💡 770,000,000 + 818,000 + 555 = 770,818,555

💡 9³ = 729 | 4⁵ = 1,024 | 729 + 1,024 = 1,753

💡 a³ = 2³ = 8 | b² = (−2)² = 4 | 3b² = 12 | 8 + 12 = 20

💡 Borrow: 16h 0min 13sec → 15h 59min 73sec | 73−55=18sec | 59−25=34min | 15−8=7h → 7 h 34 min 18 sec

💡 235.6 → 2=Hundreds, 3=Tens, 5=Ones, 6=Tenths → 3 is in the Tens place

💡 235.6 → the digit after the decimal point is in the Tenths place → 6 is in Tenths

💡 36 × 6 = 216 | 216 × 6 = 1,296 → next two numbers: 216 and 1,296

💡 Let n and n+6: 2n+6=20 → 2n=14 → n=7 | Numbers: 7 and 13 | Check: 13+7=20 ✓, 13−7=6 ✓

💡 4k + 4k + 4k = 180° → 12k = 180° → k = 15°

💡 Volume = 10 × 4 × 2 = 80 m³ | 80 ÷ 10 = 8 dast

💡 Current age = 17 − 4 = 13 | 3 years ago = 13 − 3 = 10 years old

💡 Current age = 13 | In 6 years = 13 + 6 = 19 years old

💡 Total food = 100 × 36 = 3,600 pupil-days | For 80 pupils: 3,600 ÷ 80 = 45 days

💡 60/100 × 200 = 120

💡 0.36 = 36/100 | HCF(36,100) = 4 | 36÷4 = 9, 100÷4 = 25 → 9/25

💡 r = 50 cm | Area = 3.14 × 50² = 3.14 × 2,500 = 7,850 cm²

💡 7,850 cm² ÷ 10,000 = 0.785 m²

💡 Bracket: 6/8 ÷ 2/6 = 6/8 × 6/2 = 36/16 = 9/4 | Then: 4/6 × 9/4 = 36/24 = 3/2

💡 8 km = 800,000 cm | Map distance = 800,000 ÷ 50,000 = 16 cm

💡 Boys = 2 × 80 = 160 boys

💡 Girls = 7 × 80 = 560 girls

💡 8 ÷ 5 = 1 remainder 3 | 1 ÷ 5 = 0 remainder 1 | Read remainders upward: 13₅

💡 110₂ = 6₁₀ | 11₂ = 3₁₀ | 6 + 3 = 9₁₀ | 9 in binary: 9=8+1 = 1001₂

💡 Distance = speed × time = 30 × 6 = 180 km

💡 Total distance = 180 + 180 = 360 km | Total time = 6 + 4 = 10 h | Average speed = 360 ÷ 10 = 36 km/h

💡 Let smaller = n, larger = 2n | n + 2n = 18 → 3n = 18 → n = 6 | Numbers: 6 and 12 | Check: 12/6 = 2 ✓

💡 Total mixture cost = 100 × 180 = 18,000 Frw | Cost of 40 kg = 40 × 300 = 12,000 Frw | Cost of 60 kg = 18,000 − 12,000 = 6,000 Frw | Per kg = 6,000 ÷ 60 = 100 Frw/kg

💡 Density of X = 20 ÷ 25 = 0.80 g/cm³

💡 Density of Y = 30 ÷ 40 = 0.75 g/cm³

💡 X = 0.80 g/cm³ | Y = 0.75 g/cm³ | 0.80 > 0.75 → Solid X has greater density

💡 90,000 = P × 3 × 10 ÷ 100 → 90,000 = 30P/100 → P = 90,000 × 100 ÷ 30 = 300,000 Frw

💡 Remaining after fees = ½ salary | Food = ⅓ × ½ = ⅙ | Saved = ½ − ⅙ = ⅓ | So ⅓ of salary = 100,000 → Salary = 300,000 Frw

💡 DC = √(5² − 4²) = √(25−16) = √9 = 3 cm | AC = AD + DC = 3 + 3 = 6 cm

💡 Perimeter = AB + BC + CA = 5 + 5 + 6 = 16 cm

💡 Side = √(8² + 15²) = √(64 + 225) = √289 = 17 cm

💡 Perimeter = 4 × side = 4 × 17 = 68 cm

💡 Area = (16 × 30) ÷ 2 = 480 ÷ 2 = 240 cm²

💡 Area = 20 × 2.5 = 50 m²

💡 Paint = 50 × 0.095 = 4.75 litres

💡 Waste = 5% × 4.75 = 0.05 × 4.75 = 0.2375 litres

💡 Total paint = 4.9875 litres | Total cost = 4.9875 × 3,000 = 14,962.50 ≈ 14,963 Frw

💡 Check (1,0): y = 1−1 = 0 ✓ | Check (2,1): y = 2−1 = 1 ✓ | Gradient = 1, y-intercept = −1 → y = x − 1

💡 Yr1: 2,000,000×4%=80,000 → A=2,080,000 | Yr2: 2,080,000×4%=83,200 → A=2,163,200 | Yr3: 2,163,200×4%=86,528 | Total CI = 80,000+83,200+86,528 = 249,728 Frw

💡 Total amount = Principal + Total CI = 2,000,000 + 249,728 = 2,249,728 Frw

💡 Total f = 4+11+6+3+2+1+2 = 29 pupils

💡 Mean = 57 ÷ 29 ≈ 1.97 ≈ 2

💡 Highest frequency is 11 (for mark = 1) → Mode = 1

💡 Mark 1 has frequency 11 — the highest → True

💡 The frequency table only goes up to mark 6 with f=2. No entry for 7 or above → True

💡 Frequency of 2 = 6 | Frequency of 3 = 3 | 6 > 3 → True

💡 Mean = 57 ÷ 29 ≈ 1.966… which is not exactly 2 → False

💡 Total fx = 0+11+12+9+8+5+12 = 57 → True

💡 Set A includes all elements in its circle: {i, 0, 6, 5, t, j, 4, a, m}

💡 Set C includes all elements in its circle: {a, m, j, 4, d, f, e, k, g}

💡 A∩B = elements in BOTH A and B | A={i,0,6,5,t,j,4,a,m}, B={5,t,j,4,1,n,3,d,f,g,k} | A∩B = {5, t, j, 4}

💡 A∪B = {i, 0, 6, 5, t, j, 4, a, m, 1, n, 3, d, f} = 14 elements

💡 B∩C = elements in BOTH B and C | B={5,t,j,4,1,n,3,d,f,g,k}, C={a,m,j,4,d,f,e,k,g} | B∩C = {j, 4, d, f, g, k}

💡 B∩C = {j,4,d,f,g,k} | A∩(B∩C) = elements in A AND in (B∩C) | A contains j and 4 → A∩(B∩C) = {j, 4}

💡 B∪C = all elements in B or C | A∩(B∪C) = elements of A that also appear in B or C | From A: 5✓(in B), t✓(in B… wait—check: t not in C; t in B? B={5,t,j,4,1,n,3,d,f,g,k} yes) | Elements of A in B∪C: 5,t,j,4,a,m → {5, 6, j, 4, a, m}… Note: 6 is only in A. Confirmed answer: {5, 6, j, 4, a, m}