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Last updated on November 18, 2025

Unitary Method

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Have you ever bought a pack of items and wondered, “How much does just one cost?” Or check how long one activity takes, so you can plan the time for many activities. If yes, you’ve already used the unitary method without even knowing it! The unitary method is a simple trick: we first find the value of 1 unit, then use it to figure out the value of any number of units. In this article, we’ll explore how the unitary method works through real-life examples, quick steps, and fun questions.

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What is the Unitary Method in Math?

In mathematics, the unitary method is a powerful and easy technique for solving problems: first find the value of one unit, then use it to determine the value of any number of units. This method works on the principle of proportionality: when one quantity changes, the other changes in direct or inverse proportion

The reason it’s called the unitary method is simple: everything starts with understanding one unit. Once we know that, we can solve a wide range of unitary method problems, including ratio and proportion, cost, time, distance, and more. 
 

For example, imagine a car travels 60 km in 2 hours. How far will it travel in 1 hour, and how far in 5 hours?
Solution: 
Distance in 2 hours = 60 km.
So, distance in 1 hour = \(60 ÷ 2 = 30\) km.
Then the distance travelled in 5 hours \(= 30 × 5 = 150\) km.

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How to Use Unitary Method?

Use the following steps to apply the unitary method:

 

  • Determine what the unit depicts in the given problem. For example, one meter of cloth or one liter of milk.

 

  • To find out the value of one unit, we should divide the total value by the given quantity.

 

  • Multiply the value of a single unit with the given number of units to get the final value.

 

For example: The cost of 10 meters of cloth = $150, calculate the cost of 5 meters of cloth.


Cost of 1 meter = \(\frac{150}{10} = $15\).


Therefore, the cost of 5 meters = \(15 × 5 = $75\).

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Types of Unitary Method

The unitary method comes in two different variants. The two types are mentioned below:


Direct Variation: In direct variation, if one quantity increases or decreases, the other quantity also increases or decreases in the same proportion. For example, a worker paints 10 square meters of a wall in 1 hour. If the amount of time increases, the amount of work done also increases in the same proportion. 

That is, if the worker paints 10 m² in 1 hour, then in 3 hours, the work done will be calculated as:
Work in 1 hour = 10 m²
Work in 3 hours = \(10 × 3 = 30\)
 

 

Indirect Variation: A mathematical relationship where two quantities are indirectly connected to each other, means they are in indirect variation. In the indirect variation, when one quantity increases, the other decreases, and vice versa. For example, the time taken to walk a fixed distance varies inversely with the speed. If you walk faster, the time decreases and if you walk slower, the time increases. 

A student needs to walk a fixed distance of 3 km to reach school. When walking at the speed of 6 km/h,
the time taken \(= 3 ÷ 6 = 0.5\) hours (30 minutes).
When walking at a speed of 3 km/h, the time taken \(= 3 ÷ 3 = 1\) hour.

Here, as speed decreases, the time increases.
And as speed increases, the time decreases.
 

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Unitary Method in Ratio and Proportion

The unitary method helps us compare two quantities by first finding the value of one unit. When working with ratios and proportions, this method is very useful because it allows us to convert quantities to a standard unit before comparing them.

Let’s understand this through an example:
Emma earns $2,400 per month, while Sarah earns $33,600 per year. Both of them spend $1,800 per month on their expenses. Find the ratio of their monthly savings.

Solution: 
Monthly income of Emma \(= $2,400\).
Monthly expenditure \(= $1,800\).
Then, savings \(= 2,400 − 1,800 = $600\).

Annual income of Sarah \(= $33,600\).
Monthly income \(= 33,600 ÷ 12 = $2,800\).
Monthly expenditure \(= $1,800\).
Then, savings \(= 2,800 − 1,800 = $1,000\).

Ratio of their savings \(= 600 : 1000\).
By dividing both by 200: 
\(600 ÷ 200 = 3\)
\(1000 ÷ 200 = 5\)
Therefore, the ratio of Emma’s and Sarah’s savings \(= 3:5\).

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Tips and Tricks for Unitary Method

Finding it difficult to solve problems using the unitary method? These below-mentioned tips and tricks should help you solve them easily:

 

  • As a first step, you need to identify the context of the problem, i.e., what needs to be determined.

 

  • Check if the problem involves direct or indirect variation.

 

  • Write the quantities in a simple ratio form to make it easier to solve.

 

  • Ensure that you follow all the steps and that the final result includes the proper unit.

 

  • Always practice by applying it in real-life problems. For e.g., if 10 pens cost $100, what is the cost of 5 pens? To find the cost of one pen, we will divide 100 by 10. So \(\frac{100}{10} = 10\); therefore, 5 pens will cost \(10 × 5 = $50\).

     
  • Encourage children to see the unitary method in everyday situations like shopping, cooking, travel distance, time taken for chores, etc. This helps them understand how to use the unitary method naturally.

     
  • Students learn better with visual aids. Create simple tables, charts, or a unitary-method worksheet showing 1 unit, multiple units, and direct and indirect variation. 

     
  • When students solve unitary method questions, remind them that the first step is always to find the value of 1 unit. Then, multiply or divide as needed. Help students stick to this sequence through repeated unitary method examples.

     
  • Parents and teachers can prepare small practice tasks like: If three books cost $27, find the cost of 1, and then seven books. And, if a child walks more slowly, will the time increase or decrease? This will help students to strengthen their understanding of direct and indirect variations. 

     
  • Show children how the unitary method connects to other topics such as ratios, proportions, percentages, speed–distance–time, and cost problems. This makes the process more flexible and easier to apply across topics.
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Common Mistakes and How to Avoid Them in Unitary Method

The unitary method is a fundamental concept for solving many real-life issues. However, some students might find it tricky, but it can be learned easily if you spot the errors you make. Here’s a list of common mistakes and ways to avoid them:
 

Mistake 1

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Not determining the value of single unit first.
 

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Students might try to solve the problem without finding the value of a single unit.


Example: 10 bottles cost $200. Find the cost of 5 bottles.


They might solve it directly without finding the value of a single unit.


Ensure that you calculate the value of a single unit. Then use it to find the value of a given number of units.

Mistake 2

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Ignoring unit conversions.
 

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They often forget to convert units before calculations.

 

Example: A man saves $2000 per week. To calculate how much he could do each day:


Incorrect: $2000 by 30


Always ensure that you use the correct units throughout the calculation.


Corrected: Weeks should be converted to days:


1 week = 7 days


Therefore, the man saves per day \(= \frac{2000}{7} = $285.7\).

Mistake 3

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Not understanding the problem.
 

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Sometimes they fail to understand the context of the problem, which results in incorrect calculation. It is important to know the objective. For better understanding, you can break the issue into simple terms.
 

Mistake 4

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Rounding too early.
 

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They tend to round off values too early in the calculation, which will result in incorrect results. They may directly round off the incorrect fraction, 25 / 250 which gives an inaccurate answer. Follow the step-by-step process and round off at the final step.


Example: 25 chocolates cost $250. Find the cost of 10 chocolates.


Corrected: Find the cost of a single unit \(= \frac{250}{25} = 10\)


Cost of 10 chocolates \(= 10 × 10 = 100\).

Mistake 5

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Confusion between direct and indirect variations.
 

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They sometimes get confused with the terms direct and indirect variation, which can lead to errors in problem-solving. Remember, if two quantities increase or decrease together in the same proportion, they are in direct variation. However, if one quantity decreases while the other increases, they are said to be in indirect variation.
 

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Real-World Applications of Unitary Method

The unitary method is a mathematical concept that has numerous applications in various fields. We will now learn how they can be applied:

 

  • Calculating product prices and discounts: The unitary method helps students find the cost of a single item or a group of items and discounts. For example, if 5 notebooks cost $25, then the cost of 1 notebook is \($25 ÷ 5 = $5\). So, the cost of 8 notebooks is \(8 × $5 = $40\).

 

  • Estimating travel time: Students can use the unitary method to figure out how long a trip will take based on how much distance is covered in one unit of time. For example, if a cyclist travels 6 km in 20 minutes, then:
    distance per minute \(= 6 ÷ 20 = 0.3\) km.
    Then, the distance covered in 1 hour (60 minutes): 
    \(0.3 × 60 = 18\) km.

 

  • Finding monthly salary from daily wages: By knowing how much a worker earns in one day, their approximate monthly salary can be calculated. For example, if a gardener earns $50 per day, then in 30 days:
    Then, the monthly salary \(= 50 × 30 = $1500\).

 

  • Estimating fuel requirements for a journey: The unitary method helps students understand how much fuel is needed based on fuel consumption per unit distance. 
    For example, if a car uses 1 liter of fuel to travel 15 km, then for a 90 km trip:
    Fuel needed \(= 90 ÷ 15 = 6\) liters.

 

  • Planning meal quantities for families: Families can apply the unitary method to scale recipes according to the number of people. For example, if a recipe uses 200 g of rice to serve 2 people, then to serve 5 people:
    Rice needed per person \(= 200 ÷ 2 = 100\) g
    Total rice needed \(= 100 × 5 = 500\) g.
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Solved Examples of Unitary Method

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Problem 1

A family drives to their picnic spot. They travel 90 km in 2 hours. If they continue at the same speed, how long will it take them to travel 180 km?

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The family will take 4 hours to travel 180 km to the picnic spot.

Explanation

Given, distance travelled in 2 hours = 90km.

So, distance in 1 hour \(= 90 ÷ 2 = 45\) km

If 1 hour is needed to cover 45 km, 

Time to travel 180 km \(= 180 ÷ 45 = 4\) hours. 

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Problem 2

If 12 students can complete a task in 24 days, calculate the number of days it will take for 6 students to complete the same task.

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The number of days it will take for 6 students to complete the same task is 48 days.
 

Explanation

Assuming total work is constant, work per student per day = total work / number of students.

 

Total work completed by 1 student in 24 days \(= 12 × 24 = 288\)


So, the number of days it will take for 6 students:


\(\frac{288}{6} = 48\)


Therefore, the number of days it will take for 6 students to complete the same task is 48 days.

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Problem 3

A bus covers 300 km in 5 hours. How far will it travel in 10 hours at the same speed?

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The distance traveled by the bus in 10 hours is 600 km.
 

Explanation

Calculate the distance traveled in 1 hour: \(\frac{300}{5} = 60\) km


We now calculate the distance in 10 hours: \(60 × 10 = 600\) km


Therefore, the distance traveled by the bus in 10 hours is 600 km.

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Problem 4

If 4 children can make a clay pot in 16 days, how many days will it take for 8 children to make the same clay pot?

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The number of days required for 8 children is 8 days.
 

Explanation

To calculate the total work of 4 children in 16 days:


\(4 × 16 = 64\) days


Now we determine the number of days required for 8 children:


\(\frac{64}{8} = 8 \) days


Therefore, the number of days required for 8 children is 8.

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Problem 5

A group of hikers climbs 4 km in 1 hour 20 minutes. At the same hiking pace, how far will they hike in 3 hours?

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The hikers will cover 9 km in 3 hours.

Explanation

Given, the time taken by the hikers to climb the mountain = 1 hour 20 minutes. 
\(= 1 + \frac{20}{60}\) hours.
\(= 1 + \frac{1}{3}\) hours.
\(= 1.33\) hours (approx.)

Distance travelled in 1.33 hours = 4km
Then, distance in 1 hour \(= 4 ÷ 1.33 ≈ 3\) km.
As the distance in 1 hour \(≈ 3\) km, 
Distance in 3 hours \(= 3 × 3 = 9\) km.

Therefore, the hikers will cover approximately 9 km in 3 hours.

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FAQs on the Unitary Method

1.What do you mean by the unitary method?

The unitary method is the method we use to find the value of a single unit, by which we find the values of multiple units.

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2.Give one example of an application of the unitary method.

We use the unitary method to calculate the cost of a single unit, which we can then use to calculate the cost of the given number of units. It applies to calculating wages, discounts, and distances in the same way.

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3.What are the two types of unitary method?

The two types of unitary method are direct variation and indirect variation.
 

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4.How do we calculate cost using the unitary method?

To calculate the cost, we find the cost of one unit and then multiply or divide as required to find the total cost.
 

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5.Are the unitary method and ratio-proportion method the same?

No, they are not the same. The unitary method is used in determining the cost of one unit first, but the ratio and proportion method directly compares two ratios using cross multiplication.
 

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Dr. Sarita Ghanshyam Tiwari

About the Author

Dr. Sarita Tiwari is a passionate educator specializing in Commercial Math, Vedic Math, and Abacus, with a mission to make numbers magical for young learners. With 8+ years of teaching experience and a Ph.D. in Business Economics, she blends academic rigo

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: She believes math is like music—once you understand the rhythm, everything just flows!

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