Solve Unitary Method Questions and Word Problems

The Unitary Method is a fundamental mathematical technique used to solve problems by finding the value of a single unit and then using it to determine the value of multiple units. It is widely applicable in solving real-life problems, particularly in areas like ratio and proportion, percentage calculations, and unit conversions. The Unitary method is essential for understanding how to break down complex problems into simpler, more manageable parts. This approach is often used in various types of competitive exams, including those for banking, government jobs, and entrance tests, where questions related to the Unitary Method appear frequently. Key concepts include understanding ratios, proportions, and basic arithmetic operations. This guide covers the definition, types, properties, formulas, and word problems associated with the Unitary Method, providing a comprehensive overview to enhance your problem-solving skills.

What is the Unitary Method?

The Unitary Method is a fundamental mathematical approach used to solve problems by first determining the value of a single unit and then using that value to find the value of multiple units. It operates on the principle that if the value of one unit is known, the value of any number of units can be calculated through multiplication or division.

Important Terms:

• Unit: A single quantity or measurement, such as one item, one kilogram, or one liter.
• Ratio: A relationship between two quantities, showing how many times one value is contained within another.
• Proportion: An equation that states that two ratios are equal.
• Multiplier: A factor by which a unit is multiplied to find the value of multiple units.
• Divisor: A number by which a total is divided to find the value of a single unit.
• Cost per Unit: The price of one unit of an item or service.
• Rate: A measure of a quantity relative to another quantity, such as speed (distance per unit of time) or density (mass per unit of volume).
• Percentage: A ratio or fraction out of 100, used to express proportions.

How It Works:

1. Finding the Value of One Unit:
   • If the value of multiple units is known, divide the total value by the number of units to find the value of one unit.
   • Example: If 5 apples cost $10, the cost of one apple is $10 ÷ 5 = $2.
2. Finding the Value of Multiple Units:
   • If the value of one unit is known, multiply it by the number of units to find the total value.
   • Example: If one apple costs $2, the cost of 8 apples is $2 × 8 = $16.

Types of Unitary Method

The Unitary Method can be categorized into two main types, based on the relationship between the quantities involved:

1. Direct Variation (Direct Proportion)

In direct variation, as one quantity increases, the other quantity also increases in the same proportion, and vice versa. The Unitary Method is applied here to find the value of one quantity when the other is known.

Example: If 5 kilograms of rice cost $20, then the cost of 1 kilogram (the unit) is $20 ÷ 5 = $4. If you want to find the cost of 8 kilograms of rice, you multiply the unit cost by 8: $4 × 8 = $32.

Application: This type of unitary method is used in problems involving prices, wages, and other quantities that increase or decrease in direct proportion to each other.

2. Inverse Variation (Inverse Proportion)

In inverse variation, as one quantity increases, the other quantity decreases proportionally, and vice versa. Here, the Unitary Method is used to find one quantity when the product of two quantities is constant.

Example: If 6 workers can complete a task in 12 days, then the work done by 1 worker (the unit) in 12 days is equivalent to the work done by 6 workers. To find out how long it will take for 3 workers to complete the task, divide the total work (6 × 12 = 72 worker-days) by the number of workers (3), resulting in 72 ÷ 3 = 24 days.

Application: This type is used in problems involving speed and time, work and workers, and other scenarios where an increase in one quantity leads to a proportional decrease in another.

Also Read: Reciprocal: Definition, Meaning and Solved Examples

Properties of Unitary Method

The Unitary Method has several key properties that make it an effective and versatile tool for solving a wide range of mathematical problems. Here are the main properties of the Unitary Method:

1. Proportionality

• Property: If one quantity changes, the other changes in a consistent ratio.
• Example: If 2 pens cost $4, then 4 pens will cost $8 (double the quantity, double the cost).

2. Linearity

• Property: Changes between quantities are linear, meaning direct or inverse proportionality.
• Example: If a car travels 60 miles in 1 hour, it will travel 120 miles in 2 hours (distance is directly proportional to time).

3. Scalability

• Property: The method can be applied to any number of units.
• Example: If 1 apple costs $2, then 50 apples will cost $2 × 50 = $100.

4. Versatility

• Property: Can be used for various types of problems (e.g., ratios, speed, cost).
• Example: If a worker can complete a task in 5 days, 2 workers can complete it in 2.5 days (work is inversely proportional to the number of workers).

5. Simplicity

• Property: Breaks down complex problems into simpler steps.
• Example: To find the cost of 7 chocolates when 1 chocolate costs $3, simply multiply: $3 × 7 = $21.

6. Consistency

• Property: Provides uniform and reliable results.
• Example: If 10 books cost $50, then 1 book costs $50 ÷ 10 = $5. For 20 books, it’s $5 × 20 = $100.

7. Applicability in Both Direct and Inverse Proportions

• Property: Works for both direct and inverse relationships.
• Example: Direct: 3 meters of cloth costs $30, so 6 meters cost $60. Inverse: 4 workers take 8 hours to complete a task; 8 workers will take 4 hours.

Formulas of Unitary Method

The Unitary Method primarily revolves around finding the value of one unit and then using that value to determine the value of multiple units. The formulas associated with the Unitary Method are straightforward and can be categorized based on direct and inverse proportion problems.

1. Direct Proportion Formula

In direct proportion, as one quantity increases, the other increases proportionally. The key formulas are:

• Finding the value of one unit:

Value of one unit = Total value of multiple units / Number of units

Example: If 5 apples cost $10, then the cost of one apple is:
10/5 = 2 dollars

• Finding the value of multiple units:

Value of multiple units = Value of one unit × Number of units

Example: If one apple costs $2, then the cost of 8 apples is:
2×8 = 16 dollars

2. Inverse Proportion Formula

In inverse proportion, as one quantity increases, the other decreases proportionally. The key formulas are:

• Finding the value of one unit (work or time):

Value of one unit = Total value (e.g., worker-days, total time) / Number of units (e.g., workers, tasks)​

Example: If 6 workers take 12 days to complete a task, the total worker-days is:
6×12 = 72 worker-days

• Finding the value of multiple units:

Value of multiple units = Total value / Number of units

Example: If 3 workers are to complete the same task, the time taken would be:
72/3 = 24 days

Also Read: All Perfect Cube Numbers

Types of Unitary Method Word Problems

Here are five-word problems that can be solved using the Unitary Method:

1. Cost Calculation

Problem: If 7 notebooks cost Rs 35, what is the cost of 5 notebooks? 

Solution:

• Cost of 1 notebook = Rs35 ÷ 7 = Rs5
• Cost of 5 notebooks = Rs 5 × 5 = $25

2. Time and Work

Problem: If 8 workers can complete a task in 12 days, how many days will it take for 6 workers to complete the same task? 

Solution:

• Total work = 8 workers × 12 days = 96 worker-days
• Time taken by 6 workers = 96 worker-days ÷ 6 workers = 16 days

3. Distance and Speed

Problem: A car travels 240 kilometers in 4 hours. How far will it travel in 7 hours at the same speed? 

Solution:

• Distance covered in 1 hour = 240 km ÷ 4 hours = 60 km/hour
• Distance in 7 hours = 60 km/hour × 7 hours = 420 kilometers

4. Wage Calculation

Problem: If 5 workers earn Rs 400 in a day, how much will 8 workers earn in a day? 

Solution:

• Earnings of 1 worker = Rs 400 ÷ 5 = $80
• Earnings of 8 workers = Rs 80 × 8 = $640

5. Quantity and Cost

Problem: If 10 kilograms of sugar cost $50, how much will 15 kilograms of sugar cost? 

Solution:

• Cost of 1 kilogram of sugar = Rs 50 ÷ 10 kg = $5/kg
• Cost of 15 kilograms = Rs 5/kg × 15 kg = $75

Unitary Method Questions: Step-by-Step Solution

Question 1 If 15 workers can complete a construction project in 20 days, how many days will it take for 25 workers to complete the same project, assuming the work rate is consistent?

Solution: Total work = Number of workers × Number of days

Total work = 15 × 20

Total work = 300 worker-days

Days for 25 workers = Total work / Number of workers

Days for 25 workers = 300 / 25

Days for 25 workers = 12 days

Question 2:A machine can produce 1200 units in 8 hours. If the production rate is consistent, how many units can it produce in 15 hours?

Solution: Units per hour = Total units / Number of hours

Units per hour = 1200 / 8

Units per hour = 150

Units in 15 hours = Units per hour × Number of hours

Units in 15 hours = 150 × 15

Units in 15 hours = 2250

Question 3:If 30 students can complete a group project in 10 days, how many students are required to complete the same project in 6 days?

Solution: Total work = Number of students × Number of days

Total work = 30 × 10

Total work = 300 student-days

Number of students for 6 days = Total work / Number of days

Number of students for 6 days = 300 / 6

Number of students for 6 days = 50 students

Question 4:A factory produces 5000 units of a product in 12 hours. How many units will it produce in 20 hours?

Solution: Units per hour = Total units / Number of hours

Units per hour = 5000 / 12

Units per hour ≈ 416.67

Units in 20 hours = Units per hour × Number of hours

Units in 20 hours = 416.67 × 20

Units in 20 hours ≈ 8333.33

Question 5:If 9 liters of paint cover 72 square meters, how many liters are required to cover 150 square meters?

Solution: Coverage per liter = Total coverage / Number of liters

Coverage per liter = 72 / 9

Coverage per liter = 8 square meters

Liters required = Total area / Coverage per liter

Liters required = 150 / 8

Liters required = 18.75 liters

Question 6:If a car travels 360 kilometers on 40 liters of fuel, how many liters will it need to travel 540 kilometers?

Solution: Distance per liter = Total distance / Number of liters

Distance per liter = 360 / 40

Distance per liter = 9 kilometers

Liters required = Total distance / Distance per liter

Liters required = 540 / 9

Liters required = 60 liters

Question 7:If 5 workers can finish a task in 16 days, how many days will it take for 8 workers to complete the same task?

Solution: Total work = Number of workers × Number of days

Total work = 5 × 16

Total work = 80 worker-days

Days for 8 workers = Total work / Number of workers

Days for 8 workers = 80 / 8

Days for 8 workers = 10 days

Question 8:A factory produces 2400 units of an item in 6 hours. How many hours will it take to produce 6000 units?

Solution: Production rate = Total units / Number of hours

Production rate = 2400 / 6

Production rate = 400 units per hour

Hours required = Total units / Production rate

Hours required = 6000 / 400

Hours required = 15 hours

Question 9:If 10 kg of a substance is used to prepare 25 liters of solution, how many kilograms are needed to prepare 75 liters of the same solution?

Solution: Kilograms per liter = Total kilograms / Number of liters

Kilograms per liter = 10 / 25

Kilograms per liter = 0.4 kg per liter

Kilograms required = Kilograms per liter × Number of liters

Kilograms required = 0.4 × 75

Kilograms required = 30 kg

Question 10:A contractor can paint 1500 square meters in 10 hours. If another contractor can paint at twice the speed, how many hours will it take for this second contractor to paint 1500 square meters?

Solution: First contractor’s rate = Total area / Number of hours

First contractor’s rate = 1500 / 10

First contractor’s rate = 150 square meters per hour

Second contractor’s rate = 2 × First contractor’s rate

Second contractor’s rate = 2 × 150

Second contractor’s rate = 300 square meters per hour

Hours required = Total area / Second contractor’s rate

Hours required = 1500 / 300

Hours required = 5 hours

FAQs

RELATED BLOGS

Elementary Linear Algebra	Factorisation Method
How to Solve Fraction Equations	Equations With Variables on Both Sides
Heights and Distances	Additive Inverse

This was all about the “Unitary Method”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our  Indian exams page.