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Dhaka Board · SSC Mathematics · 2024

Complete Solved Paper · Step-by-Step Solutions

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

Chapter Coverage

QuestionChapterTopicMarks
Q1Ch 3Algebraic Expressions — Reciprocal Sums10
Q2Ch 3Algebraic Expressions — Factorization & Identity Proof10
Q3Ch 5Quadratic Equations — Roots & Transformations10
Q4Ch 8Geometric Progression — Terms & Sums10
Q5Ch 9Parallelogram — Diagonal Properties10
Q6Ch 9Isosceles Triangle — Congruence & Angles10
Q7Ch 12Trigonometry — Ratios & Identities10
Q8Ch 17Statistics — Mean, Median, Mode, SD10

Creative Questions

Question 1 10 marks Ch 3 · Algebraic Expressions

STEM: Let x + 1/x = 3.

(a) Find x² + 1/x². [2 marks]

(b) Find x³ + 1/x³. [4 marks]

(c) Find x⁴ + 1/x⁴. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Square the given
(x + 1/x)² = x² + 2 + 1/x²
Substitute
9 = x² + 2 + 1/x² → x² + 1/x² = 7
Answer
x² + 1/x² = 7 ✓

(b) Solution [4 marks]

Cube identity
(x + 1/x)³ = x³ + 3x + 3/x + 1/x³ = x³ + 1/x³ + 3(x + 1/x)
Substitute
3³ = x³ + 1/x³ + 3(3) → 27 = x³ + 1/x³ + 9
Solve
x³ + 1/x³ = 18
Answer
x³ + 1/x³ = 18 ✓

(c) Solution [4 marks]

Square result from (a)
(x² + 1/x²)² = x⁴ + 2 + 1/x⁴
Substitute
7² = x⁴ + 2 + 1/x⁴ → 49 = x⁴ + 1/x⁴ + 2
Solve
x⁴ + 1/x⁴ = 47
Answer
x⁴ + 1/x⁴ = 47 ✓
Question 2 10 marks Ch 3 · Algebraic Expressions

STEM: Consider the algebraic expressions involving a, b, c, d.

(a) Factorize a⁴ + 4b⁴. [2 marks]

(b) Factorize x⁴ + x²y² + y⁴. [4 marks]

(c) Prove that (a² + b²)(c² + d²) = (ac + bd)² + (ad − bc)². [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Sophie Germain identity
a⁴ + 4b⁴ = a⁴ + 4b⁴ + 4a²b² − 4a²b²
Rewrite
= (a² + 2b²)² − (2ab)²
Difference of squares
= (a² + 2b² + 2ab)(a² + 2b² − 2ab)
Answer
a⁴ + 4b⁴ = (a² + 2ab + 2b²)(a² − 2ab + 2b²) ✓

(b) Solution [4 marks]

Add and subtract x²y²
x⁴ + x²y² + y⁴ = x⁴ + 2x²y² + y⁴ − x²y²
Perfect square
= (x² + y²)² − (xy)²
Difference of squares
= (x² + y² + xy)(x² + y² − xy)
Answer
x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² − xy + y²) ✓

(c) Solution [4 marks]

Expand RHS
RHS = (ac+bd)² + (ad−bc)²
Expand (ac+bd)²
= a²c² + 2abcd + b²d²
Expand (ad−bc)²
= a²d² − 2abcd + b²c²
Add them
RHS = a²c² + b²d² + a²d² + b²c² = a²(c²+d²) + b²(c²+d²)
Factor
= (a²+b²)(c²+d²) = LHS ✓
Conclusion
Proved: (a²+b²)(c²+d²) = (ac+bd)² + (ad−bc)² ✓
Question 3 10 marks Ch 5 · Quadratic Equations

STEM: The equation x² − 3x + 2 = 0 has roots α and β.

(a) Find α + β and αβ. [2 marks]

(b) Find α² + β². [4 marks]

(c) Form a new quadratic equation whose roots are 1/α² and 1/β². [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Vieta's formulas
For ax²+bx+c=0: sum = −b/a, product = c/a
Apply
α + β = 3, αβ = 2
Answer
α + β = 3, αβ = 2 ✓

(b) Solution [4 marks]

Identity
α² + β² = (α+β)² − 2αβ
Substitute
= 3² − 2(2) = 9 − 4 = 5
Answer
α² + β² = 5 ✓

(c) Solution [4 marks]

New roots
New roots are 1/α² and 1/β²
Sum of new roots
1/α² + 1/β² = (α²+β²)/(α²β²) = 5/(αβ)² = 5/4
Product of new roots
1/α² × 1/β² = 1/(αβ)² = 1/4
New equation
x² − (5/4)x + 1/4 = 0
Multiply by 4
4x² − 5x + 1 = 0
Answer
4x² − 5x + 1 = 0 ✓
Question 4 10 marks Ch 8 · Geometric Progression

STEM: A geometric progression (GP) has first term a = 2 and common ratio r = 3.

(a) Find the 5th term of the GP. [2 marks]

(b) Find the sum of the first 6 terms. [4 marks]

(c) Find the sum: 1 + 3 + 9 + 27 + … + 3⁹. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

GP term formula
aₙ = a × r^(n−1)
Substitute
a₅ = 2 × 3^(5−1) = 2 × 81 = 162
Answer
5th term = 162 ✓

(b) Solution [4 marks]

GP sum formula
Sₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1
Substitute n=6
S₆ = 2(3⁶ − 1)/(3 − 1) = 2(729 − 1)/2 = 728
Answer
S₆ = 728 ✓

(c) Solution [4 marks]

Identify the series
1 + 3 + 9 + ··· + 3⁹ is a GP with a = 1, r = 3
Find n
Last term = 3⁹ = a × r^(n−1) = 1 × 3^(n−1), so n − 1 = 9 → n = 10
Apply sum formula
S₁₀ = 1×(3¹⁰ − 1)/(3 − 1) = (59049 − 1)/2 = 59048/2 = 29524
Answer
Sum = 29524 ✓
Question 5 10 marks Ch 9 · Parallelogram

STEM: ABCD is a parallelogram. P is a point on diagonal BD.

(a) Prove that △APB ≅ △CPB in area. [2 marks]

(b) If the diagonals AC = 10 cm and BD = 8 cm are perpendicular, find the area of the parallelogram. [4 marks]

(c) Prove that the diagonals of a parallelogram bisect each other. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Key property
The diagonal BD divides parallelogram ABCD into two triangles of equal area: △ABD = △CBD
Adding/subtracting △BPC
Since P is on BD: Area(△APB) = Area(△ABD) − Area(△APD) and by symmetry these equal Area(△CPB)
Equal base
Triangles APB and CPB share the same base PB, and since ABCD is a parallelogram, AB ∥ CD, so perpendicular distances from A and C to BD are equal
Conclusion
△APB and △CPB have equal base PB and equal heights, so Area(△APB) = Area(△CPB) ✓

(b) Solution [4 marks]

Formula for area using perpendicular diagonals
Area = ½ × d₁ × d₂ when diagonals are perpendicular
Substitute
Area = ½ × 10 × 8 = 40 cm²
Answer
Area of parallelogram = 40 cm² ✓

(c) Solution [4 marks]

Setup
Let diagonals AC and BD intersect at O. We prove AO = OC and BO = OD
In △AOB and △COD
AB ∥ CD (opposite sides of parallelogram), so ∠OAB = ∠OCD (alternate angles) and ∠OBA = ∠ODC (alternate angles)
Equal sides
AB = CD (opposite sides of parallelogram are equal)
ASA Congruence
△AOB ≅ △COD (by ASA: ∠OAB=∠OCD, AB=CD, ∠OBA=∠ODC)
Conclusion
Therefore AO = OC and BO = OD, so diagonals bisect each other ✓
Question 6 10 marks Ch 9 · Triangle Congruence

STEM: In triangle PQR, PQ = PR (isosceles). PD is the altitude from P to QR.

(a) Prove △PDQ ≅ △PDR. [2 marks]

(b) Prove that PD bisects angle QPR. [4 marks]

(c) If ∠QPR = 36°, find all angles of the triangle. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Given
PQ = PR (isosceles), PD ⊥ QR so ∠PDQ = ∠PDR = 90°
In △PDQ and △PDR
PD = PD (common side), ∠PDQ = ∠PDR = 90°, PQ = PR (given)
RHS Congruence
△PDQ ≅ △PDR by RHS (right angle, hypotenuse, side)
Answer
△PDQ ≅ △PDR (RHS) ✓

(b) Solution [4 marks]

From congruence in (a)
Since △PDQ ≅ △PDR, corresponding angles are equal
Equal angles
∠QPD = ∠RPD (corresponding angles of congruent triangles)
Conclusion
PD bisects ∠QPR since ∠QPD = ∠RPD ✓

(c) Solution [4 marks]

Given
∠QPR = 36°
Base angles
In isosceles △PQR, PQ = PR, so base angles are equal: ∠PQR = ∠PRQ
Angle sum
∠QPR + ∠PQR + ∠PRQ = 180°
Solve
36° + 2∠PQR = 180° → ∠PQR = 72°
Answer
∠QPR = 36°, ∠PQR = ∠PRQ = 72° ✓
Question 7 10 marks Ch 12 · Trigonometry

STEM: In right triangle ABC, sin A = 3/5, with ∠C = 90°.

(a) Find all six trigonometric ratios for angle A. [2 marks]

(b) Prove that (sin A + cos A)² = 1 + 2 sin A cos A and find its numerical value. [4 marks]

(c) Verify that sin²A + cos²A = 1 using the values found. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Given sin A = 3/5
Using Pythagorean theorem: opposite = 3, hypotenuse = 5, so adjacent = √(25−9) = 4
All six ratios
sin A = 3/5, cos A = 4/5, tan A = 3/4
Reciprocals
cosec A = 5/3, sec A = 5/4, cot A = 4/3
Answer
All six trig ratios found ✓

(b) Solution [4 marks]

Expand LHS
(sin A + cos A)² = sin²A + 2 sin A cos A + cos²A
Apply identity
sin²A + cos²A = 1, so = 1 + 2 sin A cos A
Proved algebraically
(sin A + cos A)² = 1 + 2 sin A cos A = 1 + sin 2A ✓
Numerical value
= 1 + 2×(3/5)×(4/5) = 1 + 24/25 = 49/25
Check
(3/5 + 4/5)² = (7/5)² = 49/25 ✓
Answer
(sin A + cos A)² = 49/25 ✓

(c) Solution [4 marks]

LHS
sin²A + cos²A = (3/5)² + (4/5)² = 9/25 + 16/25 = 25/25 = 1
RHS
= 1
Conclusion
LHS = RHS = 1, so sin²A + cos²A = 1 is verified ✓
Question 8 10 marks Ch 17 · Statistics

STEM: The following data set represents test scores: 15, 18, 21, 24, 27, 21, 18, 24, 15, 21.

(a) Find the mean of the data. [2 marks]

(b) Find the median and mode. [4 marks]

(c) Find the standard deviation. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Sum
15+18+21+24+27+21+18+24+15+21 = 204
Mean
x̄ = 204/10 = 20.4
Answer
Mean = 20.4 ✓

(b) Solution [4 marks]

Sort data
15, 15, 18, 18, 21, 21, 21, 24, 24, 27
Median (n=10)
Median = average of 5th and 6th values = (21 + 21)/2 = 21
Mode
21 appears 3 times (most frequent)
Answer
Median = 21, Mode = 21 ✓

(c) Solution [4 marks]

Deviations from mean (x̄=20.4)
Values and (x−x̄)²: 15→29.16, 15→29.16, 18→5.76, 18→5.76, 21→0.36, 21→0.36, 21→0.36, 24→12.96, 24→12.96, 27→43.56
Sum of (x−x̄)²
Σ(x−x̄)² = 29.16+29.16+5.76+5.76+0.36+0.36+0.36+12.96+12.96+43.56 = 140.4
Variance
σ² = Σ(x−x̄)²/n = 140.4/10 = 14.04
Standard Deviation
σ = √14.04 ≈ 3.747
Answer
Standard Deviation σ ≈ 3.75 ✓

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