Welcome to AssamSchool. This article contains 30 Extra Questions from Pair of Linear Equations in Two Variables from Class 10 Mathematics Chapter 3. These Pair of Linear Equations in Two Variables Class 10 Extra Questions, are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Pair of Linear Equations in Two Variables Class 10 Extra Questions
1. Determine if the pair of equations \(2x + 3y = 5\) and \(4x + 6y = 10\) are consistent, inconsistent, or dependent by comparing ratios.
2. Solve \(3x + 2y = 11\) and \(2x – y = 5\) using substitution.
3. Solve \(5x – 3y = 8\) and \(3x + 4y = 1\) using elimination.
4. Five years ago, A’s age was three times B’s age. Ten years later, A’s age will be twice B’s age. Find their current ages.
5. For what value of \(k\) will the system \(kx + 3y = 7\) and \(2x – 6y = 14\) be inconsistent?
6. 3 pens and 5 pencils cost ₹50, while 5 pens and 3 pencils cost ₹54. Find the cost of each pen and pencil.
7. If the graphs of two equations are parallel lines, what does this imply about their solutions?
8. Show that the equations \(x + 2y = 4\) and \(2x + 4y = 12\) have no solution.
9. The sum of a two-digit number and its reverse is 99. If the digits differ by 3, find the number.
10. Solve \( \frac{x}{3} + \frac{y}{4} = 5 \) and \( \frac{x}{2} – \frac{y}{3} = 1 \)
11. Check if \( \frac{2}{3}x + \frac{5}{4}y = 7 \) and \( 8x + 15y = 28 \) are consistent.
12. A man has ₹500 in ₹10 and ₹20 notes. If the total notes are 30, find how many of each he has.
13. Solve \(0.5x + 0.7y = 3.5\) and \(0.3x – 0.2y = 0.9\)
14. The sum of the ages of a father and son is 50. Five years ago, the father was seven times as old as his son. Find their current ages.
15. The ratio of incomes of two friends is 5:4 and their expenditures are 3:2. If each saves ₹2000 monthly, find their incomes.
16. Determine if \(x – 2y = 3\) and \(3x – 6y = 9\) are dependent.
17. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it goes 40 km upstream and 55 km downstream. Find the speed of the boat and the stream.
18. Show that \(4x + 6y = 8\) and \(2x + 3y = 4\) have infinitely many solutions.
19. A fraction becomes \(1/3\) when 1 is subtracted from the numerator and 2 is added to the denominator. It becomes 1/2 when 1 is added to the numerator and 1 subtracted from the denominator. Find the fraction.
20. Solve \( \sqrt{3}x + \sqrt{2}y = 0 \) and \( \sqrt{2}x + \sqrt{3}y = 0\)
21. The perimeter of a rectangle is 36 cm. If its length is doubled and breadth tripled, the perimeter becomes 92 cm. Find original dimensions.
22. If two lines are coincident, how many solutions do they have?
23. A chemist has two acid solutions: 20% and 40%. How many liters of each should be mixed to get 10 liters of 30% acid?
24. Solve \(2x + 0y = 8\) and \(3x – y = 5\)
25. A train covers a distance at uniform speed. If speed were 10 km/h more, it would take 2 hours less. If speed were 10 km/h less, it would take 3 hours more. Find the distance.
26. After solving equations, how do you verify the solution?
27. Two pipes fill a tank in 6 hours. If one pipe alone takes 5 hours more than the other, find the time each pipe takes separately.
28. Find \(k\) if \(kx + 2y = 5\) and \(3x + y = 1\) have a unique solution.
29. ₹6000 is lent partly at 5% and partly at 8%. Total interest after a year is ₹420. Find the amounts lent at each rate.
30. Verify that \((x = 2)\), \((y = 3\)) is a solution of \(4x – 3y = -1\) and \(x + y = 5\)
Answers
- Solution:
Ratios: \( \frac{2}{4} = \frac{1}{2}, \frac{3}{6} = \frac{1}{2}, \frac{5}{10} = \frac{1}{2} \)
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Dependent (infinitely many solutions). - Solution:
From \(2x – y = 5\), \(y = 2x – 5\)
Substitute into \(3x + 2(2x – 5) = 11\)
⇒ \(7x – 10 = 11\)
⇒ \((x = 3)\), \((y = 1)\)
Solution: \(x = 3, \, y = 1\) - Solution:
Multiply first equation by 4: \(20x – 12y = 32\)
Multiply second by 3: \(9x + 12y = 3\)
Add: \(29x = 35\)
⇒ \(x = \frac{35}{29}\)
Substitute back to find \(y = -\frac{7}{29}\) - Solution:
Let A’s age = \(a\), B’s age = \(b\)
\(a – 5 = 3(b – 5)\) and \(a + 10 = 2(b + 10)\)
Solve: \(a = 50\), \(b = 20\) - Solution:
For inconsistency: \( \frac{k}{2} = \frac{3}{-6} \neq \frac{7}{14} \)
⇒ \(k = -1\) - Solution:
Let pen = ₹\(x\), pencil = ₹\(y\)
\(3x + 5y = 50\), \(5x + 3y = 54\)
Solve: \(x = 8\), \(y = 6\) - Solution:
No solution (parallel lines). - Solution:
Multiply first equation by 2: \(2x + 4y = 8\)
Compare with \(2x + 4y = 12\). Inconsistent. - Solution:
Let number = \(10a + b\).
\(10a + b + 10b + a = 99\)
⇒ \(a + b = 9\)
\(|a – b| = 3\)
Solutions: \(a=6, \, b=3\) or \(a=3, \, b=6\)
Numbers: 63, 36. - Solution:
Multiply first by 12: \(4x + 3y = 60\)
Multiply second by 6: \(3x – 2y = 6\)
Solve: \(x = 6\), \(y = 12\) - Solution:
Multiply first by 12: \(8x + 15y = 84\)
Compare with \(8x + 15y = 28\). Inconsistent. - Solution:
Let ₹10 notes = \(x\), ₹20 notes = \(y\)
\(10x + 20y = 500\), \(x + y = 30\)
Solve: (x = 10), (y = 20) - Solution:
Multiply first by 10: \(5x + 7y = 35\)
Multiply second by 10: \(3x – 2y = 9\)
Solve: \(x = 5\), \(y = 2\) - Solution:
Let father = \(f\), son = \(s\)
\(f + s = 50\), \(f – 5 = 7(s – 5)\)
Solve: \(f = 40\), \(s = 10\) - Solution:
Let incomes be \(5x, \, 4x\)
Expenditures \(3y, \, 2y\)
\(5x – 3y = 2000\), \(4x – 2y = 2000\)
Solve: \(x = 2000\), \(y = 2000\)
Incomes: ₹10,000 and ₹8,000 - Solution:
\( \frac{1}{3} = \frac{-2}{-6} = \frac{3}{9} \)
Dependent (infinitely many solutions) - Solution:
Let boat speed = \(b\), stream speed = \(s\)
\( \frac{30}{b – s} + \frac{44}{b + s} = 10 \),
\( \frac{40}{b – s} + \frac{55}{b + s} = 13 \)
Solve: \(b = 12\) km/h, \(s = 2\) km/h. - Solution:
Second equation is half of the first. Infinitely many solutions. - Solution:
Let fraction = \( \frac{n}{d} \)
\( \frac{n – 1}{d + 2} = \frac{1}{3} \)
\( \frac{n + 1}{d – 1} = \frac{1}{2} \)
Solve: \(n = 5\), \(d = 8\)
Fraction: \( \frac{5}{8} \) - Solution:
Multiply first by \( \sqrt{2} \)
second by \( \sqrt{3} \)
Subtract: \(y = 0\), \(x = 0\). Trivial solution. - Solution:
Let length = \(l\), breadth = \(b\)
\(2(l + b) = 36\), \(2(2l + 3b) = 92\)
Solve: \(l = 10\) cm, \(b = 8\) cm - Solution:
Infinitely many solutions. - Solution:
Let 20% = \(x\) L, 40% = \(10 – x\) L
\(0.2x + 0.4(10 – x) = 3\)
Solve: \(x = 5\) L - Solution:
From \(2x = 8\), \(x = 4\)
Substitute into \(3(4) – y = 5) ⇒ (y = 7\) - Solution:
Let speed = (\s\), distance = \(d\)
\( \frac{d}{s} – \frac{d}{s + 10} = 2 \)
\( \frac{d}{s – 10} – \frac{d}{s} = 3\)
Solve: \(d = 300\) km, \(s = 50\) km/h - Solution:
Substitute solutions into original equations to check validity. - Solution:
Let slower pipe = \(x\) hours, faster = \(x – 5\)
\( \frac{1}{x} + \frac{1}{x – 5} = \frac{1}{6} \)
Solve: \(x = 15\) hours, other = 10 hours. - Solution:
For unique solution: \( \frac{k}{3} \neq \frac{2}{1} \) ⇒ \(k \neq 6\) - Solution:
Let amounts be \(x\) and \(6000 – x\)
\(0.05x + 0.08(6000 – x) = 420\)
Solve: \(x = ₹2000\), other = ₹4000. - Solution:
Substitute \(x = 2, y = 3\)
\(4(2) – 3(3) = -1\), \(2 + 3 = 5\). Valid solution.
The Bottom Line
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You can also check the related MCQs for this chapter here: MCQ on Pair of Linear Equations in Two Variables
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Abdur Rohman is an Electrical Engineer from Charaideo, Assam, who wears multiple hats as a part-time teacher, blogger, entrepreneur, and digital marketer. Passionate about education, he founded The Assam School blog to provide free, comprehensive textbook solutions, MCQs (Multiple Choice Questions), and other academic content for students from Class V to XII.