Dhaka Board SSC Math 2025 · Complete Solutions

Question	Chapter	Topic	Marks
Q1	Ch 3	Algebraic Expressions — Newton's Identities	10
Q2	Ch 4	Logarithms — Properties & Proofs	10
Q3	Ch 5	Quadratic Equations — Roots & New Equations	10
Q4	Ch 8	Arithmetic Series — Terms & Sums	10
Q5	Ch 9	Triangles — Right Triangle & Circumradius	10
Q6	Ch 11	Circle — Angles & Area	10
Q7	Ch 13	Angle of Elevation — Tower Problems	10
Q8	Ch 16	Mensuration — Cylinder & Cone	10

Question 1 10 marks Ch 3 · Algebraic Expressions

STEM: Let a + b + c = 6, ab + bc + ca = 11, and abc = 6.

(a) Find the value of a² + b² + c². [2 marks]

(b) Using Newton's identity, find a³ + b³ + c³. [4 marks]

(c) Factorize a³ + b³ + c³ − 3abc completely. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Identity

We know: (a+b+c)² = a²+b²+c² + 2(ab+bc+ca)

Substitute

6² = a²+b²+c² + 2(11) → 36 = a²+b²+c² + 22

Answer

a²+b²+c² = 14 ✓

(b) Solution [4 marks]

Newton's Identity

Let p₁ = a+b+c, e₁ = a+b+c = 6, e₂ = ab+bc+ca = 11, e₃ = abc = 6

Power Sum p₃

Newton: p₃ = e₁·p₂ − e₂·p₁ + 3e₃

p₂ = a²+b²+c²

From (a): p₂ = 14

Calculate

p₃ = 6×14 − 11×6 + 3×6 = 84 − 66 + 18 = 36

Answer

a³+b³+c³ = 36 ✓

(c) Solution [4 marks]

Known identity

a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab − bc − ca)

Find a²+b²+c² − ab−bc−ca

= (a²+b²+c²) − (ab+bc+ca) = 14 − 11 = 3

Verify with given values

a³+b³+c³ − 3abc = 36 − 3(6) = 36 − 18 = 18

Also

(a+b+c)(a²+b²+c² − ab−bc−ca) = 6 × 3 = 18 ✓

Factored Form

a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca) ✓

Further factor

Note: a²+b²+c² − ab−bc−ca = ½[(a−b)² + (b−c)² + (c−a)²]

Complete Factorization

a³+b³+c³ − 3abc = ½(a+b+c)[(a−b)²+(b−c)²+(c−a)²] ✓

Question 2 10 marks Ch 4 · Logarithms

STEM: Given that log 2 = 0.3010 and log 3 = 0.4771 (base 10 throughout).

(a) Find log 6. [2 marks]

(b) Find log 12. [4 marks]

(c) Prove that log(1) + log(2) + log(3) + … + log(10) = log(10!). [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Write 6

6 = 2 × 3

Log product rule

log 6 = log 2 + log 3 = 0.3010 + 0.4771

Answer

log 6 = 0.7781 ✓

(b) Solution [4 marks]

Write 12

12 = 4 × 3 = 2² × 3

Apply rules

log 12 = log(2²) + log 3 = 2 log 2 + log 3

Substitute

= 2(0.3010) + 0.4771 = 0.6020 + 0.4771

Answer

log 12 = 1.0791 ✓

(c) Solution [4 marks]

Left side

LHS = log 1 + log 2 + log 3 + ··· + log 10

Log sum rule

By the logarithm product rule, log a + log b = log(ab)

Apply repeatedly

log 1 + log 2 + ··· + log 10 = log(1 × 2 × 3 × ··· × 10)

Definition of factorial

1 × 2 × 3 × ··· × 10 = 10!

Note

log 1 = 0, so it contributes nothing but is included for completeness

Conclusion

LHS = log(10!) = RHS ✓ (Proved)

Numerical check

10! = 3628800, log(3628800) ≈ 6.5598

Question 3 10 marks Ch 5 · Quadratic Equations

STEM: Let f(x) = x² − 5x + 6.

(a) Find the roots of f(x) = 0 by factorization. [2 marks]

(b) For what value of k does x² + (k+1)x + k = 0 have equal roots? [4 marks]

(c) If α and β are the roots of f(x) = 0, form the quadratic equation whose roots are 2α+β and α+2β. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Factorize

x² − 5x + 6 = (x − 2)(x − 3) = 0

Answer

x = 2 or x = 3 ✓

(b) Solution [4 marks]

Equal roots condition

For equal roots, discriminant Δ = b² − 4ac = 0

Identify coefficients

Here a = 1, b = (k+1), c = k

Set Δ = 0

(k+1)² − 4(1)(k) = 0

Expand

k² + 2k + 1 − 4k = 0 → k² − 2k + 1 = 0

Factorize

(k − 1)² = 0

Answer

k = 1 ✓

(c) Solution [4 marks]

Vieta's formulas for f(x)=0

α + β = 5, αβ = 6

New roots

Let p = 2α+β and q = α+2β

Sum of new roots

p + q = (2α+β) + (α+2β) = 3α + 3β = 3(α+β) = 3×5 = 15

Product of new roots

p × q = (2α+β)(α+2β) = 2α²+4αβ+αβ+2β² = 2(α²+β²)+5αβ

Find α²+β²

α²+β² = (α+β)² − 2αβ = 25 − 12 = 13

Product

p × q = 2(13) + 5(6) = 26 + 30 = 56

New Equation

x² − (p+q)x + pq = 0 → x² − 15x + 56 = 0

Answer

x² − 15x + 56 = 0 ✓

Question 4 10 marks Ch 8 · Arithmetic Series

STEM: An arithmetic progression (AP) has first term a = 7 and common difference d = 3.

(a) Find the 10th term of the AP. [2 marks]

(b) Find the sum of the first 20 terms. [4 marks]

(c) Find the sum of all terms of the AP that lie between 20 and 100 (exclusive). [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Formula

aₙ = a + (n−1)d

Substitute

a₁₀ = 7 + (10−1)×3 = 7 + 27 = 34

Answer

10th term = 34 ✓

(b) Solution [4 marks]

Sum formula

Sₙ = n/2 × [2a + (n−1)d]

Substitute n=20

S₂₀ = 20/2 × [2(7) + (19)(3)] = 10 × [14 + 57] = 10 × 71

Answer

S₂₀ = 710 ✓

(c) Solution [4 marks]

Find first term > 20

AP: 7, 10, 13, 16, 19, 22, … → first term greater than 20 is 22

Find last term < 100

aₙ = 7 + (n−1)×3 < 100 → (n−1) < 31 → n < 32, so a₃₁ = 7+90 = 97. Last term = 97

Terms from 22 to 97

These form an AP: first = 22, last = 97, d = 3

Count terms

n = (97 − 22)/3 + 1 = 75/3 + 1 = 25 + 1 = 26 terms

Sum

S = n/2 × (first + last) = 26/2 × (22 + 97) = 13 × 119 = 1547

Answer

Sum of terms between 20 and 100 = 1547 ✓

Question 5 10 marks Ch 9 · Triangles

STEM: In triangle ABC, AB = 13 cm, BC = 12 cm, AC = 5 cm.

(a) Show that triangle ABC is a right triangle. [2 marks]

(b) Find all angles of the triangle (in degrees, to 2 decimal places where needed). [4 marks]

(c) Find the circumradius of triangle ABC. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Converse of Pythagoras

Check if the square of the longest side = sum of squares of other two sides

Calculate

AB² = 13² = 169; BC² + AC² = 144 + 25 = 169

Conclusion

AB² = BC² + AC², so △ABC is a right triangle with right angle at C ✓

(b) Solution [4 marks]

Right angle

∠ACB = 90° (from part a)

sin A = opposite/hypotenuse

sin A = BC/AB = 12/13

Angle A

∠A = sin⁻¹(12/13) ≈ 67.38°

Angle B

∠B = 180° − 90° − 67.38° = 22.62°

Verify using tan

tan B = AC/BC = 5/12 → ∠B = tan⁻¹(5/12) ≈ 22.62° ✓

Answer

∠A ≈ 67.38°, ∠B ≈ 22.62°, ∠C = 90° ✓

(c) Solution [4 marks]

Key theorem

For a right triangle, the circumradius R = hypotenuse / 2

Alternatively

R = a/(2 sin A) where a is side opposite angle A

Using diameter theorem

The hypotenuse of a right triangle is the diameter of its circumscribed circle

Calculate

R = AB/2 = 13/2 = 6.5 cm

Answer

Circumradius R = 6.5 cm ✓

Question 6 10 marks Ch 11 · Circle

STEM: A circle with centre O has AB as its diameter. C is a point on the circle. ∠CAB = 35°.

(a) Find ∠ACB. [2 marks]

(b) Find ∠ABC. [4 marks]

(c) If AB = 10 cm, find the area of triangle ABC. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Angle in semicircle theorem

The angle in a semicircle (angle subtended by diameter) is always 90°

Answer

∠ACB = 90° ✓

(b) Solution [4 marks]

Angle sum in triangle

∠CAB + ∠ABC + ∠ACB = 180°

Substitute

35° + ∠ABC + 90° = 180°

Solve

∠ABC = 180° − 35° − 90° = 55°

Answer

∠ABC = 55° ✓

(c) Solution [4 marks]

AB is hypotenuse

Since ∠ACB = 90°, AB is hypotenuse. AB = 10 cm

Find BC

cos(∠CAB) = BC/AB → BC = AB × cos 35° = 10 × 0.8192 = 8.192 cm

Find AC

sin(∠CAB) = AC/AB → AC = AB × sin 35° = 10 × 0.5736 = 5.736 cm

Area of triangle

Area = ½ × AC × BC = ½ × 5.736 × 8.192

Calculate

Area = ½ × 47.00 = 23.50 cm²

Alternative formula

Area = ½ × AB² × sin(∠CAB) × cos(∠CAB) = ½ × 100 × sin35° × cos35° = 50 × sin70°/2 = 25 sin70° ≈ 23.49 cm²

Answer

Area of △ABC = 25 sin 70° ≈ 23.49 cm² ✓

Question 7 10 marks Ch 13 · Elevation & Depression

STEM: A vertical tower PQ stands on level ground. Point A is 40 m from the base P of the tower.

(a) From A, the angle of elevation of the top Q is 30°. Find the height PQ. [2 marks]

(b) From B, on the other side of the tower, the angle of elevation of Q is 60°. If PQ = 20√3 m, find the distance PB. [4 marks]

(c) Using the values from (b), find the distance AB. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Setup

In right triangle QPA: tan 30° = PQ/PA

Solve

PQ = PA × tan 30° = 40 × (1/√3) = 40/√3 = 40√3/3 ≈ 23.09 m

Answer

PQ = 40√3/3 ≈ 23.09 m ✓

(b) Solution [4 marks]

Given

PQ = 20√3 m, ∠QBP = 60°

Setup

In right triangle QPB: tan 60° = PQ/PB

Solve for PB

PB = PQ/tan 60° = 20√3/√3 = 20 m

Answer

PB = 20 m ✓

(c) Solution [4 marks]

Find PA from part (b) context

Using PQ = 20√3 and angle at A = 30°: PA = PQ/tan 30° = 20√3 × √3 = 60 m

A and B on opposite sides

Since A and B are on opposite sides of the tower P: AB = PA + PB = 60 + 20 = 80 m

Answer

AB = 80 m ✓

Question 8 10 marks Ch 16 · Mensuration

STEM: A right circular cylinder has base radius r = 7 cm and height h = 10 cm.

(a) Find the volume of the cylinder. (Use π = 22/7) [2 marks]

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

(c) A cone has the same base radius and the same height as the cylinder. Compare the volume of the cone to that of the cylinder. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Formula

V_cylinder = πr²h

Substitute

V = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cm³

Answer

Volume = 1540 cm³ ✓

(b) Solution [4 marks]

Formula

CSA = 2πrh

Substitute

CSA = 2 × (22/7) × 7 × 10 = 2 × 22 × 10 = 440 cm²

Total Surface Area (bonus)

TSA = CSA + 2πr² = 440 + 2×(22/7)×49 = 440 + 308 = 748 cm²

Answer

Curved Surface Area = 440 cm² ✓

(c) Solution [4 marks]

Cone volume formula

V_cone = (1/3)πr²h

With same r=7, h=10

V_cone = (1/3) × (22/7) × 49 × 10 = (1/3) × 1540 = 1540/3 ≈ 513.33 cm³

Comparison

V_cone/V_cylinder = [(1/3)πr²h] / [πr²h] = 1/3

Conclusion

The cone's volume is exactly 1/3 of the cylinder's volume. V_cone = 513.33 cm³, V_cylinder = 1540 cm³ ✓

Dhaka Board · SSC Mathematics · 2025

Chapter Coverage

Creative Questions

(a) Solution [2 marks]

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