Sylhet Board · SSC Mathematics · 2021
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 1 – Real Numbers | Decimal and surd classification, Haor surveys | 10 |
| 2 | Ch 3 – Algebraic Expressions | if a−b=3, ab=10, evaluate expressions | 10 |
| 3 | Ch 6 – Quadratic Equations | Stone throw from hilltop, trajectory | 10 |
| 4 | Ch 8 – Arithmetic Progression | Steps on hill staircase | 10 |
| 5 | Ch 10 – Triangles | Pythagoras on hill-river triangle | 10 |
| 6 | Ch 13 – Mensuration | Cylindrical water reservoir on hill | 10 |
| 7 | Ch 14 – Trigonometry | Radio tower on hilltop, elevation angles | 10 |
| 8 | Ch 15 – Statistics | Tea garden yield, bar chart analysis | 10 |
Classify each as rational or irrational and simplify where possible: (i) √(144/25) (ii) √(50)+√(32) (iii) π/2 (iv) 0.666…
(a) [2m] Classify (i) and (iv).
(b) [4m] Simplify (ii) and show the result is irrational.
(c) [4m] Show that √50 × √32 = 40 (rational). Why is the product of two irrationals sometimes rational?
0.666… = 2/3 → rational
Sum = 9√2 — since √2 is irrational and 9 is non-zero rational, 9√2 is irrational
This happens because √50 = 5√2 and √32 = 4√2, and their product is 20×2 = 40. When two surds are like surds (same irrational part), their product can be rational if the irrational parts cancel out.
For a hill plot: a − b = 3 and ab = 10.
(a) [2m] Find a² + b².
(b) [4m] Find a³ − b³.
(c) [4m] Find the individual values of a and b (take a > b > 0).
Since b > 0: b=2, a=5
Check: a−b=3 ✓; ab=10 ✓
A stone is thrown from a hilltop. Its height above the ground (in metres) at time t seconds is: h = −5t² + 30t + 35.
(a) [2m] Find the initial height (at t=0) and the height at t=3.
(b) [4m] Find the maximum height using completing the square.
(c) [4m] Find when the stone hits the ground (h=0). Use the quadratic formula.
h(3) = −5(9)+90+35 = −45+90+35 = 80 m
= −5(t²−6t+9−9)+35 = −5(t−3)²+45+35
= −5(t−3)²+80
Maximum at t=3: h_max = 80 m
t = 7 or t = −1 (rejected)
Stone lands at t = 7 seconds
A hill staircase has steps getting wider. The first step is 30 cm wide; each step is 5 cm wider than the previous.
(a) [2m] Write the AP and find the width of step 8.
(b) [4m] Find the total width of the first 15 steps.
(c) [4m] The staircase is 360 cm wide total. How many steps does it have?
n(55+5n) ≤ 720
5n²+55n−720 ≤ 0
n²+11n−144 ≤ 0
n = [−11+√(121+576)]/2 = [−11+√697]/2 ≈ [−11+26.4]/2 ≈ 7.7
n = 7 steps: S₇ = 7/2[60+5×6] = 3.5×90 = 315 cm; S₈ = 4×[60+35] = 4×95=380 >360
So 7 steps fit within 360 cm.
A hill H is directly north of a river bank point R. A village V is due east of R. HR = 5 km, RV = 12 km.
(a) [2m] What is angle HRV?
(b) [4m] Find the straight-line distance HV.
(c) [4m] A road is built from H to V. Find the angle that HV makes with the horizontal (east direction) at V.
∴ HR is vertical, RV is horizontal → ∠HRV = 90°
HV = 13 km
tan(∠HVR) = HR/RV = 5/12
∠HVR = tan⁻¹(5/12) ≈ 22.6° above the east direction
A cylindrical water reservoir on a Sylhet hill has diameter 3.5 m and height 6 m. It is closed at both ends. (π=22/7)
(a) [2m] Find the total surface area of the cylinder.
(b) [4m] Find the volume of the reservoir.
(c) [4m] The reservoir supplies water at 0.2 m³/min. How long will a full reservoir last? Also find cost to fill at Tk 5 per litre (1 m³ = 1000 litres).
TSA = 2πr(r+h) = 2×(22/7)×1.75×(1.75+6) = 2×(22/7)×1.75×7.75
= 2×22×0.25×7.75 = 2×42.625 = 85.25 m²
Volume in litres = 57.75×1000 = 57,750 litres
Cost = 57,750×5 = Tk 2,88,750
A radio tower stands on top of a 40 m hill. From a point on the ground 80 m from the hill's base, the angle of elevation to the tower top is 60° and to the hilltop is 45°.
(a) [2m] Find the angle of elevation to the hilltop and verify it is 45°. (Height of hill = 40 m, distance = 80 m)
(b) [4m] Find the total height (hill + tower) using the 60° angle. (tan 60°=√3)
(c) [4m] Find the height of the tower alone.
The problem states 45°, which requires height = distance. Let the hill height be h₁ and the horizontal distance be d.
If angle to hilltop = 45°: tan 45° = h₁/d = 1 → h₁ = d = 80 m. So the hill is actually 80 m high for this to work. Let us use the given data: hill=80 m (corrected for consistency).
√3 = H_total/80
H_total = 80√3 ≈ 138.56 m
Tea yield (kg/hectare) for 5 estates: 850, 920, 780, 1050, 900.
(a) [2m] Find the mean and range of yields.
(b) [4m] Find the variance and standard deviation.
(c) [4m] One estate's yield increases by 100 kg/ha next year. Find the new mean and state what happens to the standard deviation.
Mean = 4500/5 = 900 kg/ha
Range = 1050−780 = 270 kg/ha
Squared: 2500, 400, 14400, 22500, 0
Variance = 39800/5 = 7960 kg²/ha²
SD = √7960 ≈ 89.22 kg/ha
New mean = 4600/5 = 920 kg/ha
Adding a constant to one value changes the mean but also changes the variance (SD does change since the deviation of that estate changes).