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Chittagong Board · SSC Mathematics · 2024

Chittagong 2024 Total: 100 Marks 3 Hours 8 Questions

Chapter Coverage

QChapterTopicMarks
1Ch 3 · Algebraic ExpressionsSophie Germain identity — x⁴+410
2Ch 4 · LogarithmsLog properties and simplification10
3Ch 6 · Quadratic EquationsBoat speed problem10
4Ch 9 · SeriesGeometric progression — practical context10
5Ch 11 · Lines & AnglesParallel lines and transversal angle proof10
6Ch 13 · CircleCyclic quadrilateral angle problems10
7Ch 14 · TrigonometryHarbor crane height10
8Ch 15 · StatisticsFish catch data analysis10
Question 1[10 marks]Ch 3 · Algebraic Expressions

Consider the expression x⁴ + 4y⁴.

(a) Factorize x⁴ + 4 completely using the Sophie Germain identity. [2 marks]

(b) Factorize x⁴ + 4y⁴ into two quadratic factors. [4 marks]

(c) If x = 2 and y = 1, verify your factorization numerically and find the value of each factor. [4 marks]

▶ Show Solution

(a) [2 marks]

Sophie Germain identity
a⁴ + 4b⁴ = (a²+2b²+2ab)(a²+2b²−2ab). Here a = x, b = 1.
Result
x⁴+4 = (x²+2x+2)(x²−2x+2) ✓

(b) [4 marks]

Apply with b = y
x⁴+4y⁴ = (x²+2y²+2xy)(x²+2y²−2xy)
Verification
Expanding (x²+2y²+2xy)(x²+2y²−2xy) = (x²+2y²)² − (2xy)² = x⁴+4x²y²+4y⁴ − 4x²y² = x⁴+4y⁴ ✓
Answer
x⁴+4y⁴ = (x²+2y²+2xy)(x²+2y²−2xy) ✓

(c) [4 marks]

LHS (x=2, y=1)
2⁴ + 4(1)⁴ = 16 + 4 = 20
Factor 1
(4+2+4) = 10
Factor 2
(4+2−4) = 2
Check
10 × 2 = 20 = LHS ✓
Question 2[10 marks]Ch 4 · Logarithms

Use logarithm laws to simplify and evaluate the following expressions. (log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451)

(a) Prove that log(ab²) = log a + 2 log b. [2 marks]

(b) Simplify: log 12 + log 18 − log 6 + log 25 − log 75. [4 marks]

(c) Find the value of: log₅ 8 × log₂ 625. [4 marks]

▶ Show Solution

(a) [2 marks]

Product rule
log(ab²) = log a + log b²
Power rule
= log a + 2 log b ✓

(b) [4 marks]

Combine logs
log(12 × 18 / 6 × 25 / 75) = log(12×18×25 / 6×75)
Numerator
12×18×25 = 5400
Denominator
6×75 = 450
Simplify
5400/450 = 12
Answer
log 12 = log(4×3) = 2 log 2 + log 3 = 0.6020+0.4771 = 1.0791 ✓

(c) [4 marks]

Change of base
log₅ 8 = log 8/log 5, log₂ 625 = log 625/log 2
Product
log₅ 8 × log₂ 625 = (log 8/log 5) × (log 625/log 2)
Simplify
log 8 = log 2³ = 3 log 2, log 625 = log 5⁴ = 4 log 5
Cancel
= (3 log 2/log 5) × (4 log 5/log 2) = 3×4 = 12 ✓
Question 3[10 marks]Ch 6 · Quadratic Equations

A motorboat travels 36 km upstream and 36 km downstream. The total journey takes 9 hours. The speed of the river current is 3 km/h.

(a) If the boat's speed in still water is x km/h, write expressions for the upstream and downstream speeds. [2 marks]

(b) Form a quadratic equation for x and solve it. [4 marks]

(c) Find the time taken for each leg of the journey. [4 marks]

▶ Show Solution

(a) [2 marks]

Speeds
Upstream speed = (x−3) km/h, Downstream speed = (x+3) km/h ✓

(b) [4 marks]

Time equation
36/(x−3) + 36/(x+3) = 9
Multiply through
36(x+3) + 36(x−3) = 9(x−3)(x+3)
Simplify LHS
36x+108+36x−108 = 72x
Simplify RHS
9(x²−9) = 9x²−81
Equation
9x²−72x−81 = 0 → x²−8x−9 = 0
Factor
(x−9)(x+1) = 0 → x = 9 or x = −1
Answer
x = 9 km/h (rejecting negative) ✓

(c) [4 marks]

Upstream time
36/(9−3) = 36/6 = 6 hours
Downstream time
36/(9+3) = 36/12 = 3 hours
Check total
6 + 3 = 9 hours ✓
Question 4[10 marks]Ch 9 · Series

A shipping company at Chittagong Port starts with 4 vessels and doubles its fleet every 3 years.

(a) Write the geometric sequence for the number of vessels after 0, 3, 6, 9 years. [2 marks]

(b) How many vessels will the company have after 18 years? [4 marks]

(c) In how many years will the fleet first exceed 100 vessels? (log 2 = 0.3010) [4 marks]

▶ Show Solution

(a) [2 marks]

Sequence
4, 8, 16, 32 (with first term a = 4, common ratio r = 2) ✓

(b) [4 marks]

Number of terms
18 years = 6 periods of 3 years, so n = 7 (including start)
GP formula
T_n = ar^(n−1) = 4 × 2⁶ = 4 × 64
Answer
256 vessels ✓

(c) [4 marks]

Inequality
4 × 2^k > 100 where k is number of 3-year periods
Simplify
2^k > 25 → k log 2 > log 25
log 25
log 25 = log(100/4) = 2 − 2×0.3010 = 1.3980
Solve k
k > 1.3980/0.3010 = 4.644 → k = 5 periods
Answer
5 × 3 = 15 years (fleet = 4×2⁵ = 128 > 100) ✓
Question 5[10 marks]Ch 11 · Lines & Angles

Two parallel lines PQ and RS are cut by a transversal TU at points A and B respectively.

(a) State the relationship between alternate interior angles when two parallel lines are cut by a transversal. [2 marks]

(b) If ∠PAT = 65°, find all other angles formed at point A and point B. [4 marks]

(c) Prove that co-interior angles (same-side interior angles) are supplementary. [4 marks]

▶ Show Solution

(a) [2 marks]

Theorem
When two parallel lines are cut by a transversal, alternate interior angles are equal. ✓

(b) [4 marks]

∠PAT = 65°
∠QAT = 180°−65° = 115° (linear pair)
Vertically opposite
∠PAU = 115°, ∠QAU = 65°
At B (alternate)
∠RBU = ∠PAU = 65° (alternate interior angles, PQ ∥ RS)
Remaining at B
∠SBU = 65°, ∠RBT = 115°, ∠SBT = 115° ✓

(c) [4 marks]

Given
PQ ∥ RS, transversal TU. Let ∠QAB = α (interior angle at A on same side)
Alternate angles
∠ABR = ∠PAB (alternate interior angles) = 180°−α (since ∠PAB+∠QAB = 180°)
Co-interior sum
∠QAB + ∠ABR = α + (180°−α) = 180°
Conclusion
∴ Co-interior angles are supplementary ✓
Question 6[10 marks]Ch 13 · Circle

ABCD is a cyclic quadrilateral inscribed in a circle. ∠A = (3x+10)° and ∠C = (5x−30)°.

(a) State the property of opposite angles in a cyclic quadrilateral. [2 marks]

(b) Find x and then find all four angles if ∠B = (2y+15)° and ∠D = (3y−5)°. [4 marks]

(c) Find the angle ∠B subtended by arc DAB at the center, if ∠DAB = 75°. [4 marks]

▶ Show Solution

(a) [2 marks]

Property
Opposite angles of a cyclic quadrilateral are supplementary (sum = 180°). ✓

(b) [4 marks]

∠A + ∠C = 180°
(3x+10) + (5x−30) = 180 → 8x−20 = 180 → x = 25
Angles A and C
∠A = 85°, ∠C = 95°
∠B + ∠D = 180°
(2y+15)+(3y−5) = 180 → 5y+10 = 180 → y = 34
Angles B and D
∠B = 83°, ∠D = 97° ✓

(c) [4 marks]

Central angle theorem
Central angle = 2 × inscribed angle
Arc DAB
∠DAB = 75° is the inscribed angle subtended by arc DCB (the arc not containing A)
Central angle for arc DCB
= 2 × 75° = 150°
Central angle for arc DAB
= 360° − 150° = 210° ✓
Question 7[10 marks]Ch 14 · Trigonometry

A harbor crane at Chittagong Port has its base at ground level. A ship captain observes the crane's top from a point 80 m away at sea level. The angle of elevation is 51°20'.

(a) Set up the trigonometric equation to find the crane height. [2 marks]

(b) Find the height of the crane. (tan 51°20' = 1.25) [4 marks]

(c) Find the straight-line distance from the observation point to the crane's top. (sin 51°20' = 0.7808) [4 marks]

▶ Show Solution

(a) [2 marks]

Setup
Let h = crane height, d = 80 m
Equation
tan(51°20') = h/80 → h = 80 × tan(51°20') ✓

(b) [4 marks]

Calculate
h = 80 × 1.25 = 100 m
Answer
Height of crane = 100 m ✓

(c) [4 marks]

Using sin
sin(51°20') = h/L → L = h/sin(51°20')
Calculate
L = 100/0.7808 ≈ 128.1 m
Alternative (Pythagoras)
L = √(80²+100²) = √(6400+10000) = √16400 ≈ 128.1 m
Answer
Slant distance ≈ 128.1 m ✓
Question 8[10 marks]Ch 15 · Statistics

A fishing cooperative records daily catches (in tonnes) for 10 days: 12, 15, 18, 14, 20, 16, 22, 19, 17, 17.

(a) Find the mean daily catch. [2 marks]

(b) Find the median of the data. [4 marks]

(c) Calculate the mean deviation from the mean. [4 marks]

▶ Show Solution

(a) [2 marks]

Sum
12+15+18+14+20+16+22+19+17+17 = 170
Mean
x̄ = 170/10 = 17 tonnes ✓

(b) [4 marks]

Sort data
12, 14, 15, 16, 17, 17, 18, 19, 20, 22
n = 10 (even)
Median = (5th + 6th)/2 = (17+17)/2
Answer
Median = 17 tonnes ✓

(c) [4 marks]

Deviations |xᵢ − x̄|
|12−17|=5, |15−17|=2, |18−17|=1, |14−17|=3, |20−17|=3, |16−17|=1, |22−17|=5, |19−17|=2, |17−17|=0, |17−17|=0
Sum of deviations
5+2+1+3+3+1+5+2+0+0 = 22
Mean deviation
MD = 22/10 = 2.2 tonnes ✓

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