Sylhet Board · SSC Mathematics · 2024
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 1 – Real Numbers | Rationalizing hill road gradient expressions | 10 |
| 2 | Ch 4 – Logarithms | Tea production log calculations | 10 |
| 3 | Ch 5 – Simultaneous Equations | Two hill paths, speed and time | 10 |
| 4 | Ch 9 – Geometric Series | Tea leaf harvesting GP | 10 |
| 5 | Ch 11 – Circles | Circular pond in tea garden | 10 |
| 6 | Ch 12 – Similarity | Similar hill profiles, ratio problem | 10 |
| 7 | Ch 14 – Trigonometry | Cable car angle of elevation on hill | 10 |
| 8 | Ch 15 – Statistics | Tea garden worker wages distribution | 10 |
A hill road gradient is expressed as 1/(√7 + √5).
(a) [2m] Rationalize 1/(√7+√5).
(b) [4m] Simplify 1/(√7+√5) + 1/(√7−√5). Show that the result is rational.
(c) [4m] If the road elevation is √7 − √5 km, find the reciprocal and the product (√7+√5)(√7−√5).
1/(√7−√5) = (√7+√5)/2 [by similar rationalization]
Sum = (√7−√5)/2 + (√7+√5)/2 = 2√7/2 = √7
Wait — √7 is irrational. Let me verify: (√7−√5+√7+√5)/2 = 2√7/2 = √7.
Actually √7 is irrational. Correct statement: the sum equals √7 (irrational), but note that 1/(√7+√5) × 2 = √7−√5 which is irrational.
Instead note: (√7−√5)/2 + (√7+√5)/2 = √7 is the simplified form.
(√7+√5)(√7−√5) = 7−5 = 2 (rational)
A Sylhet tea garden produces 3200 kg of tea per hectare. (log 2 = 0.3010, log 5 = 0.6990)
(a) [2m] Find log 3200.
(b) [4m] If production grows by 25% next year, find the new production and its logarithm.
(c) [4m] Solve for x: log(3200) + log x = log(16000). Find x.
log 3200 = 5·log 2 + 2 = 5(0.3010)+2 = 1.5050+2 = 3.5050
4000 = 4×1000 = 2²×10³
log 4000 = 2(0.3010)+3 = 0.6020+3 = 3.6020
3200x = 16000
x = 16000/3200 = 5
Check: log 3200 + log 5 = 3.5050+0.6990 = 4.2040; log 16000 = log(16×1000) = 4·log2+3 = 4(0.3010)+3 = 4.2040 ✓
Two hikers travel hill paths. Hiker A walks at speed u km/h and covers 36 km. Hiker B walks at speed (u−3) km/h and covers 28 km. They take the same time.
(a) [2m] Write an equation for their equal times.
(b) [4m] Solve for u.
(c) [4m] Find the time each takes and the total distance covered if they walk continuously for 4 hours at their respective speeds.
8u = 108
u = 13.5 km/h
Hiker B: u−3 = 10.5 km/h
In 4 hours: A covers 4×13.5 = 54 km; B covers 4×10.5 = 42 km
A tea garden worker's weekly leaf harvest forms a GP. In week 1 she harvests 20 kg, and each subsequent week the harvest is 90% of the previous week.
(a) [2m] Write the first 4 terms and the GP formula for week n.
(b) [4m] Find the total harvest over 8 weeks.
(c) [4m] Find the sum to infinity of the weekly harvest (total over the entire season if it continues indefinitely).
20, 18, 16.2, 14.58; aₙ = 20 × 0.9^(n−1)
0.9⁸ = 0.43047
S₈ = 20(1−0.43047)/0.1 = 20 × 5.6953 = 113.91 kg
A circular ornamental pond in a tea garden has centre O and radius 7 m. Chord PQ = 7 m. (π = 22/7)
(a) [2m] Find the angle POQ at the centre (equilateral triangle check).
(b) [4m] Find the arc length PQ and area of sector POQ.
(c) [4m] Find the area of the minor segment cut off by chord PQ.
∴ ∠POQ = 60°
Area of sector = (θ/360°) × πr² = (1/6)×(22/7)×49 = (1/6)×154 = 25.67 m²
Segment area = Sector area − Triangle area = 25.67−21.22 = 4.45 m²
Two similar hill profiles △ABC ~ △PQR. In △ABC: AB = 6 cm, BC = 8 cm, AC = 10 cm. In △PQR: PQ = 9 cm.
(a) [2m] Find the scale factor (ratio AB:PQ).
(b) [4m] Find QR and PR. Also verify △ABC is right-angled.
(c) [4m] Find the ratio of areas of △ABC and △PQR.
PR = AC × (3/2) = 10 × 1.5 = 15 cm
Check △ABC: AB²+BC² = 36+64 = 100 = AC² = 10² ✓ → right angle at B
Area △ABC = ½×6×8 = 24 cm²; Area △PQR = 24×(9/4) = 54 cm²
A cable car travels from the bottom of a hill to a station at the top. The cable makes an angle of 30° with the horizontal. The horizontal ground distance is 600 m.
(a) [2m] Find the length of the cable. (cos 30° = √3/2)
(b) [4m] Find the vertical height gained. (sin 30° = 1/2)
(c) [4m] If the cable car speed along the cable is 4 m/s, find time to reach the top. Also, if a second cable at 45° covers the same horizontal distance, find its length.
√3/2 = 600/L
L = 600×2/√3 = 1200/√3 = 400√3 ≈ 692.8 m
Alternatively: tan 30° = h/600 → h = 600/√3 = 200√3 ✓
Second cable at 45°: cos 45° = 600/L₂ → L₂ = 600/(√2/2) = 600×2/√2 = 600√2 ≈ 848.5 m
Weekly wages (Tk) of 50 tea garden workers: 800–1000: 8, 1000–1200: 15, 1200–1400: 18, 1400–1600: 7, 1600–1800: 2.
(a) [2m] Find the modal class and total frequency.
(b) [4m] Calculate the mean weekly wage.
(c) [4m] Find the median weekly wage.
Σfx = 8×900+15×1100+18×1300+7×1500+2×1700
= 7200+16500+23400+10500+3400 = 61000
Mean = 61000/50 = Tk 1,220
Median at 25th value → class 1200–1400 (CF=23 before, 41 after)
Median = 1200 + [(25−23)/18] × 200 = 1200 + (2/18)×200 = 1200 + 22.22 = Tk 1,222.22