In this chapter we will learn to solve direct variation expression using unitary method.

Let us first review the basics of both the concepts.

## What is direct variation?

Direct variation is represented by below expression.

**y = k . x**

Where y and x are variables and k is constant.

The expression states that;

⟹ Increase in value of x will also proportionally increase value of y.

⟹ Decrease in value of x will decrease value of y.

## What is Unitary Method

It’s a technique which **first finds value of single unit** and **then move on to find the value of desired unit**.

**For example; **

If price of 10 apples is 60$. Find the price of 15 apples.

Here, the unitary method will first calculate the price of 1 apple and then we can find the price of 15 apples.

Price of 10 Apples = 60$

Price of 1 apples = 60/10 = 6$

Price of 15 apples = 15 x 6 = 90$

In this method we utilize simple cross multiplication & division techniques to come up with the solution.

I hope you get an idea of both the concepts. Let’s solve some questions related to direct variation.

## Solved Direct Variation Problems with Unitary Method.

**Example 01**

The total weight of 8 books is 56kg. Find the total weight of 12 books.

**Solution**

It’s a direct variation questions as the with increase in number of books, the weight will also increase.

Solving using Unitary Method.

8 Books weigh ⟹ 56 kg

1 book weigh ⟹ 56 / 8 = 7 kg

12 books weigh ⟹ 7 x 12 = 84 kg

Hence, **weight of 12 books is 84 kg**.

**Example 02**

A car travels 160 kilometers in 5 gallon. Find the distance travelled by car in 15 gallons.

**Solution**

Here the distance travelled and oil consumption are in direct relationship. As car travels more distance, the oil consumed will also increase.

5 gallon oil results in ⟹ 160 km distance

First find the distance covered by 1 gallon oil.

1 gallon oil ⟹ 160 / 5 = 32 km distance

So the distance covered by 15 gallons will be;

15 gallon oil ⟹ 32 x 15 = 480 km.

Hence in 15 gallons, **the car will travel 480 kms**.

**Example 03**

The price of 25 packet of biscuit is 62.5$. Find the cost of 13 packets of biscuit.

**Solution**

The number of packets and total price are in direct variation. As the number of packet increase, the total cost will also increase.

25 packet of Biscuits cost ⟹ 62.5$

First calculate the cost of 1 packet.

1 packet of biscuit cost ⟹ 62.5 /25 = 2.5 $

Now calculate the cost of 13 packets.

13 packets of biscuit cost ⟹ 2.5 x 13 = 32.5$

Hence, **price of 13 packets of biscuit is 32.5$**

**Example 04**

It takes 64 hours to stitch 16 trousers. Find the total time taken to stitch 26 trousers.

**Solution**

Number of trousers and time taken are in direct relationship. A the number of trousers increases, the time taken will also increase.

16 trousers took ⟹ 64 hours time.

First calculate the time taken to stitch 1 trouser.

1 trouser took ⟹ 64 / 16 = 4 hours

Now calculate time for stitching 26 trousers.

26 trouser ⟹ 4 x 26 = 104 hours

Hence, **time taken to stitch 26 trouser in 104 hours**.

**Example 05**

A software engineer gets $600 for 5 day work. In how many days will the engineer earns $2280.

**Solution**

Number of days and total earning are in direct relationship. As the working days increase, the salary earned will also increase.

600$ earned ⟹ 5 days work

Calculate the time for 1$ earning.

1$ earned ⟹ 5/600 = 1/120 days

Now calculate time for $2280

$2280 earned ⟹ 2280 x (1/120) = 19 days.

Hence **in 19 days the engineer will earn $2280**