In this chapter we will learn methods to solve direct variation equation.

Let us review the basic concept of direct variation.

## What is Direct Variation?

Direct variation is represented by below equation.

**y = k. x**

Here, x & y are variable and k is constant.

The equation says that y is directly proportional to x.

Which means that;

⟹ **If x increase, then value of y will also increase proportionally**.

⟹ **If x decrease, then value of y will also decrease**.

Now the question arises: **with increase in x value, how much value of y will increase?**

It all depend on the** value of constant k**.

⟹ If value of k is large, then increase in x value will cause large increase in value of y.

⟹ If value of k is small, then increase in value of x will cause small increase in value of y.

### Do the variables of direct variation maintain fixed ratio

**YES!!**

In the equation y = kx, the variable x and y maintain the fixed ratio k.

Consider the below equation;

y = k . x

Taking x to the left side of equation, we get;

\mathtt{\frac{y}{x} =k}

Here the entity y/x maintain the same ratio k ( called constants )

## Solving direct variation equation

Given below are different types of direct variation questions.

All the questions are fully solved with explanation. Request you to understand all the given questions.

**Example 01**

Variable x and y are direct proportional to each other.

The value of constant is 15.

Find the value of y if x = 3.

**Solution**

Since y is directly proportional to x, we can write:

y = k. x

Its given that x = 3 & k = 15.

Putting the values in above equation.

y = 15. 3

y = 45

Hence **for x = 3, the value of y is 45**.

**Example 02**y and x are in direct proportion. When x = 4 then value of y is 24. Find the value of x when y = 54.

**Solution**

Since y and x are directly proportional we can write;

y = k. x

Its given that when x = 4 then y = 24.

Putting the values in above equation.

24 = k . 4

24/4 = k

k = 6**Hence, the value of constant k = 6**.

Now the above equation can be written as;

y = 6.x

Now find the x, when y = 54.

Putting the values in equation, we get;

54 = 6 .x

x = 54 / 6

x = 9

Hence **for y = 54, the value of x = 9**

**Example 03**

The cost of 13 watches is $273. Find the cost of 6 watches.

**Solution**

Here the cost of watch and quantity is directly proportional to each other.

When number of watch increase, the total cost will also go up.

Let the cost of 1 watch be k.

y is the total cost.

x is the number of watch.

The direct equation is represented as;

k. x = y

k . 13 = 273

k = 273/13

k = 21

**The value of constant k = 21$**

i.e. cost of 1 watch = 21$

Now find cost of 6 watches.

Here x = 6 & k = 21.

The equation can be expressed as;

y = k. x

y = 21 . 6

y = 126$

**Hence, total cost of 6 watches is $126.**

**SHORTCUT Method**

Direct Variation equation question can be easily be solved using following method.

Cost of 13 watch = $273

Cost of 1 watch = 273/13

Cost of 1 watch = $21

For cost of 6 watches, simply multiply the expressing by 6.

Cost of 6 watches = 21 x 6 = 126$

Hence 126$ is the required solution.

**Example 04**

The weight of 12 water bottles is 72 kg. Find the weight of 20 water bottles.

**Solution**

Number of water bottles and total weight are in direct proportion.

As number of bottles increase, the total weight will also increase.

Let number of water bottles = x, and;

Total weight of water bottle = y

The expression can be written as;

y = k. x

72 = k . 12

k = 72 / 12

k = 6

**Hence, the value of constant is 6.**

Now find weight of 20 water bottles.

y = k. x

y = 6 . 20

y = 120 kg

**Hence, weight of 20 water bottle is 120 kg**.

**SHORTCUT Method**

Weight of 12 bottles = 72 kg

Weight of 1 bottle = 72 / 12 kg

Weight of 1 bottle = 6 kg

Now multiply the expression by 20 to find weight of 20 water bottles.

Weight of 20 bottles = 6 x 20 = 120 kgs

Hence **120 kg is the required solution**.

**Example 05**

A car runs 72 miles in 3 gallons. How many gallons required to run total of 168 miles.

**Solution**

Oil used by car and distance covered are directly proportional.

As the oil used by car increase, the distance travelled by car will also increase.

Let, Oil used = x and distance covered = y

The direct relation can be expressed as;

y = k. x

Putting the values of y and x.

72 = k . 3

k = 72 / 3

k = 24

**Hence the** **value of constant is 24**.

Now find the oil used when the distance travelled is 168 miles.

y = k. x

168 = 24 . x

x = 168 / 24

x = 7

Hence, **7 gallon oil is used to cover distance of 168 miles**

**SHORTCUT Method**

72 mile distance consumes = 3 gallon oil

1 mile distance consumes = 3 / 72 = 1/24 gallon oil.

Multiply 168 on both sides to get oil consumed by 168 miles.

168 mile distance consumes = 168 x (1 / 24) = 7 gallons

Hence, total of 7 gallons consumed while travelling 168 miles.