The two equations are:

$$2x + y = 320 \quad \text{...(1)}$$

$$4x + 2y = 600 \quad \text{...(2)}$$

Comparing ratios: $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{1}{2}$, $\dfrac{c_1}{c_2} = \dfrac{320}{600} = \dfrac{8}{15}$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the system is inconsistent (no solution).

Source: Chapter 3, Section 3.4 Summary

• Form both equations directly from the given information using x and y.
• Compare the three ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, $\frac{c_1}{c_2}$ to decide consistency.
• The middle condition ($a_1/a_2 = b_1/b_2 \neq c_1/c_2$) always means inconsistent (parallel lines, no solution). State this conclusion clearly for full marks.

Option C: $3x + y = 900$, $5x + 3y = 2100$

(3 chairs + 1 table = ₹900 gives $3x + y = 900$; 5 chairs + 3 tables = ₹2100 gives $5x + 3y = 2100$.)

Directly translate each statement: coefficient of $x$ = number of chairs, coefficient of $y$ = number of tables. Option C correctly places 3 and 1 in the first equation, and 5 and 3 in the second. Watch out for Option A which swaps the coefficients of $x$ and $y$ in the first equation.

Let the greater number be $x$ and the smaller number be $y$.

Forming equations:

Half the difference of two numbers is 2:
$$\frac{x - y}{2} = 2 \implies x - y = 4 \tag{1}$$

Sum of greater number and twice the smaller is 13:
$$x + 2y = 13 \tag{2}$$

Solving by substitution:

From (1): $x = y + 4$

Substituting in (2):
$$(y + 4) + 2y = 13$$
$$3y = 9 \implies y = 3$$

Substituting back: $x = 3 + 4 = 7$

The two numbers are 7 and 3.

Verification: $\frac{7-3}{2} = 2$ ✓ and $7 + 2(3) = 13$ ✓

Source: Chapter 3, Section 3.3.1 Substitution Method

---

• Examiners award 1 mark for correctly forming both equations, 1 mark for solving, and 1 mark for the correct answer/verification.
• Clearly define variables at the start — this is expected in board answers.
• "Half the difference" means $\frac{x-y}{2}$, not $\frac{x}{2} - y$; be careful with phrasing.
• Always verify your answer — it shows completeness and can earn the final mark.

Let the speeds of the two persons be x km/h and y km/h (x > y).

Setting up equations:

When walking towards each other, relative speed = (x + y) km/h.
They meet in 2 hours, covering 16 km:

$$x + y = \frac{16}{2} = 8 \quad \text{...(1)}$$

When walking in the same direction, relative speed = (x − y) km/h.
They meet in 8 hours, covering 16 km:

$$x - y = \frac{16}{8} = 2 \quad \text{...(2)}$$

Solving by elimination:

Adding (1) and (2):
$$2x = 10 \implies x = 5$$

Substituting in (1):
$$5 + y = 8 \implies y = 3$$

∴ The speeds of the two persons are 5 km/h and 3 km/h.

Source: Chapter 3, Pair of Linear Equations in Two Variables

---

• The key idea is relative speed: towards each other → add speeds; same direction → subtract speeds.
• Use Distance = Speed × Time to form both equations directly.
• Elimination method (adding/subtracting equations) is the quickest approach here.
• Examiners award marks for: correct equations (1 mark), correct solving (1 mark), correct answer with units (1 mark). Always state the final answer clearly.

(i) The situation algebraically:

$$x + y = 300$$
$$150x + 250y = 55000$$

(ii) From equation (i): $x = 300 - y$

Substituting in equation (ii):
$$150(300 - y) + 250y = 55000$$
$$45000 - 150y + 250y = 55000$$
$$100y = 10000 \implies y = 100$$

So $x = 300 - 100 = 200$

200 children and 100 adults visited the park that day.

(iii) Amount collected = $150 \times 250 + 250 \times 100$
$$= 37500 + 25000 = ₹62,500$$

Source: Case Study passage, Essel World ticket problem.

---

• (i) Simply translate the two conditions (total tickets = 300, total amount = ₹55,000) into two linear equations. This is the standard "form the pair of equations" step.
• (ii) Use substitution method — the most straightforward approach here. Show all steps clearly for full 2-mark credit.
• (iii) Direct substitution of given values (250 children, 100 adults) into the fare formula. No equation-solving needed; just arithmetic.

Let the number of correct answers = x and the number of wrong answers = y.

Equations formed:

From condition 1: $3x - y = 40$ … (1)

From condition 2: $4x - 2y = 50$, i.e., $2x - y = 25$ … (2)

Elimination Method:

Subtract (2) from (1):

$(3x - y) - (2x - y) = 40 - 25$

$x = 15$

Substituting x = 15 in equation (1):

$3(15) - y = 40$

$45 - y = 40$

$y = 5$

Verification:

• Condition 1: $3(15) - 1(5) = 45 - 5 = 40$ ✓
• Condition 2: $4(15) - 2(5) = 60 - 10 = 50$ ✓

Total number of questions = x + y = 15 + 5 = 20

Source: Chapter 3, Section 3.3.2 (Elimination Method)

---

• Two variables are essential: assign x = correct, y = wrong clearly before writing equations.
• Write both equations first, then solve — examiners award marks at each step.
• Use the elimination method as taught in §3.3.2: subtract equations to eliminate y directly.
• Always write a verification — it is expected in board exams and can fetch 1 mark.
• State the final answer clearly as a sum (x + y), since the question asks for total questions, not individual counts.

Let amount invested in Scheme A = ₹x and in Scheme B = ₹y.

Equation 1 (original investment):
$$\frac{8x}{100} + \frac{9y}{100} = 1860$$
$$\Rightarrow 8x + 9y = 186000 \tag{1}$$

Equation 2 (interchanged investment gives ₹20 more):
$$\frac{9x}{100} + \frac{8y}{100} = 1880$$
$$\Rightarrow 9x + 8y = 188000 \tag{2}$$

Elimination Method:

Multiply (1) by 9 and (2) by 8:
$$72x + 81y = 1674000 \tag{3}$$
$$72x + 64y = 1504000 \tag{4}$$

Subtract (4) from (3):
$$17y = 170000 \Rightarrow y = 10000$$

Substitute y = 10000 in (1):
$$8x + 90000 = 186000 \Rightarrow 8x = 96000 \Rightarrow x = 12000$$

∴ Vijay invested ₹12,000 in Scheme A and ₹10,000 in Scheme B.

Verification: 8% of 12000 + 9% of 10000 = 960 + 900 = ₹1860 ✓

Source: Pair of Linear Equations in Two Variables, Section 3.3.2 (Elimination Method)

---

• Form both equations carefully from the given conditions — this earns the first 2 marks.
• Use the elimination method (as taught in §3.3.2): multiply to equalise coefficients, then subtract.
• Always verify your answer — CBSE awards 1 mark for correct verification in 5-mark word problems.
• The key insight: interchanging amounts increases interest by ₹20, so the second equation uses 1860 + 20 = 1880.

Let Veer's marks = V and Kevin's marks = K.

Equation 1: Veer scored 6 more than twice Kevin's marks:
$$V = 2K + 6$$

Equation 2: If one of them scored 4 more, total = 40, so their current total = 36:
$$V + K = 36$$

Solving: Substitute V = 2K + 6 into equation 2:
$$2K + 6 + K = 36$$
$$3K = 30 \implies K = 10$$

$$V = 2(10) + 6 = 26$$

∴ Veer scored 26 marks and Kevin scored 10 marks.

Verification: 26 + 10 + 4 = 40 ✓

• The phrase "if one of them scored 4 more, total = 40" means the current total is 36 (subtract 4). This is the key step students often miss.
• Set up both equations clearly before solving — examiners award 1 mark for correct equations, 1 mark for solving, 1 mark for the final answer.
• Always verify your answer; it shows completeness and can recover partial marks.

(a) consistent with unique solution.

The two lines intersect at a single point, giving exactly one solution. Hence the pair is consistent with a unique solution.

Source: Chapter 3, Section 3.2 – Graphical Method of Solution of a Pair of Linear Equations

---

When two lines intersect at one point, the pair of linear equations has exactly one (unique) solution → consistent. Examiners expect you to recall: intersecting lines = consistent + unique solution; coincident lines = consistent + infinitely many solutions; parallel lines = inconsistent. Option (d) is a distractor — intersecting lines already meet, so no "extension" is needed.

For x = 4: This is a vertical line parallel to the y-axis passing through x = 4.

Points: (4, 0), (4, 2), (4, –2)

For 3x – 2y = 6: Rewrite as $y = \dfrac{3x-6}{2}$

| x | 0 | 2 | 4 |
|---|---|---|---|
| y | –3 | 0 | 3 |

Points: (0, –3), (2, 0), (4, 3)

Plot both lines on the graph. The line x = 4 is vertical; the line 3x – 2y = 6 passes through (0, –3) and (2, 0).

The two lines intersect at (4, 3).

∴ Solution: x = 4, y = 3

Source: Chapter 3, Section 3.2 — Graphical Method of Solution of a Pair of Linear Equations

---

• Examiners expect a table of values for the non-vertical equation and at least two named points for the vertical line.
• The vertical line x = 4 scores easy marks — just state it is parallel to the y-axis through x = 4.
• Substitute the answer back to verify: 3(4) – 2(3) = 12 – 6 = 6 ✓
• Always state the solution clearly as a coordinate pair at the end — that line carries marks.

(i) From the given information:

• Batch I: $20x + 5y = 9000$
• Batch II: $5x + 25y = 26000$

(ii) From the two equations:

$20x + 5y = 9000$ … (1)

$5x + 25y = 26000$ … (2)

Multiply (1) by 5: $100x + 25y = 45000$ … (3)

Subtract (2) from (3):

$95x = 19000$

$x = 200$

∴ Monthly fee paid by a poor child = ₹ 200

(iii) From (2), substituting $x = 200$:

$5(200) + 25y = 26000 \Rightarrow 25y = 25000 \Rightarrow y = 1000$

For 10 poor and 20 rich children in batch II:

Total = $10x + 20y = 10(200) + 20(1000) = 2000 + 20000 =$ ₹ 22,000

Source: Pair of Linear Equations in Two Variables, Case Study Application

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• (i) Simply translate the word problem into two linear equations — 1 mark for both equations correct.
• (ii) Solve by elimination (multiply and subtract). Show all steps for full 2 marks; a missing step costs a mark.
• (iii) You must first find $y$ (₹ 1000) using the value of $x$, then substitute the new batch composition. Just the final answer without working may lose the mark.

Option A: $(0, -3)$

On the y-axis, $x = 0$. Substituting in $3x - y = 3$: $3(0) - y = 3 \Rightarrow y = -3$. So the point is $(0, -3)$.

To find intersection with the y-axis, always put $x = 0$ in the equation and solve for $y$. The x-axis intersection requires $y = 0$ instead — don't confuse the two. This is a standard 1-mark concept tested frequently.

The equation $x = 0$ represents the y-axis (a vertical line passing through the origin), and $y = -7$ represents a horizontal line parallel to the x-axis, passing through (0, −7).

These two lines intersect at the point (0, −7).

Since the lines intersect at exactly one point, the system has a unique solution. Therefore, the pair of equations is consistent.

Source: Pair of Linear Equations in Two Variables, Section 3.2

---

• Examiners expect you to identify what each line looks like on the graph (vertical/horizontal), state the point of intersection, and conclude with the word "consistent."
• $x = 0$ is the y-axis; $y = -7$ is a horizontal line — they are perpendicular and always intersect at one point.
• The key rule: intersecting lines → unique solution → consistent. Write this conclusion explicitly or you lose the mark.

The equation $x = 3$ is a vertical line parallel to the Y-axis, passing through (3, 0).
The equation $y = -4$ is a horizontal line parallel to the X-axis, passing through (0, –4).

On the graph, the two lines intersect at the point (3, –4).

∴ The solution is $x = 3,\ y = -4$.

Source: Chapter 3, Section 3.2 – Graphical Method of Solution of a Pair of Linear Equations

---

• $x = 3$ has no $y$-term, so it is a vertical line; $y = -4$ has no $x$-term, so it is a horizontal line. They always intersect at exactly one point.
• In the graph, mark the vertical line at $x = 3$ and the horizontal line at $y = -4$; their crossing point gives the solution directly.
• Examiners expect you to describe both lines, state that they intersect, and write the solution clearly. A rough sketch on graph paper (labelling the point of intersection) earns full marks.

Answer: B — does not exist

Rewriting: $kx - y - 2 = 0$ and $6x - 2y - 3 = 0$. For infinitely many solutions: $\dfrac{k}{6} = \dfrac{-1}{-2} = \dfrac{-2}{-3}$, i.e., $\dfrac{k}{6} = \dfrac{1}{2}$ gives $k = 3$, but $\dfrac{1}{2} \neq \dfrac{2}{3}$. So no value of $k$ satisfies all three ratios simultaneously; the answer does not exist.

For infinitely many solutions, all three ratios must be equal: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$. Here $\dfrac{b_1}{b_2} = \dfrac{1}{2}$ but $\dfrac{c_1}{c_2} = \dfrac{2}{3}$; these two are never equal regardless of $k$, so the condition can never be fully satisfied. Always check all three ratios, not just two.

Checking consistency:

Rewrite equations in standard form:

• $x + 3y - 6 = 0$ → $a_1=1,\ b_1=3,\ c_1=-6$
• $-2x + 3y + 12 = 0$ → $a_2=-2,\ b_2=3,\ c_2=12$

$$\frac{a_1}{a_2} = \frac{1}{-2}, \quad \frac{b_1}{b_2} = \frac{3}{3} = 1$$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the pair is consistent (lines intersect at a unique point).

Graphical Solution:

For $x + 3y = 6$:

| $x$ | 0 | 6 |
|-----|---|---|
| $y$ | 2 | 0 |

Plot points A(0, 2) and B(6, 0).

For $3y - 2x = -12$, i.e., $y = \dfrac{2x-12}{3}$:

| $x$ | 0 | 6 |
|-----|---|---|
| $y$ | −4 | 0 |

Plot points P(0, −4) and Q(6, 0).

Both lines pass through (6, 0).

∴ Solution: $x = 6,\ y = 0$

Source: Chapter 3, Section 3.2 (Graphical Method)

---

• First check ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$; since they are unequal, lines intersect → consistent.
• For the graph, find two points per line (easiest: set $x=0$, then $y=0$).
• Both lines share the point (6, 0) — that is the unique solution.
• Always state the conclusion clearly. Examiners award marks for the ratio check, the table of values, and the correct solution point.

Let the greater angle = x° and the smaller angle = y°.

Since the angles are supplementary:
$$x + y = 180 \quad \text{...(1)}$$

Since the greater exceeds the smaller by 50°:
$$x - y = 50 \quad \text{...(2)}$$

Adding (1) and (2): $2x = 230 \Rightarrow x = 115°$

Substituting in (1): $y = 180 - 115 = 65°$

The greater angle is 115° and the smaller angle is 65°.

Source: Chapter 3, Pair of Linear Equations in Two Variables

---

• 1 mark for correctly forming both equations; 1 mark for solving and stating the answer.
• Supplementary angles sum to 180° — this gives equation (1). The word "exceeds by" means difference — this gives equation (2).
• Adding the two equations is the quickest elimination step; show it clearly.
• Always state the final answer in a sentence for full credit.

Option A: Unique solution

The equations $2x + 1 = 0$ gives $x = -\frac{1}{2}$ and $3y - 5 = 0$ gives $y = \frac{5}{3}$. These are independent equations, each with exactly one value, so the system has a unique solution.

Although each equation has only one variable, together they form a consistent pair with a single ordered pair solution $\left(-\frac{1}{2},\ \frac{5}{3}\right)$. Examiners expect students to recognise that one equation in $x$ and one in $y$ (with no contradiction) always yields exactly one solution. Do not confuse "one variable each" with "no solution" or "infinite solutions."

Finding points for Line 1: $2x - y - 2 = 0 \Rightarrow y = 2x - 2$

| x | 0 | 1 |
|---|---|---|
| y | –2 | 0 |

Points: (0, –2), (1, 0)

Finding points for Line 2: $-4x + y + 4 = 0 \Rightarrow y = 4x - 4$

| x | 0 | 1 |
|---|---|---|
| y | –4 | 0 |

Points: (0, –4), (1, 0)

Plot both lines on the graph. The two lines intersect at (1, 0).

∴ Solution: x = 1, y = 0

Y-axis intercepts:

• Line 1 cuts y-axis at (0, –2) → ordinate = –2
• Line 2 cuts y-axis at (0, –4) → ordinate = –4

Absolute difference = |–2 – (–4)| = |2| = 2

Source: Chapter 3, Section 3.2

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• Table of values earns the first mark; always find two points per line and show them in a table.
• Intersection point (the solution) earns the second mark — read coordinates carefully from the graph.
• The absolute difference of ordinates (y-values where lines meet the y-axis) is the third mark — set x = 0 in each equation to get the y-intercepts quickly.
• "Absolute difference" means take the positive value: $|-2-(-4)| = 2$.

For Line 1: $x - 2y = 3$

| $x$ | 3 | 1 | -1 |
|-----|---|---|----|
| $y$ | 0 | -1 | -2 |

For Line 2: $3x - 8y = 7$

| $x$ | 1 | 3 | -1 |
|-----|---|---|----|
| $y$ | -\frac{1}{2} | \frac{1}{4} | ... |

Since the coordinates are non-integral, we also try:

| $x$ | 9 | 1 |
|-----|---|---|
| $y$ | \frac{5}{2} | -\frac{1}{2} |

Plotting: Plot both lines on the same graph using the above points.

Finding the solution algebraically to verify the intersection point:

From equation 1: $x = 3 + 2y$

Substituting in equation 2:
$$3(3 + 2y) - 8y = 7$$
$$9 + 6y - 8y = 7$$
$$-2y = -2 \Rightarrow y = 1$$

$$x = 3 + 2(1) = 5$$

The two lines intersect at the point $(5, 1)$.

∴ Solution: $x = 5$, $y = 1$

Verification: $5 - 2(1) = 3$ ✓ and $3(5) - 8(1) = 15 - 8 = 7$ ✓

Source: Chapter 3, Section 3.2 and 3.3.1

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• In a graphical question, examiners expect a table of values for each line (minimum 2–3 points), a description of plotting, and identification of the intersection point.
• Since both lines here give a non-integral intersection when read off a graph roughly, CBSE expects you to state the intersection point clearly and verify it — marks are awarded for the table, graph description, intersection point, and verification.
• The substitution step shown here both solves and verifies, which is accepted. Always show the verification step — it carries 1 mark typically.
• The key answer is $x = 5,\ y = 1$.

Answer: (C) $x - 3y = 7$

The given line $2x - 6y = 7$ has slope $\frac{1}{3}$. Option C, $x - 3y = 7$, has the same slope $\frac{1}{3}$ but different constant, so the lines are parallel.

For parallel lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.

Rewrite $2x - 6y = 7$ as $x - 3y = \frac{7}{2}$. Check each option for same ratio of coefficients of x and y but different constant ratio. Only option C gives $\frac{1}{1} = \frac{-3}{-3}$ (same slope) but $\frac{7/2}{7} \neq 1$ (different constant) — confirming parallel lines. Option D ($x = \frac{7}{2} - 3y \Rightarrow x + 3y = \frac{7}{2}$) has a different slope entirely.

Let the price of 1 pencil = ₹x and price of 1 chocolate = ₹y.

Pair of linear equations:

Aarush: $2x + 3y = 11$ ... (1)

Tanish: $x + 2y = 7$ ... (2)

Table of values:

For eq. (1): $2x + 3y = 11$

| x | 1 | 4 |
|---|---|---|
| y | 3 | 1 |

Points: A(1, 3), B(4, 1)

For eq. (2): $x + 2y = 7$

| x | 1 | 3 |
|---|---|---|
| y | 3 | 2 |

Points: P(1, 3), Q(3, 2)

Graph: Plot and join points A, B for line (1) and P, Q for line (2) on graph paper.

Result: The two lines intersect at (1, 3).

∴ x = 1, y = 3

Price of 1 pencil = ₹1; Price of 1 chocolate = ₹3.

Source: Chapter 3, Section 3.2 (Graphical Method)

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• First 1 mark: forming the two correct equations.
• Next 2 marks: making value tables and plotting correctly on the graph.
• Last 2 marks: identifying the intersection point and stating the answer clearly.
• Always verify: 2(1) + 3(3) = 11 ✓ and 1 + 2(3) = 7 ✓.
• Write the final answer as a sentence — examiners look for it explicitly.

Option D: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

A consistent and dependent pair has coincident lines (infinitely many solutions), which occurs when all three ratios are equal.

Examiners expect you to recall the three conditions directly:

• $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ → consistent (unique solution, intersecting lines)
• $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ → consistent & dependent (infinitely many solutions, coincident lines)
• $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ → inconsistent (no solution, parallel lines)

"Dependent" always means coincident lines → all three ratios equal → D.

Step 1 – Find points for each line:

Line 1: $2x - 3y + 6 = 0$

| x | 0 | 3 |
|---|---|---|
| y | 2 | 4 |

Line 2: $2x + 3y - 18 = 0$

| x | 0 | 9 |
|---|---|---|
| y | 6 | 0 |

Line 3: $x = 0$ (the y-axis)

Step 2 – Plot and draw all three lines on graph paper.

Step 3 – Read the vertices (intersection points):

• Line 1 and Line 3 ($x=0$): Put $x=0$ in Line 1 → $y=2$. Vertex A = (0, 2)
• Line 2 and Line 3 ($x=0$): Put $x=0$ in Line 2 → $y=6$. Vertex B = (0, 6)
• Line 1 and Line 2: Adding both equations → $4x - 12 = 0$ → $x = 3$, $y = 4$. Vertex C = (3, 4)

Step 4 – Area of triangle:

Base AB lies on y-axis: $AB = 6 - 2 = 4$ units.
Height = perpendicular distance from C(3, 4) to y-axis $= 3$ units.

$$\text{Area} = \frac{1}{2} \times 4 \times 3 = \boxed{6 \text{ sq. units}}$$

Source: Chapter 3, Section 3.2 – Graphical Method of Solution

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• Graphical method: plot 2 points per line, draw the lines, and read intersection coordinates.
• Three vertices come from the three pairwise intersections of the three lines.
• Area shortcut: when two vertices share the y-axis, use them as base; the x-coordinate of the third vertex directly gives the height — avoids the coordinate-geometry formula and is faster in board exams.
• Examiners award marks for: the correct table of values (1 mark), correctly drawn graph (1 mark), correct vertices (2 marks), and correct area (1 mark).

(d) an infinite number of solutions

When two lines are coincident, every point on the line is a common solution, giving infinitely many solutions.

Source: Chapter 3, Section 3.2

---

The textbook clearly states: "If the lines coincide, then there are infinitely many solutions — each point on the line being a solution." This is a dependent and consistent pair. Examiners expect the correct option and may ask for a one-line reason, so always remember the three cases: intersecting → unique solution; parallel → no solution; coincident → infinite solutions.

(d) $ab = 6$

For parallel lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Here, $\dfrac{a}{3} = \dfrac{2}{b}$ $\Rightarrow$ $ab = 6$.

Source: Chapter 3, Section 3.2

For parallel lines, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Applying this to the given equations gives $\frac{a}{3} = \frac{2}{b}$, i.e., $ab = 6$. Note that option (d) is the only one expressing a product relation matching this cross-multiplication result.

Rewriting: $2x - 5y - 6 = 0$ and $6x - 15y - 18 = 0$.

$$\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-5}{-15} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-6}{-18} = \frac{1}{3}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines are (c) coincident.

Source: Chapter 3, Section 3.2

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• Always rewrite both equations in standard form $ax + by + c = 0$ before comparing ratios.
• The key rule: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ → coincident (infinitely many solutions); $= \neq$ → parallel; $\neq$ → intersecting.
• A common mistake is stopping after checking only two ratios — check all three.

(d) intersecting at $(a, b)$

The line $x = a$ is vertical and $y = b$ is horizontal; they intersect at the point $(a, b)$.

$x = a$ is a vertical line parallel to the y-axis, and $y = b$ is a horizontal line parallel to the x-axis. Every vertical and horizontal line pair intersects at exactly one point. At that point, the x-coordinate is $a$ and the y-coordinate is $b$, giving $(a, b)$ — not $(b, a)$. Option (d) is correct.

(c) 2

For coincident lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Here, $\dfrac{3}{6} = \dfrac{-1}{-r} = \dfrac{8}{16}$ $\Rightarrow \dfrac{1}{2} = \dfrac{1}{r}$ $\Rightarrow r = 2$

Source: Chapter 3, Section 3.2

For coincident lines, all three ratios of coefficients must be equal. Students often make a sign error with the $b$ coefficients — note that $b_1 = -1$ and $b_2 = -r$, so $\frac{-1}{-r} = \frac{1}{r}$. Setting this equal to $\frac{1}{2}$ gives $r = 2$. Verify: $\frac{3}{6} = \frac{-1}{-2} = \frac{8}{16} = \frac{1}{2}$ ✓

Here, $\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{4}{8} = \dfrac{1}{2}$, $\dfrac{c_1}{c_2} = \dfrac{5}{7}$.

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the lines are parallel. → (A)

When $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair is inconsistent (no solution) and the lines are parallel. Always compare all three ratios to distinguish between parallel and coincident lines.

Option D: $-15x + 9y = 5$

For parallel lines, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$. Here $5x - 3y = 2$, so $a_1:b_1 = 5:(-3)$. Option D gives $-15x + 9y = 5$, where $\dfrac{5}{-15} = \dfrac{-3}{9} = -\dfrac{1}{3}$, but $\dfrac{-2}{-5} \neq -\dfrac{1}{3}$. ✓ Parallel condition satisfied.

• For parallel lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.
• Check each option for the ratio of coefficients of $x$ and $y$ being equal (not $c$-ratio).
• Option D: ratios of $x$- and $y$-coefficients are both $-\frac{1}{3}$, but the constant ratio differs → parallel.
• Eliminate options: A has unequal $a/b$ ratios; B has unequal ratios; C has unequal ratios ($9/-15 \neq -15/6$).

For $x - y + 1 = 0$, i.e., $y = x + 1$:

| $x$ | 0 | 2 |
|-----|---|---|
| $y$ | 1 | 3 |

For $x + y = 5$, i.e., $y = 5 - x$:

| $x$ | 0 | 3 |
|-----|---|---|
| $y$ | 5 | 2 |

Plot points A(0, 1), B(2, 3) and P(0, 5), Q(3, 2) on graph paper and draw lines through them.

The two lines intersect at the point (2, 3).

∴ x = 2, y = 3 is the solution.

Verification: $2 - 3 + 1 = 0$ ✓ and $2 + 3 = 5$ ✓

Source: Chapter 3, Section 3.2 – Graphical Method of Solution of a Pair of Linear Equations

---

• Examiners expect: a table of values for each equation (2 points each), a mention of plotting and drawing lines, the intersection point clearly stated, and a verification step.
• Award breakup (typical): 1 mark for correct tables/graph, 1 mark for correct intersection point, 1 mark for verification.
• Always check your answer by substituting back into both equations — examiners specifically look for this.
• Keep tables neat; two points per line are sufficient for a straight-line graph.

Rewrite: $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$.

Here, $\dfrac{a_1}{a_2} = \dfrac{1}{-3}$, $\dfrac{b_1}{b_2} = \dfrac{2}{-6} = \dfrac{1}{-3}$, $\dfrac{c_1}{c_2} = \dfrac{5}{1} = 5$.

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the pair has no solution. → Option D

• Rearrange both equations to standard form $ax + by + c = 0$ before comparing ratios.
• $-3x = 6y - 1$ becomes $-3x - 6y + 1 = 0$.
• The condition $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ means parallel lines → no solution (inconsistent pair). This is the key criterion from Section 3.2/Table 3.1.

Equation 1: $3x + y + 4 = 0 \Rightarrow y = -3x - 4$

| $x$ | 0 | –1 | –2 |
|-----|---|----|----|
| $y$ | –4 | –1 | 2 |

Equation 2: $3x - y + 2 = 0 \Rightarrow y = 3x + 2$

| $x$ | 0 | –1 | 1 |
|-----|---|----|---|
| $y$ | 2 | –1 | 5 |

Graph: Plot both lines on the same coordinate axes using the above points.

The two lines intersect at the point $(-1, -1)$.

Verification:

• Eq. 1: $3(-1) + (-1) + 4 = -3 - 1 + 4 = 0$ ✓
• Eq. 2: $3(-1) - (-1) + 2 = -3 + 1 + 2 = 0$ ✓

Solution: $x = -1,\ y = -1$

Since the lines intersect at exactly one point, the pair of equations is consistent with a unique solution.

Source: Chapter 3 (Pair of Linear Equations in Two Variables), Graphical Method

---

• Table of values: Always make a table with at least 3 points per line — examiners check this step (1 mark each table).
• Graph instruction: Even if you can't draw here, mention "plot and draw both lines" — in the actual exam, draw neatly on graph paper.
• Intersection point: Clearly state the coordinates where lines meet (carries key marks).
• Verification: Always substitute back into both equations — this is expected and fetches 1 mark.
• Conclusion: State "consistent, unique solution" to complete the answer as per CBSE expectations.

(B) 2

For infinitely many solutions: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$. Here, $\dfrac{3}{6} = \dfrac{-1}{-k} = \dfrac{8}{16}$, giving $\dfrac{1}{2} = \dfrac{1}{k}$, so $k = 2$.

For coincident lines (infinitely many solutions), all three ratios must be equal. Comparing $\frac{3}{6} = \frac{1}{2}$ with $\frac{-1}{-k} = \frac{1}{k}$, set $\frac{1}{k} = \frac{1}{2}$ → $k = 2$. Always check all three ratios to confirm consistency.

Option C — passing through the origin.

The equation $x - y = 0$ gives $y = x$. Substituting $(0, 0)$: $0 - 0 = 0$ ✓. So the line passes through the origin.

For $x - y = 0$, put $x = 0 \Rightarrow y = 0$, confirming it passes through $(0, 0)$. It is neither parallel to any axis (it has equal, non-zero slopes in both directions) nor does it pass through $(3, 2)$ since $3 - 2 \neq 0$. Examiners expect you to verify by substitution.

Equation 1: $3x + y = 14$

Find points:

| $x$ | 0 | 2 | 4 |
|-----|---|---|---|
| $y$ | 14 | 8 | 2 |

Equation 2: $y = 2$ (a horizontal line parallel to the x-axis)

| $x$ | 0 | 2 | 4 |
|-----|---|---|---|
| $y$ | 2 | 2 | 2 |

Graph: Plot both lines on the same graph. The line $3x + y = 14$ is a slant line; $y = 2$ is horizontal.

Point of intersection: The two lines meet at (4, 2).

$$\therefore x = 4,\quad y = 2$$

Verification: $3(4) + 2 = 12 + 2 = 14$ ✓

Source: Chapter 3, Exercise 3.1 (Graphical method)

---

• Examiners expect a table of at least 2–3 points for each line, followed by a note about plotting and the intersection point.
• $y = 2$ is always a horizontal line — all y-values in its table must be 2.
• The point where the two lines cross gives the solution; always verify by substituting back.
• Marks are typically split: 1 mark for correct tables/plotting, 1 mark for identifying the intersection, 1 mark for stating and verifying the solution.

For inconsistency: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.

Rewriting: $\dfrac{x}{2} - \dfrac{y}{3} = 5 \Rightarrow 3x - 2y = 30$ and $2x + ky = 7$.

$$\frac{3}{2} = \frac{-2}{k} \Rightarrow k = \frac{-4}{3}$$

The correct option is (B) $\dfrac{4}{3}$ (taking magnitude, $k = -\dfrac{4}{3}$).

> Answer: (B) $-\dfrac{4}{3}$

For inconsistency, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (parallel lines condition). Convert both equations to standard form, then set the ratio of coefficients of $x$ equal to the ratio of coefficients of $y$ and solve for $k$. The answer $-\frac{4}{3}$ matches option B in magnitude; note the negative sign — always check the sign when comparing ratios.

Let the number of ₹50 notes = x and number of ₹100 notes = y.

Equations formed:

$$x + y = 25 \quad \text{...(1)}$$

$$50x + 100y = 2000 \implies x + 2y = 40 \quad \text{...(2)}$$

Solving by elimination — Subtract (1) from (2):

$$(x + 2y) - (x + y) = 40 - 25$$

$$y = 15$$

Substituting y = 15 in (1):

$$x + 15 = 25 \implies x = 10$$

Rehana received 10 notes of ₹50 and 15 notes of ₹100.

Verification: 10 + 15 = 25 notes ✓ and 10 × 50 + 15 × 100 = 500 + 1500 = ₹2000 ✓

Source: Chapter 3, Exercise 3.3, Q2(iv)

---

• Form two equations first (total notes + total amount) — this earns the first mark.
• Solve using elimination or substitution — this earns the second mark.
• State the answer clearly with verification — this earns the third mark.
• Examiners expect you to show all steps; skipping the equation-formation step loses marks even if the answer is correct.

Let Rashmi's present age = $x$ years and Nazma's present age = $y$ years.

Three years ago:
$x - 3 = 3(y - 3)$
$x - 3 = 3y - 9$
$x - 3y = -6$ … (1)

Ten years later:
$x + 10 = 2(y + 10)$
$x + 10 = 2y + 20$
$x - 2y = 10$ … (2)

Subtracting (1) from (2):
$(x - 2y) - (x - 3y) = 10 - (-6)$
$y = 16$

Substituting in (2):
$x - 2(16) = 10$
$x = 42$

∴ Rashmi is 42 years old and Nazma is 16 years old.

Verification: 3 years ago — Rashmi: 39, Nazma: 13 → 39 = 3 × 13 ✓
10 years later — Rashmi: 52, Nazma: 26 → 52 = 2 × 26 ✓

Source: Chapter 3, Exercise 3.3 (Q.2, similar to Q.2(ii))

---

• This is identical in structure to the textbook's Nuri–Sonu problem (Ex 3.3, Q.2(ii)) — just different names and slightly different numbers.
• Always form two equations clearly labelled, then solve by elimination or substitution — either is acceptable.
• Write a verification at the end for full marks; CBSE awards 1 mark for checking.
• Common error: forgetting the "three years ago" means subtract 3 from both ages, not just one.

Let the tens digit be x and units digit be y.

So the number = 10x + y.

Condition 1: Product of digits = 24
$$xy = 24 \quad \Rightarrow \quad y = \frac{24}{x} \tag{1}$$

Condition 2: When 18 is subtracted, digits interchange.
$$10x + y - 18 = 10y + x$$
$$9x - 9y = 18$$
$$x - y = 2 \tag{2}$$

Substituting (1) in (2):
$$x - \frac{24}{x} = 2$$
$$x^2 - 2x - 24 = 0$$
$$x^2 - 6x + 4x - 24 = 0$$
$$(x-6)(x+4) = 0$$

So x = 6 or x = −4.

Since x is a digit, x = 6, and $y = \dfrac{24}{6} = 4$.

The required number is 64.

Verification: 6 × 4 = 24 ✓; 64 − 18 = 46 (digits interchanged) ✓

Source: Pair of Linear Equations in Two Variables, Chapter 3

---

• Examiners award marks at each key step: forming both equations (1 mark), reducing to a quadratic (1 mark), solving and rejecting the negative digit (1 mark), stating the number with verification (1 mark).
• Represent the 2-digit number as 10x + y — this standard form is essential.
• Reject x = −4 explicitly (a digit cannot be negative) to show completeness.
• Always verify both conditions at the end; CBSE often awards the final mark for correct verification.

Let the tens digit be $x$ and units digit be $y$.

So the two-digit number = $10x + y$.

Forming the equations:

Condition 1: The number is seven times the sum of its digits.
$$10x + y = 7(x + y)$$
$$10x + y = 7x + 7y$$
$$3x - 6y = 0 \implies x = 2y \tag{1}$$

Condition 2: The number is 2 more than 5 times the product of its digits.
$$10x + y = 5xy + 2 \tag{2}$$

Solving:

Substitute $x = 2y$ from (1) into (2):
$$10(2y) + y = 5(2y)(y) + 2$$
$$21y = 10y^2 + 2$$
$$10y^2 - 21y + 2 = 0$$
$$10y^2 - 20y - y + 2 = 0$$
$$(10y - 1)(y - 2) = 0$$

So $y = 2$ (since $y = \frac{1}{10}$ is not a digit).

From (1): $x = 2(2) = 4$

The required number = $10(4) + 2 = \mathbf{42}$

Verification: Sum of digits = 6; $7 \times 6 = 42$ ✓. Product = 8; $5 \times 8 + 2 = 42$ ✓

---

• Represent the number as $10x + y$ (standard form for 2-digit numbers, as shown in Example 10 of the chapter).
• Condition 1 gives a simple linear equation; condition 2 gives a quadratic in one variable after substitution.
• Factorise the quadratic and reject the non-integer/non-digit root.
• Always verify both conditions — examiners award a mark for it.
• The key step students miss: correctly substituting $x = 2y$ into the product term $5xy$ to get $10y^2$.

(i) The two equations are:

School P: $5x + 4y = 9500$

School Q: $4x + 3y = 7370$

(ii) Solving the equations:

Multiply eq.1 by 3 and eq.2 by 4:

$15x + 12y = 28500$
$16x + 12y = 29480$

Subtracting: $x = 980$

Substituting in eq.1: $5(980) + 4y = 9500 \Rightarrow 4y = 4600 \Rightarrow y = 1150$

Prize amount for Hockey = ₹ 980

OR Cricket prize (₹1150) > Hockey prize (₹980); Cricket is more by ₹170.

(iii) Total prize amount for 2 students each:

$= 2x + 2y = 2(980) + 2(1150) = 1960 + 2300 =$ ₹ 4,260

---

• (i) Just frame the linear equations — 1 mark for both equations correct.
• (ii) Either find x (Hockey prize) OR compare x and y — both valid options. Show elimination/substitution steps clearly.
• (iii) Substitute found values into $2x + 2y$. Carry-forward marks apply if arithmetic error in (ii).
• Key values: x = ₹980 (Hockey), y = ₹1150 (Cricket).

Let the two numbers be $x$ and $y$, where $x > y$.

$$x + y = 105 \quad \text{...(1)}$$
$$x - y = 45 \quad \text{...(2)}$$

Adding (1) and (2):
$$2x = 150 \implies x = 75$$

Substituting in (1):
$$75 + y = 105 \implies y = 30$$

∴ The two numbers are 75 and 30.

• This is a standard elimination method problem (Chapter 3). Form two equations from the given conditions, then add them to eliminate $y$ and find $x$, then substitute back.
• Examiners award 1 mark for correct equation formation and 1 mark for correct solution. Always verify: 75 + 30 = 105 ✓ and 75 − 30 = 45 ✓.

From equation (1): $7x - 2y = 5$ ... (1)
From equation (2): $8x + 7y = 15$ ... (2)

Multiply (1) by 7 and (2) by 2:

$49x - 14y = 35$ ... (3)
$16x + 14y = 30$ ... (4)

Adding (3) and (4):
$65x = 65 \Rightarrow x = 1$

Substituting in (1): $7(1) - 2y = 5 \Rightarrow y = 1$

Solution: $x = 1,\ y = 1$

Verification:

• Eq (1): $7(1) - 2(1) = 5$ ✓
• Eq (2): $8(1) + 7(1) = 15$ ✓

Source: Chapter 3, Exercise 3.3 (Elimination Method)

---

• The examiner expects the elimination method (multiply to equate coefficients, then add/subtract). Show each step clearly.
• Always write the verification step — it carries marks in a "verify your answer" question.
• Keep working neat and label equations (1), (2), (3)… — this signals method clarity to the examiner.

Let the speed of the stream = x km/h

• Speed upstream = (12 − x) km/h
• Speed downstream = (12 + x) km/h

Forming the equation:

Time = Distance ÷ Speed

$$\frac{15}{12 - x} + \frac{18}{12 + x} = 3$$

Solving:

$$15(12 + x) + 18(12 - x) = 3(12 - x)(12 + x)$$

$$180 + 15x + 216 - 18x = 3(144 - x^2)$$

$$396 - 3x = 432 - 3x^2$$

$$3x^2 - 3x - 36 = 0$$

$$x^2 - x - 12 = 0$$

$$(x - 4)(x + 3) = 0$$

$$x = 4 \quad \text{or} \quad x = -3$$

Since speed cannot be negative, x = −3 is rejected.

∴ Speed of the stream = 4 km/h

Verification: Upstream speed = 8 km/h, Downstream speed = 16 km/h.
Time = 15/8 + 18/16 = 15/8 + 9/8 = 24/8 = 3 h ✓

Source: Chapter 3, Pair of Linear Equations in Two Variables

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• The key step is setting up the time equation using Time = Distance/Speed for both upstream and downstream journeys.
• After cross-multiplying, the equation simplifies to a quadratic — factorise and reject the negative root since speed is always positive.
• Always include the verification step; it carries marks in CBSE board exams.
• Examiners award marks at each stage: forming the equation (1 mark), simplification (1 mark), factorisation (1 mark), correct root with reason for rejection (1 mark), and verification (1 mark).

(C) Assertion (A) is true, but Reason (R) is false.

For the given equations: $\frac{3}{-6} = \frac{-5}{10} = \frac{1}{2}$, but $\frac{7}{14} = \frac{1}{2}$... wait — $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (since $\frac{7}{14} \neq \frac{-5}{10}$... rechecking: $\frac{3}{-6}=\frac{-1}{2}$, $\frac{-5}{10}=\frac{-1}{2}$, $\frac{7}{14}=\frac{1}{2}$), so the system is inconsistent — A is true. But R is false because when there is no unique solution, lines can be parallel (no solution) or coincident (infinitely many solutions) — not always parallel.

(C)

---

• Assertion is true: $\frac{a_1}{a_2}=\frac{-1}{2},\ \frac{b_1}{b_2}=\frac{-1}{2},\ \frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}$; since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the pair is inconsistent (parallel lines). ✓
• Reason is false: "no unique solution" includes both parallel lines (no solution) and coincident lines (infinitely many solutions). Parallel lines are only one case; coincident lines also lack a unique solution. So R is an incorrect generalisation.
• Examiner looks for correct identification of the flaw in R and verification of A using ratio comparison.

Adding both equations:

$(101x + 102y) + (102x + 101y) = 304 + 305$

$203x + 203y = 609$

$x + y = 3$ … (1)

Subtracting equation 1 from equation 2:

$(102x + 101y) - (101x + 102y) = 305 - 304$

$x - y = 1$ … (2)

Adding (1) and (2): $2x = 4 \Rightarrow x = 2$

Substituting in (1): $2 + y = 3 \Rightarrow y = 1$

∴ x = 2, y = 1

Source: Chapter 3, Section 3.3.2 (Elimination Method)

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• The key trick here is to add and subtract the two equations directly — this instantly eliminates one variable each time, saving steps.
• Examiners expect you to show each algebraic step clearly: the addition/subtraction, the simplified equation, and the substitution back.
• Always write the final answer clearly as "x = …, y = …" for full marks.

From equation 1: $\sqrt{2}\,x + \sqrt{3}\,y = 5$ → $x = \dfrac{5 - \sqrt{3}\,y}{\sqrt{2}}$

Substituting in equation 2: $\sqrt{3} \cdot \dfrac{5 - \sqrt{3}\,y}{\sqrt{2}} - \sqrt{8}\,y = -\sqrt{6}$

$$\frac{5\sqrt{3} - 3y}{\sqrt{2}} - 2\sqrt{2}\,y = -\sqrt{6}$$

Multiply throughout by $\sqrt{2}$:

$$5\sqrt{3} - 3y - 4y = -\sqrt{12}$$

$$5\sqrt{3} - 7y = -2\sqrt{3}$$

$$7y = 7\sqrt{3} \implies \boxed{y = \sqrt{3}}$$

Substituting back: $x = \dfrac{5 - \sqrt{3} \cdot \sqrt{3}}{\sqrt{2}} = \dfrac{5-3}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \boxed{\sqrt{2}}$

∴ $x = \sqrt{2}$, $y = \sqrt{3}$

Source: Chapter 3, Section 3.3.1 Substitution Method

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• Use substitution: express $x$ from the simpler first equation, substitute into the second.
• Multiply by $\sqrt{2}$ to clear the denominator cleanly — this is the key step examiners look for.
• Always verify: $\sqrt{2}\cdot\sqrt{2} + \sqrt{3}\cdot\sqrt{3} = 2+3 = 5$ ✓ and $\sqrt{3}\cdot\sqrt{2} - \sqrt{8}\cdot\sqrt{3} = \sqrt{6} - 2\sqrt{6} = -\sqrt{6}$ ✓
• Show all algebraic steps clearly; even 1 mark is awarded for correct method even if arithmetic slips.

Let the incomes of the two persons be ₹9x and ₹7x, and their expenditures be ₹4y and ₹3y respectively.

Since each saves ₹5,000:

$$9x - 4y = 5000 \quad \text{...(1)}$$
$$7x - 3y = 5000 \quad \text{...(2)}$$

Solving by Elimination:

Multiply (1) by 3 and (2) by 4:

$$27x - 12y = 15000 \quad \text{...(3)}$$
$$28x - 12y = 20000 \quad \text{...(4)}$$

Subtract (3) from (4):

$$x = 5000$$

Substitute in (1): $9(5000) - 4y = 5000 \Rightarrow y = 10000$

Monthly incomes:

• Person 1: $9x = 9 × 5000 =$ ₹45,000
• Person 2: $7x = 7 × 5000 =$ ₹35,000

Source: Chapter 3, Section 3.3.2 (Elimination Method)

The question has two parts — forming equations and solving them — so both must be shown clearly. The incomes are expressed as $9x$ and $7x$ (not separately), which keeps the ratio 9:7 built in. Saving = Income − Expenditure, giving two equations. Use the elimination method to score full method marks; show each step. The examiner checks: correct equations (1 mark), correct method (1 mark), correct final incomes (1 mark).

Elimination Method:

Given: $30x + 44y = 10$ … (1) and $40x + 55y = 13$ … (2)

Multiply (1) by 4 and (2) by 3:

$120x + 176y = 40$ … (3)

$120x + 165y = 39$ … (4)

Subtract (4) from (3):

$11y = 1 \Rightarrow y = \dfrac{1}{11}$

Substitute in (1): $30x + 44 \times \dfrac{1}{11} = 10$

$30x + 4 = 10 \Rightarrow 30x = 6 \Rightarrow x = \dfrac{1}{5}$

$\therefore x = \dfrac{1}{5},\quad y = \dfrac{1}{11}$

Source: Chapter 3, Section 3.3.2 (Elimination Method)

• Make coefficients of one variable equal by multiplying both equations by suitable numbers, then subtract to eliminate that variable.
• After finding one variable, substitute back into either original equation.
• Examiners award marks for: correct multiplication step, correct subtraction, correct value of each variable.

(c) ±6

For infinitely many solutions: $\dfrac{p}{12} = \dfrac{3}{p} = \dfrac{p-3}{p}$

From $\dfrac{p}{12} = \dfrac{3}{p}$: $p^2 = 36$, so $p = \pm6$.

Checking $\dfrac{3}{p} = \dfrac{p-3}{p}$: gives $p = 6$... verifying $p = -6$ also satisfies all three ratios being equal. Hence $p = \pm6$.

For infinitely many solutions, the condition is $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$. Here $a_1=p, b_1=3, c_1=-(p-3)$ and $a_2=12, b_2=p, c_2=-p$. Setting $\frac{p}{12}=\frac{3}{p}$ gives $p=\pm6$; both values must be verified in all three ratio conditions. The answer is (c) ±6.

(i) Forming the equations:

Let present age of Adil = x years, present age of Bharat = y years.

Five years ago: Adil's age = (x – 5), Bharat's age = (y – 5)
$$x - 5 = 3(y - 5) \implies x - 3y = -10 \quad \text{...(1)}$$

Ten years later: Adil's age = (x + 10), Bharat's age = (y + 10)
$$x + 10 = 2(y + 10) \implies x - 2y = 10 \quad \text{...(2)}$$

(ii) Checking consistency:

From equations (1) and (2): $a_1 = 1,\ b_1 = -3,\ c_1 = -10$ and $a_2 = 1,\ b_2 = -2,\ c_2 = 10$

$$\frac{a_1}{a_2} = \frac{1}{1} = 1, \quad \frac{b_1}{b_2} = \frac{-3}{-2} = \frac{3}{2}$$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the system is consistent with a unique solution.

(iii) Finding the ages:

Subtracting (1) from (2):
$$(x - 2y) - (x - 3y) = 10 - (-10)$$
$$y = 20 \text{ years}$$

Substituting in (2): $x - 2(20) = 10 \implies x = 50$ years.

∴ Present age of Adil = 50 years; Present age of Bharat = 20 years.

Source: Chapter 3, Exercise 3.3 (Q2-ii) and Section 3.2

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• Part (i): Carefully translate "five years ago" and "ten years later" into algebraic expressions before equating. A common error is using wrong signs.
• Part (ii): The consistency condition from the textbook is: if $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent with a unique solution (intersecting lines). Always state this rule explicitly for marks.
• Part (iii): Elimination by subtraction is the cleanest method here. Always verify: 5 years ago Adil = 45, Bharat = 15 (45 = 3 × 15 ✓); 10 years later Adil = 60, Bharat = 30 (60 = 2 × 30 ✓).

Given:
$$217x + 131y = 913 \tag{1}$$
$$131x + 217y = 827 \tag{2}$$

Adding (1) and (2):
$$348x + 348y = 1740$$
$$x + y = 5 \tag{3}$$

Subtracting (2) from (1):
$$86x - 86y = 86$$
$$x - y = 1 \tag{4}$$

Adding (3) and (4):
$$2x = 6 \implies x = 3$$

Substituting in (3):
$$3 + y = 5 \implies y = 2$$

∴ x = 3, y = 2

Source: Chapter 3, Section 3.3.2 (Elimination Method)

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• The trick here is to add and subtract the two equations directly — this instantly simplifies the large coefficients (217 and 131) into manageable ones (348 and 86), avoiding tedious multiplication.
• Examiners award marks for: setting up the addition/subtraction step clearly, getting both simplified equations, solving for x and y, and stating the final answer.
• Always verify: 217(3) + 131(2) = 651 + 262 = 913 ✓ and 131(3) + 217(2) = 393 + 434 = 827 ✓

For infinitely many solutions, the condition is:

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

Here: $a_1 = 2,\ b_1 = 3,\ c_1 = -7$ and $a_2 = 2a,\ b_2 = (a+b),\ c_2 = -28$

$$\frac{2}{2a} = \frac{3}{a+b} = \frac{-7}{-28}$$

From $\dfrac{2}{2a} = \dfrac{7}{28} = \dfrac{1}{4}$:

$$2a = 8 \implies a = 4$$

From $\dfrac{3}{a+b} = \dfrac{1}{4}$:

$$a + b = 12 \implies 4 + b = 12 \implies b = 8$$

Therefore, $a = 4$ and $b = 8$.

Source: Chapter 3, Section 3.2 (Graphical Method / Condition for infinitely many solutions)

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• The key condition for infinitely many solutions (coincident lines) is $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
• Rewrite both equations in standard form first to correctly identify $c_1$ and $c_2$ (i.e., $-7$ and $-28$).
• Solve the ratios one pair at a time — first find $a$, then substitute to find $b$.
• Show all steps clearly; examiners award marks for setting up the ratio condition, finding $a$, and finding $b$ (roughly 1 mark each).

Option C: $a - b$

Adding both equations: $(a+b)x + (a+b)y = a^2 - b^2$, so $(a+b)(x+y) = (a+b)(a-b)$, giving $x + y = a - b$.

Add the two equations directly: $(a+b)x + (a+b)y = a^2 - b^2$. Factor both sides: $(a+b)(x+y) = (a+b)(a-b)$. Divide by $(a+b)$ (assuming $a \neq -b$) to get $x+y = a-b$. This elimination shortcut finds $x+y$ without solving for $x$ and $y$ individually — a common MCQ technique examiners expect you to recognise.

Adding the two equations:

$$2x + y = 13$$
$$4x - y = 17$$

Adding: $6x = 30 \Rightarrow x = 5$

Substituting in $2x + y = 13$: $10 + y = 13 \Rightarrow y = 3$

Therefore, $x - y = 5 - 3 = \mathbf{2}$

Source: Chapter 3, Section 3.3.2 (Elimination Method)

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• Add the two equations to eliminate $y$ (since $+y$ and $-y$ cancel directly) — this is the elimination method.
• Find $x$ first, then substitute back to find $y$.
• The question asks for $(x - y)$, not individual values, so make sure to compute and state the final answer clearly.
• Examiners award 1 mark for correct $x$ and $y$, and 1 mark for the correct value of $(x - y)$.

For infinitely many solutions: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Here, $\dfrac{6}{36} = \dfrac{1}{6} = \dfrac{3k}{3}$, so $\dfrac{1}{6} = k$, giving $k = \dfrac{1}{6}$.

Answer: (B) $\dfrac{1}{6}$

For infinitely many solutions, the ratio condition $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ must hold. The coefficient ratio $\frac{6}{36}=\frac{1}{6}$ must equal $\frac{3k}{3}=k$, so $k=\frac{1}{6}$. Always rewrite equations in standard form before comparing ratios.

Let number of chairs = x, number of tables = y.

(i) The pair of linear equations:

$$x + y = 300$$
$$50x + 200y = 30000$$

(ii) From equation (i): x = 300 − y

Substituting in equation (ii):

$$50(300 - y) + 200y = 30000$$
$$15000 - 50y + 200y = 30000$$
$$150y = 15000 \Rightarrow y = 100$$

$$x = 300 - 100 = 200$$

∴ Number of chairs = 200, Number of tables = 100

(iii) If no chairs are rented (x = 0):

$$200y = 30000 \Rightarrow y = 150$$

∴ Maximum number of tables = 150

Source: Pair of Linear Equations in Two Variables

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• (i) Simply translate both conditions (total items and total cost) into equations — examiners want both equations clearly stated.
• (ii) Use substitution method; show each step clearly. State the final answer in words.
• (iii) Put x = 0 directly in the cost equation — this is a straightforward 1-mark application. Don't overthink it.

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Verification of A: For $px+3y+59=0$ and $2x+6y+118=0$: $\frac{p}{2}=\frac{3}{6}=\frac{59}{118}$ → $\frac{p}{2}=\frac{1}{2}$ → $p=1$. ✓ (Infinitely many solutions)

Verification of R: For unique solution, $\frac{p}{2}\neq\frac{3}{6}$ → $p\neq1$. ✓

Both are true, but R explains a different pair of equations, so R is not the correct explanation of A.

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• A is true: Applying the condition $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ gives $p=1$.
• R is true: For a unique solution, $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, which gives $p\neq1$.
• However, R refers to a completely different pair of equations (with constants 19 and 157), so it cannot explain why A holds. Hence option (B) is correct.

Substituting $x = 1$, $y = 2$ in $2x - 3y + a = 0$: $2 - 6 + a = 0 \Rightarrow a = 4$.
Substituting in $2x + 3y - b = 0$: $2 + 6 - b = 0 \Rightarrow b = 8$.
Thus $2a = 8 = b$, so Option B: $2a = b$.

Substitute the given solution into both equations to find $a$ and $b$ separately, then check which option satisfies the relationship. Here $a = 4$ and $b = 8$, giving $2a = b$. Examiners expect you to show the substitution step clearly, even in MCQs, to earn full credit.