# Direct Variation Equation

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.

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