Barisal Board · SSC Mathematics · 2020
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 2 – Sets | River region villages, set operations | 10 |
| 2 | Ch 3 – Algebraic Expressions | If x+1/x=4, find x²+1/x² and x³+1/x³ | 10 |
| 3 | Ch 6 – Quadratic Equations | Net area of fishing ground, quadratic | 10 |
| 4 | Ch 9 – Geometric Series | Savings scheme, geometric growth | 10 |
| 5 | Ch 10 – Triangles | Midpoint theorem, river embankment | 10 |
| 6 | Ch 13 – Mensuration | Rectangular embankment cross-section prism | 10 |
| 7 | Ch 14 – Trigonometry | Sin, cos, tan values for special angles | 10 |
| 8 | Ch 15 – Statistics | Tidal height data, grouped frequency | 10 |
In a river region: set A = villages with electricity (35), set B = villages with piped water (28), A∩B = 12, total villages = 60.
(a) [2m] Find n(A∪B) using the inclusion-exclusion principle.
(b) [4m] Find the number of villages with neither electricity nor piped water.
(c) [4m] Draw a Venn diagram and find the probability that a randomly chosen village has exactly one of these facilities.
Exactly one = 23+16 = 39
P(exactly one) = 39/60 = 13/20 = 0.65
If x + 1/x = 4 (represents ratio of river length to width):
(a) [2m] Find x² + 1/x².
(b) [4m] Find x³ + 1/x³.
(c) [4m] Find x⁴ + 1/x⁴.
16 = x²+2+1/x²
x²+1/x² = 14
= 4 × (14−1) = 4 × 13 = 52
14² = x⁴+2+1/x⁴
196−2 = x⁴+1/x⁴ = 194
A rectangular fishing ground has perimeter 56 m and area 192 m².
(a) [2m] Write two equations using length l and width w.
(b) [4m] Form a quadratic in one variable and solve.
(c) [4m] Verify your answer, then find the diagonal length of the fishing ground.
lw = 192
28l − l² = 192
l² − 28l + 192 = 0
(l−16)(l−12) = 0
l = 16 m, w = 12 m (or l=12, w=16)
Diagonal = √(16²+12²) = √(256+144) = √400 = 20 m
A fisherman saves money in a geometric scheme: Tk 100 in month 1, then each month he saves 1.5 times the previous month.
(a) [2m] Write the first 4 terms of the GP.
(b) [4m] Find his total savings after 6 months.
(c) [4m] Find the sum to infinity if the ratio were 0.8 instead (decreasing savings).
100, 150, 225, 337.50
= 100 × 10.3906/0.5 = 100 × 20.7813 = Tk 2,078.13
In triangle ABC, D and E are midpoints of AB and AC respectively. BC = 14 m (represents an embankment section).
(a) [2m] State the midpoint theorem.
(b) [4m] Find DE and prove that DE ∥ BC.
(c) [4m] If area of △ADE = 12 m², find area of △ABC and trapezoid BCED.
By the midpoint theorem, DE ∥ BC since D and E are midpoints. ✓
Area ratio = 1²:2² = 1:4
Area △ADE = 12 m² → Area △ABC = 12 × 4 = 48 m²
Area of trapezoid BCED = 48−12 = 36 m²
A flood embankment has a trapezoidal cross-section with parallel sides 8 m (top) and 14 m (base), height 3 m. The embankment is 500 m long.
(a) [2m] Find the cross-sectional area of the embankment.
(b) [4m] Find the total volume of earth used to build the embankment.
(c) [4m] If earth costs Tk 250 per m³ to transport, find the total cost.
Without a calculator, evaluate exactly:
(a) [2m] sin 30° + cos 60° + tan 45°
(b) [4m] sin²30° + cos²60° + tan²45° and verify sin²θ + cos²θ = 1 for θ=30°.
(c) [4m] 2 sin 60° cos 60° (= sin 120° by double angle). Also evaluate (sin 30° + cos 30°)².
Sum = 1/2 + 1/2 + 1 = 2
Sum = 1/4+1/4+1 = 3/2
sin²30°+cos²30° = (1/2)²+(√3/2)² = 1/4+3/4 = 1 ✓
(sin 30°+cos 30°)² = (1/2+√3/2)² = ((1+√3)/2)²
= (1+2√3+3)/4 = (4+2√3)/4 = (2+√3)/2
Tidal heights (m) recorded over 40 days: 1.0–1.5: 6, 1.5–2.0: 10, 2.0–2.5: 14, 2.5–3.0: 7, 3.0–3.5: 3.
(a) [2m] Find the total frequency and the modal class.
(b) [4m] Calculate the mean tidal height using midpoints.
(c) [4m] Estimate the median tidal height.
Modal class = 2.0–2.5 m (frequency 14)
Σfx = 6×1.25+10×1.75+14×2.25+7×2.75+3×3.25
= 7.5+17.5+31.5+19.25+9.75 = 85.5
Mean = 85.5/40 = 2.1375 m
Median at 20th value → class 2.0–2.5 (CF=16 before, 30 after)
Median = 2.0 + [(20−16)/14] × 0.5 = 2.0 + (4/14)×0.5 = 2.0 + 0.143 = 2.143 m