CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 31

Previous Year Board Questions

Question 1. Solve 37x + 43y = 123 and 43x + 37y = 117. OR Solve x + 6/y = 6 and 3x − 8/y = 5. [2 Marks, CBSE 2013, 2015]
Answer:
(i) Adding: \( 80(x + y) = 240 \), so x + y = 3. Subtracting: \( -6x + 6y = 6 \), so y − x = 1. Hence x = 1, y = 2.
(ii) Put \( \tfrac{1}{y} = v \): \( x + 6v = 6 \) and \( 3x - 8v = 5 \). From the first x = 6 − 6v; then \( 18 - 18v - 8v = 5 \), \( v = \tfrac{1}{2} \), so y = 2 and x = 3. x = 3, y = 2.
In simple words: Add and subtract when the coefficients are swapped.

Teacher's Note:
1. For (ii) substitute 1/y = v.
2. Check in both equations.

Question 2. A and B are friends and their ages differ by a year. A's father D is twice as old as A, and B is twice as old as his sister C. The ages of D and C differ by 40 years. Find the ages of A and B. OR Solve 5/(x − 1) + 1/(y − 2) = 2 and 6/(x − 1) + 3/(y − 2) = 1. [3 Marks, CBSE 2013]
Answer:
(i) Let A's age be x and B's age y. D = 2x and C = y/2. Since D and C differ by 40: \( 2x - \tfrac{y}{2} = 40 \), that is, \( 4x - y = 80 \). For the ages to be whole numbers, B is older by one year: \( y - x = 1 \). Then \( 4x - (x + 1) = 80 \), \( 3x = 81 \), x = 27 and y = 28. A is 27 years and B is 28 years old (D = 54, C = 14, and 54 − 14 = 40).
(ii) Put \( u = \tfrac{1}{x - 1} \), \( v = \tfrac{1}{y - 2} \): \( 5u + v = 2 \) and \( 6u + 3v = 1 \). Then \( v = 2 - 5u \), so \( 6u + 6 - 15u = 1 \), \( u = \tfrac{5}{9} \), \( v = -\tfrac{7}{9} \). So \( x - 1 = \tfrac{9}{5} \) and \( y - 2 = -\tfrac{9}{7} \): x = 14/5, y = 5/7.
In simple words: Express each unknown age in terms of A's and B's ages.

Teacher's Note:
1. If A were older, the age of A would not be a whole number.
2. For (ii) the substitution reduces it to a linear pair.

Teacher's Note:
1. The intersection point is the solution.
2. Check: 3 + 7 = 10 and 7 − 3 = 4.

Question 4. If the pair of linear equations 2x + 3y = 7 and 2αx + (α + β)y = 28 has infinitely many solutions, then the values of α and β are: [1 Mark, CBSE 2014]
(A) 3 and 5   (B) 4 and 8   (C) 4 and 7   (D) 4 and 5
Answer:
Ans. (B) 4 and 8
\( \tfrac{2\alpha}{2} = \tfrac{\alpha + \beta}{3} = \tfrac{28}{7} = 4 \). So α = 4 and α + β = 12, giving β = 8.
In simple words: 28/7 = 4 gives α directly. | Then solve for β.

Teacher's Note:
1. Equate the three ratios.

Question 5. The lines representing the linear equations 2x − y = 3 and 4x − y = 5: [1 Mark, CBSE 2014]
(A) Intersect at a point   (B) Are parallel   (C) Are coincident   (D) Intersect at exactly two points
Answer:
Ans. (A) Intersect at a point
\( \tfrac{2}{4} \ne \tfrac{-1}{-1} \), so the lines intersect at exactly one point, (1, −1).
In simple words: Two distinct lines meet in at most one point. | Option D is never possible.

Teacher's Note:
1. Compare a1/a2 with b1/b2.

Question 6. If the pair of linear equations 10x + 5y − (k − 5) = 0 and 20x + 10y − k = 0 has infinitely many solutions, then the value of k is: [1 Mark, CBSE 2014]
(A) 2   (B) 11   (C) 10   (D) 8
Answer:
Ans. (C) 10
\( \tfrac{10}{20} = \tfrac{5}{10} = \tfrac{k - 5}{k} \), so \( \tfrac{1}{2} = \tfrac{k - 5}{k} \), \( 2k - 10 = k \) and k = 10.
In simple words: (k − 5)/k = 1/2. | k = 10.

Teacher's Note:
1. All three ratios equal.

Question 7. 2 tables and 3 chairs together cost Rs. 3500, whereas 3 tables and 2 chairs together cost Rs. 4000. Find the cost of a table and a chair. [2 Marks, CBSE 2014]
Answer:
Let a table cost Rs. x and a chair Rs. y. \( 2x + 3y = 3500 \) and \( 3x + 2y = 4000 \).
Adding: 5(x + y) = 7500, so x + y = 1500. Subtracting: x − y = 500. So x = 1000 and y = 500.
A table costs Rs. 1000 and a chair Rs. 500.
In simple words: Add and subtract the equations.

Teacher's Note:
1. Check: 2000 + 1500 = 3500.
2. And 3000 + 1000 = 4000.

Question 8. Solve for x and y: 4x + y/3 = 8/3 and x/2 + 3y/4 = −5/2. OR The sum of the numerator and the denominator of a fraction is 8. If 3 is added to both the numerator and the denominator, the fraction becomes 3/4. Find the fraction. [3 Marks, CBSE 2014]
Answer:
(i) Multiply the first by 3: \( 12x + y = 8 \). Multiply the second by 4: \( 2x + 3y = -10 \). From the first y = 8 − 12x; then \( 2x + 24 - 36x = -10 \), x = 1 and y = −4. x = 1, y = −4.
(ii) Let the fraction be \( \tfrac{n}{d} \). \( n + d = 8 \) and \( \tfrac{n + 3}{d + 3} = \tfrac{3}{4} \Rightarrow 4n - 3d = -3 \). With d = 8 − n: \( 4n - 24 + 3n = -3 \), n = 3 and d = 5. The fraction is 3/5.
In simple words: Clear the fractions first.

Teacher's Note:
1. Check (i): 12(1) + (−4) = 8.
2. Check (ii): 6/8 = 3/4.

Teacher's Note:
1. The lines meet at (2, 4).
2. The triangle's base is 6 units.

Question 10. If x = a, y = b is the solution of the equations x + y = 50 and 4x + 5y = 225, then the values of a and b are respectively: [1 Mark, CBSE 2015]
(A) 10 and 40   (B) 25 and 25   (C) 23 and 27   (D) 20 and 30
Answer:
Ans. (B) 25 and 25
From x + y = 50, \( x = 50 - y \). Then \( 4(50 - y) + 5y = 225 \), so \( 200 + y = 225 \), y = 25 and x = 25.
In simple words: Check: 100 + 125 = 225. | a = b = 25.

Teacher's Note:
1. Substitute x = 50 − y.

Question 11. For what value of k will the following pair of linear equations have infinitely many solutions? kx + 3y = k − 3 and 12x + ky = k. [2 Marks, CBSE 2015]
Answer:
\( \tfrac{k}{12} = \tfrac{3}{k} = \tfrac{k - 3}{k} \). From \( \tfrac{3}{k} = \tfrac{k-3}{k} \), \( k - 3 = 3 \), so k = 6. Check \( \tfrac{k}{12} = \tfrac{1}{2} = \tfrac{3}{6} \). k = 6.
In simple words: Use all three ratios.

Teacher's Note:
1. k = −6 satisfies k² = 36 but fails the c-ratio.
2. So only k = 6.

Question 12. Solve 5/(x + y) − 2/(x − y) = −1 and 15/(x + y) + 7/(x − y) = 10. OR Ram travels 760 km to his home, partly by train and partly by car. He takes 8 hours if he travels 160 km by train and the rest by car. He takes 12 minutes more if he travels 240 km by train and the rest by car. Find the speed of the train and the car separately. [3 Marks, CBSE 2015]
Answer:
(i) Put \( u = \tfrac{1}{x + y} \), \( v = \tfrac{1}{x - y} \): \( 5u - 2v = -1 \) and \( 15u + 7v = 10 \). Multiply the first by 3: \( 15u - 6v = -3 \). Subtract: \( 13v = 13 \), v = 1 and u = 1/5. So x + y = 5 and x − y = 1: x = 3, y = 2.
(ii) Let the speeds be x and y: \( \tfrac{160}{x} + \tfrac{600}{y} = 8 \) and \( \tfrac{240}{x} + \tfrac{520}{y} = \tfrac{41}{5} \). Solving, train 80 km/h, car 100 km/h.
In simple words: Reduce to linear equations in the reciprocals.

Teacher's Note:
1. 12 minutes = 1/5 hour.
2. Check the numbers by substitution.

Question 13. The value of k for which the pair of linear equations 4x + 6y − 1 = 0 and 2x + ky − 7 = 0 represents parallel lines is: [1 Mark, CBSE 2016]
(A) 3   (B) −3   (C) 2   (D) −2
Answer:
Ans. (A) 3
\( \tfrac{4}{2} = \tfrac{6}{k} \ne \tfrac{-1}{-7} \), so k = 3.
In simple words: 2 = 6/k gives k = 3. | The c-ratio is 1/7, different from 2.

Teacher's Note:
1. Parallel: a1/a2 = b1/b2 ≠ c1/c2.

Question 14. If the pair of linear equations 3x + 2y = 1 and (2k + 1)x + (k + 2)y = k − 1 has infinitely many solutions, then the value of k is: [1 Mark, CBSE 2016]
(A) 2   (B) 4   (C) 3   (D) 5
Answer:
Ans. (B) 4
\( \tfrac{3}{2k + 1} = \tfrac{2}{k + 2} = \tfrac{1}{k - 1} \). From \( \tfrac{2}{k+2} = \tfrac{1}{k-1} \): \( 2k - 2 = k + 2 \), k = 4. Check: \( \tfrac{3}{9} = \tfrac{1}{3} \) and \( \tfrac{1}{3} \).
In simple words: Check with the first ratio. | k = 4.

Teacher's Note:
1. Use the last two ratios first.

Question 15. Solve for x and y: 5/x + 1/y = 2 and 6/x − 3/y = 1, x ≠ 0, y ≠ 0. [2 Marks, CBSE 2016]
Answer:
Put \( u = \tfrac{1}{x} \), \( v = \tfrac{1}{y} \): \( 5u + v = 2 \) and \( 6u - 3v = 1 \). Multiply the first by 3 and add: \( 21u = 7 \), u = 1/3, v = 1/3. x = 3, y = 3.
In simple words: Substitute reciprocals.

Teacher's Note:
1. Then x = 1/u.
2. Check: 5/3 + 1/3 = 2.

Question 16. A and B each have a certain number of oranges. A says to B, 'If you give me 10 of your oranges, I will have twice the number of oranges left with you.' B replies, 'If you give me 10 of your oranges, I will have the same number of oranges as are left with you.' Find the number of oranges with A and B separately. [2 Marks, CBSE 2016]
Answer:
Let A have a and B have b oranges. \( a + 10 = 2(b - 10) \) and \( b + 10 = a - 10 \).
The second gives a = b + 20. Then b + 30 = 2b − 20, so b = 50 and a = 70.
A has 70 oranges and B has 50 oranges.
In simple words: Translate each statement into an equation.

Teacher's Note:
1. After B gives 10, B has b − 10.
2. Check: 80 = 2 × 40 and 60 = 60.

Question 17. A two-digit number is obtained by either multiplying the sum of the digits by 8 and adding 1, or by multiplying the difference of the digits by 13 and adding 2. Find the number. [3 Marks, CBSE 2016]
Answer:
Let the tens digit be a and the units digit b. \( 10a + b = 8(a + b) + 1 \Rightarrow 2a - 7b = 1 \) and \( 10a + b = 13(a - b) + 2 \Rightarrow 3a - 14b = -2 \).
From the first \( a = \dfrac{1 + 7b}{2} \); in the second: \( 3 + 21b - 28b = -4 \), b = 1 and a = 4. The number is 41.
In simple words: Write the number as 10a + b.

Teacher's Note:
1. Check: 8 × 5 + 1 = 41.
2. And 13 × 3 + 2 = 41.

Question 18. Reena has pens and pencils which together are 40 in number. If she has 5 more pencils and 5 fewer pens, then the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils. [2 Marks, CBSE 2017]
Answer:
Let pens = p and pencils = c. \( p + c = 40 \) and \( c + 5 = 4(p - 5) \Rightarrow c - 4p = -25 \).
Substituting c = 40 − p: 40 − 5p = −25, so p = 13 and c = 27. Reena has 13 pens and 27 pencils.
In simple words: Form the two equations.

Teacher's Note:
1. Check: 32 = 4 × 8.
2. 27 + 5 = 32 and 13 − 5 = 8.

Question 19. Places 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? OR Solve 10/(x + y) + 2/(x − y) = 4 and 15/(x + y) + 5/(x − y) = −2. [4 Marks, CBSE 2017]
Answer:
(i) Let the speeds be x km/h (from A) and y km/h (from B), x > y. Same direction: the faster car gains 100 km in 5 hours, so \( 5(x - y) = 100 \), x − y = 20. Opposite directions: \( 1 \times (x + y) = 100 \). So x = 60 and y = 40: 60 km/h and 40 km/h.
(ii) Put \( u = \tfrac{1}{x + y} \), \( v = \tfrac{1}{x - y} \): \( 10u + 2v = 4 \) and \( 15u + 5v = -2 \). Multiply the first by 5 and the second by 2: \( 50u + 10v = 20 \) and \( 30u + 10v = -4 \). Subtract: \( 20u = 24 \), u = 6/5 and v = −4. So x + y = 5/6 and x − y = −1/4: x = 7/24, y = 13/24.
In simple words: Same direction: relative speed x − y. Opposite: x + y.

Teacher's Note:
1. In (ii) reduce to a pair in u and v.
2. Check (ii): 10 × 6/5 + 2 × (−4) = 4.