Barisal Board · SSC Mathematics · 2023
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 2 – Sets & Functions | Fishermen survey, Venn diagram | 10 |
| 2 | Ch 3 – Algebraic Expressions | Boat deck area polynomial | 10 |
| 3 | Ch 5 – Simultaneous Equations | Two rivers, flow rate system | 10 |
| 4 | Ch 8 – Arithmetic Series | Water level rise AP | 10 |
| 5 | Ch 11 – Circle Theorems | Inscribed angle in semicircle | 10 |
| 6 | Ch 13 – Mensuration | Cone-shaped river bank erosion pile | 10 |
| 7 | Ch 14 – Trigonometry | Angle of elevation from boat to bridge | 10 |
| 8 | Ch 15 – Statistics | River flow velocity data, mean & variance | 10 |
In a fishing village of 80 fishermen: 45 own nets, 30 own boats, and 15 own both nets and boats.
(a) [2m] Draw a Venn diagram for this information.
(b) [4m] Find the number who own only nets, only boats, and neither.
(c) [4m] What fraction of all fishermen own at least one of nets or boats?
N = nets, B = boats. N∩B = 15.
Only boats = 30 − 15 = 15
N∪B = 30 + 15 + 15 = 60
Neither = 80 − 60 = 20 fishermen
Fraction = 60/80 = 3/4
A boat deck has length (3x + 4) m and width (2x − 1) m. The deck area is 77 m².
(a) [2m] Expand (3x+4)(2x−1).
(b) [4m] Form and solve a quadratic equation to find x.
(c) [4m] Find the actual length and width of the deck. Also factorize 6x² + 5x − 4 completely.
6x² + 5x − 81 = 0
Using quadratic formula: x = [−5 ± √(25 + 4×6×81)] / 12
= [−5 ± √(25+1944)] / 12 = [−5 ± √1969] / 12
√1969 ≈ 44.37
x = (−5+44.37)/12 ≈ 3.28 or negative (rejected)
x ≈ 3.28
Check: try integer — 6(3)²+5(3)−81 = 54+15−81 = −12 ≠ 0; x=3.28 confirmed
Factorize 6x²+5x−4: find ac = −24, need two numbers with product −24 and sum 5 → 8 and −3
6x²+8x−3x−4 = 2x(3x+4) − 1(3x+4) = (2x−1)(3x+4)
River A and River B together supply 1200 litres/min to a delta. River A supplies 200 L/min more than three times River B's flow.
(a) [2m] Write a system of two equations using variables a and b.
(b) [4m] Solve by substitution to find individual flow rates.
(c) [4m] If River A's flow doubles during monsoon, find the new total supply.
a = 3b + 200
4b = 1000 → b = 250 L/min
a = 3(250)+200 = 950 L/min
New total = 1900 + 250 = 2150 L/min
During monsoon, the water level of a Barisal canal rises arithmetically: 0.3 m on day 1, increasing by 0.15 m each day.
(a) [2m] Find the water level rise on day 7.
(b) [4m] Find the total rise over 14 days.
(c) [4m] The danger level is 5 m total rise. On which day is this total first reached or exceeded?
n[0.6+0.15(n−1)] ≥ 10
n[0.6+0.15n−0.15] ≥ 10
n[0.45+0.15n] ≥ 10
0.15n² + 0.45n − 10 ≥ 0
n² + 3n − 66.67 ≥ 0
n = [−3 + √(9+266.67)]/2 = [−3 + √275.67]/2 ≈ [−3+16.6]/2 ≈ 6.8
Danger level reached on day 7
AB is a diameter of a circle with centre O. C is a point on the circle such that angle BAC = 35°.
(a) [2m] State the angle in a semicircle theorem and find angle ACB.
(b) [4m] Find angle ABC and angle AOC.
(c) [4m] If the circle has radius 7 cm, find the length of chord AC. (sin 35° ≈ 0.574)
∴ ∠ACB = 90°
35° + 90° + ∠ABC = 180°
∠ABC = 55°
∠AOC = 2 × ∠ABC = 2 × 55° = 110° (central angle = twice inscribed angle subtending same arc)
AC/AB = sin(∠ABC) = sin 55° ≈ 0.819
Alternatively: using ∠BAC = 35° and AB = 14:
By sine rule: AC/sin(∠ABC) = AB/sin(∠ACB)
AC/sin 55° = 14/sin 90° = 14
AC = 14 × sin 55° = 14 × 0.819 ≈ 11.47 cm
Erosion has created a conical pile of sand on a river bank with base radius 6 m and height 8 m.
(a) [2m] Find the slant height of the cone. (π = 22/7)
(b) [4m] Find the curved surface area and total surface area of the cone.
(c) [4m] Find the volume of sand in the pile.
Base area = πr² = (22/7) × 36 = 792/7 ≈ 113.14 m²
Total SA = 188.57 + 113.14 = 301.71 m²
A boat on a river looks up at a bridge. From a point 30 m away from the base of the bridge pillar, the angle of elevation of the bridge deck is 45°. From a point 20 m further back, the angle of elevation is 30°.
(a) [2m] Let the bridge height be h. Write two equations using tan 45° and tan 30°.
(b) [4m] Show that these two equations are consistent and find h.
(c) [4m] Find the distance between the two observation points using the two expressions for h. (tan 30° = 1/√3)
From 50 m: tan 30° = h/50 → h = 50/√3 = 50√3/3 ≈ 28.87
Let h = bridge height, d = distance to first point.
tan 45° = h/d → h = d
tan 30° = h/(d+20) → 1/√3 = d/(d+20)
d+20 = d√3 → 20 = d(√3−1) → d = 20/(√3−1) = 20(√3+1)/2 = 10(√3+1) ≈ 27.32 m
h = 10(√3+1) ≈ 27.32 m
Verify using h: tan 45° = 27.32/27.32 = 1 ✓; tan 30° = 27.32/47.32 = 0.577 ≈ 1/√3 ✓
River flow velocities (m/s) measured at 7 locations: 1.2, 1.8, 2.4, 1.6, 2.0, 1.4, 2.2.
(a) [2m] Find the mean velocity.
(b) [4m] Calculate the variance.
(c) [4m] Find the standard deviation and comment on consistency of flow.
Mean = 12.6/7 = 1.8 m/s
Squared deviations: 0.36, 0, 0.36, 0.04, 0.04, 0.16, 0.16
Sum = 1.12
Variance = 1.12/7 = 0.16 m²/s²
CV = σ/mean = 0.4/1.8 = 22.2%
Comment: moderate variability in river flow; reasonably consistent for navigational purposes.