CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 36

Mathematics Objective Questions and Answers: Chapter 03 Pair of Linear Equations in Two Variables

Review structured MCQ sets for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: A sailor goes 12 km downstream in 2 hours and returns to the starting point in 3 hours. Find the speed of the sailor in still water and the speed of the current.
(a) 5 km/hr, 1 km/hr
(b) 1 km/hr, 5 km/hr
(c) 4 km/hr, 1 km/hr
(d) 3 km/hr, 1 km/hr
Show Answer & Explanation

Answer: (a) 5 km/hr, 1 km/hr

Explanation:
1. Downstream speed = 6 km/hr and upstream speed = 4 km/hr.
2. Still water = \( (6 + 4)/2 = 5 \) km/hr and current = \( (6 - 4)/2 = 1 \) km/hr.

Teacher's Note:
1. Downstream = boat + current, upstream = boat − current.

Question 2: A half-ticket issued by the railway costs half the full fare but the reservation charge is the same on a half-ticket as on a full ticket. One full reserved first class ticket for a journey between two stations costs Rs. 362. One full and one half reserved first class ticket cost Rs. 554. The reservation charge is
(a) Rs. 18
(b) Rs. 22
(c) Rs. 38
(d) None of these
Show Answer & Explanation

Answer: (b) Rs. 22

Explanation:
1. Let the fare be \( F \) and the reservation charge \( R \): \( F + R = 362 \) and \( F + (F/2 + R) + R = 554 \), i.e. \( 1.5F + 2R = 554 \).
2. Substituting \( F = 362 - R \): \( 543 + 0.5R = 554 \), so \( R = 22 \).

Teacher's Note:
1. Write the half ticket as \( F/2 + R \).

Question 3: A three-digit number abc is 459 more than the sum of its digits. What is the sum of the two-digit number ab and the one-digit number a?
(a) 71
(b) 61
(c) 51
(d) Cannot be determined
Show Answer & Explanation

Answer: (c) 51

Explanation:
1. \( 100a + 10b + c - (a + b + c) = 99a + 9b = 459 \), so \( 11a + b = 51 \).
2. \( a = 4 \) and \( b = 7 \), so ab \( + a = 47 + 4 = 51 \).

Teacher's Note:
1. Digits must be single-digit integers.

Question 4: If 2x + 3y = 7 and (p + q)x + 2(p − q)y = 21 and the system has infinitely many solutions, then the values of p and q are
(a) p = 10, q = 3
(b) p = 5, q = 2
(c) p = 21/4, q = 3/4
(d) p = 10, q = 2
Show Answer & Explanation

Answer: (c) p = 21/4, q = 3/4

Explanation:
1. Infinite solutions need \( (p + q)/2 = 2(p - q)/3 = 21/7 = 3 \).
2. \( p + q = 6 \) and \( p - q = 9/2 \), so \( p = 21/4 \) and \( q = 3/4 \).

Teacher's Note:
1. Set all three ratios equal.

Question 5: In a fraction, if the numerator is multiplied by 3 and the denominator is reduced by 3, it is equal to 18/11. If the numerator is increased by 8 and the denominator is doubled, it is equal to 2/5. The fraction is
(a) 12/25
(b) 13/12
(c) 8/5
(d) 7/19
Show Answer & Explanation

Answer: (a) 12/25

Explanation:
1. \( 3n/(d - 3) = 18/11 \) gives \( 33n = 18d - 54 \), and \( (n + 8)/(2d) = 2/5 \) gives \( 5n + 40 = 4d \).
2. Solving, \( n = 12 \) and \( d = 25 \).

Teacher's Note:
1. Form two equations in the numerator and denominator.

Question 6: The solution of the equations x/4 = y/3 = z/2 and 7x + 8y + 5z = 62 is
(a) (4, 3, 2)
(b) (2, 3, 4)
(c) (3, 4, 2)
(d) (4, 2, 3)
Show Answer & Explanation

Answer: (a) (4, 3, 2)

Explanation:
1. Let each ratio be \( k \): \( x = 4k, y = 3k, z = 2k \).
2. \( 28k + 24k + 10k = 62k = 62 \), so \( k = 1 \) and \( (x, y, z) = (4, 3, 2) \).

Teacher's Note:
1. Introduce a common ratio \( k \).

Question 7: If 173x + 197y = 149 and 197x + 173y = 221, then (x, y) is
(a) (3, −2)
(b) (2, 1)
(c) (1, −2)
(d) (2, −1)
Show Answer & Explanation

Answer: (d) (2, −1)

Explanation:
1. Adding: \( 370(x + y) = 370 \), so \( x + y = 1 \). Subtracting: \( -24x + 24y = -72 \), so \( x - y = 3 \).
2. \( x = 2 \) and \( y = -1 \).

Teacher's Note:
1. Add and subtract the equations.

Question 8: A mother said to her son, 'The sum of our present ages is twice my age 12 years ago, and nine years hence the sum of our ages will be thrice my age 14 years ago.' What is the son's present age (in years)?
(a) 8
(b) 12
(c) 15
(d) 10
Show Answer & Explanation

Answer: (b) 12

Explanation:
1. Let the mother be \( M \) and the son \( S \): \( M + S = 2(M - 12) \) and \( M + S + 18 = 3(M - 14) \).
2. These give \( S = M - 24 \) and \( S = 2M - 60 \), so \( M = 36 \) and \( S = 12 \).

Teacher's Note:
1. Translate each statement into an equation.

Question 9: Dheeraj has twice as many sisters as he has brothers. Deepa, Dheeraj's sister, has the same number of brothers as she has sisters. How many brothers does Deepa have?
(a) 2
(b) 3
(c) 4
(d) Cannot be determined
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Let there be \( B \) boys and \( G \) girls. Dheeraj: \( G = 2(B - 1) \). Deepa: \( B = G - 1 \).
2. So \( G = B + 1 = 2B - 2 \), giving \( B = 3 \) and \( G = 4 \). Deepa has 3 brothers.

Teacher's Note:
1. Remember not to count the person himself or herself.

Question 10: The demand for a fertilizer product is given by p + 5q = 21 and the supply by p − 2q = 7, where p is the price of the commodity and q is the number of units supplied. If a man wants to buy q units, the amount he pays is
(a) Rs. 22
(b) Rs. 30
(c) Rs. 32
(d) Rs. 24
Show Answer & Explanation

Answer: (a) Rs. 22

Explanation:
1. Subtracting: \( 7q = 14 \), so \( q = 2 \) and \( p = 11 \).
2. Amount = \( p \times q = 11 \times 2 \) = Rs. 22.

Teacher's Note:
1. Amount = price \( x \) units.

Question 11: Ram and Mohan are friends. If Ram gives Rs. 30 to Mohan, then Mohan will have twice the money left with Ram. But if Mohan gives Rs. 10 to Ram, then Ram will have thrice as much as is left with Mohan. How much money does each have?
(a) Rs. 62, Rs. 34
(b) Rs. 6, Rs. 2
(c) Rs. 170, Rs. 124
(d) Rs. 43, Rs. 26
Show Answer & Explanation

Answer: (a) Rs. 62, Rs. 34

Explanation:
1. \( M + 30 = 2(R - 30) \) and \( R + 10 = 3(M - 10) \).
2. Solving gives \( R = 62 \) and \( M = 34 \).

Teacher's Note:
1. Write the two conditions as equations.

Question 12: In a zoo there are rabbits and pigeons. If their heads are counted there are 90, while their legs are 224. The number of pigeons is
(a) 70
(b) 68
(c) 72
(d) 22
Show Answer & Explanation

Answer: (b) 68

Explanation:
1. \( r + p = 90 \) and \( 4r + 2p = 224 \), so \( 2r = 44 \) and \( r = 22 \).
2. \( p = 68 \).

Teacher's Note:
1. Count heads and legs.

Question 13: A florist was asked to make a bouquet with exactly Rs. 1000 with 100 sticks of roses of three colours: pink, yellow and red. Pink roses cost Rs. 0.50 per stick, red roses cost Rs. 10 per stick and yellow roses cost Rs. 50 per stick. If all three colours are used, how many red roses did the florist use?
(a) 1
(b) 5
(c) 80
(d) Several combinations are possible
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. \( p + y + r = 100 \) and \( 0.5p + 50y + 10r = 1000 \), which gives \( 99y + 19r = 1900 \).
2. With \( y \) and \( r \) positive integers the only solution is \( y = 19, r = 1 \) (and \( p = 80 \)).

Teacher's Note:
1. Reduce to a single equation in two variables and test integers.

Question 14: A woman sells to the first customer half her stock of apples and half an apple, to the second customer half her remaining stock and half an apple, and so on to the third and to a fourth customer. She finds that she has now 15 apples left. How many apples did she have before she started selling?
(a) 63
(b) 127
(c) 240
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. Each sale changes \( N \) to \( N/2 - 1/2 \), so working backwards \( L \) becomes \( 2L + 1 \): 15 → 31 → 63 → 127 → 255.
2. She started with 255 apples, which is not among the options.

Teacher's Note:
1. Work backwards from the apples left.

Question 15: To a proper fraction, when six is added to the numerator and the denominator is increased by its 50%, the ratio becomes 1/2. When the numerator is multiplied by 4 and the denominator is reduced by 8, the fraction becomes 3. The fraction (simplified) is
(a) 1/3
(b) 15/28
(c) 33/52
(d) Cannot be determined
Show Answer & Explanation

Answer: (d) Cannot be determined

Explanation:
1. \( (n + 6)/(1.5d) = 1/2 \) gives \( n = 0.75d - 6 \), and \( 4n/(d - 8) = 3 \) gives \( n = 0.75d - 6 \) as well.
2. Both conditions give the same equation, so \( n \) and \( d \) are not fixed.

Teacher's Note:
1. If the two equations are the same, the system has infinitely many solutions.

Question 16: The pair of equations \( 3^{x + y} = 81 \) and \( 81^{x - y} = 3 \) has
(a) no solution
(b) the solution x = 2 1/2, y = 1 7/8
(c) the solution x = 2, y = 2
(d) the solution x = 2 1/8, y = 1 7/8
Show Answer & Explanation

Answer: (d) the solution x = 2 1/8, y = 1 7/8

Explanation:
1. \( x + y = 4 \) and \( 4(x - y) = 1 \), so \( x - y = 1/4 \).
2. \( x = 17/8 = 2 1/8 \) and \( y = 15/8 = 1 7/8 \).

Teacher's Note:
1. Convert to linear equations by comparing exponents.

Question 17: If the numerator and the denominator of a fraction are each increased by 4, the fraction becomes 2, and when the numerator and the denominator of the same fraction are each decreased by 6, the fraction becomes 12. The sum of the numerator and the denominator is
(a) 11
(b) −11
(c) 25
(d) −25
Show Answer & Explanation

Answer: (c) 25

Explanation:
1. \( n + 4 = 2(d + 4) \) and \( n - 6 = 12(d - 6) \), so \( n = 2d + 4 \) and \( 2d - 2 = 12d - 72 \).
2. \( d = 7 \) and \( n = 18 \), so the sum is 25.

Teacher's Note:
1. Set up two equations.

Question 18: There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs. 42 per day while the average expenditure per head diminishes by Re. 1. The original expenditure of the mess is
(a) Rs. 480
(b) Rs. 520
(c) Rs. 420
(d) Rs. 460
Show Answer & Explanation

Answer: (c) Rs. 420

Explanation:
1. \( E/35 - (E + 42)/42 = 1 \) gives \( 42E - 35E - 1470 = 1470 \), so \( 7E = 2940 \) and \( E = 420 \).

Teacher's Note:
1. Average = total/number of students.

Question 19: The sum of the two digits of a number is 15. If 9 is added to the number, the digits are reversed. The number is
(a) 96
(b) 87
(c) 78
(d) 69
Show Answer & Explanation

Answer: (c) 78

Explanation:
1. \( 10a + b + 9 = 10b + a \) gives \( b - a = 1 \). With \( a + b = 15 \): \( a = 7 \) and \( b = 8 \).
2. The number is 78.

Teacher's Note:
1. Form the reversed number.

Question 20: Consider the system 2x + 3y + 4z = 16, 4x + 4y + 5z = 26 and x + by + cz = 5. The system has infinitely many solutions for a suitable b if c =
(a) 3/2
(b) 1
(c) 1/2
(d) 0
Show Answer & Explanation

Answer: (c) 1/2

Explanation:
1. The third equation must be a combination λ(first) + μ(second): \( 2\lambda + 4\mu = 1 \) and \( 16\lambda + 26\mu = 5 \), giving \( \lambda = -1/2 \) and \( \mu = 1/2 \).
2. Then \( b = 3\lambda + 4\mu = 1/2 \) and \( c = 4\lambda + 5\mu = 1/2 \).

Teacher's Note:
1. A dependent third equation gives infinitely many solutions.

Question 21: In an examination there are 30 questions. 1 mark is given for each correct answer and 0.25 is deducted for every incorrect answer. Ankur attempted all the questions and scored 13.75. How many incorrect answers did he have?
(a) 10
(b) 11
(c) 12
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. \( c + w = 30 \) and \( c - 0.25w = 13.75 \), so \( 30 - 1.25w = 13.75 \) and \( w = 13 \).
2. 13 incorrect answers is not among \( (a) \) to \( (c) \).

Teacher's Note:
1. Substitute \( c = 30 - w \).

Question 22: Which condition must a, b and c satisfy so that the system m + 2n − 3r = a, 2m + 6n − 11r = b, m − 2n + 7r = c has at least one solution?
(a) 5a + 2b + c = 0
(b) 5a + 2b − c = 0
(c) 5a − 2b − c = 0
(d) 5a − 2b + c = 0
Show Answer & Explanation

Answer: (c) 5a − 2b − c = 0

Explanation:
1. Eliminating \( m, n, r \): 5(first) − 2(second) − (third) gives 0 on the left, since \( 5 - 4 - 1 = 0, 10 - 12 + 2 = 0 \) and \( -15 + 22 - 7 = 0 \).
2. So \( 5a - 2b - c = 0 \).

Teacher's Note:
1. Find the combination of the equations whose left side vanishes.

Question 23: If 2x + 3y = 78 and 3x + 2y = 72, the value of x + y is
(a) 36
(b) 32
(c) 30
(d) Cannot be determined
Show Answer & Explanation

Answer: (c) 30

Explanation:
1. Adding: \( 5(x + y) = 150 \), so \( x + y = 30 \).

Teacher's Note:
1. Add the two equations.

Question 24: If 3Y + 9X = 54 and 28X/(13Y) = 140/39, the value of Y − X is
(a) −1
(b) −2
(c) 2
(d) 1
Show Answer & Explanation

Answer: (b) −2

Explanation:
1. \( X/Y = (140 \times 13)/(39 \times 28) = 5/3 \), so \( X = 5Y/3 \).
2. \( 3Y + 15Y = 54 \) gives \( Y = 3 \) and \( X = 5 \), so \( Y - X = -2 \).

Teacher's Note:
1. Find the ratio \( X \) : \( Y \) first.

Question 25: How can the relationship between x and y best be defined if the values are: x = 2, 3, 4, 5, 6 and y = 0, 2, 6, 12, 20?
(a) y = 2x − 4
(b) y = x² − 3x + 2
(c) y = x² − 4x
(d) y = x² − 4
Show Answer & Explanation

Answer: (b) y = x² − 3x + 2

Explanation:
1. For \( x = 2, 3, 4 \): \( x^{2} - 3x + 2 \) gives 0, 2, 6, and for \( x = 5, 6 \) gives 12 and 20.

Teacher's Note:
1. Check the formula on all five pairs.

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