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Dhaka Board · SSC Mathematics · 2023

Complete Solved Paper · Step-by-Step Solutions

8 Creative Questions 30 MCQ 100 Marks Dhaka Style · Abstract Algebra

Chapter Coverage

QuestionChapterTopicMarks
Q1Ch 3Algebraic Expressions — Sum/Product Identities10
Q2Ch 4Logarithms — Simplification & Equations10
Q3Ch 5Quadratic Equations — Word Problem10
Q4Ch 8AP and GP Combined10
Q5Ch 9Triangle Congruence Proof10
Q6Ch 11Circle — Cyclic Quadrilateral10
Q7Ch 13Angle of Elevation — Two Positions10
Q8Ch 16Mensuration — Cone + Cylinder Combined10

Creative Questions

Question 1 10 marks Ch 3 · Algebraic Expressions

STEM: Let a + b = 5 and ab = 6.

(a) Find a² + b². [2 marks]

(b) Find a³ + b³. [4 marks]

(c) Find the value of a⁴ + b⁴ and hence find (a² − b²)² given that a > b. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Identity
(a+b)² = a² + 2ab + b²
Substitute
25 = a² + b² + 2(6) → a² + b² = 13
Answer
a² + b² = 13 ✓

(b) Solution [4 marks]

Sum of cubes identity
a³ + b³ = (a+b)(a²−ab+b²) = (a+b)(a²+b²−ab)
Substitute
= 5 × (13 − 6) = 5 × 7 = 35
Answer
a³ + b³ = 35 ✓

(c) Solution [4 marks]

Find a⁴+b⁴
(a²+b²)² = a⁴ + 2a²b² + b⁴ → a⁴+b⁴ = (a²+b²)² − 2(ab)² = 169 − 2(36) = 169 − 72 = 97
Find (a²−b²)²
(a²−b²)² = (a²+b²)² − 4a²b² = 169 − 4×36 = 169 − 144 = 25
So a²−b²
Since a > b, a²−b² = 5
Answer
a⁴+b⁴ = 97, (a²−b²)² = 25, a²−b² = 5 ✓
Question 2 10 marks Ch 4 · Logarithms

STEM: Work with logarithm laws (log to base 10 unless stated).

(a) Simplify: log 25 + log 4 − log 2. [2 marks]

(b) Solve for x: log(x+3) + log(x−2) = log 14. [4 marks]

(c) Prove: log√(ab) = ½(log a + log b) and hence find log√12 given log 2 = 0.3010, log 3 = 0.4771. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Combine
log 25 + log 4 − log 2 = log(25×4/2) = log(50)
Simplify
= log(100/2) = log 100 − log 2 = 2 − 0.3010 = 1.6990
Answer
log 50 = 1.6990 ✓

(b) Solution [4 marks]

Combine LHS
log[(x+3)(x−2)] = log 14
Remove logs
(x+3)(x−2) = 14
Expand
x² + x − 6 = 14 → x² + x − 20 = 0
Factorize
(x+5)(x−4) = 0 → x = −5 or x = 4
Domain check
x = −5: log(−5+3) = log(−2) undefined. So x = 4
Answer
x = 4 ✓

(c) Solution [4 marks]

Prove
log√(ab) = log(ab)^(1/2) = ½ log(ab) = ½(log a + log b) ✓
Apply to √12
12 = 4 × 3 = 2² × 3
log√12
= ½ log 12 = ½(log 4 + log 3) = ½(2 log 2 + log 3)
Substitute
= ½(2×0.3010 + 0.4771) = ½(0.6020 + 0.4771) = ½(1.0791) = 0.53955
Answer
log√12 ≈ 0.5396 ✓
Question 3 10 marks Ch 5 · Quadratic Equations

STEM: The product of two consecutive positive integers is 240.

(a) Set up the quadratic equation representing this situation. [2 marks]

(b) Solve the equation to find the two integers. [4 marks]

(c) If instead the sum of the squares of two consecutive integers is 265, find those integers. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Let integers be
Let the smaller integer = n, so the next is n+1
Product equation
n(n+1) = 240 → n² + n − 240 = 0
Answer
n² + n − 240 = 0 ✓

(b) Solution [4 marks]

Factorize
We need two numbers that multiply to −240 and add to 1: try 16 and −15
Factor
(n+16)(n−15) = 0 → n = −16 or n = 15
Take positive
Since we need positive integers: n = 15
The two integers
15 and 16
Verify
15 × 16 = 240 ✓
Answer
The integers are 15 and 16 ✓

(c) Solution [4 marks]

Let integers be n and n+1
n² + (n+1)² = 265
Expand
n² + n² + 2n + 1 = 265 → 2n² + 2n + 1 = 265 → 2n² + 2n − 264 = 0
Simplify
n² + n − 132 = 0
Factorize
(n+12)(n−11) = 0 → n = 11 (taking positive)
Verify
11² + 12² = 121 + 144 = 265 ✓
Answer
The integers are 11 and 12 ✓
Question 4 10 marks Ch 8 · AP and GP

STEM: An AP has 3rd term = 14 and 7th term = 30. A GP has first term 3 and common ratio 2.

(a) Find the first term and common difference of the AP. [2 marks]

(b) Find the sum of the first 15 terms of the AP. [4 marks]

(c) Find the 8th term and sum of first 8 terms of the GP. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

AP term equations
a₃ = a + 2d = 14 and a₇ = a + 6d = 30
Subtract
4d = 16 → d = 4
Find a
a = 14 − 2(4) = 6
Answer
First term a = 6, common difference d = 4 ✓

(b) Solution [4 marks]

Sum formula
S₁₅ = 15/2 × [2(6) + 14×4] = 15/2 × [12 + 56] = 15/2 × 68 = 510
Answer
S₁₅ = 510 ✓

(c) Solution [4 marks]

GP 8th term
a₈ = 3 × 2^7 = 3 × 128 = 384
GP sum formula
S₈ = 3(2⁸−1)/(2−1) = 3(256−1)/1 = 3×255 = 765
Answer
8th term = 384, S₈ = 765 ✓
Question 5 10 marks Ch 9 · Triangle Congruence

STEM: In triangles ABC and DEF, AB = DE, BC = EF, and ∠ABC = ∠DEF.

(a) State the congruence criterion that applies and prove △ABC ≅ △DEF. [2 marks]

(b) In △PQR and △XYZ, PQ = XY = 8 cm, QR = YZ = 6 cm, PR = XZ = 10 cm. Prove they are congruent and find ∠Q if ∠Y = 90°. [4 marks]

(c) The medians from B in △ABC and from E in △DEF are BM and EN respectively. If △ABC ≅ △DEF (as in part a), prove BM = EN. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

SAS criterion
We have AB = DE (side), ∠ABC = ∠DEF (included angle), BC = EF (side)
Conclusion
△ABC ≅ △DEF by SAS (Side-Angle-Side) ✓

(b) Solution [4 marks]

Check SSS
PQ = XY = 8, QR = YZ = 6, PR = XZ = 10 — all three sides equal
Congruence
△PQR ≅ △XYZ by SSS
Find ∠Q
Since △PQR ≅ △XYZ, corresponding angles are equal: ∠Q = ∠Y = 90°
Verify
8² + 6² = 64 + 36 = 100 = 10² ✓ confirms right angle at Q
Answer
△PQR ≅ △XYZ by SSS, ∠Q = 90° ✓

(c) Solution [4 marks]

M is midpoint of AC
AM = MC = AC/2
N is midpoint of DF
DN = NF = DF/2
Since △ABC ≅ △DEF
AC = DF, so AM = DN (half of equal sides)
In △ABM and △DEN
AB = DE, AM = DN, ∠BAM = ∠EDN (corresponding angles from congruent triangles)
△ABM ≅ △DEN by SAS
Therefore BM = EN
Answer
BM = EN (medians from B and E are equal) ✓
Question 6 10 marks Ch 11 · Circle — Cyclic Quadrilateral

STEM: ABCD is a cyclic quadrilateral (inscribed in a circle). ∠A = 75°, ∠B = 85°.

(a) Find ∠C and ∠D. [2 marks]

(b) The diagonal AC divides ABCD into △ABC and △ACD. Find ∠BAC if ∠ACD = 40°. [4 marks]

(c) Prove that opposite angles of a cyclic quadrilateral are supplementary. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Cyclic quad property
Opposite angles of a cyclic quadrilateral are supplementary (sum = 180°)
∠A + ∠C = 180°
∠C = 180° − 75° = 105°
∠B + ∠D = 180°
∠D = 180° − 85° = 95°
Answer
∠C = 105°, ∠D = 95° ✓

(b) Solution [4 marks]

In △ACD
We know ∠ADC = ∠D = 95°, ∠ACD = 40°
Angle sum
∠DAC = 180° − 95° − 40° = 45°
∠BAC is inscribed in arc BC
Angles BAC and BDC subtend the same arc BC, so they are equal
In △ABC
∠ABC = 85°, ∠BAC = ?, ∠BCA = ?
Use ∠BAD = ∠BAC + ∠CAD = 75°
∠BAC = 75° − 45° = 30°
Answer
∠BAC = 30° ✓

(c) Solution [4 marks]

Setup
Let ABCD be a cyclic quadrilateral with centre O
Inscribed angle theorem
∠A is inscribed angle subtending arc BCD; ∠C is inscribed angle subtending arc BAD
Arc relation
Arc BCD + Arc BAD = complete circle = 360°
Central angle
Each inscribed angle = ½ × (subtended arc): ∠A = ½ arc(BCD), ∠C = ½ arc(BAD)
Add
∠A + ∠C = ½(arc BCD + arc BAD) = ½(360°) = 180°
Conclusion
∠A + ∠C = 180°, similarly ∠B + ∠D = 180° (Proved) ✓
Question 7 10 marks Ch 13 · Angle of Elevation

STEM: A man standing 60 m from the base of a vertical tower observes its top at an angle of elevation of 45°. He then walks towards the tower.

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

(b) After walking closer, the angle of elevation becomes 60°. Find how far he has walked. [4 marks]

(c) Find the angle of elevation of the top of the tower from the midpoint between his two positions. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

From first position
tan 45° = h/60 → 1 = h/60 → h = 60 m
Answer
Height of tower = 60 m ✓

(b) Solution [4 marks]

Let new distance from tower = d₂
tan 60° = h/d₂ → √3 = 60/d₂ → d₂ = 60/√3 = 20√3 m
Distance walked
= 60 − 20√3 = 60 − 34.64 = 25.36 m
Exact form
Distance walked = 60 − 20√3 = 20(3 − √3) m
Answer
He walked 20(3−√3) ≈ 25.36 m ✓

(c) Solution [4 marks]

Midpoint distance from tower
Midpoint between 60 m and 20√3 m: d_mid = (60 + 20√3)/2 = 30 + 10√3 m
Find angle θ
tan θ = 60/(30 + 10√3) = 60/[10(3+√3)] = 6/(3+√3)
Rationalize
= 6(3−√3)/[(3+√3)(3−√3)] = 6(3−√3)/(9−3) = 6(3−√3)/6 = 3−√3
Calculate
tan θ = 3 − √3 ≈ 3 − 1.732 = 1.268
Answer
θ = tan⁻¹(3−√3) ≈ 51.7° ✓
Question 8 10 marks Ch 16 · Mensuration

STEM: A solid consists of a cylinder with a cone on top. The cylinder has radius 6 cm and height 8 cm. The cone has the same base radius and height 4.5 cm.

(a) Find the volume of the cylinder. (Use π = 3.14) [2 marks]

(b) Find the volume of the cone and the total volume of the solid. [4 marks]

(c) Find the total surface area of the solid (note: the base of the cone is joined to the cylinder top, not exposed; include the slant surface of the cone). [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Cylinder volume
V_cyl = πr²h = 3.14 × 6² × 8 = 3.14 × 36 × 8 = 3.14 × 288 = 904.32 cm³
Answer
V_cylinder = 904.32 cm³ ✓

(b) Solution [4 marks]

Cone volume
V_cone = (1/3)πr²h = (1/3)×3.14×36×4.5 = (1/3)×508.68 = 169.56 cm³
Total volume
V_total = 904.32 + 169.56 = 1073.88 cm³
Answer
Total volume = 1073.88 cm³ ✓

(c) Solution [4 marks]

Slant height of cone
l = √(r²+h²) = √(36+20.25) = √56.25 = 7.5 cm
Curved SA of cylinder
CSA_cyl = 2πrh = 2×3.14×6×8 = 301.44 cm²
Bottom circle of cylinder
= πr² = 3.14×36 = 113.04 cm²
Lateral SA of cone
= πrl = 3.14×6×7.5 = 141.3 cm²
Total SA
= 301.44 + 113.04 + 141.3 = 555.78 cm²
Answer
Total Surface Area = 555.78 cm² ✓

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