If two quantities are so related such that the increase in one of them leads to the decrease in the other vice versa, then the first one is said to vary inversely as the second.

For example, if a bus traveling at a given speed tends to increase such speed, the less time it will take the bus to cover a given distance.

Therefore, in an inverse variation, the product of the two variables (quantities) will result in a constant for instance if y varies inversely as x it is written as *yx 1/x*

Therefore *y = K/x*

*yx = K, * where K = constant

The following are among the many inverse variations that we may come across:

- Time travelled and a given distance varies inversely as the sped
- The volume of mass of gas varies inversely as the pressure

**Example 1:**

y is inversely proportional to x^{2}, when y = 10, x = 2, what is y when x = 10?

**Solution**

y ∞ 1/x^{2}

y = K/x^{2}where K is constant ∞ is the sign of proportionality

when y 10, x = 2

substitute into the equation

10 = K/(2)^{2}

10 = K/4

Cross multiply

10 x 4 = K

K = 40

Find y when x 10

y = 40/(10)^{2} = 40/100 = 2/5

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**Example 2:**

If y varies inversely as the square root of x and if y = 5 when x = 16,

Find;

- y if x = 100
- x if y = 60

**Solution**

y ∞ 1/

y = K/ where K is constant

if y = 5, k = 16

5 = K/

5 = K/4

K = 5 x 4

K = 20

- y if x = 100

y = K/

y = K/ = 21/ = 20/10 = 2

- x if y = 60

60 = 20/

Cross multiply

60 = 20 /60

= 20/60 = 1/3

Take square of both sides

N/B: = ½

(x)^{1/2×2} = (1/3)^{2}

x = 1^{2}/3^{2}= 1/9

therefore x = 1/9

**Example 3: **

Given that y varies inversely as the square of x and y =8 when x = 2, find the equation connecting y and x calculate the value of x when x = 6 and x when y = ½

**Solution**

y ∞ 1/x^{2} , y = K/x^{2}

y = 8, x = 2

8 = K/2^{2}

8 = K/4

K = 8 x 4 = 32

The equation connecting y and x is

y = 32/x^{2}

Find y when x = 6

* y = 32/(6) ^{2} = 32/36 = 8/9*

when*y = ½*

*½ = 32/x ^{2}*

Cross multiply

*x ^{2}*x

*1 = 32*x

*2*

*x ^{2} = 64*

take square root of both sides

*x ^{2}*

^{x1/2}

*x = 8*

**Example 4:**

The electrical resistance of a wire varies inversely as the square of the radius. If the resistance is 0.24n when the radius is 0.5cm find the resistance when the radius is 0.04

**Solution**

Let electrical resistance be R and radius be r

R ∞ 1/r^{2}

*R = K/r ^{2}*

*If R = 0.24n r = 0.5cm*

*0.024 = K/(0.5) ^{2}*

*0.24 = K/0.25*

*K = 0.24 *x *0.25*

*K = 0.06*

Find R when r = 0.04

But K = 0.06

*R = K/r ^{2}*

*R = 0.06/(0.04) ^{2} = 0.06/0.0016*

R = *37.5n*

**Example 5:**

P is inversely proportional to Q and P = 5 when Q = 4. What is the value of Q when P = 25?

Solution

P ∞ 1/Q

P = K/Q where K is constant

When P = 5, Q = 4

∞= K/4

K = 5 x 4 = 20

Find Q when P = 25

25 = 20/Q

Cross multiply

25 x Q = 20

25Q/25 = 20/25

Q = 20/25 = 4/5

Therefore Q = 4/5

**Example 6:**

A School inter house sports competition, the weight of a javelin W varies inversely as the cube root of the distance covered by each throw. When the weight is 270kg the distance covered is 8m. find the

- distance covered by a 90kg javelin
- weight of a javelin that covers 27m

Solution

Weight = W, distance = d

Letsinterpret the statement

W ∞ 1/

Removing the sign of proportionality ∞ we introduce a constant K

W = K/

When W = 270kg, d = 8

So we substitute into the formula

270 = K/

The cube root of 8 is 2

i.e 2x2x2 = 8

270 = K/2

K = 270 x 2 = 540

- find d when W = 90kg and K = 540

90 = 540/

= 540/90 = 6

= 6

Cube both sides

(d) ^{1/3 x 3} = 6^{3}

d = 6^{3} = 6 x 6 x 6 = 2/6

therefore, the distance is 2/6m

- W when d = 27m

We substitute into the formula

W = K/

W = 540/

Cube root of 27 is 3

W = 540/3 = 180

Weight = 180kg

From the problems we have solved, you can see it with me that inverse just require a correct interpretation of the logical statement.

Always remember that inverse is 1/a number

If you encounter any problem on this topic, just follow the laid down procedure we have established today.