Chittagong Board SSC Math 2022

Q	Chapter	Topic	Marks
1	Ch 1 · Real Numbers	Surds and irrational numbers	10
2	Ch 3 · Algebra	HCF and LCM of polynomials	10
3	Ch 6 · Quadratic	Discriminant and nature of roots	10
4	Ch 9 · Series	Mixed AP/GP — fishing season	10
5	Ch 11 · Geometry	Triangle area and median	10
6	Ch 13 · Circle	Chord properties and perpendicular bisector	10
7	Ch 14 · Trigonometry	Two-angle elevation problem	10
8	Ch 16 · Mensuration	Frustum of cone — fishing net float	10

Question 1[10 marks]Ch 1 · Real Numbers

Work with surds and rationalization.

(a) Simplify: √75 − √48 + √27. [2 marks]

(b) Rationalize the denominator of 5/(3+√2) and simplify. [4 marks]

(c) If x = (√5+√3)/(√5−√3), find x + 1/x. [4 marks]

▶ Show Solution

(a) [2 marks]

Simplify each

√75=5√3, √48=4√3, √27=3√3

Combine

5√3 − 4√3 + 3√3 = 4√3 ✓

(b) [4 marks]

Multiply by conjugate

5/(3+√2) × (3−√2)/(3−√2)

Denominator

(3+√2)(3−√2) = 9−2 = 7

Answer

5(3−√2)/7 = (15−5√2)/7 ✓

(c) [4 marks]

Find 1/x

1/x = (√5−√3)/(√5+√3)

x + 1/x

= [(√5+√3)² + (√5−√3)²] / [(√5+√3)(√5−√3)]

Numerator

(5+2√15+3)+(5−2√15+3) = 16

Denominator

5−3 = 2

Answer

x + 1/x = 16/2 = 8 ✓

Question 2[10 marks]Ch 3 · Algebra

Let f(x) = x³ − 7x + 6 and g(x) = x² − x − 2.

(a) Factorize g(x). [2 marks]

(b) Factorize f(x) completely. [4 marks]

(c) Find the HCF of f(x) and g(x). [4 marks]

▶ Show Solution

(a) [2 marks]

Factor g(x)

x²−x−2: find factors of −2 summing to −1: (−2)(+1)

Answer

g(x) = (x−2)(x+1) ✓

(b) [4 marks]

Try x=1

f(1) = 1−7+6 = 0 ✓, so (x−1) is a factor

Divide

x³−7x+6 = (x−1)(x²+x−6)

Factor x²+x−6

= (x+3)(x−2)

Answer

f(x) = (x−1)(x+3)(x−2) ✓

(c) [4 marks]

Factors of f(x)

(x−1), (x+3), (x−2)

Factors of g(x)

(x−2), (x+1)

Common factor

HCF = (x−2) ✓

Question 3[10 marks]Ch 6 · Quadratic Equations

Consider the quadratic equation 3x² − 10x + k = 0.

(a) Find the discriminant in terms of k. [2 marks]

(b) Find the value of k for which the roots are equal. [4 marks]

(c) If k = 3, find the two roots and verify using Vieta's formulas. [4 marks]

▶ Show Solution

(a) [2 marks]

Discriminant

Δ = b²−4ac = (−10)²−4(3)(k) = 100−12k

Answer

Δ = 100−12k ✓

(b) [4 marks]

Equal roots condition

Δ = 0 → 100−12k = 0

Solve

k = 100/12 = 25/3 ✓

(c) [4 marks]

Equation (k=3)

3x²−10x+3 = 0

Quadratic formula

x = (10 ± √(100−36))/6 = (10 ± √64)/6 = (10±8)/6

Roots

x₁ = 18/6 = 3, x₂ = 2/6 = 1/3

Vieta's check

Sum = 3+1/3 = 10/3 = −(−10)/3 ✓; Product = 3×(1/3) = 1 = 3/3 ✓

Answer

x = 3 or x = 1/3 ✓

Question 4[10 marks]Ch 9 · Series

During monsoon fishing season, a boat catches 60 kg on the first trip. Each subsequent trip catches 15 kg more than the previous trip (trips within the same week). In the second week the boat switches to a geometric pattern, starting at the last value from week 1, decreasing by a ratio of 4/5 each day.

(a) Find the catch on the 5th trip of week 1. [2 marks]

(b) Find the total catch for week 1 (7 trips). [4 marks]

(c) Find the total catch for week 2 (5 trips in GP). [4 marks]

▶ Show Solution

(a) [2 marks]

Week 1 AP: a=60, d=15

T₅ = 60 + 4×15 = 60+60 = 120 kg ✓

(b) [4 marks]

S₇ formula

S₇ = 7/2 × [2(60) + 6(15)] = 7/2 × [120+90] = 7/2 × 210

Answer

S₇ = 7 × 105 = 735 kg ✓

(c) [4 marks]

Week 2 first term

T₇ (week 1) = 60+6×15 = 150 kg = a for week 2, r = 4/5

GP sum formula

S = a(1−rⁿ)/(1−r) = 150×(1−(4/5)⁵)/(1−4/5)

(4/5)⁵

= 1024/3125

S₅

= 150×(1−1024/3125)/(1/5) = 750×(2101/3125) = 750×0.67232

Answer

S₅ ≈ 504.2 kg ✓

Question 5[10 marks]Ch 11 · Geometry

In triangle ABC, AB = 10 cm, BC = 8 cm, and AC = 6 cm.

(a) Show that triangle ABC is a right triangle and identify the right angle. [2 marks]

(b) Find the area of triangle ABC. [4 marks]

(c) Find the length of the median from A to BC. [4 marks]

▶ Show Solution

(a) [2 marks]

Check Pythagoras

AB² = 100, BC²+AC² = 64+36 = 100

Conclusion

BC²+AC² = AB², so ∠C = 90° ✓

(b) [4 marks]

Right triangle area

Area = (1/2) × BC × AC = (1/2) × 8 × 6

Answer

Area = 24 cm² ✓

(c) [4 marks]

Median formula

m_a = (1/2)√(2b²+2c²−a²) where a=BC=8, b=AC=6, c=AB=10 (median from A to midpoint of BC)

Wait — median from A to midpoint M of BC

In a right triangle with right angle at C, median from vertex A to hypotenuse... let me use coordinates: C=(0,0), B=(8,0), A=(0,6)

Midpoint M of BC

M = (4, 0)

AM

= √((0−4)²+(6−0)²) = √(16+36) = √52 = 2√13 ≈ 7.21 cm ✓

Question 6[10 marks]Ch 13 · Circle

In a circle of radius 13 cm, a chord AB has length 24 cm. O is the center.

(a) Find the perpendicular distance from O to the chord AB. [2 marks]

(b) Two chords AB and CD are equidistant from the center. If CD = 24 cm also, prove that AB = CD. [4 marks]

(c) Another chord EF is 5 cm from the center. Find the length of EF. [4 marks]

▶ Show Solution

(a) [2 marks]

Half-chord

AM = 24/2 = 12 cm

OM (Pythagoras)

OM = √(OA²−AM²) = √(169−144) = √25 = 5 cm ✓

(b) [4 marks]

Theorem

Chords equidistant from center are equal in length

Proof

Let OM⊥AB and ON⊥CD, with OM = ON. OA = OC = r. Then AM² = OA²−OM² = OC²−ON² = CN². So AM = CN.

Conclusion

AB = 2AM = 2CN = CD ✓

(c) [4 marks]

EF half-length

EM² = r²−d² = 13²−5² = 169−25 = 144 → EM = 12

Answer

EF = 2×12 = 24 cm ✓

Question 7[10 marks]Ch 14 · Trigonometry

From two points A and B on the same side of a tower, the angles of elevation of the top are 60° and 30° respectively. AB = 40 m and A, B and the base of the tower are collinear.

(a) Let the tower height be h and the distance from B to tower base be d. Write two equations involving h and d. [2 marks]

(b) Solve for h and d. (tan 60° = √3, tan 30° = 1/√3) [4 marks]

(c) Find the straight-line distance from point A to the top of the tower. [4 marks]

▶ Show Solution

(a) [2 marks]

From A (closer)

tan 60° = h/(d−40) → √3 = h/(d−40)

From B (farther)

tan 30° = h/d → 1/√3 = h/d ✓

(b) [4 marks]

From eq 2

h = d/√3

Substitute into eq 1

√3 = (d/√3)/(d−40) → √3(d−40) = d/√3 → 3(d−40) = d

Solve d

3d−120 = d → 2d = 120 → d = 60 m

Solve h

h = 60/√3 = 20√3 ≈ 34.64 m ✓

(c) [4 marks]

Distance from A to tower base

= d−40 = 60−40 = 20 m

Slant distance

= √(20²+(20√3)²) = √(400+1200) = √1600 = 40 m ✓

Question 8[10 marks]Ch 16 · Mensuration

A fishing net float is shaped like a frustum of a cone. The radii of the two circular ends are 6 cm (top) and 10 cm (bottom). The height is 12 cm. (π = 22/7)

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

(b) Find the curved surface area. [4 marks]

(c) Find the volume of the float. [4 marks]

▶ Show Solution

(a) [2 marks]

Slant height formula

l = √(h²+(R−r)²) = √(144+(10−6)²) = √(144+16) = √160

Answer

l = 4√10 ≈ 12.65 cm ✓

(b) [4 marks]

CSA formula

= π(R+r)l = (22/7) × (10+6) × 4√10

Calculate

= (22/7) × 16 × 12.65 = (22/7) × 202.4 ≈ 636.4 cm²

Answer

CSA ≈ 636.4 cm² ✓

(c) [4 marks]

Frustum volume

V = (πh/3)(R²+Rr+r²) = (22/7×12/3)(100+60+36)

Calculate

= (22/7 × 4) × 196 = (88/7) × 196 = 88 × 28 = 2464 cm³ ✓

Chittagong Board · SSC Mathematics · 2022

Chapter Coverage

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