Barisal Board · SSC Mathematics · 2025
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 3 – Algebraic Expressions | River boat dimensions, polynomial factoring | 10 |
| 2 | Ch 4 – Logarithms | Fishery production calculations | 10 |
| 3 | Ch 5 – Simultaneous Equations | Two boats, speed-current problem | 10 |
| 4 | Ch 9 – Geometric Series | Tidal water level pattern | 10 |
| 5 | Ch 11 – Circle Theorems | Riverside well circumference | 10 |
| 6 | Ch 13 – Mensuration | Boat hull volume, half-cylinder | 10 |
| 7 | Ch 14 – Trigonometry | River bank and ship mast elevation | 10 |
| 8 | Ch 15 – Statistics | Monthly rainfall data analysis | 10 |
A wooden boat has length x m and width y m such that the length is 3 m more than twice the width, and x² − 4y² = 24.
(a) [2m] Express x in terms of y using the first condition.
(b) [4m] Substitute into the equation x² − 4y² = 24 and find the width y.
(c) [4m] Factorize the expression x² − 4y² completely and verify your answer.
∴ x = 2y + 3
(2y+3)² − 4y² = 24
4y² + 12y + 9 − 4y² = 24
12y + 9 = 24
12y = 15
y = 5/4 = 1.25 m
∴ x = 2(1.25)+3 = 5.5 m
Using difference of squares: a²−b² = (a+b)(a−b)
= (x + 2y)(x − 2y)
Verify: substitute x=5.5, y=1.25
(5.5+2.5)(5.5−2.5) = 8 × 3 = 24 ✓
A fishery in Barisal produces 500 kg of fish in January. Each month production increases by a factor of 1.2. Using logarithms:
(a) [2m] Write the production in month n as a formula.
(b) [4m] Find in which month production first exceeds 2000 kg. (Use log 2 = 0.3010, log 1.2 = 0.0792)
(c) [4m] Prove that log(500 × 1.2ⁿ) = log 500 + n·log 1.2 and evaluate for n = 8.
1.2^(n−1) > 4
Take log both sides:
(n−1)·log 1.2 > log 4 = 2·log 2
(n−1) × 0.0792 > 2 × 0.3010 = 0.6020
n−1 > 0.6020 / 0.0792 = 7.60
n > 8.60
First month is n = 9 (September)
= log 500 + n·log 1.2 [power rule] ✓
For n=8: log 500 = log(5×100) = log 5 + 2 = (1−log 2)+2 = 2.6990
+ 8 × 0.0792 = 0.6336
Total = 3.3326 → P₈ = 10^3.3326 ≈ 2150 kg
Two boats travel on the Kirtankhola river. Boat A travels upstream at speed (v − c) and Boat B travels downstream at speed (v + c), where v is the boat's speed in still water and c is the current speed. Boat A covers 48 km in 4 hours; Boat B covers 72 km in 3 hours.
(a) [2m] Write two equations from the given information.
(b) [4m] Solve for v and c using elimination or substitution.
(c) [4m] Find how long Boat A takes to travel 60 km downstream.
Downstream speed = 72/3 = 24 km/h → v + c = 24
Subtract: 2c = 12 → c = 6 km/h
Time = distance/speed = 60/24 = 2.5 hours
During monsoon, tidal water levels in a Barisal estuary rise in a geometric pattern. At 6 AM the level is 2 m; each subsequent hour the level is multiplied by a ratio r = 1.5.
(a) [2m] Write the first 4 terms of this geometric sequence.
(b) [4m] Find the sum of water levels over the first 6 hours.
(c) [4m] The embankment is at height 20 m. In which hour does the water first exceed the embankment height?
Terms: 2, 3, 4.5, 6.75 m
S₆ = 2(1.5⁶−1)/(1.5−1)
1.5⁶ = 11.3906
S₆ = 2(10.3906)/0.5 = 41.5625 m
1.5^(n−1) > 10
Take log: (n−1)·log1.5 > log10 = 1
log 1.5 = log(3/2) = 0.4771−0.3010 = 0.1761
(n−1) > 1/0.1761 = 5.68
n > 6.68 → n = 7, i.e., at hour 7 (12 noon)
A circular well on the riverside has centre O. Points A and B are on the circumference. The angle AOB = 120°. A tangent at point A meets the line OB extended at point T.
(a) [2m] State the angle between radius OA and tangent AT.
(b) [4m] Find the angle OAT and angle ATB, given angle AOB = 120°.
(c) [4m] If the well radius is 3.5 m, find the circumference and the arc length AB. (π = 22/7)
∴ ∠OAT = 90°
Wait — angle AOB = 120°, so angle AOT (OT along OB extended) = 180°−120° = 60°? No:
T is on OB extended beyond B, so ∠AOT = ∠AOB = 120° (same ray from O through B to T)
In △OAT: ∠OAT + ∠AOT + ∠ATB... Let ∠ATB = α
∠OAT = 90° (radius ⊥ tangent)
∠AOT = 120° — but this makes sum > 180°.
Correction: angle AOT = angle AOB = 120°. In △OAT: 90 + 120 + α = 180 is impossible.
T must be outside: ∠OAT = 90°, ∠OTA = ?, ∠AOT = 180°−120° = 60°
(OB extended means T is beyond B; angle between OA and OT = 180°−120° = 60°)
∠OTA = 180°−90°−60° = 30°
∠ATB = 30° (same angle)
∠OAB: OA=OB (radii), so △OAB is isosceles, base angles = (180°−120°)/2 = 30°
∠OAB = 30°
Arc AB: fraction of circle = 120/360 = 1/3
Arc AB = (1/3) × 22 = 7.33 m
A wooden boat hull is approximated as a half-cylinder of length 12 m and radius 1.4 m. A layer of paint covers the entire outer curved surface and the two semicircular ends.
(a) [2m] Find the curved surface area of the full cylinder.
(b) [4m] Find the total outer surface area of the half-cylinder hull (curved + two half-circles). (π = 22/7)
(c) [4m] Find the volume of water displaced (= volume of half-cylinder) and the weight of water displaced if 1 m³ of water weighs 1000 kg.
= 2 × (22/7) × 16.8
= 2 × 22 × 2.4 = 105.6 m²
Two semicircular ends = 2 × (1/2)πr² = πr² = (22/7) × 1.4² = (22/7) × 1.96 = 6.16 m²
Total = 52.8 + 6.16 = 58.96 m²
= (1/2) × 73.92 = 36.96 m³
Weight of water displaced = 36.96 × 1000 = 36,960 kg
From the top of a river bank 15 m high, a sailor observes a ship's mast. The angle of elevation of the top of the mast is 30° and the angle of depression of the base of the ship (at water level) is 45°.
(a) [2m] Draw and label a diagram showing the bank, water level, and ship.
(b) [4m] Find the horizontal distance from the bank to the ship.
(c) [4m] Find the height of the ship's mast. (tan 30° = 1/√3, tan 45° = 1)
AB = 15 m (vertical). D = base of ship (water level), E = top of mast.
BD = horizontal distance = d. Angle of depression to D = 45°. Angle of elevation to E = 30°.
tan 45° = AB/BD = 15/d
1 = 15/d
d = 15 m
Height of E above A's level = h − 15 (if h > 15, else below)
Angle of elevation of E from A = 30°
tan 30° = (h − 15)/d = (h−15)/15
1/√3 = (h−15)/15
h − 15 = 15/√3 = 15√3/3 = 5√3 ≈ 8.66
h = 15 + 5√3 ≈ 23.66 m
Monthly rainfall (mm) in Barisal for 8 months: 120, 145, 200, 310, 420, 380, 250, 160.
(a) [2m] Find the range and median of the data.
(b) [4m] Calculate the mean monthly rainfall.
(c) [4m] Calculate the mean absolute deviation (MAD) from the mean.
Range = 420 − 120 = 300 mm
Median = average of 4th and 5th = (200+250)/2 = 225 mm
Mean = 1985/8 = 248.125 mm
|120−248.125| = 128.125
|145−248.125| = 103.125
|200−248.125| = 48.125
|310−248.125| = 61.875
|420−248.125| = 171.875
|380−248.125| = 131.875
|250−248.125| = 1.875
|160−248.125| = 88.125
Sum = 735
MAD = 735/8 = 91.875 mm