Pair of Linear Equations in Two Variables · Class 10 Mathematics: 220 MCQs with answers

Showing 200 of 220 multiple-choice questions from this kit.

Warm-up · 61 questions

Q1. A linear equation in two variables has the form

Beginner
A $ax + by + c = 0$ with a, b not both zero B $ax^2 + by + c = 0$ C $\frac{a}{x} + by = c$ D $ax + b = 0$
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$ax + by + c = 0$ with a, b not both zero

Both variables appear to the first power, and at least one of a, b is non-zero.

Q2. A solution of a pair of linear equations is a pair (x, y) that

Beginner
A satisfies at least one equation B makes both sides zero C satisfies both equations D lies on the x-axis
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satisfies both equations

It must work in both equations at the same time.

Q3. The graph of a linear equation in two variables is

Beginner
A a parabola B a straight line C a pair of lines D a circle
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a straight line

That is why they are called linear.

Q4. If two lines cross at exactly one point, the pair of equations has

Beginner
A no solution B infinitely many solutions C two solutions D exactly one solution
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exactly one solution

The crossing point is the only pair satisfying both.

Q5. If the two lines are parallel, the pair has

Beginner
A one solution B no solution C infinitely many solutions D exactly two solutions
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no solution

Parallel lines never meet, so no point satisfies both.

Q6. If the two lines coincide, the number of solutions is

Beginner
A one B infinitely many C none D two
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infinitely many

Every point of the line satisfies both equations.

Q7. The point where the line $x + 3y = 6$ meets the y-axis is

Beginner
A $(2, 0)$ B $(0, 6)$ C $(0, 2)$ D $(6, 0)$
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$(0, 2)$

On the y-axis $x = 0$, so $3y = 6$ and $y = 2$.

Q8. The point where $2x - 3y = 12$ meets the x-axis is

Beginner
A $(0, 6)$ B $(0, -4)$ C $(-4, 0)$ D $(6, 0)$
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$(6, 0)$

On the x-axis $y = 0$, so $2x = 12$ and $x = 6$.

Q9. A pair of equations that has at least one solution is called

Beginner
A inconsistent B dependent only C consistent D parallel
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consistent

Consistent covers both the crossing and the coinciding cases.

Q10. A pair with no solution is called

Beginner
A consistent B dependent C equivalent D inconsistent
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inconsistent

Its lines are parallel.

Q11. Which point lies on the line $2x + 3y = 12$?

Beginner
A $(2, 3)$ B $(1, 4)$ C $(3, 2)$ D $(4, 2)$
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$(3, 2)$

$2(3) + 3(2) = 12$ ✓ The others give 13, 14 and 14.

Q12. Assertion (A): One linear equation in two variables has infinitely many solutions. Reason (R): Every point on its graph, which is a straight line, satisfies the equation.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

A line holds infinitely many points, and each is a solution — R explains A.

Q13. How many solutions does the single equation $x + y = 7$ have?

Beginner
A One B Infinitely many C Two D None
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Infinitely many

Every point on its line is a solution, and a line holds infinitely many points.

Q14. The line $y = 4$ is

Beginner
A horizontal, through $(0, 4)$ B vertical, through $(4, 0)$ C through the origin D not a straight line
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horizontal, through $(0, 4)$

Every point with height 4 satisfies it, so the line is horizontal.

Q15. Which equation is satisfied by the point $(0, 0)$?

Beginner
A $3x - 5y = 2$ B $x + y = 1$ C $2x - y = -4$ D $3x - 5y = 0$
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$3x - 5y = 0$

Putting both coordinates to 0 gives $0 = 0$ only in the first.

Q16. Form a pair of linear equations: the cost of 2 kg of wheat and 3 kg of rice is ₹140, and the cost of 4 kg of wheat and 5 kg of rice is ₹260.

Beginner
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Q17. Find two solutions of the equation $2x + y = 8$ and use them to describe its line.

Beginner
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Q18. The sum of two numbers is 20 and their difference is 4. Form a pair of equations and solve it.

Beginner
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Q19. A pair of linear equations has exactly one solution when

Beginner
A $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ B $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ C $\frac{a_1}{a_2} = \frac{c_1}{c_2}$ D $\frac{b_1}{b_2} = \frac{c_1}{c_2}$
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$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Unequal a-to-b ratios mean different slopes, so the lines must cross once.

Q20. A pair has infinitely many solutions when

Beginner
A $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ B $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ C $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ D $a_1 = a_2$ and $b_1 = b_2$
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$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

All three ratios equal means one equation is a multiple of the other.

Q21. A pair has no solution when

Beginner
A all three ratios are equal B $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ C $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ D $c_1 = c_2$
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$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Matching slopes with different constants give parallel lines.

Q22. Coincident lines correspond to

Beginner
A infinitely many solutions B one solution C no solution D two solutions
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infinitely many solutions

Every point of the shared line satisfies both equations.

Q23. Parallel lines correspond to a pair that is

Beginner
A consistent B dependent C unique D inconsistent
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inconsistent

No solution means inconsistent.

Q24. Written in standard form, $3x + 2y = 7$ has c equal to

Beginner
A −7 B 7 C 3 D 2
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−7

$3x + 2y - 7 = 0$, so $c = -7$.

Q25. For the pair $x + y - 5 = 0$ and $2x + 2y - 10 = 0$, the ratio $\frac{c_1}{c_2}$ is

Beginner
A $\frac{1}{2}$ B 2 C $-\frac{1}{2}$ D 5
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$\frac{1}{2}$

$\frac{-5}{-10} = \frac12$ — the minus signs cancel.

Q26. A dependent pair of linear equations is

Beginner
A always inconsistent B sometimes inconsistent C always consistent D never solvable
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always consistent

Dependent means the equations are equivalent, so they have infinitely many solutions.

Q27. Which case does NOT need the ratio of the constants?

Beginner
A no solution B one solution C infinitely many solutions D all three need it
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one solution

If the a and b ratios already differ, the lines cross whatever the constants are.

Q28. Assertion (A): The pair $2x + 3y - 5 = 0$ and $4x + 6y - 10 = 0$ has infinitely many solutions. Reason (R): All three ratios of the coefficients are equal.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

$\frac24 = \frac36 = \frac{-5}{-10} = \frac12$, which is exactly the condition for coincident lines.

Q29. A pair of equations represents two lines that meet at exactly one point. The pair is

Beginner
A consistent with a unique solution B inconsistent C dependent D without solution
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consistent with a unique solution

One crossing point means one solution, so the pair is consistent.

Q30. Without drawing, say how many solutions $x - 2y = 0$ and $3x + 4y = 20$ have, and name the pair.

Beginner
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Q31. Classify the pair $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$, giving all three ratios.

Beginner
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Q32. Show that $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$ is an inconsistent pair.

Beginner
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Q33. State the conditions on the ratios for a pair to be (i) consistent with a unique solution (ii) inconsistent.

Beginner
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Q34. The first step of the substitution method is to

Beginner
A add the two equations B make one variable the subject in one equation C multiply both equations by a constant D draw the graphs
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make one variable the subject in one equation

Substitution begins by expressing one variable in terms of the other.

Q35. Rearranging $x + 2y = 3$ for x gives

Beginner
A $x = 3 - 2y$ B $x = 3 + 2y$ C $x = 2y - 3$ D $x = \frac{3}{2y}$
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$x = 3 - 2y$

The term $2y$ changes sign as it crosses the equals sign.

Q36. Substituting $x = 3 - 2y$ into $7x - 15y = 2$ gives

Beginner
A $7(3 - 2y) - 15y = 2$ B $7x - 15(3 - 2y) = 2$ C $(3 - 2y) - 15y = 2$ D $7(3 - 2y) + 15y = 2$
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$7(3 - 2y) - 15y = 2$

x is replaced by its expression, brackets included.

Q37. $7(3 - 2y)$ equals

Beginner
A $21 - 14y$ B $21 - 2y$ C $10 - 14y$ D $21 + 14y$
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$21 - 14y$

Multiply both terms inside the bracket by 7.

Q38. You must substitute the expression into

Beginner
A the other equation B the same equation C either equation D a new equation
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the other equation

The equation you rearranged would give an identity such as 3 = 3.

Q39. Solving $y = 2x$ and $x + y = 9$ by substitution gives

Beginner
A $x = 6,\ y = 3$ B $x = 4.5,\ y = 4.5$ C $x = 9,\ y = 18$ D $x = 3,\ y = 6$
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$x = 3,\ y = 6$

$x + 2x = 9$ gives $3x = 9$, so $x = 3$ and $y = 6$.

Q40. Which variable is easiest to make the subject in $3x + 4y = 10$ and $x - 2y = 1$?

Beginner
A y, from the first equation B x, from the first equation C y, from the second equation D x, from the second equation
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x, from the second equation

Its coefficient is 1, so $x = 1 + 2y$ brings no fractions.

Q41. After substituting, the variables cancel and you are left with $5 = 5$. The pair has

Beginner
A no solution B infinitely many solutions C exactly one solution D exactly two solutions
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infinitely many solutions

A true statement means the equations are equivalent.

Q42. After substituting, you are left with $4 = 9$. The pair has

Beginner
A infinitely many solutions B one solution C two solutions D no solution
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no solution

A false statement means no values can satisfy both equations.

Q43. Where should the final answer be checked?

Beginner
A in the rearranged equation B in both original equations C in one equation only D on the graph
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in both original equations

The rearranged equation cannot fail, so it proves nothing.

Q44. Assertion (A): Substituting $x = 3 - 2y$ back into $x + 2y = 3$ is useless. Reason (R): It gives the identity $3 = 3$, which contains no new information.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

The rearranged equation is the same equation, so nothing new can come out of it.

Q45. Making y the subject of $5x - y = 12$ gives

Beginner
A $y = 12 - 5x$ B $y = 5x + 12$ C $y = \frac{12}{5x}$ D $y = 5x - 12$
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$y = 5x - 12$

Move y to the right and 12 to the left: $5x - 12 = y$.

Q46. Which line of working shows Step 2 of the method?

Beginner
A adding the two equations B replacing x by its expression in the second equation C subtracting the two equations D plotting both lines
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replacing x by its expression in the second equation

Step 2 is the substitution itself; adding or subtracting belongs to elimination.

Q47. Solve by substitution: $x + y = 14$ and $x - y = 4$.

Beginner
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Q48. The sum of two numbers is 45 and one is four times the other. Find them by substitution.

Beginner
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Q49. The first step of the elimination method is to

Beginner
A make one variable the subject B multiply so that one variable's coefficients become equal C draw both lines D divide the equations
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multiply so that one variable's coefficients become equal

Only matching coefficients allows a variable to cancel.

Q50. If the matched terms are $+5y$ and $+5y$, you should

Beginner
A subtract the equations B add the equations C multiply the equations D divide the equations
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subtract the equations

Same Sign → Subtract: $5y - 5y = 0$.

Q51. If the matched terms are $+4x$ and $-4x$, you should

Beginner
A subtract the equations B multiply by 4 C swap the equations D add the equations
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add the equations

Opposite signs cancel on addition.

Q52. Multiplying $2x + 3y = 13$ by 3 gives

Beginner
A $6x + 9y = 13$ B $6x + 3y = 39$ C $6x + 9y = 39$ D $2x + 9y = 39$
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$6x + 9y = 39$

Every term, including the constant, is multiplied.

Q53. Adding $x + y = 7$ and $x - y = 3$ gives

Beginner
A $2y = 10$ B $2x = 4$ C $2x = 10$ D $x = 10$
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$2x = 10$

The y-terms are $+y$ and $-y$, so they cancel on addition.

Q54. Solving $x + y = 7$ and $x - y = 3$ gives

Beginner
A $x = 2,\ y = 5$ B $x = 5,\ y = 2$ C $x = 4,\ y = 3$ D $x = 3,\ y = 4$
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$x = 5,\ y = 2$

Adding gives $2x = 10$, so $x = 5$ and then $y = 2$.

Q55. To eliminate y from $2x + 3y = 13$ and $5x - 2y = 4$, the multipliers are

Beginner
A 2 and 3 B 3 and 2 C 5 and 2 D 1 and 1
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2 and 3

The LCM of 3 and 2 is 6, so multiply the first by 2 and the second by 3.

Q56. Reversing the digits of the number $10x + y$ produces

Beginner
A $10x + y$ B $x + y$ C $xy$ D $10y + x$
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$10y + x$

After reversing, y sits in the tens place and x in the units place.

Q57. Elimination leaves $0 = 0$. The pair has

Beginner
A no solution B one solution C infinitely many solutions D two solutions
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infinitely many solutions

A true statement means the two equations are equivalent.

Q58. Elimination leaves $0 = 5$. The pair has

Beginner
A infinitely many solutions B no solution C one solution D five solutions
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no solution

A false statement can never be satisfied.

Q59. Assertion (A): An equation may be multiplied throughout by 3 during elimination. Reason (R): Multiplying every term by a non-zero number gives an equivalent equation with the same solutions.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

That equivalence is exactly what makes Step 1 legitimate.

Q60. Subtracting $x - y = 2$ from $x + y = 8$ gives

Beginner
A $2x = 6$ B $2y = 10$ C $2y = 6$ D $y = 6$
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$2y = 6$

The x-terms cancel and $8 - 2 = 6$ remains on the right.

Q61. Which is NOT a step of the elimination method?

Beginner
A multiplying an equation by a constant B adding the two equations C making one variable the subject of a formula D substituting back at the end
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making one variable the subject of a formula

Making a variable the subject belongs to substitution; elimination matches coefficients instead.

Building up · 86 questions

Q62. Two pencils and three erasers cost ₹9. With ₹x per pencil and ₹y per eraser, the equation is

Intermediate
A $3x + 2y = 9$ B $2x + 3y = 18$ C $2x + 3y = 9$ D $x + y = 9$
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$2x + 3y = 9$

Two lots of x and three lots of y come to 9.

Q63. 'The number of skirts is two less than twice the number of pants' becomes

Intermediate
A $y = 2x + 2$ B $y = 2(x - 2)$ C $y = 2x - 2$ D $x = 2y - 2$
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$y = 2x - 2$

Twice the pants is $2x$; two less is $2x - 2$.

Q64. The lines $y = 2x - 2$ and $y = 4x - 4$ meet at

Intermediate
A $(0, 1)$ B $(2, 2)$ C $(1, 0)$ D $(-1, 0)$
Show answer & explanation

$(1, 0)$

$2x - 2 = 4x - 4$ gives $2x = 2$, so $x = 1$ and $y = 0$.

Q65. The solution of $x + 3y = 6$ and $2x - 3y = 12$ is

Intermediate
A $x = 0,\ y = 2$ B $x = 6,\ y = 0$ C $x = 3,\ y = 1$ D $x = 6,\ y = 2$
Show answer & explanation

$x = 6,\ y = 0$

Adding gives $3x = 18$, so $x = 6$ and then $y = 0$.

Q66. Which pair of points can be used to draw $3x + 4y = 12$?

Intermediate
A $(0, 4)$ and $(3, 0)$ B $(0, 3)$ and $(4, 0)$ C $(0, 12)$ and $(12, 0)$ D $(3, 4)$ and $(4, 3)$
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$(0, 3)$ and $(4, 0)$

$x = 0$ gives $y = 3$; $y = 0$ gives $x = 4$.

Q67. A father is 30 years older than his daughter. With ages x and y, the equation is

Intermediate
A $x + y = 30$ B $x = 30y$ C $y - x = 30$ D $x - y = 30$
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$x - y = 30$

Taking x as the father's age, his age minus hers is 30.

Q68. If $(2, 3)$ is a solution of $ax + 3y = 13$, then a is

Intermediate
A 2 B 3 C 1 D 4
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2

$2a + 9 = 13$ gives $2a = 4$, so $a = 2$.

Q69. The pair $2x + 3y = 9$ and $4x + 6y = 18$ represents

Intermediate
A two parallel lines B two crossing lines C a line and a curve D the same line drawn twice
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the same line drawn twice

The second equation is twice the first, so both draw the same line.

Q70. Which of these pairs is inconsistent?

Intermediate
A $x + 2y = 4$ and $2x + 4y = 8$ B $x + 2y = 4$ and $2x + 4y = 12$ C $x - 2y = 0$ and $3x + 4y = 20$ D $x + y = 5$ and $x - y = 1$
Show answer & explanation

$x + 2y = 4$ and $2x + 4y = 12$

Doubling the first gives $2x + 4y = 8$, which contradicts 12 — parallel lines, no solution.

Q71. The solution of $x + y = 5$ and $x - y = 1$ is

Intermediate
A $x = 3,\ y = 2$ B $x = 2,\ y = 3$ C $x = 4,\ y = 1$ D $x = 1,\ y = 4$
Show answer & explanation

$x = 3,\ y = 2$

Adding gives $2x = 6$, so $x = 3$ and $y = 2$.

Q72. A line passes through $(0, 5)$ and $(5, 0)$. Its equation is

Intermediate
A $x - y = 5$ B $x + y = 5$ C $x + y = 0$ D $5x + 5y = 1$
Show answer & explanation

$x + y = 5$

Both points satisfy $x + y = 5$.

Q73. If a pair of lines coincides, the pair of equations is

Intermediate
A inconsistent B consistent with one solution C dependent and consistent D not a pair at all
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dependent and consistent

Equivalent equations give infinitely many solutions, and a dependent pair is always consistent.

Q74. Which situation needs a PAIR of equations?

Intermediate
A Finding the cost of a pencil and of an eraser from two bills B Finding one number whose double is 10 C Finding the area of a rectangle with known sides D Finding 15% of 200
Show answer & explanation

Finding the cost of a pencil and of an eraser from two bills

Two unknown prices need two facts about them.

Q75. Assertion (A): The pair $x + y = 5$ and $2x + 2y = 10$ has exactly one solution. Reason (R): The second equation is twice the first, so both represent the same line.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is false but R is true

There are infinitely many, so the assertion is false. The reason explains why: the two equations describe the same line.

Q76. Assertion (A): The point $(2, 3)$ is a solution of the pair $x + y = 5$ and $x - y = 1$. Reason (R): A solution of a pair must satisfy both equations.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true. A is false: $2 + 3 = 5$ ✓ but $2 - 3 = -1 \neq 1$. The solution is $(3, 2)$.

Q77. The equations $x + y = 6$ and $x + y = 9$ represent

Intermediate
A two parallel lines B the same line C two crossing lines D a line and a point
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two parallel lines

The same expression cannot equal both 6 and 9, so the lines never meet.

Q78. A pair of equations is formed from two facts about a situation. If the second fact repeats the first, the pair will have

Intermediate
A no solution B infinitely many solutions C exactly one solution D exactly two solutions
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infinitely many solutions

A repeated fact gives an equivalent equation, so both lines coincide.

Q79. A two-digit number has tens digit x and units digit y. The number is

Intermediate
A $10x + y$ B $x + y$ C $10y + x$ D $xy$
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$10x + y$

The tens digit is worth ten times its face value.

Q80. Solve graphically (describing the two lines): $x + y = 7$ and $x - y = 1$.

Intermediate
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Q81. Champa buys two fewer skirts than twice the number of pairs of trousers, and also four fewer than four times that number. Form the equations and solve them.

Intermediate
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Q82. Show that the pair $2x + 3y = 9$ and $4x + 6y = 18$ has infinitely many solutions.

Intermediate
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Q83. Show that the pair $x + 2y = 4$ and $2x + 4y = 12$ has no solution.

Intermediate
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Q84. Explain why the graphical method is unsuitable for $7x - 15y = 2$ and $x + 2y = 3$.

Intermediate
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Q85. (a) Write the general form of a pair of linear equations in two variables and say what a solution of the pair means. (b) Solve graphically (describing the lines): $x + 3y = 6$ and $2x - 3y = 12$. (c) Verify the solution in both equations and state whether the pair is consistent.

Intermediate
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1 mark — (a) general form. 1 mark — (a) meaning of a solution. 1 mark — (b) the two tables of points. 1 mark — (b) crossing point (6, 0). 1 mark — (c) verification and 'consistent'.

Q86. A shopkeeper sells 2 pens and 3 notebooks for ₹80, and 3 pens and 5 notebooks for ₹130. (a) Form a pair of linear equations, stating clearly what your letters mean. (b) Solve the pair. (c) Find the cost of 1 pen and 2 notebooks together.

Intermediate
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1½ marks — (a) letters named and both equations. 2 marks — (b) solving to x = 10, y = 20. 1½ marks — (c) ₹50.

Q87. The pair $x - 2y = 0$ and $3x + 4y - 20 = 0$ has

Intermediate
A no solution B infinitely many solutions C exactly one solution D exactly two solutions
Show answer & explanation

exactly one solution

$\frac13 \neq \frac{-2}{4}$, so the lines intersect once.

Q88. The pair $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$ has

Intermediate
A no solution B one solution C infinitely many solutions D exactly two solutions
Show answer & explanation

no solution

$\frac12 = \frac24$ but $\frac{-4}{-12} = \frac13$ — parallel lines.

Q89. For $3x + 2y = 5$ and $6x + 4y = 11$, the number of solutions is

Intermediate
A 1 B 0 C infinitely many D 2
Show answer & explanation

0

$\frac36 = \frac24 = \frac12$ but $\frac{-5}{-11} \neq \frac12$ — no solution.

Q90. For $2x + 3y - 8 = 0$ and $4x + 6y - 16 = 0$, the lines are

Intermediate
A parallel B intersecting C perpendicular D coincident
Show answer & explanation

coincident

All three ratios equal ½, so the two equations describe the same line.

Q91. For which value of k does $2x + ky = 5$ and $4x + 6y = 11$ have a unique solution?

Intermediate
A $k = 3$ B $k = 6$ C $k \neq 3$ D $k \neq 6$
Show answer & explanation

$k \neq 3$

A unique solution needs $\frac24 \neq \frac{k}{6}$, i.e. $k \neq 3$.

Q92. For which k does $x + 2y = 3$ and $5x + ky + 7 = 0$ have no solution?

Intermediate
A 5 B 2 C −10 D 10
Show answer & explanation

10

Need $\frac15 = \frac{2}{k}$, so $k = 10$; then $\frac{-3}{7} \neq \frac15$ ✓ no solution.

Q93. If a pair of equations is consistent with a unique solution, then

Intermediate
A $a_1b_2 = a_2b_1$ B $c_1 = c_2$ C $a_1b_2 \neq a_2b_1$ D $a_1 = b_1$
Show answer & explanation

$a_1b_2 \neq a_2b_1$

Cross-multiplying $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ gives exactly this.

Q94. The pair $5x - 8y + 1 = 0$ and $3x - \frac{24}{5}y + \frac{3}{5} = 0$ has

Intermediate
A infinitely many solutions B no solution C exactly one solution D exactly two solutions
Show answer & explanation

infinitely many solutions

Multiplying the second by $\frac53$ gives the first, so all three ratios agree.

Q95. Two lines have the same slope but different intercepts. The pair is

Intermediate
A consistent B dependent C unique D inconsistent
Show answer & explanation

inconsistent

Same slope and different position means parallel, so there is no solution.

Q96. For $9x - 4y = 2000$ and $7x - 3y = 2000$, the number of solutions is

Intermediate
A 0 B infinitely many C 1 D 2
Show answer & explanation

1

$\frac97 \neq \frac{-4}{-3}$, so the lines cross exactly once.

Q97. Which pair represents intersecting lines?

Intermediate
A $x + y = 5$ and $2x + 2y = 10$ B $x + y = 5$ and $x - y = 1$ C $x + y = 5$ and $x + y = 9$ D $2x + 4y = 6$ and $x + 2y = 3$
Show answer & explanation

$x + y = 5$ and $x - y = 1$

Only the first has different a-to-b ratios: $\frac11 \neq \frac{1}{-1}$.

Q98. Assertion (A): The pair $x + y = 3$ and $2x + 2y = 7$ is inconsistent. Reason (R): All three of the coefficient ratios are equal.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is true but R is false

The third ratio differs, which is exactly what makes the pair inconsistent rather than coincident. The assertion stands, the reason is false.

Q99. Assertion (A): If $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, the pair has infinitely many solutions. Reason (R): Equal ratios of a and b mean the lines have the same slope.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true, but A is false: with the same slope the lines may also be parallel. The c ratio decides.

Q100. A pair of equations that is consistent may have

Intermediate
A only one solution B no solution C exactly two solutions D one or infinitely many solutions
Show answer & explanation

one or infinitely many solutions

Consistent means at least one solution, which covers both cases.

Q101. Which pair of lines is coincident?

Intermediate
A $x - y = 2$ and $3x - 3y = 9$ B $x - y = 2$ and $3x - 3y = 6$ C $x - y = 2$ and $x + y = 2$ D $x - y = 2$ and $2x - y = 4$
Show answer & explanation

$x - y = 2$ and $3x - 3y = 6$

Tripling the first gives $3x - 3y = 6$ exactly, so both equations describe the same line.

Q102. If both equations of a pair are multiplied by 3, the number of solutions

Intermediate
A does not change B becomes three times as many C becomes zero D becomes infinite
Show answer & explanation

does not change

Multiplying an equation by a non-zero number gives an equivalent equation, so the ratios and the lines are unchanged.

Q103. For $x + y = 5$ and $3x + 3y = 15$, the value of $\frac{c_1}{c_2}$ is

Intermediate
A 3 B $-\frac13$ C $\frac13$ D 5
Show answer & explanation

$\frac13$

In standard form the constants are −5 and −15, so the ratio is $\frac{-5}{-15} = \frac13$, matching the other two ratios.

Q104. Assertion (A): For $4x + 6y = 12$ and $2x + 3y = 6$ the lines coincide. Reason (R): Dividing the first equation by 2 gives the second.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

The two equations are equivalent, so they draw the same line and the pair is dependent.

Q105. Find the value of k for which $2x + ky = 5$ and $4x + 6y = 11$ have no solution.

Intermediate
Show answer & explanation

—

Q106. The pair $3x + 2y = 5$ and $6x + 4y = 11$ is given. Decide the number of solutions and explain what the graph looks like.

Intermediate
Show answer & explanation

—

Q107. Why is it not enough to check only $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ before answering 'infinitely many solutions'?

Intermediate
Show answer & explanation

—

Q108. Find a value of c for which $2x + 3y = 7$ and $4x + 6y = c$ has (i) infinitely many solutions (ii) no solution.

Intermediate
Show answer & explanation

—

Q109. (a) State the three conditions on the ratios of coefficients, with the picture and the number of solutions in each case. (b) Apply them to: (i) $x - 2y = 0$, $3x + 4y = 20$ (ii) $2x + 3y = 9$, $4x + 6y = 18$ (iii) $x + 2y = 4$, $2x + 4y = 12$.

Intermediate
Show answer & explanation

—

1½ marks — (a) the three conditions. ½ mark — (a) the pictures and names. 1 mark each — (b) (i), (ii), (iii) with the ratios shown.

Q110. Two friends compare mobile plans. Plan A costs ₹x fixed plus ₹y per GB and Plan B is described by the equation $2x + 6y = 700$; Plan A's bill for 3 GB is given by $x + 3y = 350$. (a) Write both equations in standard form and find the three ratios. (b) How many solutions does the pair have, and what does that mean about the two descriptions? (c) Give one change to the second equation that would make the pair inconsistent, and say what that would mean.

Intermediate
Show answer & explanation

—

1½ marks — (a) standard form and ratios. 1½ marks — (b) infinitely many, with the meaning. 2 marks — (c) a valid change and its meaning.

Q111. Solving $x + y = 14$ and $x - y = 4$ by substitution gives

Intermediate
A $x = 5,\ y = 9$ B $x = 10,\ y = 4$ C $x = 7,\ y = 7$ D $x = 9,\ y = 5$
Show answer & explanation

$x = 9,\ y = 5$

$x = 14 - y$, so $14 - 2y = 4$, giving $y = 5$ and $x = 9$.

Q112. The solution of $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$ is

Intermediate
A $s = 6,\ t = 3$ B $s = 9,\ t = 6$ C $s = 12,\ t = 9$ D $s = 3,\ t = 0$
Show answer & explanation

$s = 9,\ t = 6$

$s = 3 + t$, so $\frac{3+t}{3} + \frac{t}{2} = 6$; multiplying by 6: $2(3+t) + 3t = 36$, giving $5t = 30$, $t = 6$ and $s = 9$.

Q113. Solving $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$ gives

Intermediate
A $x = 3,\ y = 2$ B $x = 2,\ y = 3$ C $x = 1,\ y = 3$ D $x = 2,\ y = 1$
Show answer & explanation

$x = 2,\ y = 3$

Multiplying by 10: $2x + 3y = 13$ and $4x + 5y = 23$. Then $x = \frac{13 - 3y}{2}$ leads to $y = 3$, $x = 2$.

Q114. For $2x + 3y = 9$ and $4x + 6y = 18$, substitution ends with

Intermediate
A a true statement, so infinitely many solutions B a false statement, so no solution C $y = 0$ D $x = 0$
Show answer & explanation

a true statement, so infinitely many solutions

The second equation is twice the first, so both variables cancel and $18 = 18$ remains.

Q115. For $2x + 3y = 9$ and $4x + 6y = 20$, substitution ends with

Intermediate
A a true statement, so infinitely many B $y = 1$ C $x = 2$ D a false statement, so no solution
Show answer & explanation

a false statement, so no solution

The working leaves $18 = 20$, which is never true.

Q116. Aftab's ages give $x - 7y = -42$ and $x - 3y = 6$. His present age is

Intermediate
A 12 B 42 C 35 D 45
Show answer & explanation

42

$x = 6 + 3y$ gives $-4y = -48$, so $y = 12$ and $x = 42$.

Q117. A two-digit number has digits adding to 9, and it stands 27 above the number its digits make in reverse. The number is

Intermediate
A 36 B 63 C 72 D 54
Show answer & explanation

63

$x + y = 9$ and $x - y = 3$ give $x = 6$, $y = 3$, so the number is 63.

Q118. If $x = 2y$ and $3x - y = 10$, then y is

Intermediate
A 4 B 5 C 2 D $\frac{10}{3}$
Show answer & explanation

2

$3(2y) - y = 10$ gives $5y = 10$, so $y = 2$ (and $x = 4$).

Q119. A number is three times another and their sum is 48. The smaller number is

Intermediate
A 16 B 24 C 12 D 36
Show answer & explanation

12

With $x = 3y$ and $x + y = 48$: $4y = 48$, so $y = 12$.

Q120. Which pair is best suited to substitution?

Intermediate
A $3x + 4y = 7$ and $5x - 6y = 2$ B $7x + 9y = 11$ and $4x - 5y = 3$ C $6x + 5y = 1$ and $9x - 7y = 4$ D $y = 3x - 2$ and $2x + y = 8$
Show answer & explanation

$y = 3x - 2$ and $2x + y = 8$

y is already the subject in the first equation, so substitution costs nothing.

Q121. Solving $y = 3x - 2$ and $2x + y = 8$ gives

Intermediate
A $x = 4,\ y = 2$ B $x = 1,\ y = 1$ C $x = 2,\ y = 4$ D $x = 3,\ y = 7$
Show answer & explanation

$x = 2,\ y = 4$

$2x + 3x - 2 = 8$ gives $5x = 10$, so $x = 2$ and $y = 4$.

Q122. Assertion (A): A fractional answer such as $x = \frac{49}{29}$ means a mistake has been made. Reason (R): The solution of a pair of linear equations is always a whole number.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

Both are false as stated; the correct choice is that A is false — fractional solutions are perfectly normal, as NCERT's own Example 4 shows.

Q123. Assertion (A): In the substitution method, the expression must be put into the equation that was NOT rearranged. Reason (R): Substituting into the rearranged equation gives an identity.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

R explains A exactly: the rearranged equation cannot produce new information.

Q124. Solving $x - y = 1$ and $2x + y = 8$ by substitution gives

Intermediate
A $x = 2,\ y = 3$ B $x = 1,\ y = 0$ C $x = 4,\ y = 3$ D $x = 3,\ y = 2$
Show answer & explanation

$x = 3,\ y = 2$

$x = 1 + y$, so $2(1 + y) + y = 8$ gives $3y = 6$, $y = 2$ and $x = 3$.

Q125. For $x + y = 6$ and $2x + 2y = k$, the pair has NO solution when k is

Intermediate
A 12 B 8 C 6 D 3
Show answer & explanation

8

$k = 12$ would make the equations identical; any other value, such as 8, leaves a false statement.

Q126. Five pens and three pencils cost ₹44, and one pen costs ₹4 more than one pencil. The cost of a pen is

Intermediate
A ₹3 B ₹7 C ₹5 D ₹9
Show answer & explanation

₹7

Let a pencil cost ₹y, so a pen costs ₹(y + 4). Then $5(y + 4) + 3y = 44$ gives $8y = 24$, so $y = 3$ and a pen costs ₹7.

Q127. The sum of two numbers is 35 and their difference is 13. The larger number is

Intermediate
A 22 B 11 C 24 D 13
Show answer & explanation

24

$x = 35 - y$, so $35 - 2y = 13$ gives $y = 11$ and $x = 24$.

Q128. Solve by substitution: $3x - y = 3$ and $9x - 3y = 9$.

Intermediate
Show answer & explanation

—

Q129. Solve by substitution: $x + 2y = 5$ and $2x + 4y = 15$.

Intermediate
Show answer & explanation

—

Q130. Solve by substitution: $7x - 15y = 2$ and $x + 2y = 3$.

Intermediate
Show answer & explanation

—

Q131. Solve by substitution: $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$.

Intermediate
Show answer & explanation

—

Q132. The larger of two supplementary angles exceeds the smaller by 18°. Find both angles by substitution.

Intermediate
Show answer & explanation

—

Q133. Explain what it means if, while solving by substitution, both variables disappear and you are left with $0 = 0$.

Intermediate
Show answer & explanation

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Q134. (a) Write the three steps of the substitution method. (b) Use it to solve $2x + 3y = 11$ and $2x - 4y = -24$. (c) Verify your answer in both original equations.

Intermediate
Show answer & explanation

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1½ marks — (a) the three steps. 2½ marks — (b) working to x = −2, y = 5. 1 mark — (c) both checks.

Q135. (a) Solve by substitution: $3x - y = 3$, $9x - 3y = 9$, and say what your result means. (b) Solve by substitution: $x + 2y = 5$, $2x + 4y = 15$, and say what your result means. (c) What picture do the lines make in each case?

Intermediate
Show answer & explanation

—

2 marks — (a) working and meaning. 2 marks — (b) working and meaning. 1 mark — (c) both pictures.

Q136. Solving $x + y = 5$ and $2x - 3y = 4$ by elimination gives

Intermediate
A $x = \frac{19}{5},\ y = \frac{6}{5}$ B $x = 3,\ y = 2$ C $x = 2,\ y = 3$ D $x = 4,\ y = 1$
Show answer & explanation

$x = \frac{19}{5},\ y = \frac{6}{5}$

Multiplying the first by 3 and adding gives $5x = 19$, so $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

Q137. Solving $3x + 4y = 10$ and $2x - 2y = 2$ gives

Intermediate
A $x = 2,\ y = 1$ B $x = 1,\ y = 2$ C $x = 3,\ y = 2$ D $x = 2,\ y = 2$
Show answer & explanation

$x = 2,\ y = 1$

Doubling the second gives $4x - 4y = 4$; adding removes y and leaves $7x = 14$.

Q138. Solving $3x - 5y = 4$ and $9x = 2y + 7$ gives

Intermediate
A $x = \frac{9}{13},\ y = -\frac{5}{13}$ B $x = 1,\ y = 1$ C $x = \frac{5}{13},\ y = \frac{9}{13}$ D $x = -1,\ y = -1$
Show answer & explanation

$x = \frac{9}{13},\ y = -\frac{5}{13}$

Writing the second as $9x - 2y = 7$ and tripling the first gives $9x - 15y = 12$; subtracting leaves $13y = -5$.

Q139. Two people's incomes are in the ratio 9 : 7 and their spendings in the ratio 4 : 3, each saving ₹2000 a month. Their incomes are

Intermediate
A ₹9,000 and ₹7,000 B ₹18,000 and ₹14,000 C ₹36,000 and ₹28,000 D ₹2,000 and ₹2,000
Show answer & explanation

₹18,000 and ₹14,000

$9x - 4y = 2000$ and $7x - 3y = 2000$ give $x = 2000$, so the incomes are $9x$ and $7x$.

Q140. Adding a two-digit number to the number its digits make in reverse gives 66, and the two digits are 2 apart. One such number is

Intermediate
A 31 B 45 C 57 D 24
Show answer & explanation

24

$11(x + y) = 66$ gives $x + y = 6$, and with $y - x = 2$ the digits are 2 and 4.

Q141. Five years ago a father was seven times as old as his son; in five years he will be three times as old. The son's present age is

Intermediate
A 12 years B 15 years C 10 years D 8 years
Show answer & explanation

10 years

$x - 5 = 7(y - 5)$ and $x + 5 = 3(y + 5)$ give $y = 10$ and $x = 40$.

Q142. Which pair is better suited to elimination than to substitution?

Intermediate
A $y = 2x + 1$ and $3x + y = 6$ B $x = 5 - y$ and $2x + 3y = 8$ C $x + y = 4$ and $x - y = 2$ D $7x + 9y = 11$ and $4x - 5y = 3$
Show answer & explanation

$7x + 9y = 11$ and $4x - 5y = 3$

No coefficient is 1, so making a variable the subject would bring fractions immediately.

Q143. Half the perimeter of a rectangular garden is 36 m and its length is 4 m more than its width. Its dimensions are

Intermediate
A 18 m by 18 m B 22 m by 14 m C 24 m by 12 m D 20 m by 16 m
Show answer & explanation

20 m by 16 m

$x + y = 36$ and $x - y = 4$ give $x = 20$ and $y = 16$.

Q144. Subtracting $(5x + 3y = 13)$ from $(5x - 2y = 4)$ gives

Intermediate
A $-5y = 9$ B $-5y = -9$ C $y = 9$ D $5y = -9$
Show answer & explanation

$-5y = -9$

$(-2y) - (3y) = -5y$ and $4 - 13 = -9$; the constant's sign changes too.

Q145. Two kilograms of apples and one of grapes cost ₹160, while four kilograms of apples and two of grapes cost ₹300. The pair has

Intermediate
A one solution B infinitely many solutions C no solution D exactly two solutions
Show answer & explanation

no solution

Doubling the first bill gives ₹320, not ₹300, so elimination leaves a false statement.

Q146. Assertion (A): In elimination you add the equations when the matched coefficients have the same sign. Reason (R): Adding two equal positive terms gives zero.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

Both statements are false: with the same sign you SUBTRACT, and adding equal positive terms doubles them rather than cancelling.

Q147. Assertion (A): Elimination and substitution give the same solution for any pair. Reason (R): Both methods only ever replace an equation by an equivalent one.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Each step preserves the solution set, so the two routes must end in the same place.

Topper level · 53 questions

Q148. For what value of k do the lines $x + 2y = 3$ and $2x + ky = 6$ coincide?

Advanced
A 2 B 3 C 6 D 4
Show answer & explanation

4

Doubling the first gives $2x + 4y = 6$, so $k = 4$ makes the equations identical.

Q149. The lines $x = 3$ and $y = -2$ meet at

Advanced
A $(-2, 3)$ B $(3, 2)$ C $(3, -2)$ D they do not meet
Show answer & explanation

$(3, -2)$

$x = 3$ is a vertical line and $y = -2$ a horizontal one; they cross at $(3, -2)$.

Q150. A pair of equations has the solution $(4, 1)$. Which pair could it be?

Advanced
A $x + y = 5$ and $x - y = 1$ B $x + y = 4$ and $x - y = 1$ C $x + 2y = 5$ and $x - y = 3$ D $x + y = 5$ and $x - y = 3$
Show answer & explanation

$x + y = 5$ and $x - y = 3$

$4 + 1 = 5$ ✓ and $4 - 1 = 3$ ✓ Only the first pair holds for both.

Q151. The cost of 3 kg apples and 2 kg grapes is ₹180, and of 2 kg apples and 4 kg grapes is ₹200. Taking prices x and y per kg, the pair is

Advanced
A $2x + 3y = 180$, $4x + 2y = 200$ B $3x + 2y = 200$, $2x + 4y = 180$ C $3x + 2y = 180$, $2x + 4y = 200$ D $x + y = 180$, $x + y = 200$
Show answer & explanation

$3x + 2y = 180$, $2x + 4y = 200$

Each bill is one equation, with the quantities as coefficients.

Q152. Two lines meet at $(2, 5)$. How many pairs of linear equations have this solution?

Advanced
A Exactly one B Infinitely many C Exactly two D None
Show answer & explanation

Infinitely many

Any two different lines through $(2, 5)$ give such a pair, and there are infinitely many lines through a point.

Q153. If a pair of lines has exactly one common point, which is NOT possible?

Advanced
A The equations are equivalent B The pair is consistent C The lines have different slopes D The solution is unique
Show answer & explanation

The equations are equivalent

Equivalent equations draw the same line, which shares every point, not one.

Q154. Assertion (A): Two distinct straight lines can meet at exactly two points. Reason (R): Two points determine a unique straight line.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true, and it is precisely why A is false: lines sharing two points must be the same line.

Q155. A boat's speed downstream and upstream gives the pair $x + y = 10$ and $x - y = 6$, where x is the boat's speed and y the stream's. The stream's speed is

Advanced
A 2 km/h B 8 km/h C 4 km/h D 16 km/h
Show answer & explanation

2 km/h

Subtracting gives $2y = 4$, so $y = 2$ (and $x = 8$).

Q156. Which pair of equations is NOT a pair of linear equations in two variables?

Advanced
A $x + y = 6$ and $x - y = 2$ B $xy = 6$ and $x + y = 5$ C $2x + 3y = 1$ and $x = 4$ D $x = 2y$ and $3x + y = 7$
Show answer & explanation

$xy = 6$ and $x + y = 5$

$xy = 6$ multiplies the variables, so it is not linear.

Q157. If $(3, 2)$ is a solution of $2x + ky = 12$, find k and write one more solution of that equation.

Advanced
Show answer & explanation

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Q158. A father is three times as old as his son. Five years later he will be two and a half times as old. Form the pair of equations (do not solve).

Advanced
Show answer & explanation

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Q159. (a) Describe the three ways in which the lines of a pair of linear equations can be placed, and give the number of solutions in each case. (b) For each case, give one example of a pair of equations. (c) Which of your three pairs are consistent?

Advanced
Show answer & explanation

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1½ marks — (a) three cases with counts. 2 marks — (b) one valid example each. 1½ marks — (c) the two consistent pairs, with the reason.

Q160. The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later she buys another bat and 3 more balls of the same kind for ₹1300. (a) Represent this situation algebraically. (b) Find the cost of one bat and one ball by any method. (c) If she now buys 2 bats and 5 balls, what will she pay?

Advanced
Show answer & explanation

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1½ marks — (a) both equations with the letters named. 2 marks — (b) the solution, with the comment on the data. 1½ marks — (c) ₹2600.

Q161. For what value of k do $2x + 3y - 7 = 0$ and $(k-1)x + (k+2)y - 3k = 0$ have infinitely many solutions?

Advanced
A 3 B 5 C −7 D 7
Show answer & explanation

7

$2(k+2) = 3(k-1)$ gives $k = 7$, and then $\frac{2}{6} = \frac{-7}{-21}$ ✓

Q162. For which value of k does $kx + 2y = 5$ and $3x + y = 1$ have a unique solution?

Advanced
A $k \neq 6$ B $k = 6$ C $k \neq 3$ D $k = 3$
Show answer & explanation

$k \neq 6$

A unique solution needs $\frac{k}{3} \neq \frac21$, i.e. $k \neq 6$.

Q163. The pair $2x + 3y = 7$ and $(a - b)x + (a + b)y = 3a + b - 2$ has infinitely many solutions. Then

Advanced
A $a = 1,\ b = 5$ B $a = 5,\ b = -1$ C $a = 5,\ b = 1$ D $a = -5,\ b = 1$
Show answer & explanation

$a = 5,\ b = 1$

From $\frac{2}{a-b} = \frac{3}{a+b}$: $2a + 2b = 3a - 3b$, so $a = 5b$. The constant ratio then gives $b = 1$, $a = 5$; checking, the second equation becomes $4x + 6y = 14$ ✓

Q164. If the lines $3x + 2ky = 2$ and $2x + 5y + 1 = 0$ are parallel, then k is

Advanced
A $\frac{4}{15}$ B $\frac{5}{2}$ C $\frac{2}{5}$ D $\frac{15}{4}$
Show answer & explanation

$\frac{15}{4}$

Parallel needs $\frac32 = \frac{2k}{5}$, so $4k = 15$ and $k = \frac{15}{4}$. (The constants then differ, as required.)

Q165. A pair with $a_1b_2 - a_2b_1 = 0$ and $b_1c_2 - b_2c_1 \neq 0$ has

Advanced
A one solution B infinitely many solutions C no solution D exactly two solutions
Show answer & explanation

no solution

The first says the slopes match; the second says the constants do not — parallel lines.

Q166. Assertion (A): A pair of linear equations can have exactly two solutions. Reason (R): Two straight lines can cross at no point, one point, or infinitely many.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true and rules A out: exactly two crossings is impossible for straight lines.

Q167. For $x + 2y = 3$ and $2x + 4y = k$, the pair has infinitely many solutions when k is

Advanced
A 3 B 12 C 6 D any value
Show answer & explanation

6

All three ratios must agree: $\frac12 = \frac24 = \frac{-3}{-k}$ needs $k = 6$.

Q168. Which change to $x + y = 4$ and $2x + 2y = 8$ would make the pair inconsistent?

Advanced
A Change 4 to 2 B Change 2x to 3x C Change both signs D Change 8 to 9
Show answer & explanation

Change 8 to 9

With 9 the a and b ratios still match while the c ratio does not, giving parallel lines.

Q169. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, then the second equation is

Advanced
A half the first B twice the first C unrelated to the first D the first with signs changed
Show answer & explanation

twice the first

Each coefficient of the second is twice the matching one in the first.

Q170. Two equations with $a_1 = 2$, $b_1 = -3$, $a_2 = -4$, $b_2 = 6$ represent lines that are

Advanced
A always intersecting B parallel or coincident C always perpendicular D always coincident
Show answer & explanation

parallel or coincident

$\frac{2}{-4} = \frac{-3}{6} = -\frac12$, so the slopes match; only the constants can tell parallel from coincident.

Q171. Which of these pairs is consistent?

Advanced
A $x + y = 2$ and $x + y = 7$ B $2x + y = 5$ and $3x - y = 5$ C $3x + 6y = 9$ and $x + 2y = 5$ D $2x - 4y = 3$ and $x - 2y = 7$
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$2x + y = 5$ and $3x - y = 5$

Only the first has unequal a-to-b ratios, so its lines cross; the other three are parallel pairs.

Q172. For what value of k does $kx + 3y = k - 3$ and $12x + ky = k$ have infinitely many solutions?

Advanced
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Q173. Two lines are given by $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ with $a_2 = 0$. Explain how to test for a unique solution without dividing by zero.

Advanced
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Q174. (a) Find the value of k for which the pair $3x + y = 1$ and $(2k - 1)x + (k - 1)y = 2k + 1$ has no solution. (b) Show that for this k the lines are parallel and not coincident. (c) For what values of k does the pair have a unique solution?

Advanced
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2 marks — (a) the condition and k = 2. 1½ marks — (b) the ratios shown for k = 2. 1½ marks — (c) k ≠ 2, with the reasoning.

Q175. (a) For what value of k does $x + 2y = 3$ and $5x + ky + 7 = 0$ have no solution? (b) For that k, write both equations in standard form and give all three ratios. (c) Is there any value of k for which this pair has infinitely many solutions? Justify your answer.

Advanced
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1½ marks — (a) k = 10. 1½ marks — (b) the three ratios. 2 marks — (c) 'no', with the reason that c is fixed.

Q176. If $2x + 3y = 11$ and $2x - 4y = -24$, then the value of $y$ is

Advanced
A −5 B 2 C 5 D 3
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5

Subtracting the equations gives $7y = 35$, so $y = 5$; by substitution $2x = 11 - 15$, giving $x = -2$.

Q177. For the pair $x + y = 5$ and $2x + 2y = k$, substitution leaves a true statement when k is

Advanced
A 5 B 2 C any value D 10
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10

The second equation must be exactly twice the first, so $k = 10$.

Q178. If (x, y) is the solution of $3x + 2y = 11$ and $2x + 3y = 4$, then $x - y$ equals

Advanced
A 5 B −5 C 7 D 3
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7

Solving the pair gives $x = 5$ and $y = -2$, so $x - y = 5 - (-2) = 7$.

Q179. A fraction becomes $\frac{4}{5}$ if 1 is added to both numerator and denominator, and $\frac{1}{2}$ if 5 is subtracted from both. The fraction is

Advanced
A $\frac{3}{4}$ B $\frac{7}{9}$ C $\frac{9}{11}$ D $\frac{5}{7}$
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$\frac{7}{9}$

With numerator x and denominator y: $5(x+1) = 4(y+1)$ and $2(x-5) = y-5$. Solving gives $x = 7$, $y = 9$.

Q180. Two numbers differ by 26 and one is three times the other. The larger number is

Advanced
A 13 B 26 C 39 D 52
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39

With $x = 3y$ and $x - y = 26$: $2y = 26$, so $y = 13$ and $x = 39$.

Q181. The larger of two supplementary angles exceeds the smaller by 18°. The larger angle is

Advanced
A 81° B 108° C 99° D 90°
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99°

$x + y = 180$ and $x - y = 18$ give $x = 99$, $y = 81$.

Q182. Assertion (A): The pair $y = 2x + 1$ and $y = 2x + 5$ has no solution. Reason (R): Substituting gives $2x + 1 = 2x + 5$, i.e. $1 = 5$, which is false.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

The false statement is exactly the sign of parallel lines and no solution.

Q183. For which value of m does $y = mx + 3$ and $y = 2x + 3$ have infinitely many solutions?

Advanced
A 2 B 3 C 0 D −2
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2

The two right-hand sides must be identical, so $m = 2$.

Q184. The cost of 2 kg of apples and 1 kg of grapes is ₹160, and of 4 kg of apples and 2 kg of grapes is ₹300. The pair has

Advanced
A one solution B infinitely many solutions C no solution D two solutions
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no solution

Doubling the first gives ₹320, not ₹300, so the two statements contradict each other.

Q185. Assertion (A): Substitution works on any pair of linear equations in two variables. Reason (R): One variable can always be written in terms of the other when its coefficient is not zero.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

In every linear equation at least one coefficient is non-zero, so a variable can always be made the subject; R is exactly why A holds.

Q186. In solving $3x + y = 10$ and $x + y = 6$ by substitution, the cleanest first move is

Advanced
A $x = \frac{10 - y}{3}$ from the first B adding the two equations C $y = 6 - x$ from the second equation D $y = 10 - 3x$ then substituting into the first
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$y = 6 - x$ from the second equation

Both y-coefficients are 1, and taking $y = 6 - x$ keeps the arithmetic whole-numbered; substituting a rearranged equation back into itself does nothing.

Q187. Aftab is seven times as old as his daughter was seven years ago, and three times as old as she will be in three years. Find their present ages.

Advanced
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Q188. A fraction becomes $\frac{4}{5}$ when 1 is added to both its numerator and denominator, and $\frac{1}{2}$ when 5 is subtracted from both. Find the fraction.

Advanced
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Q189. A two-digit number is such that the sum of its digits is 9, and the number is 27 more than the number obtained by reversing its digits. (a) Taking the tens digit as x and the units digit as y, write both equations. (b) Solve them by substitution. (c) State the number and verify both conditions.

Advanced
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1½ marks — (a) both equations, with the simplification. 2 marks — (b) substitution and both digits. 1½ marks — (c) the number and both verifications.

Q190. A lending library charges a fixed amount for the first three days and an extra charge for each further day. Saritha paid ₹27 for a book kept for seven days, and Susy paid ₹21 for a book kept for five days. (a) Taking the fixed charge as ₹x and the charge per extra day as ₹y, write the two equations. (b) Solve them by substitution. (c) What would a book kept for ten days cost?

Advanced
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1½ marks — (a) both equations with the 'extra days' counted correctly. 2 marks — (b) x = 15, y = 3. 1½ marks — (c) ₹36.

Q191. Twice a number added to three times a second gives 92, while four times the first minus the second gives 16. The first number is

Advanced
A 24 B 12 C 8 D 10
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10

$2x + 3y = 92$ and $4x - y = 16$. Tripling the second and adding removes y: $14x = 140$.

Q192. A boat covers 30 km upstream and 44 km downstream in 10 hours, and 40 km upstream and 55 km downstream in 13 hours. The boat's speed in still water is

Advanced
A 3 km/h B 8 km/h C 11 km/h D 5 km/h
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8 km/h

With $p = \frac{1}{u-v}$ and $q = \frac{1}{u+v}$ the times give $p = \frac15$, $q = \frac1{11}$, so $u - v = 5$ and $u + v = 11$.

Q193. For what value of k do $2x + 3y = 7$ and $4x + 6y = k$ have infinitely many solutions?

Advanced
A 7 B 2 C 14 D no such value
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14

The second equation must be exactly twice the first, so $k = 14$.

Q194. Eliminating x from $ax + by = c$ and $a'x + b'y = c'$ requires multiplying by

Advanced
A $b$ and $b'$ respectively B $a'$ and $a$ respectively C $c$ and $c'$ respectively D 1 and 1
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$a'$ and $a$ respectively

Multiplying the first by $a'$ and the second by $a$ makes both x-coefficients $aa'$.

Q195. Two complementary angles differ by 20°. Eliminating gives the larger angle as

Advanced
A 35° B 55° C 70° D 45°
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55°

Complementary angles sum to 90°, so $x + y = 90$ and $x - y = 20$; adding gives $2x = 110$.

Q196. Five books and seven pens together cost ₹79, while seven books and five pens cost ₹77. The cost of one book is

Advanced
A ₹6 B ₹7 C ₹5 D ₹8
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₹6

Adding gives $12(x + y) = 156$, and subtracting gives $2(x - y) = -2$; so $x + y = 13$ and $y - x = 1$, making a book ₹6.

Q197. A two-digit number is four times the sum of its digits, and twice the product of its digits. The number is

Advanced
A 24 B 36 C 48 D 12
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36

$10x + y = 4(x + y)$ gives $2x = y$, and $10x + y = 2xy$ then gives $x = 3$, $y = 6$.

Q198. Assertion (A): For $3x + 2y = 8$ and $6x + 4y = 16$, elimination leaves $0 = 0$. Reason (R): The two equations represent parallel lines that never meet.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is true but R is false

They represent the same line, which is why elimination collapses to $0 = 0$ instead of a contradiction. The assertion holds, the reason is false.

Q199. In a 3-mark word problem, the mark most often lost is for

Advanced
A not drawing a graph B using elimination instead of substitution C not checking the discriminant D not answering in a sentence with units
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not answering in a sentence with units

The final statement carries its own mark, and 'x = 42' does not earn it.

Q200. The coach of a team buys 7 bats and 6 balls for ₹3800, and later 3 bats and 5 balls for ₹1750. The cost of one bat is

Advanced
A ₹500 B ₹50 C ₹350 D ₹300
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₹500

$7x + 6y = 3800$ and $3x + 5y = 1750$; eliminating y gives $17x = 8500$.

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