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Pair of Linear Equations in Two Variables | Class 10 Board Questions

Learn how to solve Class 10 Mathematics word problems using the Substitution Method in Pair of Linear Equations in Two Variables. This video covers ag

Linear Equations in Two Variables – Word Problems with Solutions | Class 10 Maths

Word problems based on Linear Equations in Two Variables are among the most important topics in Class 10 Mathematics. These questions help students understand how algebra is used in everyday life situations such as taxi fares, age problems, fractions, shopping, and geometry.

In this article, we solve some of the most frequently asked Class 10 word problems using the Substitution Method. Every solution is explained step by step so that students can easily understand the complete process.

Pair of Linear Equations in Two Variables | Class 10 Board Questions


Question 1: Taxi Fare Problem

Question

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered.

  • For a journey of 10 km, the fare is ₹105.
  • For a journey of 15 km, the fare is ₹155.

Find:

  • Fixed charge
  • Charge per kilometre
  • Fare for a journey of 25 km

Solution

Let,

  • x = Fixed charge (₹)
  • y = Charge per kilometre (₹)

According to the question,

x + 10y = 105

x + 15y = 155

Step 1: Express x

x = 105 − 10y

Step 2: Substitute into the second equation

105 − 10y + 15y = 155

5y = 50

y = 10

Step 3: Find x

x = 105 − 10(10)

x = 5

Step 4: Find the fare for 25 km

Fare = x + 25y

= 5 + 25 × 10

= ₹255


Answer

  • Fixed Charge = ₹5
  • Charge per kilometre = ₹10
  • Fare for 25 km = ₹255

Question 2: Fraction Problem

Question

A fraction becomes 9/11 if 2 is added to both its numerator and denominator. If 3 is added to both the numerator and denominator, it becomes 5/6. Find the original fraction.


Solution

Let the fraction be

x/y

Then,

(x + 2)/(y + 2) = 9/11

11x − 9y = −4

x = (9y − 4)/11

Also,

(x + 3)/(y + 3) = 5/6

6x − 5y = −3

Substitute the value of x

6(9y − 4)/11 − 5y = −3

54y − 24 − 55y = −33

−y = −9

y = 9

Now,

x = (9 × 9 − 4)/11

= 77/11

= 7


Answer

The required fraction is 7/9.


Question 3: Difference Between Two Numbers

Question

The difference between two numbers is 26 and one number is three times the other. Find the two numbers.


Solution

Let the numbers be x and y.

x − y = 26

x = 3y

Substitute x = 3y into the first equation.

3y − y = 26

2y = 26

y = 13

Therefore,

x = 3 × 13 = 39


Answer

The required numbers are 39 and 13.

Pair of Linear Equations in Two Variables | Class 10 Board Questions


Question 4: Supplementary Angles

Question

The larger of two supplementary angles exceeds the smaller by 18°. Find both angles.


Solution

Let the angles be x and y.

Since they are supplementary,

x + y = 180°

Also,

x = y + 18°

Substitute into the first equation.

y + 18 + y = 180

2y = 162

y = 81°

Hence,

x = 81 + 18 = 99°


Answer

  • Smaller Angle = 81°
  • Larger Angle = 99°

Question 5: Cost of Bats and Balls

Question

A coach buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of one bat and one ball.


Solution

Let,

  • x = Cost of one bat
  • y = Cost of one ball

Then,

7x + 6y = 3800

3x + 5y = 1750

From the second equation,

x = (1750 − 5y)/3

Substitute into the first equation.

7(1750 − 5y)/3 + 6y = 3800

After simplification,

y = 50

Substitute into the second equation.

3x + 250 = 1750

3x = 1500

x = 500


Answer

  • Cost of one Bat = ₹500
  • Cost of one Ball = ₹50
Pair of Linear Equations in Two Variables | Class 10 Board Questions

Question 6: Age Problem

Question

Five years hence, Jacob's age will be three times the age of his son. Five years ago, Jacob's age was seven times the age of his son. Find their present ages.


Solution

Let,

  • x = Present age of Jacob
  • y = Present age of his son

Five years hence,

x + 5 = 3(y + 5)

x − 3y = 10

x = 10 + 3y

Five years ago,

x − 5 = 7(y − 5)

x − 7y = −30

Substitute x = 10 + 3y.

10 + 3y − 7y = −30

−4y = −40

y = 10

Therefore,

x = 10 + 3(10)

x = 40


Answer

  • Jacob's Present Age = 40 years
  • Son's Present Age = 10 years
Pair of Linear Equations in Two Variables | Class 10 Board Questions

Important Exam Tips

  • Always define the variables before forming equations.
  • Translate the word problem carefully into mathematical equations.
  • Use the substitution method systematically.
  • Substitute the obtained values back to verify your answer.
  • Write the final answer with correct units such as ₹, years, km, or degrees.

Key Takeaways

  • Linear equations in two variables model many real-life situations.
  • The substitution method is one of the easiest techniques to solve such equations.
  • Accuracy in forming equations is more important than lengthy calculations.
  • These questions are frequently asked in CBSE Class 10 board examinations.

Conclusion

These solved examples demonstrate how Linear Equations in Two Variables can be applied to everyday situations like taxi fares, fractions, shopping, age calculations, and geometry. By carefully defining variables, forming equations from the given information, and solving them using the substitution method, students can confidently tackle similar questions in board examinations.

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