(i) Let boys = x, girls = y Given: Total students = 10 → x + y = 10 Girls are 4 more than boys → y = x + 4 Rewrite y = x + 4 as x - y = -4 Pair of equations: x + y = 10 x - y = -4 Solution by adding: 2x = 6 → x = 3 boys y = 3 + 4 = 7 girls (ii) Let cost of pencil = x, pen = y 5x + 7y = 50 7x + 5y = 46 Multiply first by 7, second by 5 and subtract: 35x + 49y = 350 35x + 25y = 230 Subtract: 24y = 120 → y = 5 Put y=5 in first eq: 5x + 35 = 50 → 5x = 15 → x = 3 Cost: Pencil = ₹3, Pen = ₹5

(i) 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0 a₁/a₂ = 5/7, b₁/b₂ = -4/6 = -2/3, c₁/c₂ = 8/-9 Ratios not equal → Lines intersect (ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0 a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2 All ratios equal → Lines coincident (iii) 6x – 3y + 10 = 0 and 2x – y + 9 = 0 a₁/a₂ = 6/2 = 3, b₁/b₂ = -3/-1 = 3, c₁/c₂ = 10/9 Ratios a and b equal, c different → Lines parallel

(i) 3x + 2y = 5 ; 2x – 3y = 7 Ratios a: 3/2, b: 2/-3, c: 5/7 → Not equal → Consistent (intersect) (ii) 2x – 3y = 8 ; 4x – 6y = 9 Ratios a: 2/4=1/2, b: -3/-6=1/2, c: 8/9 ≠ 1/2 → Inconsistent (parallel) (iii) 3x + 5y = 7 ; 2x – 3y = 14 Ratios differ → Consistent (iv) 5x – 3y = 11 ; -10x + 6y = -22 Multiply first by 2: 10x - 6y = 22 Second: -10x + 6y = -22 Adding both: 0 = 0 → Coincident (consistent) (v) 4x + 2y = 8 ; 3x + 2y = 12 Ratios differ → Consistent

(i) x + y = 5, 2x + 2y = 10 Second eq. is just twice first → Coincident (consistent) (ii) x – y = 8, 3x – 3y = 16 Second eq. thrice first would be 3x - 3y = 24 ≠ 16 → Intersect (consistent) (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 Check ratios: Not equal → Intersect (consistent) (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 Ratios of a and b equal but c different → Parallel (inconsistent)

Let width = x m, length = x + 4 m Half perimeter = 36 → (Length + Width) = 36 x + (x + 4) = 36 2x + 4 = 36 → 2x = 32 → x = 16 Width = 16 m, Length = 20 m

(i) Intersecting lines: Example 3x – y + 1 = 0 (different slope) (ii) Parallel lines: 4x + 6y – 10 = 0 (multiple of first but different c) (iii) Coincident lines: 4x + 6y – 16 = 0 (exact multiple of first eq.)

Find intercepts: For x – y + 1 = 0 x-intercept: Put y=0 → x + 1 = 0 → x = -1 y-intercept: Put x=0 → -y + 1=0 → y=1 For 3x + 2y – 12 = 0 x-intercept: y=0 → 3x=12 → x=4 y-intercept: x=0 → 2y=12 → y=6 Find intersection of two lines: From first line: y = x + 1 Put in second: 3x + 2(x +1) –12=0 → 3x + 2x + 2 –12=0 → 5x –10=0 → x=2 y = 2 +1=3 Vertices of triangle formed with x-axis are: (-1,0), (4,0), (2,3) Shade region bounded by these points.

(i) x + y = 14 x – y = 4 Substitute y = 14 – x in second: x – (14 – x) = 4 → 2x – 14 = 4 → 2x = 18 → x = 9 y = 14 – 9 = 5 (ii) s – t = 3 (6/3)s + (2/3)t = 1 → 2s + (2/3)t = 1 (rewrite second) Rewrite second: Multiply whole eq by 3: 6s + 2t = 3 Express s = t + 3 from first eq: Substitute: 6(t+3) + 2t = 3 → 6t + 18 + 2t = 3 → 8t = -15 → t = -15/8 s = t + 3 = -15/8 + 24/8 = 9/8 (iii) 3x – y = 3 9x – 3y = 9 From first: y = 3x – 3 Put in second: 9x – 3(3x – 3) = 9 → 9x – 9x + 9 = 9 → 9=9 true → Infinite solutions (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 From first: 0.2x = 1.3 – 0.3y → x = (1.3 – 0.3y)/0.2 = 6.5 – 1.5y Put in second: 0.4(6.5 – 1.5y) + 0.5y = 2.3 → 2.6 – 0.6y + 0.5y = 2.3 –0.1y = –0.3 → y = 3 x = 6.5 – 1.5(3) = 6.5 – 4.5 = 2 (v) √2 x + √3 y = 0 √3 x – √8 y = 0 From first: √2 x = –√3 y → x = –(√3/√2) y Put in second: √3 (–(√3/√2) y) – √8 y = 0 – (3/√2) y – 2√2 y = 0 Multiply all by √2: –3 y – 2*2 y = 0 → –3 y –4 y = 0 → –7 y = 0 → y = 0 Then x = 0 (vi) (3x)/2 – (5y)/3 = –2 x/3 + y/2 = 13/6 Multiply second eq by 6: 2x + 3y = 13 Multiply first eq by 6: 9x – 10y = –12 From second: 2x = 13 – 3y → x = (13 – 3y)/2 Put in first: 9(13 – 3y)/2 – 10y = –12 Multiply both sides by 2: 9(13 – 3y) – 20y = –24 117 – 27y – 20y = –24 → –47y = –141 → y = 3 x = (13 – 3*3)/2 = (13 – 9)/2 = 2

From first: 2x + 3y = 11 From second: 2x – 4y = –24 Subtract second from first: (2x + 3y) – (2x – 4y) = 11 – (–24) 2x + 3y – 2x + 4y = 35 7y = 35 → y = 5 Put y = 5 in first: 2x + 15 = 11 → 2x = –4 → x = –2 Equation y = mx + 3 passes through (–2, 5): 5 = m(–2) + 3 → 5 – 3 = –2m → 2 = –2m → m = –1

(i) The difference between two numbers is 26 and one number is three times the other. Find them. Let numbers be x and y: x – y = 26 x = 3y Substitute x = 3y in first: 3y – y = 26 → 2y = 26 → y = 13 x = 3*13 = 39 (ii) Larger supplementary angle exceeds smaller by 18° Let smaller = x, larger = y x + y = 180 y – x = 18 Add: 2y = 198 → y = 99 x = 180 – 99 = 81 (iii)The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball. 7 bats + 6 balls = ₹3800 3 bats + 5 balls = ₹1750 Let cost of bat = x, ball = y 7x + 6y = 3800 3x + 5y = 1750 From second: 3x = 1750 – 5y → x = (1750 – 5y)/3 Put in first: 7((1750 – 5y)/3) + 6y = 3800 Multiply all by 3: 7(1750 – 5y) + 18y = 11400 12250 – 35y + 18y = 11400 –17y = 11400 – 12250 = –850 → y = 50 x = (1750 – 5*50)/3 = (1750 – 250)/3 = 1500/3 = 500 Cost: Bat = ₹500, Ball = ₹50 (iv) Fixed charge = x, charge per km = y 10x + 10y = 105 15x + 15y = 155 Since fixed charge + (distance × rate) = total Let fixed charge = f, rate = r So: f + 10r = 105 f + 15r = 155 Subtract: (f + 15r) – (f + 10r) = 155 – 105 → 5r = 50 → r = 10 Put r in first: f + 10*10 = 105 → f = 5 For 25 km: Charge = f + 25r = 5 + 250 = ₹255 (v) Let fraction be x/y (x + 2)/(y + 2) = 9/11 (x + 3)/(y + 3) = 5/6 Cross multiply both: 11(x + 2) = 9(y + 2) → 11x + 22 = 9y + 18 → 11x – 9y = –4 ...(1) 6(x + 3) = 5(y + 3) → 6x + 18 = 5y + 15 → 6x – 5y = –3 ...(2) Multiply (1) by 5: 55x – 45y = –20 Multiply (2) by 9: 54x – 45y = –27 Subtract: 55x – 54x = –20 + 27 → x = 7 Put x = 7 in (2): 6*7 – 5y = –3 → 42 – 5y = –3 → –5y = –45 → y = 9 Fraction = 7/9 (vi)Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? Let present age of Jacob = J Let present age of son = S Five years hence: Jacob's age = J + 5 Son's age = S + 5 Given: J + 5 = 3(S + 5) → J + 5 = 3S + 15 → J - 3S = 10 ...(1) Five years ago: Jacob's age = J - 5 Son's age = S - 5 Given: J - 5 = 7(S - 5) → J - 5 = 7S - 35 → J - 7S = -30 ...(2) Subtract (2) from (1): (J - 3S) - (J - 7S) = 10 - (-30) J - 3S - J + 7S = 40 4S = 40 S = 10 Put S = 10 in (1): J - 3(10) = 10 J - 30 = 10 J = 40 **Answer:** Jacob’s present age = 40 years Son’s present age = 10 years

Let the fraction = x/y From first condition: (x + 1) / (y - 1) = 1 → x + 1 = y - 1 → x - y = -2 ...(1) From second condition: x / (y + 1) = 1/2 → 2x = y + 1 → 2x - y = 1 ...(2) Multiply (1) by 2: 2x - 2y = -4 Subtract (2) from this: (2x - 2y) - (2x - y) = -4 - 1 2x - 2y - 2x + y = -5 -y = -5 y = 5 Put y = 5 in (1): x - 5 = -2 x = 3 Answer: The fraction is 3/5.

Let present age of Nuri = N Let present age of Sonu = S Five years ago: N - 5 = 3(S - 5) → N - 5 = 3S - 15 → N - 3S = -10 ...(1) Ten years later: N + 10 = 2(S + 10) → N + 10 = 2S + 20 → N - 2S = 10 ...(2) Subtract (2) from (1): (N - 3S) - (N - 2S) = -10 - 10 N - 3S - N + 2S = -20 -S = -20 S = 20 Put S = 20 in (2): N - 2(20) = 10 N - 40 = 10 N = 50 Answer: Nuri is 50 years old, Sonu is 20 years old.

Let the digits be x (tens) and y (ones). From sum of digits: x + y = 9 ...(1) Number = 10x + y Reversed number = 10y + x Given: 9 × (10x + y) = 2 × (10y + x) → 90x + 9y = 20y + 2x → 90x - 2x = 20y - 9y → 88x = 11y → 8x = y ...(2) Put y = 8x in (1): x + 8x = 9 9x = 9 x = 1 Then y = 8 × 1 = 8 Answer: The number is 18.

Let number of ₹ 50 notes = x Let number of ₹ 100 notes = y Total notes: x + y = 25 ...(1) Total amount: 50x + 100y = 2000 Divide by 50: x + 2y = 40 ...(2) Subtract (1) from (2): (x + 2y) - (x + y) = 40 - 25 x + 2y - x - y = 15 y = 15 Put y = 15 in (1): x + 15 = 25 x = 10 Answer: Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Let fixed charge = x Charge for each extra day = y Saritha kept for 7 days: Fixed charge + 4 extra days = 27 x + 4y = 27 ...(1) Susy kept for 5 days: Fixed charge + 2 extra days = 21 x + 2y = 21 ...(2) Subtract (2) from (1): (x + 4y) - (x + 2y) = 27 - 21 2y = 6 y = 3 Put y = 3 in (2): x + 2(3) = 21 x + 6 = 21 x = 15 Answer: Fixed charge = ₹ 15, charge for each extra day = ₹ 3.