Chittagong Board SSC Math 2020

Q	Chapter	Topic	Marks
1	Ch 1 · Real Numbers	Properties and number line	10
2	Ch 4 · Exponents & Log	Exponent laws — storage capacity	10
3	Ch 6 · Quadratic	Word problem — dock length	10
4	Ch 9 · Series	Savings plan — port workers	10
5	Ch 11 · Geometry	Exterior angle theorem	10
6	Ch 13 · Circle	Inscribed angle and arc	10
7	Ch 14 · Trigonometry	Karnafuli Bridge elevation	10
8	Ch 16 · Mensuration	Rectangular prism warehouse	10

Question 1[10 marks]Ch 1 · Real Numbers

Classify and work with real numbers.

(a) Arrange in ascending order: √7, π, 8/3, 2.6̄ (2.666...). [2 marks]

(b) Show that 0.142857142857... = 1/7 using the algebraic method. [4 marks]

(c) Between any two real numbers there is another real number. Find two real numbers between 3/11 and 5/11, and express them as decimals. [4 marks]

▶ Show Solution

(a) [2 marks]

Decimal values

√7 ≈ 2.646, π ≈ 3.1416, 8/3 ≈ 2.667, 2.6̄ = 2.666...

Ascending order

2.6̄ < √7 < 8/3 < π ✓

(b) [4 marks]

Let x = 0.142857142857...

x = 0.̄142857̄

Multiply by 10⁶

1000000x = 142857.142857...

Subtract

999999x = 142857 → x = 142857/999999

Simplify

142857/999999 = 1/7 ✓

(c) [4 marks]

3/11 and 5/11

3/11 ≈ 0.2727..., 5/11 ≈ 0.4545...

Two numbers between

4/11 ≈ 0.3636... and 7/22 = 0.3181...

Verify

0.2727 < 0.3181 < 0.3636 < 0.4545 ✓

Question 2[10 marks]Ch 4 · Exponents & Logarithms

Apply exponent and logarithm laws.

(a) Simplify: (2³ × 4² × 8) / (16 × 2⁵). [2 marks]

(b) If 2ˣ = 3ʸ = 6ᶻ, prove that z = xy/(x+y). [4 marks]

(c) Solve for x: log(x+3) + log(x−3) = log 7. [4 marks]

▶ Show Solution

(a) [2 marks]

Convert to powers of 2

= (2³ × 2⁴ × 2³) / (2⁴ × 2⁵) = 2^(3+4+3) / 2^(4+5) = 2^10 / 2^9

Answer

= 2¹ = 2 ✓

(b) [4 marks]

Let 2ˣ=3ʸ=6ᶻ=k

2=k^(1/x), 3=k^(1/y), 6=k^(1/z)

Since 6=2×3

k^(1/z) = k^(1/x) × k^(1/y) = k^(1/x+1/y)

Equate exponents

1/z = 1/x + 1/y = (x+y)/(xy)

Answer

z = xy/(x+y) ✓

(c) [4 marks]

Combine logs

log[(x+3)(x−3)] = log 7 → x²−9 = 7

Solve

x² = 16 → x = ±4

Check domain

x+3>0 and x−3>0 requires x>3; so x=4 is valid, x=−4 is rejected

Answer

x = 4 ✓

Question 3[10 marks]Ch 6 · Quadratic Equations

A rectangular dock at Chittagong Port has a perimeter of 60 m. The area of the dock is 200 m².

(a) Let the length be x m. Write expressions for width and area in terms of x. [2 marks]

(b) Form and solve a quadratic equation for x. [4 marks]

(c) Find both possible dimensions and explain which is the valid solution. [4 marks]

▶ Show Solution

(a) [2 marks]

Perimeter constraint

2(length + width) = 60 → width = 30−x

Area

Area = x(30−x) = 200 ✓

(b) [4 marks]

Expand

30x − x² = 200 → x²−30x+200 = 0

Factor

(x−20)(x−10) = 0

Roots

x = 20 or x = 10 ✓

(c) [4 marks]

If x = 20

length = 20 m, width = 30−20 = 10 m

If x = 10

length = 10 m, width = 30−10 = 20 m

Both valid

Both give 20 m × 10 m dock (same rectangle, convention: length ≥ width, so length = 20 m, width = 10 m) ✓

Question 4[10 marks]Ch 9 · Series

A port worker saves 500 taka in the first month, then increases savings by 200 taka each month.

(a) Write the first four terms of this arithmetic sequence. [2 marks]

(b) Find total savings after 12 months. [4 marks]

(c) After how many months will cumulative savings exceed 50,000 taka? [4 marks]

▶ Show Solution

(a) [2 marks]

Sequence

500, 700, 900, 1100 (a=500, d=200) ✓

(b) [4 marks]

S₁₂

= 12/2 × [2(500)+11(200)] = 6 × [1000+2200] = 6 × 3200

Answer

S₁₂ = 19,200 taka ✓

(c) [4 marks]

Inequality

S_n = n/2[2(500)+(n−1)(200)] > 50000

Simplify

n[1000+200n−200]/2 > 50000 → n(800+200n) > 100000

Simplify

200n²+800n−100000 > 0 → n²+4n−500 > 0

Solve n²+4n−500=0

n = (−4+√(16+2000))/2 = (−4+√2016)/2 ≈ (−4+44.9)/2 ≈ 20.45

Answer

n = 21 months ✓

Question 5[10 marks]Ch 11 · Geometry

In triangle ABC, side BC is extended to point D. ∠BAC = 45° and ∠ABC = 75°.

(a) State the exterior angle theorem. [2 marks]

(b) Find ∠ACD (exterior angle). [4 marks]

(c) Find ∠ACB and verify all angles sum to 180°. [4 marks]

▶ Show Solution

(a) [2 marks]

Theorem

The exterior angle of a triangle equals the sum of the two non-adjacent interior angles. ✓

(b) [4 marks]

Exterior angle

∠ACD = ∠BAC + ∠ABC = 45°+75°

Answer

∠ACD = 120° ✓

(c) [4 marks]

∠ACB

∠ACB = 180°−∠ACD = 180°−120° = 60°

Verify

∠BAC+∠ABC+∠ACB = 45°+75°+60° = 180° ✓

Question 6[10 marks]Ch 13 · Circle

Points A, B, C, D lie on a circle. ∠BAC = 35° and ∠ABD = 42°.

(a) Find ∠BDC (angles in the same segment). [2 marks]

(b) Find the central angle ∠BOC (O is center). [4 marks]

(c) If arc BC has length 11 cm and radius is 10 cm, find ∠BOC in degrees. (π = 22/7) [4 marks]

▶ Show Solution

(a) [2 marks]

Angles in same segment

∠BAC and ∠BDC are in the same segment (both subtended by arc BC)

Answer

∠BDC = ∠BAC = 35° ✓

(b) [4 marks]

Central angle theorem

∠BOC = 2 × ∠BAC (central angle = 2 × inscribed angle on same arc)

Answer

∠BOC = 2 × 35° = 70° ✓

(c) [4 marks]

Arc length formula

Arc = (θ/360°) × 2πr → 11 = (θ/360) × 2 × (22/7) × 10

Simplify RHS

(θ/360) × (440/7) = 11 → θ/360 = 11×7/440 = 77/440 = 7/40

Answer

θ = 360 × 7/40 = 63° ✓

Question 7[10 marks]Ch 14 · Trigonometry

From the top of the Karnafuli Bridge (height H m), an observer sees a boat upstream at an angle of depression of 30° and a boat downstream at an angle of depression of 45°. The two boats are 200 m apart (along the river).

(a) Express the horizontal distance to each boat in terms of H. [2 marks]

(b) Using the 200 m distance, find H. (tan 30° = 1/√3, tan 45° = 1) [4 marks]

(c) Find the actual slant distance from the bridge top to the upstream boat. [4 marks]

▶ Show Solution

(a) [2 marks]

Upstream boat

tan 30° = H/d₁ → d₁ = H/tan30° = H√3

Downstream boat

tan 45° = H/d₂ → d₂ = H/1 = H ✓

(b) [4 marks]

Total distance

d₁ + d₂ = 200 → H√3 + H = 200

Solve

H(√3+1) = 200 → H = 200/(√3+1)

Rationalize

H = 200(√3−1)/((√3+1)(√3−1)) = 200(√3−1)/2 = 100(√3−1)

Answer

H = 100(√3−1) ≈ 73.2 m ✓

(c) [4 marks]

d₁ (upstream)

= H√3 = 100(√3−1)×√3 = 100(3−√3) ≈ 100×1.268 = 126.8 m

Slant distance

= √(d₁²+H²) = √((H√3)²+H²) = H√(3+1) = 2H

Answer

Slant = 2H = 200(√3−1) ≈ 146.4 m ✓

Question 8[10 marks]Ch 16 · Mensuration

A port warehouse is 50 m long, 20 m wide, and 8 m high. It has a pitched roof shaped like a triangular prism on top, with the triangular face having base 20 m and height 4 m.

(a) Find the volume of the rectangular warehouse section. [2 marks]

(b) Find the volume of the roof (triangular prism). [4 marks]

(c) Find the total external surface area of the structure (excluding the floor). [4 marks]

▶ Show Solution

(a) [2 marks]

Rectangular section

V = l×w×h = 50×20×8 = 8000 m³ ✓

(b) [4 marks]

Triangular prism volume

V = (1/2 × base × height) × length = (1/2 × 20 × 4) × 50

Answer

V = 40 × 50 = 2000 m³ ✓

(c) [4 marks]

4 walls of rectangle

2(50×8) + 2(20×8) = 800+320 = 1120 m²

2 triangular end faces of roof

2×(1/2×20×4) = 80 m²

2 sloping roof faces

Slant height = √(10²+4²) = √116 ≈ 10.77 m; each face = 50×10.77 = 538.5 m²; total = 1077 m²

Total SA

1120+80+1077 = 2277 m² ✓

Chittagong Board · SSC Mathematics · 2020

Chapter Coverage

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