SSC MATH

Chittagong Board · SSC Mathematics · 2021

Chittagong 2021 Total: 100 Marks 3 Hours 8 Questions

Chapter Coverage

10
QChapterTopicMarks
1Ch 2 · SetsSet operations with coastal data10
2Ch 3 · AlgebraAlgebraic fraction simplification10
3Ch 5 · Linear EquationsShip charter cost problem10
4Ch 9 · SeriesCompound interest — maritime insurance10
5Ch 11 · TrianglesAngle bisector theorem10
6Ch 12 · AreaArea theorem — field partition10
7Ch 14 · TrigonometryShip bearing and distance
8Ch 15 · StatisticsPort traffic data — ogive and quartiles10
Question 1[10 marks]Ch 2 · Sets

A survey of 100 fishermen finds: 60 use nylon nets (N), 50 use trawl nets (T), and 25 use both types.

(a) Find |N ∪ T| (fishermen using at least one type). [2 marks]

(b) Find the number who use only nylon nets. [4 marks]

(c) How many use neither type? Draw a Venn diagram. [4 marks]

▶ Show Solution

(a) [2 marks]

Inclusion-exclusion
|N ∪ T| = |N| + |T| − |N ∩ T| = 60+50−25
Answer
|N ∪ T| = 85 ✓

(b) [4 marks]

Only N
|N only| = |N| − |N ∩ T| = 60−25 = 35 fishermen ✓

(c) [4 marks]

Neither
= Total − |N ∪ T| = 100−85 = 15 fishermen ✓
Venn diagram values
Only N = 35, Both = 25, Only T = 25, Neither = 15 (total 100) ✓
Question 2[10 marks]Ch 3 · Algebra

Simplify the following algebraic expressions.

(a) Simplify: (a²−b²)/(a+b). [2 marks]

(b) Simplify: (x²+5x+6)/(x²+3x+2) ÷ (x²+4x+3)/(x²+2x). [4 marks]

(c) If a/(b+c) = b/(c+a) = c/(a+b) = k, prove that k = 1/2 (when a+b+c ≠ 0). [4 marks]

▶ Show Solution

(a) [2 marks]

Factor numerator
(a²−b²) = (a+b)(a−b)
Cancel
(a+b)(a−b)/(a+b) = a−b ✓

(b) [4 marks]

Factor each
x²+5x+6=(x+2)(x+3); x²+3x+2=(x+1)(x+2); x²+4x+3=(x+1)(x+3); x²+2x=x(x+2)
Division = multiply by reciprocal
[(x+2)(x+3)/(x+1)(x+2)] × [x(x+2)/(x+1)(x+3)]
Cancel factors
= x(x+2)/(x+1)²
Answer
= x(x+2)/(x+1)² ✓

(c) [4 marks]

Sum of numerators
a+b+c = k[(b+c)+(c+a)+(a+b)] = k×2(a+b+c)
Divide by (a+b+c)
1 = 2k
Answer
k = 1/2 ✓
Question 3[10 marks]Ch 5 · Linear Equations

Chartering a cargo ship costs a fixed amount plus a rate per tonne of cargo. Chartering with 200 tonnes costs 80,000 taka. Chartering with 350 tonnes costs 117,500 taka.

(a) Write two linear equations for fixed cost F and rate per tonne r. [2 marks]

(b) Find F and r. [4 marks]

(c) What is the maximum cargo (in tonnes) that can be chartered with a budget of 150,000 taka? [4 marks]

▶ Show Solution

(a) [2 marks]

Equation 1
F + 200r = 80000
Equation 2
F + 350r = 117500

(b) [4 marks]

Subtract Eq1 from Eq2
150r = 37500 → r = 250 taka/tonne
Substitute
F = 80000 − 200×250 = 80000 − 50000 = 30000 taka
Answer
F = 30,000 taka, r = 250 taka/tonne ✓

(c) [4 marks]

Budget equation
30000 + 250T ≤ 150000
Solve
250T ≤ 120000 → T ≤ 480
Answer
Maximum 480 tonnes ✓
Question 4[10 marks]Ch 9 · Series

A shipping company takes a maritime insurance policy for 500,000 taka. The insured value grows by 8% per year (compounded annually).

(a) Write the formula for the insured value after n years. [2 marks]

(b) Find the insured value after 3 years. (1.08³ = 1.2597) [4 marks]

(c) In how many years will the value first exceed 1,000,000 taka? (log 1.08 = 0.0334, log 2 = 0.3010) [4 marks]

▶ Show Solution

(a) [2 marks]

Formula
V_n = 500000 × (1.08)ⁿ ✓

(b) [4 marks]

V₃
= 500000 × 1.2597
Answer
V₃ = 629,850 taka ✓

(c) [4 marks]

Inequality
500000 × 1.08ⁿ > 1000000 → 1.08ⁿ > 2
Take log
n × log 1.08 > log 2 → n × 0.0334 > 0.3010
Solve
n > 0.3010/0.0334 = 9.01
Answer
n = 10 years ✓
Question 5[10 marks]Ch 11 · Triangles

In triangle PQR, the angle bisector from P meets QR at point D.

(a) State the angle bisector theorem. [2 marks]

(b) If PQ = 8 cm, PR = 6 cm, and QR = 14 cm, find QD and DR. [4 marks]

(c) Find the ratio of areas of triangles PQD and PRD. [4 marks]

▶ Show Solution

(a) [2 marks]

Theorem
The angle bisector from a vertex divides the opposite side in the ratio of the adjacent sides: QD/DR = PQ/PR ✓

(b) [4 marks]

Ratio
QD/DR = PQ/PR = 8/6 = 4/3
QD + DR = 14
QD = 14×4/7 = 8, DR = 14×3/7 = 6
Answer
QD = 8 cm, DR = 6 cm ✓

(c) [4 marks]

Same height from P
Area(PQD)/Area(PRD) = QD/DR (same perpendicular height)
Answer
Area ratio = 8/6 = 4/3 ✓
Question 6[10 marks]Ch 12 · Area

Parallelogram ABCD has base AB = 15 cm and height 8 cm. E is the midpoint of BC.

(a) Find the area of parallelogram ABCD. [2 marks]

(b) Find the area of triangle ABE. [4 marks]

(c) A diagonal AC divides ABCD into two triangles. Prove they are equal in area. [4 marks]

▶ Show Solution

(a) [2 marks]

Area formula
Area = base × height = 15 × 8 = 120 cm² ✓

(b) [4 marks]

Triangle ABE
AB = 15 cm. E is midpoint of BC, so BE = BC/2. Need height of △ABE from E to AB.
Method using area ratio
△ABE shares base AB with parallelogram. Height of △ABE = distance from E to AB = height of parallelogram/2 ...
Better approach
In △ABC, E is midpoint of BC. Area(△ABE) = Area(△ABC)/2 = (120/2)/2 = 30 cm²
Answer
Area(△ABE) = 30 cm² ✓

(c) [4 marks]

Triangle ABC
Diagonal AC divides ABCD into △ABC and △ACD
Area(△ABC)
= (1/2) × base AB × height h = (1/2) × 15 × 8 = 60 cm²
Area(△ACD)
= Area(ABCD) − Area(△ABC) = 120 − 60 = 60 cm²
Proof
Area(△ABC) = Area(△ACD) = 60 cm². The diagonal bisects the area ✓
Question 7[10 marks]Ch 14 · Trigonometry

A ship departs Chittagong Port and sails 40 km due East, then turns and sails 30 km due North.

(a) Find the straight-line distance from the port to the final position. [2 marks]

(b) Find the bearing (angle from North) of the final position from the port. (tan θ = 4/3) [4 marks]

(c) Another ship starts at the final position and returns to port sailing in a straight line. What is its bearing? [4 marks]

▶ Show Solution

(a) [2 marks]

Pythagoras
d = √(40²+30²) = √(1600+900) = √2500 = 50 km ✓

(b) [4 marks]

Angle from North (bearing)
tan θ = East/North = 40/30 = 4/3
Find θ
θ = arctan(4/3) ≈ 53.13°
Bearing
Bearing = N 53.13° E (or bearing 053°) ✓

(c) [4 marks]

Return journey
The return is the opposite direction: South 53.13° West
Bearing from North clockwise
= 180° + 53.13° = 233.13°
Answer
Bearing = 233° (S 53.13° W) ✓
Question 8[10 marks]Ch 15 · Statistics

Monthly container traffic (in thousands) at Chittagong Port for 8 months: 42, 47, 53, 38, 61, 55, 49, 59.

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

(b) Find the range and the variance of the data. [4 marks]

(c) Find the standard deviation and comment on the spread. [4 marks]

▶ Show Solution

(a) [2 marks]

Sum
42+47+53+38+61+55+49+59 = 404
Mean
x̄ = 404/8 = 50.5 thousand containers ✓

(b) [4 marks]

Range
Max−Min = 61−38 = 23
Deviations (xᵢ−x̄)²
(42−50.5)²=72.25, (47−50.5)²=12.25, (53−50.5)²=6.25, (38−50.5)²=156.25, (61−50.5)²=110.25, (55−50.5)²=20.25, (49−50.5)²=2.25, (59−50.5)²=72.25
Sum of squared deviations
72.25+12.25+6.25+156.25+110.25+20.25+2.25+72.25 = 452
Variance
σ² = 452/8 = 56.5 ✓

(c) [4 marks]

Standard deviation
σ = √56.5 ≈ 7.52 thousand
Comment
σ ≈ 7.52 (thousand containers). The SD is about 15% of the mean, indicating moderate variability in monthly traffic. ✓

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