Solving Problems Using Unitary Method

Unitary method problems are problems where you are required to find the value of one thing based on the given values of something else. Let us take an example of this.

Problem 1:

The weight of $10$ identical books is $500$ gms. What is the weight of $18$ such books?

We know that since the books are identical, their weights are same, and the total weight is dependent on the number of books.

In this case since we need to find the value of the weight of books, the weight of books is called the dependent value, and the since the number of books is given, we call that the independent value.

We also know that the weight total weight increases with the number of books.

So, the steps we follow is as follows:

1. Calculate the value of the dependent thing for a value of 1 of the independent value. The name unitary method comes from the number 1 (unit).

2. Calculate the dependent value for the given independent value.

In our example the dependent value is the weight of the books, and the independent value is the number of books.

Here is how we go about writing the statement.

The weight of $10$ books is $500$ gms.

The weight of $1$ book is $500 \div 10 = \dfrac{500}{10}\ gms$

Therefore, the weight of $18$ books is $\dfrac{500}{10} \times {18} = 50 \times 18 = 900\ gms$

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It is very important to identify whether the dependent value increases or decreases with the increase of the independent value.

Our previous example example the dependent value (weight) increases with the independent value (number of books).

We will see another example where the dependent value decreases with any increase in the independent value.

Problem 2:

Driving at 60 km/hr it takes 20 minutes to reach from your house to your school. How long will it take when driving at 40 km/hr?

Here the dependent thing is time which you are supposed to find out, and the independent thing is the speed which is given. And we also know, that if you drive at a higher speed the time taken will be less. So let us find the answer to our question:

Driving at $60$ km/hr it takes $20$ mins to school.

Driving at $1$ km/hr it takes $20 \times 60\ mins$ to school.

Driving at $40$ km/hr, it takes $\dfrac{20 \times 60}{40} = 30\ mins$ to school

Notice, that in our first example, in step 2, we divided the value of the independent value by the dependent value, and in step 3, we multiplied them. But in our second example we multiplied the independent value by the dependent value in step 2. This is because in first case the dependent value will increase and in the second case the dependent value decreased with an increase in the independent value.

You will get a better feel of this concept once you have gone through some practice problems.

Note that it is always a good practice to do the calculations at the last step. If you do the calculations earlier you may end up with much larger numbers, and the chances of committing error will be higher.