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CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Model Questions - 1 Marks - Part 2

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SECTION A — Multiple Choice Questions [1 Mark Each]

51. 1 Mark Q51. If $\frac{1}{x+y} + \frac{1}{x-y} = \frac{1}{4}$ and $\frac{5}{x+y} - \frac{2}{x-y} = -\frac{3}{2}$, then the values of $x$ and $y$ are:
    (a) $x = 5, y = 3$

    (b) $x = 3, y = 5$

    (c) $x = 4, y = 2$

    (d) $x = 2, y = 4$
    Solution:
    Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.
    $u + v = \frac{1}{4}$ and $5u - 2v = -\frac{3}{2}$.
    Solving this system gives $u = \frac{1}{8} \implies x + y = 8$, and $v = \frac{1}{8} \implies x - y = 2$.
    Solving $x + y = 8$ and $x - y = 2$ yields $x = 5, y = 3$.
    Answer: (a) $x = 5, y = 3$
52. 1 Mark Q52. The value of $k$ for which the lines $2x + 3y = 7$ and $(k+1)x + (2k-1)y = 4k+1$ are coincident is:
    (a) $k = 5$

    (b) $k = 3$

    (c) $k = 2$

    (d) $k = 4$
    Solution:
    For coincident lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$:
    $\frac{2}{k+1} = \frac{3}{2k-1} = \frac{7}{4k+1}$
    Taking the first two ratios: $2(2k-1) = 3(k+1) \implies 4k - 2 = 3k + 3 \implies k = 5$.
    Answer: (a) $k = 5$
53. 1 Mark Q53. A train covered a certain distance at a uniform speed. If the train had been $10\text{ km/h}$ faster, it would have taken $2\text{ hours}$ less. If it were slower by $10\text{ km/h}$, it would have taken $3\text{ hours}$ more. The distance covered is:
    (a) $600\text{ km}$

    (b) $300\text{ km}$

    (c) $500\text{ km}$

    (d) $400\text{ km}$
    Solution:
    Let speed be $S$ and time be $T$. Distance $D = S \times T$.
    $(S + 10)(T - 2) = ST \implies -2S + 10T = 20 \implies -S + 5T = 10$.
    $(S - 10)(T + 3) = ST \implies 3S - 10T = 30$.
    Solving gives $S = 50\text{ km/h}$ and $T = 12\text{ hours}$. Distance = $50 \times 12 = 600\text{ km}$.
    Answer: (a) $600\text{ km}$
54. 1 Mark Q54. In $\triangle ABC$, $\angle C = 3 \angle B = 2(\angle A + \angle B)$. The angles $\angle A, \angle B, \angle C$ are respectively:
    (a) $20^\circ, 40^\circ, 120^\circ$

    (b) $30^\circ, 60^\circ, 90^\circ$

    (c) $45^\circ, 45^\circ, 90^\circ$

    (d) $20^\circ, 60^\circ, 100^\circ$
    Solution:
    Let angles be $A, B, C$. $C = 3B$ and $3B = 2(A + B) \implies 3B = 2A + 2B \implies B = 2A \implies A = \frac{B}{2}$.
    Also, $A + B + C = 180^\circ \implies \frac{B}{2} + B + 3B = 180^\circ \implies \frac{9B}{2} = 180^\circ \implies B = 40^\circ$.
    Then $A = 20^\circ$ and $C = 120^\circ$.
    Answer: (a) $20^\circ, 40^\circ, 120^\circ$
55. 1 Mark Q55. If the system $x + 2y = 3$ and $a x + b y = c$ has infinitely many solutions, then which relation must hold?
    (a) $a = \frac{b}{2} = \frac{c}{3}$

    (b) $2a = b$ and $3a = c$

    (c) $a = 2b = 3c$

    (d) Both (A) and (B)
    Solution:
    For infinitely many solutions, $\frac{1}{a} = \frac{2}{b} = \frac{3}{c} \implies a = \frac{b}{2} = \frac{c}{3}$ (Option A).
    Cross-multiplying gives $b = 2a$ and $c = 3a$ (Option B). Thus both hold.
    Answer: (d) Both (A) and (B)
56. 1 Mark Q56. The coordinates of the vertices of a triangle formed by the lines $y = x$, $y = 2x$, and $y = 3$ are:
    (a) $(0,0), (3,3), (1.5, 3)$

    (b) $(0,0), (3,3), (3,6)$

    (c) $(0,0), (1,2), (3,3)$

    (d) $(0,0), (2,3), (3,3)$
    Solution:
    Intersection of $y = x$ and $y = 3$ is $(3,3)$. Intersection of $y = 2x$ and $y = 3$ is $(1.5, 3)$. Intersection of $y = x$ and $y = 2x$ is $(0,0)$.
    Answer: (a) $(0,0), (3,3), (1.5, 3)$
57. 1 Mark Q57. A shopkeeper sells a saree at $8\%$ profit and a sweater at $10\%$ discount, thereby getting an amount of ₹$1008$. If she had sold the saree at $10\%$ profit and the sweater at $8\%$ discount, she would have got ₹$1028$. The cost price of the saree is:
    (a) ₹$600$

    (b) ₹$400$

    (c) ₹$500$

    (d) ₹$800$
    Solution:
    Let cost price of saree be $x$ and sweater be $y$.
    $1.08x + 0.90y = 1008$ and $1.10x + 0.92y = 1028$.
    Solving this system gives $x = 600$.
    Answer: (a) ₹$600$
58. 1 Mark Q58. The pair of linear equations $2x - y - 4 = 0$ and $x + y + 1 = 0$ intersect at a point in which quadrant?
    (a) First quadrant

    (b) Fourth quadrant

    (c) Second quadrant

    (d) Third quadrant
    Solution:
    Adding equations: $3x - 3 = 0 \implies x = 1$. Substituting $x = 1$ gives $1 + y + 1 = 0 \implies y = -2$.
    The point is $(1, -2)$, which lies in the fourth quadrant.
    Answer: (b) Fourth quadrant
59. 1 Mark Q59. The value of $a$ for which the pair of equations $ax + y = a^2$ and $x + ay = 1$ has infinitely many solutions is:
    (a) $a = 1$

    (b) $a = -1$

    (c) $a = 2$

    (d) $a = 0$
    Solution:
    For infinitely many solutions, $\frac{a}{1} = \frac{1}{a} = \frac{a^2}{1}$.
    $\frac{a}{1} = \frac{1}{a} \implies a^2 = 1 \implies a = \pm 1$. But $\frac{a^2}{1} = \frac{1}{1} \implies a^2 = 1$. Combining requirements gives $a = 1$.
    Answer: (a) $a = 1$
60. 1 Mark Q60. A father is 3 times as old as his son. In 12 years, his age will be twice the age of his son. The present age of the father is:
    (a) $36\text{ years}$

    (b) $42\text{ years}$

    (c) $48\text{ years}$

    (d) $30\text{ years}$
    Solution:
    Let son's age be $S$ and father's age be $F = 3S$.
    In 12 years: $3S + 12 = 2(S + 12) \implies 3S + 12 = 2S + 24 \implies S = 12$.
    Father's present age $F = 3(12) = 36\text{ years}$.
    Answer: (a) $36\text{ years}$
61. 1 Mark Q61. If $217x + 131y = 913$ and $131x + 217y = 827$, then the value of $x + y$ is:
    (a) $5$

    (b) $6$

    (c) $7$

    (d) $8$
    Solution:
    Adding both equations: $(217 + 131)x + (131 + 217)y = 913 + 827$
    $348x + 348y = 1740 \implies 348(x + y) = 1740 \implies x + y = 5$.
    Answer: (a) $5$
62. 1 Mark Q62. A fraction becomes $\frac{4}{5}$ if $1$ is added to both numerator and denominator. If, however, $5$ is subtracted from both numerator and denominator, the fraction becomes $\frac{1}{2}$. The fraction is:
    (a) $\frac{7}{9}$

    (b) $\frac{5}{9}$

    (c) $\frac{7}{11}$

    (d) $\frac{9}{11}$
    Solution:
    Let the fraction be $\frac{x}{y}$. $\frac{x+1}{y+1} = \frac{4}{5} \implies 5x - 4y = -1$.
    $\frac{x-5}{y-5} = \frac{1}{2} \implies 2x - y = 5$.
    Solving yields $x = 7, y = 9$, so the fraction is $\frac{7}{9}$.
    Answer: (a) $\frac{7}{9}$
63. 1 Mark Q63. If a pair of linear equations is consistent and independent, then the lines representing them will be:
    (a) Intersecting at a single point

    (b) Parallel

    (c) Coincident

    (d) Horizontal
    Solution:
    Consistent and independent systems have exactly one unique solution, meaning the lines intersect at a single point.
    Answer: (a) Intersecting at a single point
64. 1 Mark Q64. A motorboat can travel $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10\text{ hours}$. It can travel $40\text{ km}$ upstream and $55\text{ km}$ downstream in $13\text{ hours}$. The speed of the boat in still water is:
    (a) $8\text{ km/h}$

    (b) $3\text{ km/h}$

    (c) $11\text{ km/h}$

    (d) $5\text{ km/h}$
    Solution:
    Let upstream speed be $u$ and downstream speed be $v$.
    $\frac{30}{u} + \frac{44}{v} = 10$ and $\frac{40}{u} + \frac{55}{v} = 13$.
    Solving gives $u = 5\text{ km/h}$ and $v = 11\text{ km/h}$.
    Speed in still water = $\frac{11 + 5}{2} = 8\text{ km/h}$.
    Answer: (a) $8\text{ km/h}$
65. 1 Mark Q65. The area of the triangle formed by the line $x + y = 4$ with the coordinate axes is:
    (a) $8\text{ sq units}$

    (b) $16\text{ sq units}$

    (c) $4\text{ sq units}$

    (d) $2\text{ sq units}$
    Solution:
    The line intercepts axes at $(4,0)$ and $(0,4)$.
    Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8\text{ sq units}$.
    Answer: (a) $8\text{ sq units}$
66. 1 Mark Q66. The values of $p$ and $q$ for which $2x + 3y = 7$ and $(p+q)x + (2p-q)y = 21$ have infinitely many solutions are:
    (a) $p = 5, q = 1$

    (b) $p = 1, q = 5$

    (c) $p = 3, q = 2$

    (d) $p = 4, q = 2$
    Solution:
    $\frac{2}{p+q} = \frac{3}{2p-q} = \frac{7}{21} = \frac{1}{3}$.
    $p + q = 6$ and $2p - q = 9$. Adding gives $3p = 15 \implies p = 5$, and $q = 1$.
    Answer: (a) $p = 5, q = 1$
67. 1 Mark Q67. If $3x + y = 1$ and $(2k-1)x + (k-1)y = 2k+1$ are inconsistent, then $k$ equals:
    (a) $2$

    (b) $-2$

    (c) $0$

    (d) $1$
    Solution:
    For inconsistent lines, $\frac{3}{2k-1} = \frac{1}{k-1} \neq \frac{1}{2k+1}$.
    $3(k-1) = 2k-1 \implies 3k - 3 = 2k - 1 \implies k = 2$.
    Answer: (a) $2$
68. 1 Mark Q68. A 2-digit number is 4 times the sum of its digits and twice the product of its digits. The number is:
    (a) $36$

    (b) $24$

    (c) $48$

    (d) $12$
    Solution:
    Testing option (a) $36$: Sum = $3+6=9$, $4 \times 9 = 36$. Product = $3 \times 6 = 18$, $2 \times 18 = 36$. Both conditions are satisfied.
    Answer: (a) $36$
69. 1 Mark Q69. If $x = a$ and $y = b$ is the solution of the system $x - y = 2$ and $x + y = 4$, then the values of $a$ and $b$ are:
    (a) $a = 3, b = 1$

    (b) $a = 1, b = 3$

    (c) $a = 2, b = 2$

    (d) $a = 4, b = 0$
    Solution:
    Adding equations: $2x = 6 \implies x = 3$. Substituting gives $3 + y = 4 \implies y = 1$.
    Answer: (a) $a = 3, b = 1$
70. 1 Mark Q70. If the system $2x + 3y = 5$ and $4x + k y = 10$ has infinitely many solutions, then $k$ is equal to:
    (a) $6$

    (b) $3$

    (c) $4$

    (d) $5$
    Solution:
    $\frac{2}{4} = \frac{3}{k} = \frac{5}{10} \implies \frac{1}{2} = \frac{3}{k} \implies k = 6$.
    Answer: (a) $6$
71. 1 Mark Q71. The area of the triangle formed by the lines $x = 3$, $y = 4$, and $x = y$ is:
    (a) $0.5\text{ sq units}$

    (b) $1\text{ sq unit}$

    (c) $2\text{ sq units}$

    (d) $2.5\text{ sq units}$
    Solution:
    Vertices are $(3,3)$, $(4,4)$, and $(3,4)$. Area = $\frac{1}{2} \times 1 \times 1 = 0.5\text{ sq units}$.
    Answer: (a) $0.5\text{ sq units}$
72. 1 Mark Q72. A taxi charges a fixed charge together with the charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is ₹$105$, and for a journey of $15\text{ km}$, the charge paid is ₹$155$. What is the charge for traveling $25\text{ km}$?
    (a) ₹$255$

    (b) ₹$235$

    (c) ₹$245$

    (d) ₹$260$
    Solution:
    Let fixed charge be $F$ and per km charge be $C$.
    $F + 10C = 105$ and $F + 15C = 155$.
    Subtracting gives $5C = 50 \implies C = 10$, and $F = 5$.
    Charge for $25\text{ km} = 5 + 25(10) = 255$.
    Answer: (a) ₹$255$
73. 1 Mark Q73. If $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ represent intersecting lines, then:
    (a) $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

    (b) $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

    (c) $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

    (d) $\frac{a_1}{a_2} = \frac{c_1}{c_2}$
    Solution:
    Intersecting lines have a unique solution, which requires $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
    Answer: (a) $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
74. 1 Mark Q75. The pair of equations $x = a$ and $y = b$ graphically represents lines which are:
    (a) Intersecting at $(a, b)$

    (b) Parallel

    (c) Intersecting at $(b, a)$

    (d) Coincident
    Solution:
    $x = a$ is a vertical line parallel to the Y-axis, and $y = b$ is a horizontal line parallel to the X-axis. They intersect each other perpendicularly at the point $(a, b)$.
    Answer: (a) Intersecting at $(a, b)$
75. 1 Mark Q76. If $3x + 2y = 13$ and $3x - 2y = 5$, then the value of $x + y$ is:
    (a) $5$

    (b) $3$

    (c) $2$

    (d) $7$
    Solution:
    Adding the two equations: $(3x + 2y) + (3x - 2y) = 13 + 5 \implies 6x = 18 \implies x = 3$.
    Substituting $x = 3$ into the first equation: $3(3) + 2y = 13 \implies 9 + 2y = 13 \implies 2y = 4 \implies y = 2$.
    Therefore, $x + y = 3 + 2 = 5$.
    Answer: (a) $5$
76. 1 Mark Q77. The value of $c$ for which the pair of equations $c x - y = 2$ and $6x - 2y = 3$ will have infinitely many solutions is:
    (a) No value of $c$ exists

    (b) $c = 3$

    (c) $c = -3$

    (d) $c = 12$
    Solution:
    For infinitely many solutions, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$:
    $\frac{c}{6} = \frac{-1}{-2} = \frac{2}{3} \implies \frac{c}{6} = \frac{1}{2}$ and $\frac{1}{2} = \frac{2}{3}$ (which is false).
    Since the consistency conditions for infinitely many solutions conflict ($\frac{1}{2} \neq \frac{2}{3}$), no such value of $c$ can exist.
    Answer: (a) No value of $c$ exists
77. 1 Mark Q78. In a competitive exam, 3 marks are awarded for every correct answer and 1 mark is deducted for every wrong answer. Jay scored 40 marks. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Jay would have scored 50 marks. How many questions were there in the test if Jay attempted all?
    (a) $20$

    (b) $15$

    (c) $25$

    (d) $30$
    Solution:
    Let $x$ be correct answers and $y$ be incorrect answers.
    $3x - y = 40$
    $4x - 2y = 50 \implies 2x - y = 25$
    Subtracting the second from the first gives $x = 15$. Substituting gives $3(15) - y = 40 \implies y = 5$.
    Total questions = $x + y = 15 + 5 = 20$.
    Answer: (a) $20$
78. 1 Mark Q79. For what value of $k$, do the equations $3x - y + 8 = 0$ and $6x - k y + 16 = 0$ represent coincident lines?
    (a) $2$

    (b) $-2$

    (c) $\frac{1}{2}$

    (d) $-\frac{1}{2}$
    Solution:
    For coincident lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$:
    $\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16} \implies \frac{1}{2} = \frac{1}{k} \implies k = 2$.
    Answer: (a) $2$
79. 1 Mark Q80. If the pair of linear equations $x - y = 1$ and $x + y = 3$ is solved, the triangle formed by these two lines and the Y-axis has an area equal to:
    (a) $2\text{ sq units}$

    (b) $3\text{ sq units}$

    (c) $4\text{ sq units}$

    (d) $1\text{ sq unit}$
    Solution:
    Solving the equations yields intersection point $(2, 1)$.
    For $x - y = 1$, Y-intercept (where $x=0$) is $(0, -1)$.
    For $x + y = 3$, Y-intercept (where $x=0$) is $(0, 3)$.
    Base on the Y-axis = distance between $(0, -1)$ and $(0, 3)$ which is $4$.
    Height = X-coordinate of intersection point = $2$.
    Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4\text{ sq units}$.
    Answer: (c) $4\text{ sq units}$
80. 1 Mark Q81. If $148x + 231y = 527$ and $231x + 148y = 610$, then the value of $x - y$ is:
    (a) $1$

    (b) $2$

    (c) $3$

    (d) $4$
    Solution:
    Subtracting the second equation from the first equation:
    $(148 - 231)x + (231 - 148)y = 527 - 610$
    $-83x + 83y = -83 \implies -83(x - y) = -83 \implies x - y = 1$.
    Answer: (a) $1$
81. 1 Mark Q82. The pair of equations $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$ has:
    (a) A unique solution

    (b) Exactly two solutions

    (c) Infinitely many solutions

    (d) No solution
    Solution:
    $\frac{a_1}{a_2} = \frac{1}{-3}$, $\frac{b_1}{b_2} = \frac{2}{-6} = \frac{1}{-3}$, and $\frac{c_1}{c_2} = \frac{5}{1}$.
    Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel, meaning there is no solution.
    Answer: (d) No solution
82. 1 Mark Q83. If $x = a, y = b$ is the solution of the equations $x - y = 2$ and $x + y = 4$, then the values of $a$ and $b$ are respectively:
    (a) $3$ and $1$

    (b) $3$ and $1$

    (c) $5$ and $3$

    (d) $4$ and $2$
    Solution:
    Adding the equations: $2x = 6 \implies x = 3$. Substituting gives $y = 1$.
    Therefore, $a = 3, b = 1$.
    Answer: (b) $3$ and $1$
83. 1 Mark Q84. The value of $k$ for which the pair of equations $kx - y = 2$ and $6x - 2y = 3$ has no solution is:
    (a) $k = 3$

    (b) $k \neq 3$

    (c) $k = 4$

    (d) $k = 2$
    Solution:
    For no solution, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:
    $\frac{k}{6} = \frac{-1}{-2} \implies \frac{k}{6} = \frac{1}{2} \implies k = 3$.
    Answer: (a) $k = 3$
84. 1 Mark Q85. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old is Nuri now?
    (a) $50\text{ years}$

    (b) $20\text{ years}$

    (c) $60\text{ years}$

    (d) $40\text{ years}$
    Solution:
    Let Nuri's age be $N$ and Sonu's age be $S$.
    $N - 5 = 3(S - 5) \implies N - 3S = -10$
    $N + 10 = 2(S + 10) \implies N - 2S = 10$
    Subtracting the first from the second gives $S = 20$, and $N = 50$.
    Answer: (a) $50\text{ years}$
85. 1 Mark Q86. The area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is:
    (a) $ab\text{ sq units}$

    (b) $\frac{1}{2}ab\text{ sq units}$

    (c) $2ab\text{ sq units}$

    (d) $\frac{1}{4}ab\text{ sq units}$
    Solution:
    The line intercepts the X-axis at $(a, 0)$ and the Y-axis at $(0, b)$.
    Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}ab\text{ sq units}$.
    Answer: (b) $\frac{1}{2}ab\text{ sq units}$
86. 1 Mark Q87. For what value of $p$ does the system of equations $px + 2y = 5$ and $3x + y = 1$ have a UNIQUE solution?
    (a) $p = 6$

    (b) $p \neq 6$

    (c) $p = 3$

    (d) $p \neq 3$
    Solution:
    For a unique solution, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \frac{p}{3} \neq \frac{2}{1} \implies p \neq 6$.
    Answer: (b) $p \neq 6$
87. 1 Mark Q88. If $x + y = 10$ and $x - y = 4$, then the value of $x^2 - y^2$ is:
    (a) $40$

    (b) $14$

    (c) $6$

    (d) $24$
    Solution:
    $x^2 - y^2 = (x + y)(x - y) = 10 \times 4 = 40$.
    Answer: (a) $40$
88. 1 Mark Q89. Aruna has only ₹$1$ and ₹$2$ coins with her. If the total number of coins that she has is $50$ and the total amount of money with her is ₹$75$, then the number of ₹$1$ and ₹$2$ coins are respectively:
    (a) $25$ and $25$

    (b) $35$ and $15$

    (c) $15$ and $35$

    (d) $30$ and $20$
    Solution:
    Let ₹$1$ coins be $x$ and ₹$2$ coins be $y$.
    $x + y = 50$
    $1x + 2y = 75$
    Subtracting the first from the second gives $y = 25$, and $x = 25$.
    Answer: (a) $25$ and $25$
89. 1 Mark Q90. If the system of linear equations $2x + 3y = 7$ and $2ax + (a+b)y = 28$ has infinitely many solutions, then the values of $a$ and $b$ are:
    (a) $a = 4, b = 8$

    (b) $a = 8, b = 4$

    (c) $a = 4, b = 4$

    (d) $a = 2, b = 6$
    Solution:
    $\frac{2}{2a} = \frac{3}{a+b} = \frac{7}{28} = \frac{1}{4}$
    $\frac{1}{a} = \frac{1}{4} \implies a = 4$
    $\frac{3}{4+b} = \frac{1}{4} \implies 4+b = 12 \implies b = 8$.
    Answer: (a) $a = 4, b = 8$
90. 1 Mark Q91. If $\frac{x}{a} + \frac{y}{b} = 2$ and $ax - by = a^2 - b^2$, then the values of $x$ and $y$ are:
    (a) $x = a, y = b$

    (b) $x = b, y = a$

    (c) $x = -a, y = -b$

    (d) $x = a^2, y = b^2$
    Solution:
    Testing $x = a$ and $y = b$:
    $\frac{a}{a} + \frac{b}{b} = 1 + 1 = 2$ (satisfied).
    $a(a) - b(b) = a^2 - b^2$ (satisfied).
    Answer: (a) $x = a, y = b$
91. 1 Mark Q92. For what value of $k$ do the equations $3x - y + 8 = 0$ and $6x - ky = -16$ represent coincident lines?
    (a) $k = 2$

    (b) $k = -2$

    (c) $k = 1/2$

    (d) $k = -1/2$
    Solution:
    Rewrite second equation in standard form: $6x - ky + 16 = 0$.
    For coincident lines, $\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16} \implies \frac{1}{2} = \frac{1}{k} \implies k = 2$.
    Answer: (a) $k = 2$
92. 1 Mark Q93. A boat goes $12\text{ km}$ upstream and $40\text{ km}$ downstream in $8\text{ hours}$. It can go $16\text{ km}$ upstream and $32\text{ km}$ downstream in the same time. The speed of the boat in still water is:
    (a) $6\text{ km/h}$

    (b) $8\text{ km/h}$

    (c) $2\text{ km/h}$

    (d) $10\text{ km/h}$
    Solution:
    Let upstream speed be $u$ and downstream speed be $v$.
    $\frac{12}{u} + \frac{40}{v} = 8$ and $\frac{16}{u} + \frac{32}{v} = 8$.
    Solving this gives $u = 4\text{ km/h}$ and $v = 12\text{ km/h}$.
    Speed in still water = $\frac{12 + 4}{2} = 8\text{ km/h}$.
    Answer: (b) $8\text{ km/h}$
93. 1 Mark Q94. The area of the triangle formed by the line $2x + 3y = 12$ and the coordinate axes is:
    (a) $12\text{ sq units}$

    (b) $24\text{ sq units}$

    (c) $6\text{ sq units}$

    (d) $18\text{ sq units}$
    Solution:
    X-intercept ($y=0$): $2x = 12 \implies x = 6$.
    Y-intercept ($x=0$): $3y = 12 \implies y = 4$.
    Area = $\frac{1}{2} \times 6 \times 4 = 12\text{ sq units}$.
    Answer: (a) $12\text{ sq units}$
94. 1 Mark Q95. The pair of linear equations $x + 2y - 5 = 0$ and $3x + 12y - 10 = 0$ has:
    (a) Unique solution

    (b) No solution

    (c) Infinitely many solutions

    (d) Exactly two solutions
    Solution:
    $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{2}{12} = \frac{1}{6}$.
    Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines have a unique solution.
    Answer: (a) Unique solution
95. 1 Mark Q96. The sum of a two-digit number and the number obtained by reversing the digits is $66$. If the digits of the number differ by $2$, how many such numbers are there?
    (a) $2$

    (b) $1$

    (c) $3$

    (d) $4$
    Solution:
    Let number be $10x + y$. Sum with reversed number is $11(x+y) = 66 \implies x + y = 6$.
    Given $|x - y| = 2$.
    Solving $x+y=6, x-y=2 \implies x=4, y=2$ (Number: $42$).
    Solving $x+y=6, y-x=2 \implies x=2, y=4$ (Number: $24$).
    Both numbers ($42$ and $24$) satisfy the conditions, so there are $2$ such numbers.
    Answer: (a) $2$
96. 1 Mark Q97. If $2x + y = 23$ and $4x - y = 19$, then the value of $5y - 2x$ is:
    (a) $31$

    (b) $37$

    (c) $29$

    (d) $33$
    Solution:
    Adding the equations: $6x = 42 \implies x = 7$.
    Substituting $x = 7$: $2(7) + y = 23 \implies y = 9$.
    Now evaluate $5y - 2x = 5(9) - 2(7) = 45 - 14 = 31$.
    Answer: (a) $31$
97. 1 Mark Q98. For what value of $a$ do the equations $3x + y = 1$ and $(2a-1)x + (a-1)y = 2a+1$ have NO solution?
    (a) $a = 2$

    (b) $a = -2$

    (c) $a = 1$

    (d) $a = 0$
    Solution:
    For no solution, $\frac{3}{2a-1} = \frac{1}{a-1} \neq \frac{1}{2a+1}$.
    $3(a-1) = 2a-1 \implies 3a - 3 = 2a - 1 \implies a = 2$.
    Answer: (a) $a = 2$
98. 1 Mark Q99. The perimeter of a rectangle is $44\text{ cm}$. Its length is $2\text{ cm}$ more than twice its breadth. The area of the rectangle is:
    (a) $96\text{ cm}^2$

    (b) $88\text{ cm}^2$

    (c) $108\text{ cm}^2$

    (d) $112\text{ cm}^2$
    Solution:
    Let breadth be $b$ and length be $l$.
    $2(l + b) = 44 \implies l + b = 22$
    $l = 2b + 2 \implies 2b - l = -2$
    Solving gives $b = 8\text{ cm}$ and $l = 14\text{ cm}$.
    Area = $l \times b = 14 \times 8 = 112\text{ cm}^2$.
    Answer: (d) $112\text{ cm}^2$
99. 1 Mark Q100. If the system $x + y = 2$ and $2x + 2y = k$ has infinitely many solutions, then $k$ is equal to:
    (a) $4$

    (b) $2$

    (c) $1$

    (d) $8$
    Solution:
    For infinitely many solutions, $\frac{1}{2} = \frac{1}{2} = \frac{2}{k} \implies \frac{1}{2} = \frac{2}{k} \implies k = 4$.
    Answer: (a) $4$

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