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Chittagong Board · SSC Mathematics · 2025

Chittagong 2025 Total: 100 Marks 3 Hours 8 Questions

Chapter Coverage

QChapterTopicMarks
1Ch 3 · Algebraic ExpressionsCube sum/difference identity10
2Ch 4 · Exponents & LogarithmsLog properties, navigation speed10
3Ch 5 · Simultaneous EquationsTwo ships meeting problem10
4Ch 9 · Finite SeriesAP with harbor loading context10
5Ch 13 · Circle TheoremsLighthouse arc angles10
6Ch 12 · Triangle SimilarityMap scale — Chittagong geography10
7Ch 14 · TrigonometryHill height — Chittagong Hill Tracts10
8Ch 16 · MensurationCylindrical water tower volume10
Question 1[10 marks]Ch 3 · Algebraic Expressions

Let a = 3 and b = 2.

(a) Show that (a+b)³ − (a−b)³ = 2b(3a² + b²). [2 marks]

(b) Compute (a+b)³ + (a−b)³ and express it in its simplest factored form. [4 marks]

(c) A port storage container has dimensions where the length exceeds the width by (a+b) metres and the width exceeds the height by (a−b) metres. If the height is 2 m, find the volume. [4 marks]

▶ Show Solution

(a) [2 marks]

Identity
We know (a+b)³ − (a−b)³ = [(a+b) − (a−b)][(a+b)² + (a+b)(a−b) + (a−b)²]
Simplify bracket 1
(a+b) − (a−b) = 2b
Expand bracket 2
(a+b)² + (a+b)(a−b) + (a−b)² = a²+2ab+b² + a²−b² + a²−2ab+b² = 3a²+b²
Result
∴ (a+b)³ − (a−b)³ = 2b(3a²+b²) ✓

(b) [4 marks]

Substitute a=3, b=2
(3+2)³ + (3−2)³ = 5³ + 1³ = 125 + 1 = 126
Factored form identity
(a+b)³ + (a−b)³ = [(a+b)+(a−b)][(a+b)²−(a+b)(a−b)+(a−b)²]
First factor
(a+b)+(a−b) = 2a = 6
Second factor
a²+2ab+b² − (a²−b²) + a²−2ab+b² = a²+3b² = 9+12 = 21
Answer
126 = 2a(a²+3b²) = 6×21 = 126 ✓

(c) [4 marks]

Dimensions
height h = 2, width = h + (a−b) = 2 + 1 = 3, length = width + (a+b) = 3 + 5 = 8
Volume
V = l × w × h = 8 × 3 × 2 = 48 m³ ✓
Question 2[10 marks]Ch 4 · Exponents & Logarithms

A cargo ship travels from Chittagong Port to Singapore. The navigator uses logarithms for calculations.

(a) Prove that log(1 + 1/n) = log(n+1) − log n. [2 marks]

(b) If the ship's speed S satisfies log S = (log 1250 − 2 log 5) / log 10, find S. [4 marks]

(c) The journey distance is 1800 nautical miles. If the ship travels the first half at speed S knots and the second half at (S+10) knots, find the total travel time in hours. (log 2 = 0.3010, log 3 = 0.4771) [4 marks]

▶ Show Solution

(a) [2 marks]

LHS
log(1 + 1/n) = log((n+1)/n)
Log quotient rule
= log(n+1) − log n ✓

(b) [4 marks]

Numerator
log 1250 − 2 log 5 = log 1250 − log 25 = log(1250/25) = log 50
log 50
log 50 = log(100/2) = log 100 − log 2 = 2 − 0.3010 = 1.6990
Denominator
log 10 = 1
Therefore
log S = 1.6990 / 1 = 1.6990 = log 50
Answer
S = 50 knots ✓

(c) [4 marks]

Each half
Distance per half = 1800/2 = 900 nautical miles
Time for first half
t₁ = 900/50 = 18 hours
Time for second half
t₂ = 900/(50+10) = 900/60 = 15 hours
Total time
T = 18 + 15 = 33 hours ✓
Question 3[10 marks]Ch 5 · Simultaneous Equations

Two ships leave Chittagong Port simultaneously. Ship A heads northeast at speed u km/h, Ship B heads northwest at speed v km/h. After 3 hours Ship A has traveled 15 km more than Ship B. Together they cover 165 km.

(a) Write two simultaneous equations in u and v. [2 marks]

(b) Solve the system to find u and v. [4 marks]

(c) If the two ships then travel towards each other from their positions and need to meet in 2 hours, what must their new combined closing speed be? [4 marks]

▶ Show Solution

(a) [2 marks]

Equation 1 (difference)
3u − 3v = 15 → u − v = 5
Equation 2 (total)
3u + 3v = 165 → u + v = 55

(b) [4 marks]

Add equations
2u = 60 → u = 30
Substitute
30 + v = 55 → v = 25
Answer
u = 30 km/h (Ship A), v = 25 km/h (Ship B) ✓

(c) [4 marks]

Distance A traveled
d_A = 30 × 3 = 90 km
Distance B traveled
d_B = 25 × 3 = 75 km
Gap between ships
Since ships went opposite directions, gap = d_A + d_B = 90 + 75 = 165 km
Required closing speed
Speed = Distance/Time = 165/2 = 82.5 km/h ✓
Question 4[10 marks]Ch 9 · Finite Series

The Chittagong Port Authority loads containers in a systematic schedule. On day 1 they load 40 containers. Each subsequent day they load 8 more containers than the previous day.

(a) Find the number of containers loaded on day 10. [2 marks]

(b) Find the total number of containers loaded in 15 days. [4 marks]

(c) On which day will the daily load first exceed 200 containers? [4 marks]

▶ Show Solution

(a) [2 marks]

AP formula
a = 40, d = 8, n = 10
T₁₀
T_n = a + (n−1)d = 40 + 9×8 = 40 + 72 = 112 containers ✓

(b) [4 marks]

Sum formula
S_n = n/2 × [2a + (n−1)d]
S₁₅
S₁₅ = 15/2 × [2×40 + 14×8] = 15/2 × [80 + 112] = 15/2 × 192
Answer
S₁₅ = 15 × 96 = 1440 containers ✓

(c) [4 marks]

Inequality
T_n > 200 → 40 + (n−1)×8 > 200
Solve
(n−1)×8 > 160 → n−1 > 20 → n > 21
Verify n=22
T₂₂ = 40 + 21×8 = 40 + 168 = 208 > 200 ✓
Answer
Day 22 is the first day the load exceeds 200 containers ✓
Question 5[10 marks]Ch 13 · Circle Theorems

A lighthouse at point L stands on a circular island. Three boats are anchored at points A, B, and C on the circumference of a circle of radius 7 km centered at L. The arc AB subtends an angle of 80° at the center.

(a) Find the angle ∠ACB inscribed in the major arc. [2 marks]

(b) If ∠AOC = 110° (O is center), find ∠ABC where A, B, C lie on the circle. [4 marks]

(c) Find the length of arc AB. (π = 22/7) [4 marks]

▶ Show Solution

(a) [2 marks]

Inscribed angle theorem
Inscribed angle = ½ × central angle
Angle ACB
∠ACB = 80°/2 = 40° ✓

(b) [4 marks]

Reflex ∠AOC
Reflex ∠AOC = 360° − 110° = 250°
Arc ABC
The arc AC not containing B subtends 110° at center. The arc AC containing B subtends 250°.
∠ABC (inscribed in major arc)
∠ABC = ½ × reflex ∠AOC = 250°/2 = 125° ✓

(c) [4 marks]

Arc length formula
Arc = (θ/360°) × 2πr
Substitute
Arc AB = (80/360) × 2 × (22/7) × 7
Calculate
= (2/9) × 44 = 88/9 ≈ 9.78 km ✓
Question 6[10 marks]Ch 12 · Triangle Similarity

On a map of Chittagong, the distance between Patenga Beach and Foy's Lake is 6 cm, while the actual distance is 12 km. The map shows a triangular region with vertices at Patenga (P), Agrabad (A), and Halishahar (H). On the map, PA = 4 cm, AH = 5 cm, and PH = 6 cm.

(a) Find the map scale as a ratio. [2 marks]

(b) Find the actual distances PA, AH, and PH in km. [4 marks]

(c) Show that the triangle PAH is a scalene triangle and find its perimeter in km. [4 marks]

▶ Show Solution

(a) [2 marks]

Scale calculation
6 cm on map = 12 km actual = 1,200,000 cm
Scale ratio
Scale = 1 : 200,000 ✓

(b) [4 marks]

PA actual
4 cm × 200,000 = 800,000 cm = 8 km
AH actual
5 cm × 200,000 = 1,000,000 cm = 10 km
PH actual
6 cm × 200,000 = 1,200,000 cm = 12 km
Answer
PA = 8 km, AH = 10 km, PH = 12 km ✓

(c) [4 marks]

Check sides
PA = 8 km, AH = 10 km, PH = 12 km — all three sides are different
Scalene proof
Since 8 ≠ 10 ≠ 12, triangle PAH is scalene ✓
Perimeter
P = 8 + 10 + 12 = 30 km ✓
Question 7[10 marks]Ch 14 · Trigonometry

A surveyor in the Chittagong Hill Tracts stands 500 m from the base of a hill. She observes the top of the hill at an angle of elevation of 36°52'. The base of the hill is at the same level as the surveyor.

(a) Write down the trigonometric relationship needed to find the hill's height. [2 marks]

(b) Find the height of the hill if tan 36°52' = 0.75. [4 marks]

(c) From the top of the hill, a river valley is seen at an angle of depression of 18°26'. Find the horizontal distance from the hilltop to the valley. (tan 18°26' = 1/3) [4 marks]

▶ Show Solution

(a) [2 marks]

Setup
Let h = height of hill, d = 500 m horizontal distance
Relationship
tan(elevation angle) = h/d → h = d × tan(36°52') ✓

(b) [4 marks]

Formula
h = 500 × tan 36°52'
Substitute
h = 500 × 0.75 = 375 m
Answer
Height of hill = 375 m ✓

(c) [4 marks]

Depression angle
tan(depression) = vertical drop / horizontal distance
Setup
tan 18°26' = h/x → 1/3 = 375/x
Solve
x = 375 × 3 = 1125 m ✓
Question 8[10 marks]Ch 16 · Mensuration

Chittagong City Corporation plans to build a cylindrical water tower. The tower has a base radius of 3.5 m and a total height of 10 m. The tower is topped with a hemispherical dome of the same radius.

(a) Find the volume of the cylindrical part. (π = 22/7) [2 marks]

(b) Find the volume of the hemispherical dome. [4 marks]

(c) Find the total surface area of the structure (curved surface of cylinder + curved surface of hemisphere + base circle only). [4 marks]

▶ Show Solution

(a) [2 marks]

Cylinder volume
V = πr²h = (22/7) × (3.5)² × 10
Calculate
= (22/7) × 12.25 × 10 = (22/7) × 122.5 = 22 × 17.5 = 385 m³ ✓

(b) [4 marks]

Hemisphere volume
V = (2/3)πr³ = (2/3) × (22/7) × (3.5)³
Calculate r³
(3.5)³ = 42.875
Final
V = (2/3) × (22/7) × 42.875 = (2/3) × (22 × 6.125) = (2/3) × 134.75 = 89.83 m³ ✓

(c) [4 marks]

CSA of cylinder
= 2πrh = 2 × (22/7) × 3.5 × 10 = 220 m²
CSA of hemisphere
= 2πr² = 2 × (22/7) × 12.25 = 77 m²
Base circle
= πr² = (22/7) × 12.25 = 38.5 m²
Total SA
220 + 77 + 38.5 = 335.5 m² ✓

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