SSC MATH

Sylhet Board · SSC Mathematics · 2022

Full Paper8 Questions80 MarksSylhet Board

Chapter Coverage

QChapterTopicMarks
1Ch 1 – Real NumbersSurd operations, hill measurements10
2Ch 3 – Algebraic Expressionsif x+y=7, xy=12, find algebraic values10
3Ch 5 – Simultaneous EquationsTwo hill villages, distance and population10
4Ch 9 – Geometric SeriesPopulation of hill town, exponential growth10
5Ch 11 – CirclesTangent-secant from hill observation10
6Ch 13 – MensurationHemispherical hill cave: surface and volume10
7Ch 14 – TrigonometryAngle of depression from hill top to road10
8Ch 15 – StatisticsRainfall data, cumulative frequency10
Jump to: Q1Q2Q3Q4 Q5Q6Q7Q8
Question 1[10 marks]Ch 1

Two hill trail lengths are √180 m and √45 m.

(a) [2m] Simplify each surd fully.

(b) [4m] Find 2√180 − 3√45 and √180 × √45.

(c) [4m] Show that √(180/45) = 2 and explain in terms of surd laws.

▶ Show Solution
Part (a) — Simplify
√180 = √(36×5) = 6√5
√45 = √(9×5) = 3√5
Part (b) — Operations
2√180−3√45 = 2(6√5)−3(3√5) = 12√5−9√5 = 3√5
√180×√45 = 6√5×3√5 = 18×5 = 90
Part (c) — Quotient surd law
√(180/45) = √4 = 2
Using surd law: √(a/b) = √a/√b = 6√5/(3√5) = 2 ✓
This is the surd quotient law: √(m/n) = √m/√n for positive m, n.
Answer
√180=6√5; √45=3√5; 2√180−3√45=3√5; Product=90; √(180/45)=2 ✓
Question 2[10 marks]Ch 3

The length and width of a hill plot satisfy: x + y = 7 and xy = 12.

(a) [2m] Find x² + y².

(b) [4m] Find x³ + y³.

(c) [4m] Find the values of x and y individually. Then evaluate x⁴ + y⁴.

▶ Show Solution
Part (a) — x²+y²
(x+y)² = x²+2xy+y² → 49 = x²+24+y² → x²+y² = 25
Part (b) — x³+y³
x³+y³ = (x+y)(x²−xy+y²) = 7×(25−12) = 7×13 = 91
Part (c) — x, y values and x⁴+y⁴
x+y=7, xy=12 → roots of t²−7t+12=0 → (t−3)(t−4)=0
x=4, y=3 (or x=3, y=4)
x⁴+y⁴ = (x²+y²)²−2(xy)² = 25²−2×144 = 625−288 = 337
Answer
x²+y²=25; x³+y³=91; x=4, y=3; x⁴+y⁴=337 ✓
Question 3[10 marks]Ch 5

Village A and Village B are in the Sylhet hills. The sum of their populations is 4200. Village A has 3 times the population of Village B minus 600.

(a) [2m] Set up a system of equations.

(b) [4m] Solve for the individual populations.

(c) [4m] If Village A grows by 10% and Village B shrinks by 5%, find the new total population.

▶ Show Solution
Part (a) — Equations
A + B = 4200
A = 3B − 600
Part (b) — Solve
Substitute: (3B−600)+B = 4200 → 4B = 4800 → B = 1200
A = 3(1200)−600 = 3000
Part (c) — New total
New A = 3000×1.1 = 3300; New B = 1200×0.95 = 1140
New total = 3300+1140 = 4440
Answer
A=3000, B=1200; New total after growth/shrinkage = 4440 ✓
Question 4[10 marks]Ch 9

A Sylhet hill town has population 10,000 and grows at 5% per year (geometric growth).

(a) [2m] Write the population as a GP with first term and ratio.

(b) [4m] Find the population after 4 years. (1.05⁴ = 1.2155)

(c) [4m] How many complete years until the population doubles? (log 2 = 0.3010, log 1.05 = 0.0212)

▶ Show Solution
Part (a) — GP
a = 10000, r = 1.05
Pₙ = 10000 × 1.05^(n−1)
Part (b) — After 4 years
P₅ = 10000 × 1.05⁴ = 10000 × 1.2155 = 12,155
Part (c) — Doubling time
10000 × 1.05ⁿ = 20000
1.05ⁿ = 2
n·log 1.05 = log 2
n = 0.3010/0.0212 = 14.2 → 15 years (complete years)
Answer
a=10000, r=1.05; After 4 yrs: 12,155; Doubles in 15 complete years ✓
Question 5[10 marks]Ch 11

From an external point T on a hilltop, a tangent TA touches a circle at A, and a secant TBC passes through the circle (B and C on circle). TB = 4 cm, TC = 9 cm.

(a) [2m] State the tangent-secant theorem.

(b) [4m] Find the length of tangent TA using the theorem.

(c) [4m] Find BC (the chord). If the circle radius is r and OA⊥TA, find r given OT=10 cm.

▶ Show Solution
Part (a) — Theorem
If a tangent and a secant are drawn from the same external point, then (tangent)² = (external segment) × (whole secant). TA² = TB × TC.
Part (b) — TA length
TA² = TB × TC = 4 × 9 = 36
TA = 6 cm
Part (c) — BC and radius r
BC = TC−TB = 9−4 = 5 cm
OA ⊥ TA (radius to tangent point), OT = 10, TA = 6
OA² = OT²−TA² = 100−36 = 64 → r = OA = 8 cm
Answer
TA=6 cm; BC=5 cm; Radius r=8 cm ✓
Question 6[10 marks]Ch 13

A hemispherical cave entrance in a Sylhet hill has radius 3.5 m. (π = 22/7)

(a) [2m] Find the curved surface area of the hemisphere.

(b) [4m] Find the total surface area (including flat circular base).

(c) [4m] Find the volume of the hemisphere.

▶ Show Solution
Part (a) — Curved SA
Curved SA = 2πr² = 2×(22/7)×3.5² = 2×(22/7)×12.25 = 2×38.5 = 77 m²
Part (b) — Total SA
Base area = πr² = (22/7)×12.25 = 38.5 m²
Total SA = 77+38.5 = 115.5 m²
Part (c) — Volume
V = (2/3)πr³ = (2/3)×(22/7)×3.5³ = (2/3)×(22/7)×42.875
= (2/3)×134.75 = 89.83 m³
Answer
Curved SA=77 m²; Total SA=115.5 m²; Volume≈89.83 m³ ✓
Question 7[10 marks]Ch 14

From the top of a 100 m hill, a person sees a road below. The angle of depression to the near end of the road is 60° and to the far end is 30°.

(a) [2m] Draw and label the diagram.

(b) [4m] Find the horizontal distances to both ends of the road from the base of the hill. (tan 30°=1/√3, tan 60°=√3)

(c) [4m] Find the length of the road section visible.

▶ Show Solution
Part (a) — Diagram
Hill of height 100 m. Near end of road: angle of depression 60°. Far end: angle of depression 30°. Both on same horizontal level as base.
Part (b) — Distances
Near end: tan 60° = 100/d₁ → d₁ = 100/√3 = 100√3/3 ≈ 57.74 m
Far end: tan 30° = 100/d₂ → d₂ = 100√3 ≈ 173.2 m
Part (c) — Road length
Length = d₂−d₁ = 100√3−100/√3 = 100√3−100√3/3 = 100√3(1−1/3) = 200√3/3 ≈ 115.47 m
Answer
Near=100/√3≈57.74 m; Far=100√3≈173.2 m; Road length=200√3/3≈115.47 m ✓
Question 8[10 marks]Ch 15

Annual rainfall (mm) in Sylhet over 6 years: 3200, 3650, 2900, 4100, 3500, 3750.

(a) [2m] Find the mean annual rainfall.

(b) [4m] Find the variance.

(c) [4m] Find the standard deviation. Compare with Barisal mean of 248 mm/month (multiply Sylhet mean by appropriate factor for comparison).

▶ Show Solution
Part (a) — Mean
Sum = 3200+3650+2900+4100+3500+3750 = 21100
Mean = 21100/6 = 3516.67 mm/year
Part (b) — Variance
Deviations from 3516.67: −316.67, 133.33, −616.67, 583.33, −16.67, 233.33
Squared: 100,279, 17,769, 380,282, 340,274, 278, 54,443
Sum ≈ 893,325
Variance = 893325/6 ≈ 148,888 mm²
Part (c) — SD and comparison
SD = √148888 ≈ 385.9 mm/year
Sylhet monthly mean = 3516.67/12 ≈ 293 mm/month vs Barisal 248 mm/month
Sylhet receives about 18% more monthly rainfall than Barisal.
Answer
Mean=3516.67 mm/yr; Variance≈148,888; SD≈385.9 mm/yr; Sylhet∼18% wetter than Barisal ✓

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