ADO - Linear Equations (Multi-Variable Word Problems)

Linear equations in multi-variable word problems involve forming and solving equations with two or more unknowns based on real-life situations. These problems test your ability to translate verbal information into mathematical relationships and then solve them logically.

1. Understanding the Concept

A linear equation is an equation in which the highest power of the variable is 1. When a problem has two or more unknowns, you need the same number of independent equations to find a unique solution.

For example:
x + y = 10
2x − y = 4

Here, x and y are variables, and the equations are linear because the variables are not raised to powers higher than 1.

2. Common Types of Word Problems

a) Age Problems

These involve present, past, or future ages.

Example:
The sum of ages of A and B is 40. After 10 years, A will be twice B’s age.

Let A = x and B = y
Equation 1: x + y = 40
After 10 years: x + 10 = 2(y + 10)

Solve:
x + 10 = 2y + 20
x = 2y + 10

Substitute into first equation:
2y + 10 + y = 40
3y = 30 → y = 10
x = 30

b) Mixture Problems

Used when combining two or more substances with different values (price, concentration, etc.).

Example:
A mixture contains milk and water. Total volume is 20 liters, and milk is 12 liters.

Let milk = x, water = y
x + y = 20
x = 12 → y = 8

More complex cases include cost or ratio conditions.

c) Work Problems

Based on efficiency or rates of work.

Example:
A and B together complete a task in 6 days. A alone takes 10 days.

Let B take y days
1/10 + 1/y = 1/6

Solve:
1/y = 1/6 − 1/10 = (5 − 3)/30 = 2/30 = 1/15
y = 15 days

d) Speed, Time, Distance Problems

These involve relationships like distance = speed × time.

Example:
Two trains start from the same point. One travels at 60 km/h and another at 40 km/h. After how many hours will they be 100 km apart?

Let time = t
Distance difference = 60t − 40t = 20t
20t = 100 → t = 5 hours

e) Cost and Profit Problems

Involve selling price, cost price, and profit.

Example:
Total cost of two items is 500. One costs 100 more than the other.

Let costs be x and y
x + y = 500
x = y + 100

Substitute:
y + 100 + y = 500
2y = 400 → y = 200
x = 300

3. Methods to Solve

Substitution Method

Solve one equation for one variable and substitute into another.

Elimination Method

Add or subtract equations to eliminate one variable.

Comparison Method

Solve both equations for the same variable and compare.

4. Key Steps to Solve Word Problems

  1. Read the problem carefully and identify unknowns

  2. Assign variables clearly

  3. Convert statements into equations

  4. Solve using a suitable method

  5. Verify the answer in the original context

5. Important Tips

  • Always check if the number of equations equals the number of variables

  • Pay attention to units (years, liters, km, etc.)

  • Translate words like “more than,” “less than,” “twice,” correctly into equations

  • Practice forming equations, not just solving them

6. Why This Topic is Important for ADO

  • Frequently appears in quantitative aptitude

  • Helps in solving case-based and real-life problems

  • Improves logical thinking and interpretation skills

  • Often combined with data interpretation questions