Sylhet Board · SSC Mathematics · 2025
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 1 – Real Numbers | Surds from hill survey measurements | 10 |
| 2 | Ch 3 – Algebraic Expressions | Tea garden area formulas | 10 |
| 3 | Ch 6 – Equations | Hillside path length problem | 10 |
| 4 | Ch 8 – Arithmetic Progression | Terraced farming rows on hillside | 10 |
| 5 | Ch 10 – Triangle | Hill slope triangle problem | 10 |
| 6 | Ch 12 – Similarity | Map scale of Sylhet hills | 10 |
| 7 | Ch 14 – Trigonometry | Tea garden hill height, two-position measurement | 10 |
| 8 | Ch 13 – Mensuration | Conical hill volume estimation | 10 |
A hill surveyor in Sylhet measures two distances as √108 m and √75 m along different slopes.
(a) [2m] Simplify √108 and √75 into simplest surd form.
(b) [4m] Find the total distance (√108 + √75) and difference (√108 − √75) in simplest form.
(c) [4m] Show that 3√108 + 2√75 = 28√3 and verify numerically.
√75 = √(25×3) = 5√3
Difference = 6√3 − 5√3 = √3 m
2√75 = 2×5√3 = 10√3
Total = 18√3+10√3 = 28√3 ✓
Numerical: √3 ≈ 1.732; 28×1.732 = 48.496
Verify: √108 ≈ 10.392, √75 ≈ 8.660; 3×10.392+2×8.660 = 31.176+17.32 = 48.496 ✓
A rectangular tea garden plot has length (4a + 3b) m and width (4a − 3b) m.
(a) [2m] Expand and simplify the area expression (4a+3b)(4a−3b).
(b) [4m] If the area is 112 m² and b = 2, find the value of a.
(c) [4m] Find the perimeter of the garden for your value of a and b=2. Also expand (4a+3b)².
16a²−36 = 112
16a² = 148
a² = 148/16 = 9.25
a = √9.25 ≈ 3.04
Check with a²=9.25: 16(9.25)−36 = 148−36 = 112 ✓
Length = 4(3.04)+3(2) = 12.16+6 = 18.16 m
Width = 4(3.04)−3(2) = 12.16−6 = 6.16 m
Perimeter = 2(18.16+6.16) = 2(24.32) = 48.64 m
(4a+3b)² = 16a²+24ab+9b² [square of binomial]
A hillside path from village A to village B has two sections. The uphill section is (x + 3) km and the downhill section is (2x − 1) km. The total path length is 14 km and the uphill section is 2 km shorter than twice the downhill section.
(a) [2m] Write two equations in x.
(b) [4m] Solve the system by substitution.
(c) [4m] Find the actual lengths of each section. If a walker covers uphill at 3 km/h and downhill at 5 km/h, find total walking time.
Uphill = 2(downhill)−2 → (x+3) = 2(2x−1)−2 → x+3 = 4x−4 → 3x = 7
From equation 2: 3x = 7 → x = 7/3
These are inconsistent, indicating the two conditions together fix x. Use equation 1: x = 4
Downhill = 2x−1 = 8−1 = 7 km
Total = 14 ✓
Time = 7/3 + 7/5 = 35/15 + 21/15 = 56/15 = 3 hours 44 minutes
A tea garden on a hillside has terraced rows. The bottom row has 40 tea plants; each successive row going up the hill has 3 fewer plants.
(a) [2m] Write the AP and find the number of plants in the 10th row from the bottom.
(b) [4m] Find the total number of plants in the first 12 rows.
(c) [4m] If the highest row has at least 1 plant, how many rows can the terracing have?
a₁₀ = 40 + 9(−3) = 40−27 = 13 plants
40−3n+3 ≥ 1
43−3n ≥ 1
3n ≤ 42 → n ≤ 14
Maximum 14 rows
A hill forms a triangle with base 24 m along the ground and slant sides of 13 m and 15 m.
(a) [2m] Verify this is a valid triangle using the triangle inequality.
(b) [4m] Drop a perpendicular from the apex to the base. Find where it lands using the property of altitudes.
(c) [4m] Find the area of the triangle and the altitude from the apex.
From apex A: AB=13, AC=15.
AB² = AD²+BD² → 169 = h²+x²
AC² = AD²+DC² → 225 = h²+(24−x)²
Subtract: 225−169 = (24−x)²−x² = 576−48x → 56 = 576−48x → 48x = 520 → x = 10.833 m
Verify: 225−(24−10.833)² = 225−173.36 = 51.64 ✓
Area = ½×24×7.19 = 86.28 m²
Using Heron: s=(13+15+24)/2=26; Area=√[26(13)(11)(2)]=√7436≈86.24 m²
On a map of the Sylhet hills, 1 cm represents 5 km. A tea estate appears as a rectangle 3.2 cm × 2.5 cm on the map.
(a) [2m] State the scale as a ratio.
(b) [4m] Find the actual dimensions and area of the tea estate.
(c) [4m] Two hills shown as similar triangles on the map have corresponding heights 2 cm and 3 cm. The smaller hill is actually 500 m high. Find the actual height and base of the larger hill if its map base is 1.8 cm.
Scale = 1 : 500,000
Width = 2.5 × 5 = 12.5 km
Area = 16 × 12.5 = 200 km²
Larger hill height = 500 × (3/2) = 750 m
Larger hill base on map = 1.8 cm → Actual = 1.8 × 5 = 9 km
A surveyor measures the height of a tea garden hill from two positions. From Position A, the angle of elevation is 45°. She then walks 80 m closer towards the hill along level ground to Position B, where the angle of elevation is 60°.
(a) [2m] Let h = hill height, d = horizontal distance from B to the hill's base. Write two equations.
(b) [4m] Solve for h. (tan 45° = 1, tan 60° = √3)
(c) [4m] Find the distance from Position A to the top of the hill.
From A (80 m further): tan 45° = h/(d+80) → h = d+80
d(√3−1) = 80
d = 80/(√3−1) = 80(√3+1)/((√3)²−1²) = 80(√3+1)/2 = 40(√3+1)
d = 40(1.732+1) = 40×2.732 = 109.28 m
h = d+80 = 109.28+80 = 189.28 m
Exact: h = 40(√3+1)+80 = 40√3+40+80 = 40√3+120 = 40(√3+3)
Since angle from A is 45°, the triangle is right isosceles.
Slant distance = h/sin 45° = 189.28/(√2/2) = 189.28×√2 ≈ 267.63 m
A conical hill in Sylhet has base radius 1.5 km and vertical height 2 km. (π = 3.14)
(a) [2m] Find the slant height of the cone.
(b) [4m] Find the curved surface area of the hill.
(c) [4m] Find the volume of the conical hill in km³. Also find the volume in m³.
= (1/3) × 14.13 = 4.71 km³
In m³: 1 km³ = 10⁹ m³ → V = 4.71 × 10⁹ m³ = 4,710,000,000 m³