NCERT Solutions for Class 10 Maths Ex 3.7: Linear Equations in Two Variables

Strategy for Tackling Optional Exercise 3.7

The questions in Exercise 3.7 can seem intimidating, but they all follow a logical pattern. The main skill is not solving the equations, but forming them correctly from the given text. Here’s a solid strategy to follow:

1. Read and Re-read: Read the problem at least twice. The first time, get a general idea of the story. The second time, identify the key pieces of information and what you need to find.
2. Define Your Variables: Clearly identify the two unknown quantities in the problem and assign them variables, usually x and y. For example, if the problem is about ages, you might write, "Let the present age of the father be x years and the present age of the son be y years."
3. Translate Words to Equations: This is the most critical step. A word problem will always contain two distinct conditions that you can translate into two linear equations. Look for keywords:

• 'Sum', 'total', 'altogether' means addition (+).
• 'Difference', 'more than', 'less than' means subtraction (-).
• 'Times', 'product', 'of' often means multiplication (×).
• 'Is', 'are', 'was', 'will be' translates to equals (=).

1. Solve and Verify: Once you have your two equations, use the most suitable algebraic method (elimination is often best for complex problems) to solve for x and y. Finally, and most importantly, substitute your answers back into the original word problem to ensure they make logical sense.

Step-by-Step Guide: From Word Problem to Solution

1. Step 1: Analyze the Problem and Define Variables — Let's take a sample problem: "The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number." Here, the unknowns are the digits of the number. Let the digit in the tens place be x. Let the digit in the units place be y. Therefore, the original number can be written as 10x + y. The number obtained by reversing the digits is 10y + x.
2. Step 2: Formulate the Linear Equations — Now, translate the two conditions from the problem into equations. Condition 1: "The sum of the digits... is 9." Equation 1: x + y = 9 Condition 2: "Nine times this number is twice the number obtained by reversing the order of the digits." Equation 2: 9(10x + y) = 2(10y + x) Simplifying Equation 2: 90x + 9y = 20y + 2x 90x - 2x = 20y - 9y 88x = 11y 8x = y or 8x - y = 0
3. Step 3: Solve the System of Equations — We have two simple equations: 1) x + y = 9 2) 8x - y = 0 Using the elimination method is easy here. Add Equation 1 and Equation 2: (x + y) + (8x - y) = 9 + 0 9x = 9 x = 1 Now, substitute x = 1 into Equation 1: 1 + y = 9 y = 8
4. Step 4: State and Verify the Final Answer — The digits are x = 1 and y = 8. The original number is 10x + y = 10(1) + 8 = 18. Let's verify: - Sum of digits: 1 + 8 = 9. (Correct) - Nine times the number: 9 18 = 162. - Number reversed is 81. Twice the reversed number: 2 81 = 162. (Correct) Thus, the solution is verified. The number is 18.

Common Mistakes to Avoid in Ex 3.7

These problems require precision. A small error at the beginning can lead to a completely wrong answer. Here are the most common pitfalls to watch out for:

• Incorrectly Representing a Two-Digit Number: A common mistake is writing a number with digits x and y as xy. Remember, the correct algebraic representation is 10x + y, where x is the tens digit.
• Mixing Up Upstream and Downstream Speeds: In boat problems, always remember:
• Downstream speed = Speed of boat in still water + Speed of stream (x + y)
• Upstream speed = Speed of boat in still water - Speed of stream (x - y)

Getting this reversed is a very frequent error.

• Errors in Framing 'Age' Problems: When dealing with ages 'n' years ago or 'n' years later, make sure to apply the change to both people. If their current ages are x and y, then 5 years ago, their ages were x-5 and y-5, not just one of them.
• Forgetting to Check for Practicality: Your final answer should make sense. Age cannot be negative, the number of objects cannot be a fraction, and speed must be positive. If you get an illogical answer, it's a sign that you need to recheck your equations or calculations.

Practice Questions with Solutions

• Q: A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction. A: Step 1: Define variables. Let the numerator be x and the denominator be y. The fraction is x/y. Step 2: Formulate equations from the given conditions. Condition 1: (x-1)/y = 1/3 => 3(x-1) = y => 3x - 3 = y => 3x - y = 3 (Equation 1) Condition 2: x/(y+8) = 1/4 => 4x = y+8 => 4x - y = 8 (Equation 2) Step 3: Solve the system of equations. Subtract Equation 1 from Equation 2: (4x - y) - (3x - y) = 8 - 3 x = 5 Substitute x=5 into Equation 1: 3(5) - y = 3 => 15 - y = 3 => y = 12. Final answer: The fraction is 5/12.
• Q: Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test? A: Step 1: Define variables. Let the number of right answers be x and the number of wrong answers be y. Step 2: Formulate equations. Condition 1: 3x - y = 40 (Equation 1) Condition 2: 4x - 2y = 50. We can simplify this by dividing by 2: 2x - y = 25 (Equation 2) Step 3: Solve the equations. Subtract Equation 2 from Equation 1: (3x - y) - (2x - y) = 40 - 25 x = 15 Substitute x=15 into Equation 1: 3(15) - y = 40 => 45 - y = 40 => y = 5. Step 4: Find the total number of questions. Total questions = Right answers + Wrong answers = x + y. 15 + 5 = 20. Final answer: There were 20 questions in the test.
• Q: The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle. A: Step 1: Define variables. Let the length be l and the breadth be b. The original area is A = lb. Step 2: Formulate equations. Condition 1: (l-5)(b+3) = lb - 9 => lb + 3l - 5b - 15 = lb - 9 => 3l - 5b = 6 (Equation 1) Condition 2: (l+3)(b+2) = lb + 67 => lb + 2l + 3b + 6 = lb + 67 => 2l + 3b = 61 (Equation 2) Step 3: Solve the system. Multiply Eq 1 by 2 and Eq 2 by 3 to eliminate l. 6l - 10b = 12 6l + 9b = 183 Subtracting the first from the second: 19b = 171 => b = 9. Substitute b=9 into Eq 1: 3l - 5(9) = 6 => 3l - 45 = 6 => 3l = 51 => l = 17. Final answer: The length is 17 units and the breadth is 9 units.
• Q: Points A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? A: Step 1: Define variables. Let the speed of the car from A be x km/hr and the speed of the car from B be y km/hr. Assume x > y. Step 2: Formulate equations using Distance = Speed × Time. Case 1 (Same direction): Car A travels 5x km and car B travels 5y km. The difference in distance is 100 km. 5x - 5y = 100 => x - y = 20 (Equation 1) Case 2 (Opposite direction): Car A travels 1x km and car B travels 1y km. The sum of their distances is 100 km. x + y = 100 (Equation 2) Step 3: Solve the equations. Add Equation 1 and Equation 2: (x - y) + (x + y) = 20 + 100 2x = 120 => x = 60. Substitute x=60 into Equation 2: 60 + y = 100 => y = 40. Final answer: The speed of the car from A is 60 km/hr and the speed of the car from B is 40 km/hr.

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