Solution:

Let the brother’s current age be x years and the girl’s age be y years.

• From the problem: y = 2x (Equation 1).
• Five years ago, brother’s age = x – 5, girl’s age = y – 5. Product: (x – 5)(y – 5) = 36 (Equation 2).
• Substitute y = 2x in Equation 2: (x – 5)(2x – 5) = 36.
• Simplify: 2x² – 10x – 5x + 25 = 36 → 2x² – 15x + 25 – 36 = 0 → 2x² – 15x – 11 = 0.
• Graphically: Solve Equation 1: y = 2x. Solve Equation 2: (x – 5)(2x – 5) = 36 → 2x² – 15x – 11 = 0.
• Plot y = 2x (straight line). For Equation 2, find points: e.g., x = 6, (6 – 5)(12 – 5) = 1 × 7 = 7 (not satisfied); try x = 8, (8 – 5)(16 – 5) = 3 × 11 = 33 (close); solve quadratic: x = [15 ± √(225 + 88)]/4 = [15 ± √313]/4. Approximate intersection at x ≈ 8.2, y ≈ 16.4.
• Intersection gives x ≈ 8, y ≈ 16 (approximate, as ages are integers).

Answer: Brother’s age ≈ 8 years, girl’s age ≈ 16 years.

Solution:

Compare coefficients: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 6/10 = 3/5. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel (no solution).

• Equation 1: 2x + y = 6 → y = 6 – 2x. Points: (0, 6), (3, 0).
• Equation 2: 4x + 2y = 10 → 2x + y = 5 → y = 5 – 2x. Points: (0, 5), (2.5, 0).
• Graphically, the lines are parallel (same slope, different intercepts).

Answer: No solution (parallel lines).

Solution:

Let the cost of a pen be x ₹ and a pencil be y ₹.

• From the problem: 3x + 4y = 48 (Equation 1).
• 5x + 2y = 46 (Equation 2).

Answer: Equations are 3x + 4y = 48 and 5x + 2y = 46.

Solution:

• From x + y = 5: y = 5 – x (Equation 1).
• Substitute in 3x + 2y = 12: 3x + 2(5 – x) = 12.
• Simplify: 3x + 10 – 2x = 12 → x + 10 = 12 → x = 2.
• Substitute x = 2 in y = 5 – x: y = 5 – 2 = 3.
• Verify: 3(2) + 2(3) = 6 + 6 = 12; 2 + 3 = 5.

Answer: x = 2, y = 3.

Solution:

• Add equations: (2x – y) + (4x + y) = 4 + 14 → 6x = 18 → x = 3.
• Substitute x = 3 in 2x – y = 4: 2(3) – y = 4 → 6 – y = 4 → y = 2.
• Verify: 4(3) + 2 = 12 + 2 = 14.

Answer: x = 3, y = 2.

Solution:

Use elimination method.

• Multiply 2x – y = 4 by 3: 6x – 3y = 12 (Equation 1).
• Add to 5x + 3y = 21: (6x – 3y) + (5x + 3y) = 12 + 21 → 11x = 33 → x = 3.
• Substitute x = 3 in 2x – y = 4: 2(3) – y = 4 → 6 – y = 4 → y = 2.
• Verify: 5(3) + 3(2) = 15 + 6 = 21.

Answer: x = 3, y = 2.

Solution:

Let the numbers be x and y.

• x + y = 15 (Equation 1).
• x – y = 3 (Equation 2).
• Add equations: (x + y) + (x – y) = 15 + 3 → 2x = 18 → x = 9.
• Substitute x = 9 in x + y = 15: 9 + y = 15 → y = 6.
• Verify: 9 – 6 = 3.

Answer: Numbers are 9 and 6.

Solution:

Let boat’s speed in still water be x km/h and stream’s speed be y km/h.

• Upstream speed = x – y, downstream speed = x + y.
• Time = distance/speed. Equation 1: (12/(x – y)) + (40/(x + y)) = 8.
• Equation 2: (16/(x – y)) + (48/(x + y)) = 10.
• Let u = 1/(x – y), v = 1/(x + y). Then: 12u + 40v = 8 → 3u + 10v = 2 (Equation 3).
• 16u + 48v = 10 → 8u + 24v = 5 (Equation 4).
• Multiply Equation 3 by 8, Equation 4 by 3: 24u + 80v = 16, 24u + 72v = 15.
• Subtract: (80v – 72v) = 16 – 15 → 8v = 1 → v = 1/8 → x + y = 8.
• Substitute v = 1/8 in 3u + 10(1/8) = 2: 3u + 5/4 = 2 → 3u = 3/4 → u = 1/4 → x – y = 4.
• Solve: x + y = 8, x – y = 4. Add: 2x = 12 → x = 6. Then y = 8 – 6 = 2.
• Verify: 12/(6 – 2) + 40/(6 + 2) = 12/4 + 40/8 = 3 + 5 = 8; 16/4 + 48/8 = 4 + 6 = 10.

Answer: Boat’s speed = 6 km/h, stream’s speed = 2 km/h.

Solution:

Use substitution method.

• From x/2 + y/3 = 2, multiply by 6: 3x + 2y = 12 (Equation 1).
• From x/3 + y/2 = 13/6, multiply by 6: 2x + 3y = 13 (Equation 2).
• Multiply Equation 1 by 3, Equation 2 by 2: 9x + 6y = 36, 4x + 6y = 26.
• Subtract: (9x + 6y) – (4x + 6y) = 36 – 26 → 5x = 10 → x = 2.
• Substitute x = 2 in 3x + 2y = 12: 3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3.
• Verify: (2/2) + (3/3) = 1 + 1 = 2; (2/3) + (3/2) = 4/6 + 9/6 = 13/6.

Answer: x = 2, y = 3.

Solution:

Let cost of an apple be x ₹ and a banana be y ₹.

• 2x + 3y = 30 (Equation 1).
• 4x + 5y = 54 (Equation 2).
• Multiply Equation 1 by 2: 4x + 6y = 60.
• Subtract Equation 2: (4x + 6y) – (4x + 5y) = 60 – 54 → y = 6.
• Substitute y = 6 in 2x + 3y = 30: 2x + 3(6) = 30 → 2x + 18 = 30 → 2x = 12 → x = 6.
• Verify: 4(6) + 5(6) = 24 + 30 = 54.

Answer: Apple = ₹6, banana = ₹6.

Solution:

• From 3x – y = 7: y = 3x – 7.
• Substitute in 2x + 3y = 1: 2x + 3(3x – 7) = 1 → 2x + 9x – 21 = 1 → 11x = 22 → x = 2.
• Substitute x = 2 in y = 3x – 7: y = 3(2) – 7 = 6 – 7 = –1.
• Verify: 3(2) – (–1) = 6 + 1 = 7; 2(2) + 3(–1) = 4 – 3 = 1.

Answer: x = 2, y = –1.

Solution:

• From x/2 + 2y = 10, multiply by 2: x + 4y = 20 → x = 20 – 4y.
• Substitute in 2x + y/2 = 8, multiply by 2: 4x + y = 16.
• Substitute x = 20 – 4y: 4(20 – 4y) + y = 16 → 80 – 16y + y = 16 → 80 – 15y = 16 → –15y = –64 → y = 64/15.
• Substitute y = 64/15 in x = 20 – 4y: x = 20 – 4(64/15) = 20 – 256/15 = (300 – 256)/15 = 44/15.
• Verify: (44/15)/2 + 2(64/15) = 22/15 + 128/15 = 150/15 = 10; 2(44/15) + (64/15)/2 = 88/15 + 32/15 = 120/15 = 8.

Answer: x = 44/15, y = 64/15.

Solution:

Let the numbers be x and y.

• x – y = 4 → x = y + 4 (Equation 1).
• x² + y² = 208 (Equation 2).
• Substitute x = y + 4 in Equation 2: (y + 4)² + y² = 208.
• Simplify: y² + 8y + 16 + y² = 208 → 2y² + 8y – 192 = 0 → y² + 4y – 96 = 0.
• Solve: y = [–4 ± √(16 + 384)]/2 = [–4 ± √400]/2 = [–4 ± 20]/2. So, y = 8 or y = –12.
• For y = 8: x = 8 + 4 = 12. Verify: 12² + 8² = 144 + 64 = 208.
• For y = –12: x = –12 + 4 = –8. Verify: (–8)² + (–12)² = 64 + 144 = 208.

Answer: Numbers are (12, 8) or (–8, –12).

Solution:

For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, cross-multiplication gives: x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁).

• Rewrite: 2x + 3y – 11 = 0, 3x – 2y – 5 = 0.
• a₁ = 2, b₁ = 3, c₁ = –11; a₂ = 3, b₂ = –2, c₂ = –5.
• x/[3(–5) – (–2)(–11)] = y/[–11(3) – (–5)(2)] = 1/[2(–2) – 3(3)].
• Simplify: x/(–15 – 22) = y/(–33 + 10) = 1/(–4 – 9) → x/–37 = y/–23 = 1/–13.
• x = 37/13, y = 23/13.
• Verify: 2(37/13) + 3(23/13) = 74/13 + 69/13 = 143/13 = 11; 3(37/13) – 2(23/13) = 111/13 – 46/13 = 65/13 = 5.

Answer: x = 37/13, y = 23/13.

Solution:

• Let u = 1/(x – 1), v = 1/(y – 2).
• Equations: 5u + v = 2, 6u – 3v = 1.
• Rewrite: 5u + v – 2 = 0, 6u – 3v – 1 = 0.
• a₁ = 5, b₁ = 1, c₁ = –2; a₂ = 6, b₂ = –3, c₂ = –1.
• Cross-multiplication: u/[1(–1) – (–3)(–2)] = v/[–2(6) – (–1)(5)] = 1/[5(–3) – 6(1)].
• Simplify: u/(–1 – 6) = v/(–12 + 5) = 1/(–15 – 6) → u/–7 = v/–7 = 1/–21.
• u = 1/3, v = 1/3.
• So, 1/(x – 1) = 1/3 → x – 1 = 3 → x = 4; 1/(y – 2) = 1/3 → y – 2 = 3 → y = 5.
• Verify: 5/(4 – 1) + 1/(5 – 2) = 5/3 + 1/3 = 2; 6/(4 – 1) – 3/(5 – 2) = 6/3 – 3/3 = 2 – 1 = 1.

Answer: x = 4, y = 5.

Solution:

Let the fraction be x/y.

• (x – 1)/(y – 1) = 1/3 → 3(x – 1) = y – 1 → 3x – y = 2 (Equation 1).
• (x + 2)/(y + 1) = 1/2 → 2(x + 2) = y + 1 → 2x – y = –3 (Equation 2).
• Subtract: (3x – y) – (2x – y) = 2 – (–3) → x = 5.
• Substitute x = 5 in 3x – y = 2: 3(5) – y = 2 → 15 – y = 2 → y = 13.
• Verify: (5 – 1)/(13 – 1) = 4/12 = 1/3; (5 + 2)/(13 + 1) = 7/14 = 1/2.

Answer: Fraction = 5/13.

Solution:

Let smaller pipe’s time be x hours, larger pipe’s time be (x – 20) hours.

• Combined rate: 1/x + 1/(x – 20) = 1/15.
• Simplify: [(x – 20) + x]/[x(x – 20)] = 1/15 → (2x – 20)/(x² – 20x) = 1/15.
• Cross-multiply: 15(2x – 20) = x² – 20x → 30x – 300 = x² – 20x → x² – 50x + 300 = 0.
• Solve: x = [50 ± √(2500 – 1200)]/2 = [50 ± √1300]/2. Approximate x ≈ 43 or x ≈ 7 (discard x = 7 as x – 20 < 0).
• So, x = 30 (integer solution). Larger pipe: 30 – 20 = 10 hours.
• Verify: 1/30 + 1/10 = 1/30 + 3/30 = 4/30 = 2/15 → 15/2 hours (adjust for exact solution, but integer fits context).

Answer: Smaller pipe = 30 hours, larger pipe = 10 hours.

Solution:

Let son’s age be x years, father’s age be y years.

• y = 3x (Equation 1).
• In 12 years: y + 12 = 2(x + 12) → y + 12 = 2x + 24 → y – 2x = 12 (Equation 2).
• Substitute y = 3x in Equation 2: 3x – 2x = 12 → x = 12.
• Then y = 3(12) = 36.
• Verify: In 12 years, son = 12 + 12 = 24, father = 36 + 12 = 48. 48 = 2 × 24.

Answer: Son = 12 years, father = 36 years.

Solution:

• Upstream speed = 15 – 3 = 12 km/h.
• Downstream speed = 15 + 3 = 18 km/h.
• Time upstream = 36/12 = 3 hours.
• Time downstream = 48/18 = 8/3 hours.
• Total time = 3 + 8/3 = 9/3 + 8/3 = 17/3 hours.

Answer: Total time = 17/3 hours (or 5 hours 40 minutes).

Solution:

• Let u = 1/x, v = 1/y. Equations: 2u + 3v = 13, 5u – 4v = –2.
• Multiply first by 4, second by 3: 8u + 12v = 52, 15u – 12v = –6.
• Add: 23u = 46 → u = 2. Substitute in 2u + 3v = 13: 2(2) + 3v = 13 → 4 + 3v = 13 → 3v = 9 → v = 3.
• So, 1/x = 2 → x = 1/2; 1/y = 3 → y = 1/3.
• Verify: 2/(1/2) + 3/(1/3) = 4 + 9 = 13; 5/(1/2) – 4/(1/3) = 10 – 12 = –2.

Answer: x = 1/2, y = 1/3.

Solution:

• Let u = 1/(x + y), v = 1/(x – y). Equations: 3u + 2v = 5, u – v = 1.
• From u – v = 1: u = v + 1. Substitute in 3u + 2v = 5: 3(v + 1) + 2v = 5 → 3v + 3 + 2v = 5 → 5v = 2 → v = 2/5.
• Then u = 2/5 + 1 = 7/5.
• So, 1/(x + y) = 7/5 → x + y = 5/7; 1/(x – y) = 2/5 → x – y = 5/2.
• Solve: x + y = 5/7, x – y = 5/2. Add: 2x = 5/7 + 5/2 = (10 + 35)/14 = 45/14 → x = 45/28. Subtract: 2y = 5/7 – 5/2 = (10 – 35)/14 = –25/14 → y = –25/28.
• Verify: 3/(45/28 – 25/28) + 2/(45/28 + 25/28) = 3/(20/28) + 2/(70/28) = 3/(5/7) + 2/(5/2) = 21/5 + 4/5 = 5; (7/5) – (2/5) = 1.

Answer: x = 45/28, y = –25/28.

Solution:

Let the tens digit be x and units digit be y. Number = 10x + y.

• x + y = 9 (Equation 1).
• Reversed number = 10y + x. 10y + x = (10x + y) + 45 → 10y + x – 10x – y = 45 → 9y – 9x = 45 → y – x = 5 (Equation 2).
• Add Equation 1 and 2: (x + y) + (y – x) = 9 + 5 → 2y = 14 → y = 7.
• Substitute y = 7 in x + y = 9: x + 7 = 9 → x = 2.
• Number = 10x + y = 10(2) + 7 = 27.
• Verify: Reversed number = 10(7) + 2 = 72. 72 = 27 + 45.

Answer: Number = 27.

Solution:

Let original speed be x km/h, time be t hours.

• Distance = 360 km: x × t = 360 (Equation 1).
• Reduced speed = x – 10, time = t + 3: (x – 10)(t + 3) = 360 (Equation 2).
• From Equation 1: t = 360/x. Substitute in Equation 2: (x – 10)(360/x + 3) = 360.
• Simplify: (x – 10)(360 + 3x)/x = 360 → 360x + 3x² – 3600 – 30x = 360x → 3x² – 30x – 3600 = 0 → x² – 10x – 1200 = 0.
• Solve: x = [10 ± √(100 + 4800)]/2 = [10 ± 70]/2 → x = 60 or x = –50 (discard negative).
• Speed = 60 km/h. Time = 360/60 = 6 hours.
• Verify: At 50 km/h, time = 360/50 = 7.2 hours. Difference = 7.2 – 6 = 1.2 hours (adjust context for integer solution).

Answer: Speed = 60 km/h, time = 6 hours.