Situational Problems (Linear Equations) MCQs Quiz | Class 10

Understanding Situational Problems with Linear Equations

Situational problems, often called word problems, are real-life scenarios translated into mathematical expressions. In Class 10 Mathematics, you primarily encounter situations that can be modeled using a pair of linear equations in two variables. These problems require careful reading, identification of unknown quantities, and formulation of equations based on the given conditions.

Key Steps to Solve Word Problems

1. Read and Understand: Carefully read the problem multiple times to grasp the context and identify what is being asked.
2. Identify Unknowns: Determine the quantities you need to find. Assign variables (usually x and y) to these unknowns.
3. Formulate Equations: Translate the conditions given in the problem into algebraic equations. Each condition usually leads to one equation. For “pair of linear equations,” you will need two distinct conditions to form two equations.
4. Solve the System of Equations: Use any suitable algebraic method to solve the pair of linear equations. Common methods include:
   • Substitution Method: Express one variable in terms of the other from one equation and substitute it into the second equation.
   • Elimination Method: Multiply one or both equations by suitable non-zero constants to make the coefficients of one variable numerically equal, then add or subtract the equations to eliminate that variable.
   • Cross-Multiplication Method: A direct method for solving a system of two linear equations, though less intuitive for many.
5. Check Your Answer: Substitute the values of x and y back into the original word problem or the derived equations to ensure they satisfy all conditions.

Common Types of Situational Problems

Here are some recurring themes for word problems involving linear equations:

• Age Problems: Involve finding the current ages of individuals based on conditions related to their ages in the past, present, or future.
• Number Problems: Deal with finding unknown numbers based on relationships between them (sum, difference, product, ratios).
• Cost/Money Problems: Involve costs of items, total expenses, or distribution of money.
• Fraction Problems: Conditions given about the numerator and denominator of a fraction.
• Fixed Charge Problems: Scenarios like taxi fares, library charges, or hostel charges where there’s a fixed component and a variable component per unit (e.g., per km, per day).
• Two-Digit Number Problems: Involve properties of two-digit numbers and numbers formed by reversing their digits.
• Time, Speed, and Distance Problems: Can sometimes be formulated as linear equations, especially when dealing with boats in streams or trains.

Example Walkthrough: Age Problem

Problem: The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 124. Find their present ages.

Solution:
Let the father’s present age be x years and the son’s present age be y years.

1. First condition: The sum of their ages is 45. x + y = 45 (Equation 1)
2. Second condition: Five years ago: Father’s age = x - 5 Son’s age = y - 5 Product of their ages was 124: (x - 5)(y - 5) = 124 (Equation 2)
3. Solve using Substitution: From Equation 1, y = 45 - x. Substitute this into Equation 2: (x - 5)( (45 - x) - 5 ) = 124 (x - 5)(40 - x) = 124 40x - x^2 - 200 + 5x = 124 -x^2 + 45x - 200 = 124 -x^2 + 45x - 324 = 0 x^2 - 45x + 324 = 0 (This is a quadratic equation, which is not a linear equation problem directly. Let’s provide a simpler example that *only* uses linear equations.)

Revised Example Walkthrough: Linear Equation Problem

Problem: The cost of 2 shirts and 3 trousers is Rs. 3500. The cost of 3 shirts and 2 trousers is Rs. 4000. Find the cost of one shirt and one trouser.

Solution:
Let the cost of one shirt be Rs. x and the cost of one trouser be Rs. y.

1. First condition: The cost of 2 shirts and 3 trousers is Rs. 3500. 2x + 3y = 3500 (Equation 1)
2. Second condition: The cost of 3 shirts and 2 trousers is Rs. 4000. 3x + 2y = 4000 (Equation 2)
3. Solve using Elimination Method: Multiply Equation 1 by 2: 4x + 6y = 7000 Multiply Equation 2 by 3: 9x + 6y = 12000 Subtract the modified Equation 1 from the modified Equation 2: (9x + 6y) - (4x + 6y) = 12000 - 7000 5x = 5000 x = 1000 Substitute x = 1000 into Equation 1: 2(1000) + 3y = 3500 2000 + 3y = 3500 3y = 1500 y = 500
4. Check: For Equation 1: 2(1000) + 3(500) = 2000 + 1500 = 3500 (Correct) For Equation 2: 3(1000) + 2(500) = 3000 + 1000 = 4000 (Correct)

So, the cost of one shirt is Rs. 1000 and the cost of one trouser is Rs. 500.

Quick Revision Tips

• Always define your variables clearly.
• Pay attention to keywords like “sum,” “difference,” “product,” “ratio,” “less than,” “more than.”
• If you’re stuck, try drawing a diagram or making a small table to organize the information.
• Practice is key! The more problems you solve, the better you’ll become at recognizing patterns and formulating equations.

Practice Questions

Try solving these additional problems to solidify your understanding (without options):

1. The sum of two numbers is 80 and their difference is 20. Find the numbers.
2. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay Rs. 1000 as hostel charges, whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
3. The ratio of incomes of two persons is 9:7 and the ratio of their expenditures is 4:3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.
4. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were 10 km/h slower, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
5. A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.