Barisal Board · SSC Mathematics · 2021
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 1 – Real Numbers | Rationalizing river depth fractions | 10 |
| 2 | Ch 3 – Algebraic Expressions | Polynomial factorization, boat cargo | 10 |
| 3 | Ch 5 – Simultaneous Equations | River supply and irrigation allocation | 10 |
| 4 | Ch 8 – Arithmetic Series | Fish auction bid series | 10 |
| 5 | Ch 11 – Circle Theorems | Cyclic quadrilateral in river region | 10 |
| 6 | Ch 13 – Mensuration | Sphere-shaped buoy volume and surface | 10 |
| 7 | Ch 14 – Trigonometry | Tidal bore height, two-angle problem | 10 |
| 8 | Ch 15 – Statistics | Fish weight grouped data, mean deviation | 10 |
A river depth measurement gives d = 5/(2+√3) metres.
(a) [2m] Rationalize the denominator of 5/(2+√3).
(b) [4m] Show that 5/(2+√3) + 5/(2−√3) = 20. Find this value.
(c) [4m] If the river has two branches with depths 5/(2+√3) and 5/(2−√3), find the difference in their depths.
5/(2−√3) = 5(2+√3)/(4−3) = 10+5√3
Sum = (10−5√3)+(10+5√3) = 20 ✓
A boat's cargo hold has polynomial volume expression: V = x³ − 6x² + 11x − 6 cubic metres.
(a) [2m] Show that (x−1) is a factor by substituting x = 1.
(b) [4m] Factorize V completely using synthetic or long division.
(c) [4m] Find all values of x for which V = 0, and determine which gives a physically meaningful (positive) volume when x = 4.
x³−6x²+11x−6 = (x−1)(x²−5x+6)
x²−5x+6 = (x−2)(x−3)
V = (x−1)(x−2)(x−3)
At x=4: V = (4−1)(4−2)(4−3) = 3×2×1 = 6 m³ (positive, meaningful)
A river supplies 800 m³/day to two irrigation channels. Channel A receives twice as much as Channel B plus 50 m³.
(a) [2m] Write a system of equations. Let a = Channel A flow, b = Channel B flow.
(b) [4m] Solve the system.
(c) [4m] In the monsoon season, total supply doubles. If the ratio a:b remains the same, find new individual flows.
a = 2b + 50
a = 2(250)+50 = 550 m³/day
b_new = 1600 × 5/16 = 500 m³/day
a_new = 1600 × 11/16 = 1100 m³/day
Bids at a fish auction start at Tk 1000, each subsequent bid increases by Tk 250.
(a) [2m] List the first 5 bids and state the common difference.
(b) [4m] Find the 15th bid amount and the sum of first 15 bids.
(c) [4m] The auctioneer stops after a bid exceeds Tk 5000. What is the bid number and amount?
S₁₅ = 15/2 × [1000+4500] = 7.5 × 5500 = Tk 41,250
(n−1)×250 > 4000 → n−1 > 16 → n > 17
n = 18th bid
Amount = 1000+17×250 = 1000+4250 = Tk 5,250
ABCD is a cyclic quadrilateral inscribed in a circle. Angle A = 3x° and angle C = (x + 40)°.
(a) [2m] State the property of opposite angles in a cyclic quadrilateral.
(b) [4m] Find x, angle A, and angle C.
(c) [4m] If angle B = 95°, find angle D. Also find the arc BD in terms of inscribed angle.
4x + 40 = 180 → 4x = 140 → x = 35°
∠A = 3×35 = 105°; ∠C = 35+40 = 75°
Arc BD (not containing A or C) subtends inscribed ∠A = 105° at the circle.
Central angle for arc BD = 2×105° = 210° (reflex arc)
A spherical buoy floating on the Barisal river has a diameter of 0.7 m. (π = 22/7)
(a) [2m] Find the radius and the surface area of the sphere.
(b) [4m] Find the volume of the buoy.
(c) [4m] If the buoy is to be painted, and one tin covers 1.5 m², how many tins are needed?
SA = 4πr² = 4 × (22/7) × 0.35² = 4 × (22/7) × 0.1225
= 4 × 22 × 0.0175 = 4 × 0.385 = 1.54 m²
= (4/3) × (22/7) × 0.042875
= (4/3) × 0.13475 = 0.1797 ≈ 0.18 m³
A tidal bore wave is observed from two riverbank points 100 m apart in a straight line. From Point P, the angle of elevation of the wave crest is 45°. From Point Q (100 m further from the wave), the angle is 30°.
(a) [2m] Let the wave height be h and distance from Q to the wave base be d. Write two equations.
(b) [4m] Solve for h. (tan 45° = 1, tan 30° = 1/√3)
(c) [4m] Find the distance PQ and verify the 100 m separation.
From Q (farther): tan 30° = h/d → h = d/√3
d − d/√3 = 100
d(1 − 1/√3) = 100
d × (√3−1)/√3 = 100
d = 100√3/(√3−1) = 100√3(√3+1)/((√3−1)(√3+1)) = 100√3(√3+1)/2
= 50√3(√3+1) = 50(3+√3) ≈ 50(3+1.732) = 50 × 4.732 = 236.6 m
h = d−100 = 136.6 m
Alternatively: h = d/√3 = 236.6/1.732 ≈ 136.6 m
Separation = 236.6−136.6 = 100 m ✓
Weight of fish (kg) caught in 30 hauls: 2–4: 4 hauls, 4–6: 8, 6–8: 10, 8–10: 6, 10–12: 2.
(a) [2m] Find the total number of hauls and the modal class.
(b) [4m] Calculate the mean weight per haul using midpoints.
(c) [4m] Calculate the mean absolute deviation (MAD) from the mean.
Modal class = 6–8 kg (highest frequency = 10)
Σfx = 4×3+8×5+10×7+6×9+2×11 = 12+40+70+54+22 = 198
Mean = 198/30 = 6.6 kg
Σf|d| = 4×3.6+8×1.6+10×0.4+6×2.4+2×4.4
= 14.4+12.8+4.0+14.4+8.8 = 54.4
MAD = 54.4/30 = 1.813 kg