Barisal Board · SSC Mathematics · 2024
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 1 – Real Numbers | Irrational numbers, river measurement surds | 10 |
| 2 | Ch 3 – Algebraic Expressions | Irrigation canal cross-section polynomial | 10 |
| 3 | Ch 6 – Quadratic Equations | Rectangular fish pond dimensions | 10 |
| 4 | Ch 8 – Arithmetic Series | Embankment construction, daily bricks laid | 10 |
| 5 | Ch 10 – Triangles | River delta triangle area | 10 |
| 6 | Ch 13 – Mensuration | Cylindrical water tank on riverside | 10 |
| 7 | Ch 14 – Trigonometry | Ferry crossing angle of depression | 10 |
| 8 | Ch 15 – Statistics | Fish catch frequency distribution | 10 |
A surveyor measures a river width as √200 m and its depth as √72 m.
(a) [2m] Simplify √200 and √72 into simplest surd form.
(b) [4m] Find the ratio width : depth in simplified form.
(c) [4m] Show that (√200 + √72)(√200 − √72) = 128 and classify the result.
√72 = √(36×2) = 6√2
Width = 10√2 m, Depth = 6√2 m
= 200 − 72 = 128 ✓
128 is a rational number (it is a whole number), so the product of these two irrational surds is rational.
An irrigation canal has a trapezoidal cross-section. The bottom width is b m, top width is (b + 4) m, and depth is 3 m. The cross-sectional area equals 3b + 6 m².
(a) [2m] Write the area formula for a trapezoid with these dimensions.
(b) [4m] Form an equation and solve for b.
(c) [4m] Expand and simplify: (b+4)² − b². What does this represent geometrically?
= ½(b + b+4)(3) = ½ × (2b+4) × 3 = 3(b+2) = 3b+6
The equation is consistent for any b. The problem states area = 3b+6, which is exactly what we derived. So b is determined by an additional condition. If area = 24 m² (given separately):
3b + 6 = 24 → 3b = 18 → b = 6 m
Geometrically: this is the difference in squares of top and bottom widths, representing the extra area strip at the top.
A rectangular fish pond has area 60 m². Its length is 7 m more than its width.
(a) [2m] Let the width be x m. Write a quadratic equation for the area.
(b) [4m] Solve the equation using the quadratic formula.
(c) [4m] Find the perimeter of the pond and the cost of fencing at Tk 120 per metre.
Area = x(x+7) = 60
x² + 7x − 60 = 0
x = (−7+17)/2 = 5 or x = (−7−17)/2 = −12 (rejected, negative)
x = 5 m (width), length = 12 m
Cost = 34 × 120 = Tk 4,080
Workers building a flood embankment lay bricks in an arithmetic pattern: 50 bricks on day 1, increasing by 15 bricks each day.
(a) [2m] Write the AP and find the number of bricks on day 10.
(b) [4m] Find the total bricks laid in the first 20 days.
(c) [4m] On which day will exactly 500 bricks be laid? Verify.
aₙ = a + (n−1)d
a₁₀ = 50 + 9×15 = 50+135 = 185 bricks
= 10 × [100 + 285] = 10 × 385 = 3,850 bricks
(n−1)×15 = 450
n−1 = 30 → n = 31st day
Verify: 50 + 30×15 = 50+450 = 500 ✓
A triangular river delta has sides 13 m, 14 m, and 15 m.
(a) [2m] Find the semi-perimeter s.
(b) [4m] Use Heron's formula to find the area.
(c) [4m] Find the height drawn to the longest side (14 m base).
= √[21(21−13)(21−14)(21−15)]
= √[21 × 8 × 7 × 6]
= √[21 × 8 × 42]
= √7056 = 84 m²
84 = ½ × 14 × h
84 = 7h → h = 12 m
A cylindrical water storage tank on a riverside has diameter 2.8 m and height 4 m. It is open at the top.
(a) [2m] Find the radius and base area. (π = 22/7)
(b) [4m] Find the total surface area of the tank (lateral + base, no top).
(c) [4m] Find the volume and the time to fill the tank if water flows in at 0.5 m³/min.
Base area = πr² = (22/7) × 1.96 = 6.16 m²
Total (no top) = 35.2 + 6.16 = 41.36 m²
Time = Volume/rate = 24.64/0.5 = 49.28 minutes
A lighthouse stands on a straight riverbank. A ferry boat at water level sees the top of the lighthouse at an angle of elevation of 60°. The boat is 20 m from the base of the lighthouse.
(a) [2m] Sketch the situation labelling all known values.
(b) [4m] Find the height of the lighthouse. (tan 60° = √3)
(c) [4m] Find the direct distance from the boat to the top of the lighthouse (the hypotenuse).
√3 = h/20
h = 20√3 ≈ 34.64 m
d = 40 m
Alternatively: cos 60° = 20/d → ½ = 20/d → d = 40 m ✓
Daily fish catch (kg) by 40 fishermen grouped as follows: 10–20: 5 fishermen, 20–30: 10, 30–40: 12, 40–50: 8, 50–60: 5.
(a) [2m] Verify total frequency and find the modal class.
(b) [4m] Calculate the mean catch using midpoints.
(c) [4m] Find the median class and estimate the median.
Modal class = class with highest frequency = 30–40 (12 fishermen)
Σfx = 5×15 + 10×25 + 12×35 + 8×45 + 5×55
= 75 + 250 + 420 + 360 + 275 = 1380
Mean = 1380/40 = 34.5 kg
Median is at (40/2) = 20th value → falls in class 30–40
Median = L + [(n/2 − cf)/f] × h
= 30 + [(20−15)/12] × 10 = 30 + (5/12)×10 = 30 + 4.167 = 34.17 kg