Solving Direct Variation using Using Unitary Method

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