linear function that can be written in the form xy = k, where k ≠ 0.

(2) (0.25) = 0.5      (- 3) (- 0.375) = 1.125      (4) (0.5) = 2.
[Find the product xy .]

The product xy is not constant for every relation. So, x does not vary inversely with y.

4.
Write an equation to model the data given in the table:
 $x$ 2.4 2.6 2.8 $y$ 4.55 4.2 3.9
 a. $\frac{x}{y}$ = 10.92 b. $\mathrm{xy}$ = 10.92 c. there is no relation between $x$ and $y$. d. $\frac{y}{x}$ = 10.92

Solution:

As x increases, y decreases. So, check the inverse variation.

An inverse variation is a linear function that can be written in the form xy = k, where k ≠ 0.

(2.4) (4.55) = 10.92      (2.6) (4.2) = 10.92      (2.8) (3.9) = 10.92
[Find the product xy .]

The product xy is constant, so x varies inversely with y and k = 10.92

The equation is xy = 10.92.

5.
A gas exerts a pressure of 76 pounds per square inch when its volume is 40 cubic feet. What is the pressure exerted by the gas when its volume is 95 cubic feet assuming that the product of pressure and volume is constant?
 a. 37 pounds per square inch b. 32 pounds per square inch c. 96 pounds per square inch d. 34 pounds per square inch

Solution:

The relation between volume V and pressure P of a gas is PV = K .

(76)(40) = K
[Substitute the values.]

So, K = 3040

Now the equation is P = 3040V

P = 3040 / 95
[Substitute V = 95 cubic feet.]

P = 32

The gas at a volume of 95 cubic feet exerts a pressure of 32 pounds per square inch.

6.
Given $\mathrm{xy}$ = 19, then find the values of $a$, $b$ and $c$ in the table.
 $x$ $y$ 10 $a$ $b$ 19 9 $c$
 a. $a$ = 2.1, $b$ = 1.9, $c$ = 1 b. $a$ = 1.9, $b$ = 1, $c$ = 9 c. $a$ = 1.9, $b$ = 19, $c$ = 2.1 d. $a$ = 1.9, $b$ = 1, $c$ = 2.1

Solution:

xy = 19

From the table, substitute x, y values in the above equation.

 10a = 19 (b)(19) = 19 9c = 19 a = 19 / 10 b = 19 / 19 c = 19 / 9 a = 1.9 b = 1 c = 2.1

7.
If the point (2.76, 7.29) is from an inverse variation, then write an equation to model the data.
 a. $\mathrm{xy}$ = 20.71 b. $\mathrm{xy}$ = 21.12 c. $\frac{x}{y}$ = 19.12 d. $\mathrm{xy}$ = 20.12

Solution:

An equation in two variables of the form xy = k, where k ≠ 0, is an inverse variation. k is constant of variation.

So, the constant of variation k = (2.76)(7.29) = 20.12

The equation is xy = 20.12.

8.
If ($x$, $g$) and ($t$, $d$) both satisfy an inverse variation, then which of the following relation is true?
 a. $\frac{x}{t}$ = $\frac{d}{g}$ b. $\frac{t}{x}$ = $\frac{g}{d}$ c. $\mathrm{xg}$ = $\mathrm{td}$ d. All the above

Solution:

The formula for an inverse variation is xy = k, where k ≠ 0.

Substitute (x, g) in the equation.

xg = k

Substitute (t, d) in the equation.

td = k

So, xg = td

xgt = d
[Divide throughout by t .]

xt = dg
[Divide throughout by g.]

9.
If the pair of points (8, 2) and (- 4, $y$) are from an inverse variation, then find the value of $y$.
 a. - $\frac{17}{4}$ b. - $\frac{15}{4}$ c. - 4 d. - $\frac{16}{3}$

Solution:

The formula for an inverse variation is xy = k, where k ≠ 0.

So, x1 · y1 = x2 · y2

(8)(2) = (- 4)(y)

16 = - 4y

y = - 4

10.
Given $\mathrm{xy}$ = 2.5, then find the values of $a$, $b$ and $c$ in the table:
 $x$ $y$ $a$ 0.149 0.215 $b$ $c$ 5

 a. $a$ = 16.8, $b$ = 11.6, $c$ = 0.5 b. $a$ = 0.149, $b$ = 0.215, $c$ = 5 c. $a$ = 0.1, $b$ = 0.1, $c$ = 0.5 d. $a$ = 16.8, $b$ = 11.6, $c$ = 5

Solution:

xy = 2.5

From the table, substitute x, y values in the above equation.

 (a) (0.149) = 2.5 (0.215) (b) = 2.5 (c) (5) = 2.5 a = 2.5 / 0.149 b = 2.5 / 0.215 c = 2.5 / 5 a = 16.8 b = 11.6 c = 0.5