Dhaka Board SSC Math 2022 · Complete Solutions

Question	Chapter	Topic	Marks
Q1	Ch 3	Algebraic Identities — (a+b+c)²	10
Q2	Ch 4	Exponent Laws — Simplification	10
Q3	Ch 7	Simultaneous Equations — Word Problem	10
Q4	Ch 8	Series — Sum of Squares	10
Q5	Ch 9	Parallel Lines — Angle Proofs	10
Q6	Ch 11	Circle — Tangent from External Point	10
Q7	Ch 12	Trigonometric Identity Proofs	10
Q8	Ch 17	Statistics — Grouped Frequency, Mean & SD	10

Question 1 10 marks Ch 3 · Algebraic Identities

STEM: Let a + b + c = 4, a² + b² + c² = 20.

(a) Find ab + bc + ca. [2 marks]

(b) Prove (a+b+c)² = a² + b² + c² + 2(ab+bc+ca) and use it to find (a−b−c)² given b+c = 3. [4 marks]

(c) If additionally abc = 5, find a³ + b³ + c³. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

From identity

(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)

Substitute

16 = 20 + 2(ab+bc+ca) → ab+bc+ca = −2

Answer

ab + bc + ca = −2 ✓

(b) Solution [4 marks]

Prove (a+b+c)²

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca = a²+b²+c²+2(ab+bc+ca) ✓

Find (a−b−c)²

Note a−b−c = a−(b+c). With b+c = 3: a = 4−(b+c) = 4−3 = 1

Calculate

(a−b−c)² = (1−3)² = (−2)² = 4

Answer

(a−b−c)² = 4 ✓

(c) Solution [4 marks]

Newton's formula

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

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

= 4 × (20−(−2)) = 4 × 22 = 88

So

a³+b³+c³ = 88 + 3abc = 88 + 15 = 103

Answer

a³ + b³ + c³ = 103 ✓

Question 2 10 marks Ch 4 · Exponent Laws

STEM: Simplify using laws of exponents.

(a) Simplify: (x²y³)⁴ ÷ (x³y²)³. [2 marks]

(b) Simplify: (a^m · b^n)^p · (a^p · b^m)^n ÷ (a^n · b^p)^m. [4 marks]

(c) Prove: aˣ/aʸ · aʸ/aᶻ · aᶻ/aˣ = 1. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Expand numerator

(x²y³)⁴ = x⁸y¹²

Expand denominator

(x³y²)³ = x⁹y⁶

Divide

x⁸y¹²/x⁹y⁶ = x^(8−9) · y^(12−6) = x⁻¹y⁶ = y⁶/x

Answer

y⁶/x ✓

(b) Solution [4 marks]

Expand each factor

(a^m·b^n)^p = a^(mp)·b^(np)

Second factor

(a^p·b^m)^n = a^(pn)·b^(mn)

Third factor (denominator)

(a^n·b^p)^m = a^(nm)·b^(pm)

Multiply numerators, divide by denominator

a^(mp+pn−nm) · b^(np+mn−pm)

Simplify exponents

For a: mp+pn−nm = p(m+n)−nm; for b: np+mn−pm = n(p+m)−pm

When m=n=p, check

Exponents: mp+pn−nm = m²+m²−m² = m², similarly for b. Final result = a^(mp+np−mn) · b^(np+mn−mp)

Answer

= a^(p(m+n)−mn) · b^(n(m+p)−mp) ✓

(c) Solution [4 marks]

Expand each term

aˣ/aʸ = a^(x−y), aʸ/aᶻ = a^(y−z), aᶻ/aˣ = a^(z−x)

Multiply

a^(x−y) · a^(y−z) · a^(z−x) = a^((x−y)+(y−z)+(z−x))

Simplify exponent

(x−y)+(y−z)+(z−x) = x−y+y−z+z−x = 0

Conclusion

= a⁰ = 1 ✓ (Proved)

Question 3 10 marks Ch 7 · Simultaneous Equations

STEM: A shopkeeper sells two types of pens. Type A costs Tk 15 each and Type B costs Tk 25 each.

(a) If he sells 5 Type A and 3 Type B for a total of Tk 150, write the equation. [2 marks]

(b) He also knows that if he sells 4 Type A and 5 Type B, total = Tk 185. Solve simultaneously for the prices (verify they match). [4 marks]

(c) A different day: 3x + 2y = 16 and 5x − 3y = 9. Solve this system. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Let price of A = a, price of B = b

5a + 3b = 150

Answer

5a + 3b = 150 ✓

(b) Solution [4 marks]

Two equations

Eq(1): 5a + 3b = 150; Eq(2): 4a + 5b = 185

Eliminate a

4×Eq(1): 20a+12b=600; 5×Eq(2): 20a+25b=925

Subtract

13b = 325 → b = 25

Find a

5a + 75 = 150 → 5a = 75 → a = 15

Verify

4(15)+5(25) = 60+125 = 185 ✓

Answer

Type A = Tk 15, Type B = Tk 25 ✓

(c) Solution [4 marks]

Equations

Eq(1): 3x+2y=16; Eq(2): 5x−3y=9

Eliminate y

3×Eq(1): 9x+6y=48; 2×Eq(2): 10x−6y=18

Add

19x = 66 → x = 66/19

Hmm, try elimination of x instead

5×Eq(1): 15x+10y=80; 3×Eq(2): 15x−9y=27

Subtract

19y = 53 → y = 53/19

Find x from Eq(1)

3x = 16−2(53/19) = (304−106)/19 = 198/19 → x = 66/19

Answer

x = 66/19 ≈ 3.47, y = 53/19 ≈ 2.79 ✓

Question 4 10 marks Ch 8 · Series — Sum of Squares

STEM: Work with the sum of squares formula: 1² + 2² + 3² + ··· + n² = n(n+1)(2n+1)/6.

(a) Find 1² + 2² + 3² + ··· + 10². [2 marks]

(b) Find 11² + 12² + ··· + 20². [4 marks]

(c) Find the sum: 2² + 4² + 6² + ··· + 20² and compare with part (a). [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Formula

S₁₀ = 10×11×21/6 = 2310/6 = 385

Answer

1² + 2² + ··· + 10² = 385 ✓

(b) Solution [4 marks]

Strategy

11²+12²+···+20² = (1²+···+20²) − (1²+···+10²)

Find S₂₀

S₂₀ = 20×21×41/6 = 17220/6 = 2870

Subtract

2870 − 385 = 2485

Answer

11² + 12² + ··· + 20² = 2485 ✓

(c) Solution [4 marks]

Factor out 2² = 4

2²+4²+···+20² = 4(1²+2²+···+10²) = 4×385 = 1540

Comparison

The sum of even squares up to 20 is 4 times the sum of squares up to 10

Answer

2²+4²+···+20² = 1540 = 4×385 ✓

Question 5 10 marks Ch 9 · Parallel Lines

STEM: Lines AB and CD are parallel. EF is a transversal cutting AB at P and CD at Q. ∠APQ = 65°.

(a) Find the alternate interior angle ∠PQD. [2 marks]

(b) Find all eight angles formed at P and Q, labelling each type. [4 marks]

(c) A second transversal GH crosses AB at R and CD at S such that ∠ARS = 70°. Find ∠RSC and ∠PRS (the angle between the two transversals at their crossing, if they meet). [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Alternate interior angles

AB ∥ CD, EF is transversal: alternate interior angles are equal

Answer

∠PQD = ∠APQ = 65° (alternate interior angles) ✓

(b) Solution [4 marks]

At point P (on AB)

∠APQ = 65° (given); ∠BPQ = 180°−65° = 115° (supplementary)

Vertically opposite

∠EPB = 65° (vert. opp. to ∠APQ); ∠EPA = 115° (vert. opp. to ∠BPQ)

At point Q (on CD)

∠PQD = 65° (alt. interior); ∠PQC = 115° (supplementary)

Remaining at Q

∠FQD = 115° (vert. opp. to ∠PQC); ∠FQC = 65° (vert. opp. to ∠PQD)

Answer

Angles at P: 65°,115°,65°,115°. At Q: 65°,115°,65°,115° ✓

(c) Solution [4 marks]

Co-interior angles

∠ARS = 70°, and ∠RSC are co-interior (same-side interior) angles

Co-interior sum

∠ARS + ∠RSC = 180° → ∠RSC = 110°

Angle PRS

∠APQ = 65°, ∠ARS = 70°. The angle between EF and GH at their intersection (if they meet on line AB): ∠PRS = ∠ARS − ∠APR. Since P and R are different points on AB, ∠PRS is measured as the angle at R between RP (along AB) and RS: ∠PRS = 180°−70° = 110° (since ∠ARS + ∠BRS = 180°, and ∠BRS = ∠PRS as P is on line AB on side B)

Answer

∠RSC = 110°, ∠BRS = 110° ✓

Question 6 10 marks Ch 11 · Circle — Tangent

STEM: From an external point T, two tangents TA and TB are drawn to a circle with centre O and radius 5 cm. OT = 13 cm.

(a) Find the length of tangent TA. [2 marks]

(b) Find ∠OAT and ∠AOT. [4 marks]

(c) Find the area of quadrilateral OATB. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Tangent-radius perpendicularity

OA ⊥ TA, so △OAT is right-angled at A

Pythagoras

TA² = OT² − OA² = 13² − 5² = 169 − 25 = 144

Answer

TA = 12 cm ✓

(b) Solution [4 marks]

∠OAT

Since OA ⊥ TA: ∠OAT = 90°

Find ∠AOT

tan(∠AOT) = TA/OA = 12/5

Or use sin

sin(∠OTA) = OA/OT = 5/13, so ∠OTA = sin⁻¹(5/13) ≈ 22.62°

Angle AOT

∠AOT = 90° − 22.62° = 67.38°

Answer

∠OAT = 90°, ∠AOT ≈ 67.38° ✓

(c) Solution [4 marks]

Symmetry

OATB is a kite (OA = OB = 5, TA = TB = 12). By symmetry, it can be split into 2 congruent right triangles.

Area of △OAT

= ½ × OA × TA = ½ × 5 × 12 = 30 cm²

Area of OATB

= 2 × 30 = 60 cm²

Answer

Area of quadrilateral OATB = 60 cm² ✓

Question 7 10 marks Ch 12 · Trigonometric Identities

STEM: Prove the following trigonometric identities.

(a) Prove: (sin θ + cos θ)² + (sin θ − cos θ)² = 2. [2 marks]

(b) Prove: (1 − sin²A)/cos A = cos A. [4 marks]

(c) Prove: (tan θ + cot θ) sin θ cos θ = 1. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Expand

LHS = sin²θ+2sinθcosθ+cos²θ + sin²θ−2sinθcosθ+cos²θ

Simplify

= 2sin²θ + 2cos²θ = 2(sin²θ+cos²θ) = 2×1 = 2 = RHS ✓

Answer

Proved ✓

(b) Solution [4 marks]

LHS

(1−sin²A)/cos A

Use identity

1 − sin²A = cos²A

Substitute

= cos²A/cos A = cos A = RHS ✓

Answer

Proved ✓

(c) Solution [4 marks]

Write tan and cot

tan θ = sin θ/cos θ, cot θ = cos θ/sin θ

LHS

(sin θ/cos θ + cos θ/sin θ) · sin θ cos θ

Simplify bracket

= (sin²θ+cos²θ)/(sin θ cos θ) · sin θ cos θ

Cancel

= (sin²θ+cos²θ) = 1 = RHS ✓

Answer

Proved ✓

Question 8 10 marks Ch 17 · Statistics — Grouped Data

STEM: The marks of 30 students are given in the grouped frequency table:

Marks	Frequency
10–20	4
20–30	8
30–40	10
40–50	6
50–60	2

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

(b) Find the modal class and estimate the mode. [4 marks]

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

▶ Show Solution

(a) Solution [2 marks]

Midpoints and fx

Midpoints: 15,25,35,45,55. fx: 4×15=60, 8×25=200, 10×35=350, 6×45=270, 2×55=110

Sum

Σfx = 60+200+350+270+110 = 990; Σf = 30

Mean

x̄ = 990/30 = 33

Answer

Mean = 33 marks ✓

(b) Solution [4 marks]

Modal class

Class with highest frequency: 30–40 (f=10)

Mode formula

Mode = L + [(f₁−f₀)/(2f₁−f₀−f₂)] × h

Values

L=30, f₁=10, f₀=8, f₂=6, h=10

Calculate

Mode = 30 + [(10−8)/(20−8−6)] × 10 = 30 + [2/6]×10 = 30 + 3.33 = 33.33

Answer

Modal class: 30–40, Mode ≈ 33.33 ✓

(c) Solution [4 marks]

Calculate (x−x̄)² for each class

Using x̄=33: (15−33)²=324, (25−33)²=64, (35−33)²=4, (45−33)²=144, (55−33)²=484

f×(x−x̄)²

4×324=1296, 8×64=512, 10×4=40, 6×144=864, 2×484=968

Σf(x−x̄)²

= 1296+512+40+864+968 = 3680

Variance

σ² = 3680/30 = 122.67

SD

σ = √122.67 ≈ 11.08

Answer

Standard Deviation ≈ 11.08 marks ✓

Dhaka Board · SSC Mathematics · 2022

Chapter Coverage

Creative Questions

(a) Solution [2 marks]

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