Sylhet Board SSC Math 2023

Q	Chapter	Topic	Marks
1	Ch 2 – Sets	Hill villages, tea/rice growers Venn diagram	10
2	Ch 3 – Algebraic Expressions	Hill plot polynomial identity	10
3	Ch 6 – Quadratic Equations	Tea garden plot dimensions	10
4	Ch 8 – Arithmetic Series	Savings plan for hill school	10
5	Ch 10 – Triangles	Altitude and median in hill terrain triangle	10
6	Ch 13 – Mensuration	Frustum-shaped hill section volume	10
7	Ch 14 – Trigonometry	Waterfall height measurement	10
8	Ch 15 – Statistics	Elevation data of hill peaks, grouped stats	10

Question 1[10 marks]Ch 2

In Sylhet hill villages: 60 households grow tea (set T), 40 grow rice (set R), 20 grow both, total households = 90.

(a) [2m] Find n(T∪R) and households growing neither.

(b) [4m] Find the probability that a randomly chosen household grows exactly one crop.

(c) [4m] If 10 more households begin growing tea, update the set sizes and find new n(T∪R).

▶ Show Solution

Part (a) — Union and neither

n(T∪R) = 60+40−20 = 80
Neither = 90−80 = 10 households

Part (b) — Exactly one crop

Only T = 60−20 = 40; Only R = 40−20 = 20
Exactly one = 60; P = 60/90 = 2/3

Part (c) — After 10 more tea growers

New n(T) = 70; n(R) = 40; n(T∩R) = 20 (unchanged, assuming new ones don't grow rice)
New n(T∪R) = 70+40−20 = 90 (entire village)

Answer

n(T∪R)=80; Neither=10; P(exactly one)=2/3; After addition: n(T∪R)=90 ✓

Question 2[10 marks]Ch 3

If a + b = 5 and ab = 6 (representing dimensions of a hill plot in km):

(a) [2m] Find a² + b².

(b) [4m] Find a³ + b³ using the identity a³+b³ = (a+b)(a²−ab+b²).

(c) [4m] Find (a−b)² and hence |a−b|. Also determine the actual values of a and b.

▶ Show Solution

Part (a) — a²+b²

(a+b)² = a²+2ab+b² → 25 = a²+12+b² → a²+b² = 13

Part (b) — a³+b³

a³+b³ = (a+b)(a²−ab+b²) = 5×(13−6) = 5×7 = 35

Part (c) — (a−b)² and values

(a−b)² = (a+b)²−4ab = 25−24 = 1
|a−b| = 1
a+b=5, a−b=1 → a = 3, b = 2 (or a=2, b=3)

Answer

a²+b²=13; a³+b³=35; |a−b|=1; a=3, b=2 ✓

Question 3[10 marks]Ch 6

A tea garden plot is rectangular. Its area is 84 m² and perimeter is 38 m.

(a) [2m] Write equations for length l and width w.

(b) [4m] Form a quadratic equation and solve for l and w.

(c) [4m] Find the diagonal length. Also find the new area if both dimensions increase by 2 m.

▶ Show Solution

Part (a) — Equations

l+w = 19 (half perimeter)
lw = 84

Part (b) — Quadratic

l+w=19, lw=84 → l and w are roots of: t²−19t+84=0
(t−7)(t−12)=0
l=12 m, w=7 m

Part (c) — Diagonal and new area

Diagonal = √(12²+7²) = √(144+49) = √193 ≈ 13.89 m
New dimensions: 14 m × 9 m → New area = 14×9 = 126 m²

Answer

l=12 m, w=7 m; Diagonal=√193≈13.89 m; New area=126 m² ✓

Question 4[10 marks]Ch 8

Parents in a hill village save for their children's school fees in an AP. Year 1: Tk 2000; each year they save Tk 500 more.

(a) [2m] Find the amount saved in year 8.

(b) [4m] Find the total savings after 10 years.

(c) [4m] In which year do total savings first exceed Tk 50,000?

▶ Show Solution

Part (a) — Year 8

a₈ = 2000+7×500 = 2000+3500 = Tk 5,500

Part (b) — Total after 10 years

S₁₀ = 10/2×[2(2000)+9(500)] = 5×[4000+4500] = 5×8500 = Tk 42,500

Part (c) — First exceed 50,000

Sₙ = n/2[4000+(n−1)500] = n/2[3500+500n] = n(1750+250n)
= 250n²+1750n ≥ 50000
n²+7n ≥ 200
n²+7n−200 ≥ 0
n = [−7+√(49+800)]/2 = [−7+√849]/2 ≈ [−7+29.14]/2 ≈ 11.07
After year 12: S₁₂ = 250(144)+1750(12) = 36000+21000 = 57000 > 50000 ✓

Answer

Year 8: Tk 5,500; S₁₀=Tk 42,500; Total exceeds Tk 50,000 in year 12 ✓

Question 5[10 marks]Ch 10

In △ABC, AB = 10 cm, BC = 8 cm, AC = 6 cm. M is the midpoint of BC.

(a) [2m] Verify this is a right triangle and identify the right angle.

(b) [4m] Find the length of median AM.

(c) [4m] Find the altitude from A to BC.

▶ Show Solution

Part (a) — Right triangle check

BC²+AC² = 64+36 = 100 = AB² ✓
Right angle is at C (since BC and AC are the legs)

Part (b) — Median AM

M = midpoint of BC. BM = 4.
In right △ABC with right angle at C: median from A to hypotenuse AB = AB/2 = 5.
But median AM is to BC (not the hypotenuse).
Place C at origin, B at (8,0), A at (0,6) [since right angle at C].
M = midpoint of BC = (4,0)
AM = √((0−4)²+(6−0)²) = √(16+36) = √52 = 2√13 ≈ 7.21 cm

Part (c) — Altitude from A to BC

Area = ½×BC×AC = ½×8×6 = 24 cm²
Also Area = ½×BC×h_A where h_A is altitude from A to BC
24 = ½×8×h_A → h_A = 6 cm (= AC, since right angle at C means AC ⊥ BC)

Answer

Right angle at C; Median AM = 2√13 ≈ 7.21 cm; Altitude from A to BC = 6 cm ✓

Question 6[10 marks]Ch 13

A section of a Sylhet hill is modelled as a frustum (truncated cone) with top radius 30 m, bottom radius 80 m, and vertical height 50 m. (π = 3.14)

(a) [2m] Find the slant height of the frustum.

(b) [4m] Find the curved surface area of the frustum. (CSA = π(r₁+r₂)l)

(c) [4m] Find the volume of the frustum. (V = πh(r₁²+r₁r₂+r₂²)/3)

▶ Show Solution

Part (a) — Slant height

l = √(h²+(r₂−r₁)²) = √(50²+(80−30)²) = √(2500+2500) = √5000 = 50√2 ≈ 70.71 m

Part (b) — Curved surface area

CSA = π(r₁+r₂)l = 3.14×(30+80)×50√2
= 3.14×110×70.71 = 3.14×7778.1 = 24,423 m²

Part (c) — Volume

V = πh/3 × (r₁²+r₁r₂+r₂²)
= 3.14×50/3 × (900+2400+6400)
= (157/3) × 9700 = 52.333×9700 = 507,633 m³ ≈ 5.076×10⁵ m³

Answer

Slant h=50√2≈70.71 m; CSA≈24,423 m²; Volume≈507,633 m³ ✓

Question 7[10 marks]Ch 14

A tourist stands 60 m from the base of a waterfall. The angle of elevation of the top of the waterfall is 60° and the angle of depression of the base of the waterfall pool is 30°.

(a) [2m] Sketch the situation with all labels.

(b) [4m] Find the height of the waterfall above the tourist's eye level. (tan 60° = √3)

(c) [4m] Find the depth of the pool below the tourist's eye level, and the total height from pool to waterfall top. (tan 30° = 1/√3)

▶ Show Solution

Part (a) — Sketch

Tourist at point T, horizontal distance 60 m to waterfall base B. Waterfall top W above B, pool floor P below B. Angle up to W = 60°, angle down to pool = 30°.

Part (b) — Height above eye level

tan 60° = h_above/60
√3 = h/60
h_above = 60√3 ≈ 103.92 m

Part (c) — Pool depth and total

tan 30° = h_below/60
1/√3 = h_below/60
h_below = 60/√3 = 20√3 ≈ 34.64 m
Total height = 60√3+20√3 = 80√3 ≈ 138.56 m

Answer

Above eye = 60√3≈103.92 m; Pool depth = 20√3≈34.64 m; Total = 80√3≈138.56 m ✓

Question 8[10 marks]Ch 15

Elevations (m) of 10 hill peaks near Sylhet: 450, 620, 380, 710, 550, 490, 670, 420, 580, 630.

(a) [2m] Find the mean and range of these elevations.

(b) [4m] Find the median and inter-quartile range.

(c) [4m] Calculate the variance of the elevations.

▶ Show Solution

Part (a) — Mean and range

Sum = 450+620+380+710+550+490+670+420+580+630 = 5500
Mean = 5500/10 = 550 m
Range = 710−380 = 330 m

Part (b) — Median and IQR

Sorted: 380, 420, 450, 490, 550, 580, 620, 630, 670, 710
Median = (550+580)/2 = 565 m
Q1 = median of lower half = (450+490)/2 = 470
Q3 = median of upper half = (620+630)/2 = 625
IQR = 625−470 = 155 m

Part (c) — Variance

Mean = 550; deviations: −100,70,−170,160,0,−60,120,−130,30,80
Squared: 10000,4900,28900,25600,0,3600,14400,16900,900,6400
Sum = 111600; Variance = 111600/10 = 11,160 m²
SD = √11160 ≈ 105.6 m

Answer

Mean=550 m; Range=330 m; Median=565 m; IQR=155 m; Variance=11,160 m² ✓

Sylhet Board · SSC Mathematics · 2023

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