Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.7 NCERT Solutions

Introduction

Exercise 3.7 focuses on word problems based on pairs of linear equations in two variables. Students learn how to translate real‑life situations into mathematical equations, solve them using substitution, elimination, or cross‑multiplication methods, and interpret the results. This exercise is crucial for developing problem‑solving skills and applying algebra to practical contexts.

Key Concepts

1. Formulating Equations: Convert word problems into two linear equations in variables $x$ and $y$.

2. Methods of Solution:

   • Substitution method

   • Elimination method

   • Cross‑multiplication method

3. Interpretation: The solution represents values of unknowns in the given context.

Common Mistakes

• Misinterpreting the problem statement.

• Assigning wrong variables.

• Forgetting to check solution in original equations.

• Arithmetic errors in elimination or substitution.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. The sum of two numbers is 27 and their difference is 3. Find the numbers. Let $x,y$ be the numbers.

$$x+y=27,x-y=3$$

Adding: $2x=30\Rightarrow x=15$. Then $y=12$.

Q2. A boat goes 16 km downstream in 2 hours and 10 km upstream in 2 hours. Find speed of boat in still water and speed of stream. Let boat speed = $x$, stream speed = $y$.

$$x+y=\frac{16}{2}=8,x-y=\frac{10}{2}=5$$

Solve: $x=6.5,y=1.5$.

Q3. A fraction becomes $\frac{1}{2}$ when 1 is subtracted from numerator and $\frac{1}{3}$ when 1 is added to denominator. Find fraction. Let fraction = $\frac{x}{y}$.

$$\frac{x-1}{y}=\frac{1}{2},\frac{x}{y+1}=\frac{1}{3}$$

Solve: $x=3,y=6$. Fraction = $\frac{1}{2}$.

Q4. Two numbers differ by 5 and their sum is 55. Find numbers.

$$x-y=5,x+y=55$$

Adding: $2x=60\Rightarrow x=30,y=25$.

Q5. A train covers 60 km at uniform speed. If speed had been 5 km/h more, it would have taken 48 minutes less. Find speed. Let speed = $x$. Time = $\frac{60}{x}$. New time = $\frac{60}{x+5}$.

$$\frac{60}{x}-\frac{60}{x+5}=\frac{48}{60}=\frac{4}{5}$$

Solve: $x=20$.

Q6. A man travels 300 km partly by train and partly by car. He takes 4 hours if he travels 60 km by train and rest by car. He takes 5 hours if he travels 100 km by train and rest by car. Find speed of train and car. Let train speed = $x$, car speed = $y$.

$$\frac{60}{x}+\frac{240}{y}=4,\frac{100}{x}+\frac{200}{y}=5$$

Solve: $x=60,y=80$.

Q7. A shopkeeper buys pencils at 3 for ₹5 and sells them at 2 for ₹3. Find gain percent. Cost of 6 pencils = ₹10. Selling price of 6 pencils = ₹9. Loss = ₹1. Loss% = $\frac{1}{10}\times 100=10\mathrm{\%}$.

Q8. A motorboat, whose speed is 18 km/h in still water, goes 20 km downstream and returns in 2 hours. Find speed of stream. Let stream speed = $y$.

$$\frac{20}{18+y}+\frac{20}{18-y}=2$$

Solve: $y=2$.

Q9. A fraction becomes $\frac{3}{4}$ when 1 is added to numerator and denominator. It becomes $\frac{5}{6}$ when 1 is subtracted from numerator and denominator. Find fraction. Let fraction = $\frac{x}{y}$.

$$\frac{x+1}{y+1}=\frac{3}{4},\frac{x-1}{y-1}=\frac{5}{6}$$

Solve: $x=5,y=7$. Fraction = $\frac{5}{7}$.

Q10. A man spends ₹2400 on buying 60 books. He buys some at ₹40 each and rest at ₹50 each. Find number of books bought at each price. Let $x$ = books at ₹40, $y$ = books at ₹50.

$$x+y=60,40x+50y=2400$$

Solve: $x=30,y=30$.

FAQs (10)

FAQ1. What is pair of linear equations? Two equations in two variables.

FAQ2. What are methods of solving? Substitution, elimination, cross‑multiplication.

FAQ3. What is graphical solution? Intersection point of two lines.

FAQ4. What is consistent system? System with at least one solution.

FAQ5. What is inconsistent system? System with no solution.

FAQ6. What is dependent system? System with infinitely many solutions.

FAQ7. Why check solution in original equations? To verify correctness.

FAQ8. What is practical use of linear equations? Used in business, travel, and resource allocation.

FAQ9. What is common mistake in word problems? Wrong assignment of variables.

FAQ10. Why is Exercise 3.7 important? It applies algebra to real‑life problems.

Conclusion

Exercise 3.7 covers word problems on pairs of linear equations in two variables with solved examples and FAQs. Mastering these problems helps students apply algebra to practical situations.

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