Chittagong Board SSC Math 2023

Q	Chapter	Topic	Marks
1	Ch 3 · Algebra	Polynomial division and remainder	10
2	Ch 4 · Logarithms	Change of base and natural contexts	10
3	Ch 5 · Simultaneous Equations	Coastal trade route problem	10
4	Ch 9 · Series	Harbor tidal pattern — AP	10
5	Ch 12 · Geometry	Triangle congruence — coastal survey	10
6	Ch 13 · Circle	Tangent and chord properties	10
7	Ch 14 · Trigonometry	Coastal elevation — cliff height	10
8	Ch 16 · Mensuration	Composite solid — dock bollard	10

Question 1[10 marks]Ch 3 · Algebra

Let p(x) = 2x³ − 5x² + 3x − 7.

(a) Find p(2). [2 marks]

(b) Divide p(x) by (x−2) and find the remainder. [4 marks]

(c) Show that (2x−1) is not a factor of p(x) by evaluating p(1/2). [4 marks]

▶ Show Solution

(a) [2 marks]

Substitute x=2

p(2) = 2(8) − 5(4) + 3(2) − 7 = 16 − 20 + 6 − 7

Answer

p(2) = −5 ✓

(b) [4 marks]

Remainder theorem

When p(x) is divided by (x−2), remainder = p(2)

Long division check

2x³−5x²+3x−7 ÷ (x−2): quotient = 2x²−x+1, remainder = p(2)

Verify: (x−2)(2x²−x+1)

= 2x³−x²+x−4x²+2x−2 = 2x³−5x²+3x−2

Remainder

Remainder = p(2) = −5 ✓

(c) [4 marks]

Substitute x = 1/2

p(1/2) = 2(1/8) − 5(1/4) + 3(1/2) − 7

Calculate

= 1/4 − 5/4 + 3/2 − 7 = 0.25 − 1.25 + 1.5 − 7 = −6.5

Conclusion

p(1/2) = −13/2 ≠ 0, so (2x−1) is NOT a factor ✓

Question 2[10 marks]Ch 4 · Logarithms

A navigator uses logarithms to compute distances. (log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451)

(a) Evaluate log₇ 49 without tables. [2 marks]

(b) If log x = 3 log 2 + 2 log 3 − log 6, find x. [4 marks]

(c) Prove that: log(a/b) + log(b/c) + log(c/a) = 0. [4 marks]

▶ Show Solution

(a) [2 marks]

log₇ 49

= log₇ 7² = 2 × log₇ 7 = 2 × 1

Answer

log₇ 49 = 2 ✓

(b) [4 marks]

Simplify

log x = log 2³ + log 3² − log 6 = log 8 + log 9 − log 6

Combine

= log(8×9/6) = log(72/6) = log 12

Answer

x = 12 ✓

(c) [4 marks]

LHS

log(a/b) + log(b/c) + log(c/a)

Product rule

= log[(a/b) × (b/c) × (c/a)]

Simplify

= log[(abc)/(abc)] = log 1 = 0

Conclusion

∴ LHS = 0 = RHS ✓

Question 3[10 marks]Ch 5 · Simultaneous Equations

Two coastal traders meet at Cox's Bazar. Trader A has 5 kg of dried fish and 3 kg of prawns worth 1300 taka. Trader B has 4 kg of dried fish and 7 kg of prawns worth 2100 taka.

(a) Set up simultaneous equations for the price of dried fish (x taka/kg) and prawns (y taka/kg). [2 marks]

(b) Solve the equations to find x and y. [4 marks]

(c) Trader C wants 2 kg of dried fish and 5 kg of prawns. How much should he pay? [4 marks]

▶ Show Solution

(a) [2 marks]

Equation 1

5x + 3y = 1300

Equation 2

4x + 7y = 2100

(b) [4 marks]

Multiply Eq1 by 7

35x + 21y = 9100

Multiply Eq2 by 3

12x + 21y = 6300

Subtract

23x = 2800 → x = 2800/23 ≈ 121.7 ... let me recheck

Correct values

From 5x+3y=1300 and 4x+7y=2100: multiply first by 7: 35x+21y=9100; multiply second by 3: 12x+21y=6300; subtract: 23x=2800 → x=200 taka (rounding: use exact)

Re-solve cleanly

Use elimination: 7(5x+3y)−3(4x+7y)=7×1300−3×2100 → 35x+21y−12x−21y=9100−6300 → 23x=2800 → x≈121.7; substitute: 5(121.7)+3y=1300 → 3y=691.5 → y=230.5

Answer

x ≈ 121.7 taka/kg (dried fish), y ≈ 230.5 taka/kg (prawns) ✓

(c) [4 marks]

Cost for Trader C

2x + 5y = 2(121.7) + 5(230.5)

Calculate

= 243.4 + 1152.5 = 1395.9

Answer

≈ 1396 taka ✓

Question 4[10 marks]Ch 9 · Series

Tidal water levels at Chittagong Port form an arithmetic pattern. At 6 AM the water is 2.4 m deep and it rises by 0.3 m every hour.

(a) Find the water depth at 12 PM (noon). [2 marks]

(b) At what time will the depth first reach 4.5 m? [4 marks]

(c) Find the sum of hourly depth readings from 6 AM to 6 PM (inclusive, 13 readings). [4 marks]

▶ Show Solution

(a) [2 marks]

Hours from 6 AM to noon

6 hours, so n = 7 (term number)

T₇

= 2.4 + 6×0.3 = 2.4 + 1.8 = 4.2 m ✓

(b) [4 marks]

Inequality

2.4 + (n−1)×0.3 ≥ 4.5

Solve

(n−1)×0.3 ≥ 2.1 → n−1 ≥ 7 → n ≥ 8

n=8 corresponds to

6 AM + 7 hours = 1 PM ✓

(c) [4 marks]

AP sum

S = n/2 × (first + last); first = 2.4, last = 2.4 + 12×0.3 = 6.0, n = 13

Calculate

S = 13/2 × (2.4 + 6.0) = 13/2 × 8.4 = 54.6 m ✓

Question 5[10 marks]Ch 12 · Geometry

Two coastal survey teams triangulate a lighthouse L from two base points A and B on the shore. AB = 6 km, ∠LAB = 70°, ∠LBA = 65°.

(a) Find ∠ALB. [2 marks]

(b) In triangle ALB, if a line DE is drawn parallel to AB with D on AL and E on BL, and AD:DL = 2:3, find DE. [4 marks]

(c) Find the ratio of area of triangle DLE to triangle ALB. [4 marks]

▶ Show Solution

(a) [2 marks]

Angle sum

∠ALB = 180° − 70° − 65° = 45° ✓

(b) [4 marks]

Ratio AD:DL = 2:3

So DL/AL = 3/(2+3) = 3/5

Similar triangles

DE ∥ AB, so △DLE ~ △ALB (AA similarity)

Scale factor

DE/AB = DL/AL = 3/5

DE

DE = 6 × 3/5 = 3.6 km ✓

(c) [4 marks]

Area ratio theorem

If triangles are similar with ratio k, area ratio = k²

Scale factor

k = 3/5

Area ratio

Area(DLE)/Area(ALB) = (3/5)² = 9/25 ✓

Question 6[10 marks]Ch 13 · Circle

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

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

(b) Find ∠AOT and ∠AOB. [4 marks]

(c) Find the area of quadrilateral AOTB (where O is the center). [4 marks]

▶ Show Solution

(a) [2 marks]

Right angle

OA ⊥ TA (radius to tangent point)

Pythagoras

TA² = OT² − OA² = 169 − 25 = 144

Answer

TA = 12 cm ✓

(b) [4 marks]

sin(∠AOT)

sin(∠AOT) = TA/OT = 12/13 ... actually sin(∠OTA) = OA/OT = 5/13

∠AOT

cos(∠AOT) = OA/OT = 5/13 → ∠AOT = cos⁻¹(5/13) ≈ 67.38°

∠AOB

By symmetry ∠BOT = ∠AOT, so ∠AOB = 2×67.38° ≈ 134.76° ✓

(c) [4 marks]

Area of △OAT

= (1/2)×OA×TA = (1/2)×5×12 = 30 cm²

Area of △OBT

= 30 cm² (by symmetry)

Total area AOTB

= 30+30 = 60 cm² ✓

Question 7[10 marks]Ch 14 · Trigonometry

A lighthouse stands on top of a 30 m coastal cliff. From a boat at sea, the angle of elevation of the top of the lighthouse is 45° and the angle of elevation of the base of the lighthouse (top of cliff) is 30°.

(a) Set up expressions for the horizontal distance to the cliff. [2 marks]

(b) Find the horizontal distance from the boat to the base of the cliff. (tan 30° = 1/√3, tan 45° = 1) [4 marks]

(c) Find the height of the lighthouse. [4 marks]

▶ Show Solution

(a) [2 marks]

Let d = horizontal distance, h = lighthouse height

tan 30° = 30/d and tan 45° = (30+h)/d

Equations

d = 30/tan 30° and d = (30+h)/tan 45° ✓

(b) [4 marks]

From tan 30° = 1/√3

d = 30 ÷ (1/√3) = 30√3

Answer

d = 30√3 ≈ 51.96 m ✓

(c) [4 marks]

From tan 45° equation

1 = (30+h)/30√3 → 30+h = 30√3

Lighthouse height

h = 30√3 − 30 = 30(√3−1) = 30(1.732−1) = 30×0.732

Answer

h = 21.96 m ≈ 22 m ✓

Question 8[10 marks]Ch 16 · Mensuration

A dock bollard is shaped like a cylinder topped with a cone. The cylinder has radius 15 cm and height 80 cm. The cone has the same base radius and height 30 cm. (π = 3.14)

(a) Find the volume of the cylindrical part. [2 marks]

(b) Find the total volume of the bollard. [4 marks]

(c) Find the total surface area (base + lateral cylinder + lateral cone). [4 marks]

▶ Show Solution

(a) [2 marks]

Cylinder volume

V = πr²h = 3.14 × 225 × 80 = 3.14 × 18000

Answer

V_cyl = 56,520 cm³ ✓

(b) [4 marks]

Cone volume

V = (1/3)πr²h = (1/3)×3.14×225×30 = 3.14×2250 = 7065 cm³

Total volume

56520 + 7065 = 63,585 cm³ ✓

(c) [4 marks]

Base circle

= πr² = 3.14×225 = 706.5 cm²

Lateral cylinder

= 2πrh = 2×3.14×15×80 = 7536 cm²

Slant height of cone

l = √(15²+30²) = √(225+900) = √1125 = 15√5 ≈ 33.54 cm

Lateral cone

= πrl = 3.14×15×33.54 ≈ 1579 cm²

Total SA

706.5+7536+1579 ≈ 9821.5 cm² ✓

Chittagong Board · SSC Mathematics · 2023

Chapter Coverage

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