Word Problems on Pair of Linear Equations in Two Variables
Word problems on pairs of linear equations form one of the most practically oriented and frequently examined topics in CBSE Class 10 Mathematics. While students may be comfortable solving equations given in algebraic form, the real challenge lies in translating a written description of a situation into two equations and then solving them. This skill bridges the gap between abstract algebra and real-life problem solving. Word problems on linear equations cover a rich variety of contexts — ages, numbers, fractions, speeds and distances, areas, costs, work and time, and many more. The NCERT textbook dedicates substantial attention to this topic, and board examination papers almost always feature at least one word problem on linear equations. In this topic, we will explore the systematic approach to forming equations from word problems, solve a wide range of example types, and develop the critical thinking skills needed to interpret verbal statements mathematically. The ability to convert real-world scenarios into algebraic equations is a cornerstone of mathematical literacy that extends well beyond the classroom.
What is Word Problems on Linear Equations?
A word problem on a pair of linear equations is a problem stated in natural language (words) that describes a situation involving two unknown quantities and two conditions (relationships) linking those quantities. The task is to:
- Identify the two unknown quantities and assign variables (usually x and y) to them.
- Translate the two verbal conditions into two linear equations in x and y.
- Solve the pair of equations using any suitable method (substitution, elimination, or cross-multiplication).
- Interpret the solution in the context of the original problem and verify it makes sense.
The key characteristic of these problems is that the equations are not given directly — students must read the problem carefully, identify what is unknown, and set up the equations themselves. This makes word problems more challenging than pure algebraic problems, but also more rewarding because they demonstrate the power of algebra to solve practical questions.
Common Categories of Word Problems:
- Number Problems: Finding two numbers given conditions about their sum, difference, ratio, or digit properties.
- Age Problems: Finding present ages given conditions about past or future ages.
- Fraction Problems: Finding the numerator and denominator of a fraction given conditions about modifications to it.
- Speed, Distance, and Time: Finding speeds or times for upstream/downstream, with/against wind, or travel scenarios.
- Geometry Problems: Finding dimensions of shapes given perimeter, area, or angle conditions.
- Money and Cost Problems: Finding unit prices or quantities given total cost conditions.
Word Problems on Pair of Linear Equations in Two Variables Formula
General Strategy for Word Problems:
Step 1: Read the problem carefully — at least twice.
Step 2: Identify what is to be found. Assign variables x and y.
Step 3: Translate each condition into a linear equation.
Step 4: Solve the system of equations.
Step 5: Interpret the answer and check it against the original problem.
Key Formulae Used in Various Problem Types:
| Problem Type | Key Relationship |
|---|---|
| Speed-Distance-Time | Distance = Speed x Time, or Time = Distance / Speed |
| Upstream/Downstream | Upstream speed = boat speed - stream speed; Downstream speed = boat speed + stream speed |
| Two-digit number | Number = 10 x (tens digit) + (units digit) |
| Fraction | Fraction = numerator / denominator |
| Work problems | 1/x + 1/y = combined work rate |
| Ages | Future age = present age + years; Past age = present age - years |
Methods
Detailed Approach for Each Problem Type:
Number Problems: Let the numbers be x and y. Translate conditions like 'sum is 50' as x + y = 50, 'one is 4 more than twice the other' as x = 2y + 4. Solve using substitution or elimination.
Age Problems: Let present ages be x and y. Use 'k years ago' as (x - k) and 'k years hence' as (x + k). Form two equations from the given conditions about the relationship between ages at different times.
Fraction Problems: Let the numerator be x and denominator be y. The fraction is x/y. Conditions like 'adding 1 to numerator makes fraction 1/2' become (x + 1)/y = 1/2, which gives 2(x + 1) = y, or 2x - y = -2.
Speed-Distance-Time Problems: Common setup: A person travels a distance d1 at speed s1 and d2 at speed s2, with total time t. This gives d1/s1 + d2/s2 = t. For upstream/downstream: let boat speed = x, stream speed = y, then upstream speed = x - y, downstream speed = x + y.
Geometry Problems: For rectangles, perimeter = 2(length + breadth). For triangles, sum of angles = 180 degrees. Translate geometric constraints into linear equations.
Money Problems: Let the price of item A be x and item B be y. '3 items of A and 5 items of B cost Rs. 200' translates to 3x + 5y = 200.
Tips for Success:
- Define variables clearly at the start. Write 'Let x = ...' and 'Let y = ...'.
- Read each sentence separately and convert it into an equation.
- Check units: ensure both sides of the equation use the same units.
- Verify your answer by substituting back into the word problem, not just the equations.
- Reject solutions that do not make physical sense (e.g., negative ages, negative speeds).
Solved Examples
Example 1: Number Problem — Sum and Difference
Problem: The sum of two numbers is 74 and their difference is 18. Find the numbers.
Solution:
Let the two numbers be x and y (x > y).
Condition 1: x + y = 74 ... (i)
Condition 2: x - y = 18 ... (ii)
Adding (i) and (ii): 2x = 92, so x = 46
Substituting in (i): 46 + y = 74, so y = 28
Verification: 46 + 28 = 74 ✓ and 46 - 28 = 18 ✓
Answer: The two numbers are 46 and 28.
Example 2: Age Problem — Present and Past Ages
Problem: Ravi's father is 3 times as old as Ravi. Four years ago, the father was 4 times as old as Ravi. Find their present ages.
Solution:
Let Ravi's present age = x years and his father's present age = y years.
Condition 1: y = 3x → y - 3x = 0 ... (i)
Condition 2: Four years ago, father was 4 times Ravi's age:
(y - 4) = 4(x - 4)
y - 4 = 4x - 16
y - 4x = -12 ... (ii)
Subtracting (ii) from (i):
(y - 3x) - (y - 4x) = 0 - (-12)
x = 12
From (i): y = 3(12) = 36
Verification: Present: 36 = 3 × 12 ✓. Four years ago: 32 = 4 × 8 ✓
Answer: Ravi is 12 years old and his father is 36 years old.
Example 3: Fraction Problem
Problem: A fraction becomes 4/5 if 1 is added to both the numerator and the denominator. If 5 is subtracted from both, it becomes 1/2. Find the fraction.
Solution:
Let the fraction be x/y.
Condition 1: (x + 1)/(y + 1) = 4/5
5(x + 1) = 4(y + 1)
5x + 5 = 4y + 4
5x - 4y = -1 ... (i)
Condition 2: (x - 5)/(y - 5) = 1/2
2(x - 5) = y - 5
2x - 10 = y - 5
2x - y = 5 ... (ii)
From (ii): y = 2x - 5. Substituting in (i):
5x - 4(2x - 5) = -1
5x - 8x + 20 = -1
-3x = -21
x = 7
y = 2(7) - 5 = 9
Verification: (7+1)/(9+1) = 8/10 = 4/5 ✓. (7-5)/(9-5) = 2/4 = 1/2 ✓
Answer: The fraction is 7/9.
Example 4: Speed-Distance-Time (Upstream and Downstream)
Problem: A boat goes 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream.
Solution:
Let the speed of the boat in still water = x km/h and speed of the stream = y km/h.
Upstream speed = (x - y) km/h, Downstream speed = (x + y) km/h
Using Time = Distance / Speed:
Condition 1: 30/(x - y) + 44/(x + y) = 10 ... (i)
Condition 2: 40/(x - y) + 55/(x + y) = 13 ... (ii)
Let u = 1/(x - y) and v = 1/(x + y):
30u + 44v = 10 ... (iii)
40u + 55v = 13 ... (iv)
Multiply (iii) by 4 and (iv) by 3:
120u + 176v = 40 ... (v)
120u + 165v = 39 ... (vi)
Subtracting (vi) from (v): 11v = 1, so v = 1/11
From (iii): 30u + 44(1/11) = 10 → 30u + 4 = 10 → 30u = 6 → u = 1/5
So x - y = 5 and x + y = 11
Adding: 2x = 16, x = 8. Subtracting: 2y = 6, y = 3.
Verification: 30/5 + 44/11 = 6 + 4 = 10 ✓. 40/5 + 55/11 = 8 + 5 = 13 ✓
Answer: Speed of boat in still water = 8 km/h, speed of stream = 3 km/h.
Example 5: Two-Digit Number Problem
Problem: A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.
Solution:
Let the tens digit = x and units digit = y.
The number = 10x + y. The reversed number = 10y + x.
Condition 1: 10x + y = 6(x + y) + 4
10x + y = 6x + 6y + 4
4x - 5y = 4 ... (i)
Condition 2: (10x + y) - 18 = 10y + x
9x - 9y = 18
x - y = 2 ... (ii)
From (ii): x = y + 2. Substituting in (i):
4(y + 2) - 5y = 4
4y + 8 - 5y = 4
-y = -4
y = 4
x = 4 + 2 = 6
Number = 10(6) + 4 = 64
Verification: 6(6 + 4) + 4 = 60 + 4 = 64 ✓. 64 - 18 = 46 (digits reversed) ✓
Answer: The number is 64.
Example 6: Cost and Quantity Problem
Problem: 5 pencils and 7 pens together cost Rs. 195, while 7 pencils and 5 pens together cost Rs. 153. Find the cost of each pencil and each pen.
Solution:
Let the cost of one pencil = Rs. x and one pen = Rs. y.
Condition 1: 5x + 7y = 195 ... (i)
Condition 2: 7x + 5y = 153 ... (ii)
Adding (i) and (ii): 12x + 12y = 348 → x + y = 29 ... (iii)
Subtracting (i) from (ii): 2x - 2y = -42 → x - y = -21 ... (iv)
From (iii) and (iv):
Adding: 2x = 8, x = 4
Subtracting: 2y = 50, y = 25
Verification: 5(4) + 7(25) = 20 + 175 = 195 ✓. 7(4) + 5(25) = 28 + 125 = 153 ✓
Answer: Cost of each pencil = Rs. 4 and each pen = Rs. 25.
Example 7: Geometry Problem — Rectangle
Problem: The perimeter of a rectangle is 52 cm. If the length is 2 cm more than 3 times the breadth, find the dimensions.
Solution:
Let the length = x cm and breadth = y cm.
Condition 1: 2(x + y) = 52 → x + y = 26 ... (i)
Condition 2: x = 3y + 2 ... (ii)
Substituting (ii) in (i):
(3y + 2) + y = 26
4y = 24
y = 6
x = 3(6) + 2 = 20
Verification: 2(20 + 6) = 2(26) = 52 ✓. 20 = 3(6) + 2 = 20 ✓
Answer: Length = 20 cm and breadth = 6 cm.
Example 8: Age Problem — Future Ages
Problem: The sum of the ages of a mother and daughter is 50 years. Five years hence, the mother's age will be twice the daughter's age. Find their present ages.
Solution:
Let the mother's age = x years and daughter's age = y years.
Condition 1: x + y = 50 ... (i)
Condition 2: (x + 5) = 2(y + 5)
x + 5 = 2y + 10
x - 2y = 5 ... (ii)
Subtracting (ii) from (i):
(x + y) - (x - 2y) = 50 - 5
3y = 45
y = 15
x = 50 - 15 = 35
Verification: 35 + 15 = 50 ✓. After 5 years: 40 = 2 × 20 ✓
Answer: Mother is 35 years old and daughter is 15 years old.
Example 9: Fixed Charge and Rate Problem
Problem: A lending library charges a fixed registration fee plus a charge per day. Saritha paid Rs. 105 for a book kept for 7 days, and Suresh paid Rs. 145 for a book kept for 12 days. Find the fixed charge and the charge per day.
Solution:
Let fixed charge = Rs. x and charge per day = Rs. y.
Condition 1: x + 7y = 105 ... (i)
Condition 2: x + 12y = 145 ... (ii)
Subtracting (i) from (ii): 5y = 40, so y = 8
From (i): x + 56 = 105, so x = 49
Verification: 49 + 7(8) = 49 + 56 = 105 ✓. 49 + 12(8) = 49 + 96 = 145 ✓
Answer: Fixed charge = Rs. 49, charge per day = Rs. 8.
Example 10: Work and Efficiency Problem
Problem: 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone to finish the work.
Solution:
Let 1 woman's one day work = 1/x and 1 man's one day work = 1/y.
Condition 1: 2/x + 5/y = 1/4 ... (i)
Condition 2: 3/x + 6/y = 1/3 ... (ii)
Let u = 1/x and v = 1/y:
2u + 5v = 1/4 ... (iii)
3u + 6v = 1/3 ... (iv)
Multiply (iii) by 3 and (iv) by 2:
6u + 15v = 3/4 ... (v)
6u + 12v = 2/3 ... (vi)
Subtracting (vi) from (v): 3v = 3/4 - 2/3 = 9/12 - 8/12 = 1/12
v = 1/36, so y = 36
From (iii): 2u + 5(1/36) = 1/4
2u = 1/4 - 5/36 = 9/36 - 5/36 = 4/36 = 1/9
u = 1/18, so x = 18
Verification: 2/18 + 5/36 = 1/9 + 5/36 = 4/36 + 5/36 = 9/36 = 1/4 ✓
3/18 + 6/36 = 1/6 + 1/6 = 2/6 = 1/3 ✓
Answer: One woman alone takes 18 days and one man alone takes 36 days.
Real-World Applications
Word problems on pairs of linear equations model countless real-life scenarios:
Business and Commerce: Determining the cost of individual items when only combined costs are known. For instance, finding the unit price of two different products from total bills at a store.
Travel Planning: Computing speeds of vehicles and wind/current when given journey times for different directions. This applies to aviation (headwind/tailwind), boating (upstream/downstream), and road travel (with/against gradient).
Demographics and Census: Determining population groups when given total population and ratios or differences. Schools use this to plan resources based on student distributions.
Financial Planning: Fixed and variable cost analysis in business — determining the fixed overhead and per-unit production cost from total expenditure data at different production levels.
Mixture Problems: Finding the quantities of two solutions to mix to achieve a desired concentration. Pharmacists, chemists, and food scientists routinely solve such systems.
Engineering: Balancing forces, currents, or flows in networks where two unknowns are connected by two constraint equations.
Key Points to Remember
- Read the problem at least twice before attempting to form equations.
- Clearly define what x and y represent, including units.
- Each sentence or condition in the problem usually translates to one equation.
- For two-digit numbers: number = 10 × (tens digit) + (units digit).
- For fractions: fraction = numerator / denominator.
- For upstream/downstream: use (boat speed - stream speed) and (boat speed + stream speed).
- For age problems: 'k years ago' means subtract k, 'k years hence' means add k.
- Always verify the answer against the original word problem, not just the equations.
- Reject physically impossible answers (negative ages, negative prices, etc.).
- The substitution method often works best when one equation is already solved for a variable.
- The elimination method is efficient when coefficients are easily made equal by multiplication.
Practice Problems
- The sum of two numbers is 100. If one number is 20 more than the other, find the numbers.
- A father's age is three times his son's age. Ten years from now, the father will be twice as old as the son. Find their current ages.
- A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
- A train covered 300 km at a certain speed. If the speed had been 10 km/h more, the journey would have taken 1 hour less. Find the speed.
- The length of a rectangle exceeds its breadth by 9 cm. If the length and breadth are each increased by 3 cm, the area increases by 84 sq cm. Find the original dimensions.
- 8 men and 12 women can finish a work in 10 days, while 6 men and 8 women finish it in 14 days. Find the time taken by one man alone and one woman alone.
Frequently Asked Questions
Q1. How do I start a word problem on linear equations?
Start by reading the problem carefully, identify the two unknown quantities, assign variables x and y to them, and then translate each condition in the problem into a linear equation. You should get exactly two equations.
Q2. What if the problem gives more than two conditions?
For a pair of linear equations, you need exactly two independent conditions. If more are given, some may be redundant (derivable from others) or you may need to identify the two most useful ones. Extra conditions can serve as verification.
Q3. Which method should I use — substitution, elimination, or cross-multiplication?
Use whichever is most convenient. Substitution works well when one equation is already solved for a variable. Elimination is efficient when coefficients can be easily matched. Cross-multiplication gives direct results. For exam speed, choose the method that minimizes calculation.
Q4. How do I handle fraction problems?
Let numerator = x and denominator = y. Write the fraction as x/y. Convert each condition into a linear equation by cross-multiplying after forming the fraction equation. Be careful to cross-multiply correctly and handle signs.
Q5. What if the answer comes out negative in an age problem?
A negative age is physically impossible, so check your equation setup for sign errors. In some problems, the equations are set up correctly but the question's conditions cannot be satisfied — in that case, the problem has no valid solution. However, in CBSE exams, problems are designed to have positive solutions.
Q6. How do I handle upstream and downstream problems?
Let the speed of the boat (or swimmer) in still water be x and the speed of the stream be y. Upstream speed = x - y, downstream speed = x + y. Use Time = Distance / Speed to form the two equations from the given journey data.
Q7. Why do some word problems require substitution like u = 1/x?
When the unknowns appear in denominators (like distance/speed problems), the equations are not linear in x and y. By substituting u = 1/x and v = 1/y, we convert them into linear equations in u and v, solve for u and v, then find x and y.
Q8. Are word problems important for CBSE Board exams?
Yes, word problems carry significant marks in CBSE Class 10 board exams. They typically appear as 3-mark or 5-mark questions. Practicing a variety of problem types is essential for scoring well.
Q9. How do I verify my answer in a word problem?
Substitute your values of x and y back into the original word problem (not just the equations). Check that every condition stated in the problem is satisfied. Also check that the answers are reasonable (positive ages, sensible speeds, valid digits, etc.).
Q10. What is a two-digit number word problem?
In these problems, you find the digits of a two-digit number. If the tens digit is x and units digit is y, the number is 10x + y and the reversed number is 10y + x. Conditions about the number and its digits give two equations.
Related Topics
- Pair of Linear Equations in Two Variables
- Substitution Method
- Elimination Method
- Word Problems on Linear Equations
- Graphical Method for Linear Equations
- Cross-Multiplication Method
- Consistency of Linear Equations
- Word Problems Solved Graphically
- Reducing Equations to Linear Form
- Conditions for Solvability
- Linear Equations in Real Life
