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Jessore Board · SSC Mathematics · 2022

Jessore 2022 Total: 100 Marks 3 Hours 8 Questions

Chapter Coverage

QChapterTopicMarks
1Ch 2 · SetsSet operations and universal set10
2Ch 4 · LogarithmsLog equations and properties10
3Ch 6 · QuadraticNature of roots — farming land10
4Ch 9 · SeriesAP with agricultural data10
5Ch 11 · GeometryIsosceles triangle properties10
6Ch 12 · AreaQuadrilateral area decomposition10
7Ch 15 · StatisticsGrouped data mean and mode10
8Ch 16 · MensurationRectangular box and diagonal10
Question 1[10 marks]Ch 2 · Sets

Universal set U = {1,2,3,4,5,6,7,8,9,10}. A = {2,4,6,8,10}, B = {1,2,3,4,5}.

(a) Find A ∩ B and A ∪ B. [2 marks]

(b) Find A' (complement of A) and verify |A|+|A'| = |U|. [4 marks]

(c) Find (A ∪ B)' and verify De Morgan's law: (A ∪ B)' = A' ∩ B'. [4 marks]

▶ Show Solution

(a) [2 marks]

A ∩ B
= {2,4} (elements in both)
A ∪ B
= {1,2,3,4,5,6,8,10} ✓

(b) [4 marks]

A'
= U − A = {1,3,5,7,9}
Verify
|A|=5, |A'|=5, 5+5=10=|U| ✓

(c) [4 marks]

(A∪B)'
A∪B = {1,2,3,4,5,6,8,10}, so (A∪B)' = {7,9}
A'
= {1,3,5,7,9}
B'
= U−B = {6,7,8,9,10}
A'∩B'
= {7,9}
De Morgan's verified
(A∪B)' = A'∩B' = {7,9} ✓
Question 2[10 marks]Ch 4 · Logarithms

(log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451)

(a) Find the value of log 0.0084. [2 marks]

(b) Solve: log₂(x+4) + log₂(x−4) = log₂ 48. [4 marks]

(c) If log_a(bc) = x and log_b(ac) = y and log_c(ab) = z, prove that 1/(x+1)+1/(y+1)+1/(z+1) = 1. [4 marks]

▶ Show Solution

(a) [2 marks]

0.0084 = 84/10000 = 4×21/10000
log 0.0084 = log 84 − log 10000 = log(4×3×7) − 4
Calculate
= 2log2+log3+log7 − 4 = 0.6020+0.4771+0.8451−4 = 1.9242−4 = −2.0758 ✓

(b) [4 marks]

Combine
log₂[(x+4)(x−4)] = log₂ 48 → x²−16 = 48
Solve
x² = 64 → x = ±8
Check domain
x+4>0 and x−4>0 requires x>4, so x=8 valid, x=−8 rejected
Answer
x = 8 ✓

(c) [4 marks]

Note
x+1 = log_a(bc)+1 = log_a(bc)+log_a(a) = log_a(abc)
Similarly
y+1=log_b(abc), z+1=log_c(abc)
Let log(abc)=k
1/(x+1)=log(a)/k, 1/(y+1)=log(b)/k, 1/(z+1)=log(c)/k
Sum
(log a+log b+log c)/k = log(abc)/log(abc) = 1 ✓
Question 3[10 marks]Ch 6 · Quadratic

A farmer has a rectangular plot with area 180 m². The length exceeds width by 3 m.

(a) Form a quadratic equation for the width w. [2 marks]

(b) Solve the equation. [4 marks]

(c) Find the perimeter and diagonal of the plot. [4 marks]

▶ Show Solution

(a) [2 marks]

Area equation
w(w+3) = 180 → w²+3w−180 = 0 ✓

(b) [4 marks]

Factor
(w+15)(w−12) = 0 → w = 12 or w = −15
Answer
w = 12 m (rejecting negative); length = 15 m ✓

(c) [4 marks]

Perimeter
= 2(12+15) = 2×27 = 54 m
Diagonal
= √(12²+15²) = √(144+225) = √369 ≈ 19.2 m
Answer
Perimeter = 54 m, Diagonal ≈ 19.2 m ✓
Question 4[10 marks]Ch 9 · Series

In an AP, the 4th term is 14 and the 9th term is 34.

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

(b) Find the 25th term. [4 marks]

(c) Find the sum of the first 20 terms. [4 marks]

▶ Show Solution

(a) [2 marks]

Setup equations
a+3d=14 and a+8d=34
Subtract
5d=20 → d=4; a=14−12=2
Answer
a=2, d=4 ✓

(b) [4 marks]

T₂₅
= a+24d = 2+96 = 98 ✓

(c) [4 marks]

S₂₀
= 20/2 × [2a+19d] = 10×[4+76] = 10×80 = 800 ✓
Question 5[10 marks]Ch 11 · Geometry

Triangle ABC is isosceles with AB = AC = 13 cm and BC = 10 cm.

(a) Find the height from A to BC. [2 marks]

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

(c) D is a point on BC such that BD = 3 cm. Find the length of AD. [4 marks]

▶ Show Solution

(a) [2 marks]

Height from A
Foot M is midpoint of BC, BM=5. h=√(AB²−BM²)=√(169−25)=√144=12 cm
Answer
Height = 12 cm ✓

(b) [4 marks]

Area
= (1/2)×BC×h = (1/2)×10×12 = 60 cm² ✓

(c) [4 marks]

Coordinates
B=(0,0), C=(10,0), M=(5,0), A=(5,12)
D on BC with BD=3
D=(3,0)
AD
= √((5−3)²+(12−0)²) = √(4+144) = √148 = 2√37 ≈ 12.17 cm ✓
Question 6[10 marks]Ch 12 · Area

Quadrilateral ABCD has diagonals AC = 24 cm and BD = 18 cm. The diagonals are perpendicular.

(a) State the formula for area of a quadrilateral with perpendicular diagonals. [2 marks]

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

(c) A point E on AC divides it such that AE = 8 cm. Find the area of triangle ABE if BE ⊥ AC and BE = 10 cm. [4 marks]

▶ Show Solution

(a) [2 marks]

Formula
Area = (1/2) × d₁ × d₂ where d₁, d₂ are perpendicular diagonals ✓

(b) [4 marks]

Area
= (1/2) × 24 × 18 = 216 cm² ✓

(c) [4 marks]

Triangle ABE
Base AE = 8 cm, height = BE = 10 cm (since BE ⊥ AC)
Area
= (1/2) × 8 × 10 = 40 cm² ✓
Question 7[10 marks]Ch 15 · Statistics

Mustard yield (kg/plot) in Jessore — 30 plots:

YieldFreq
20–303
30–407
40–5012
50–606
60–702

(a) Find the mode. [2 marks]

(b) Find the mean using assumed mean 45. [4 marks]

(c) Find the median. [4 marks]

▶ Show Solution

(a) [2 marks]

Modal class
Highest freq = 12 in class 40–50
Mode
= 40 + [(12−7)/(24−7−6)]×10 = 40 + (5/11)×10 = 40+4.55 = 44.55 kg ✓

(b) [4 marks]

d = x−45 (midpoints 25,35,45,55,65)
d: −20,−10,0,10,20; f: 3,7,12,6,2
Σfd
3(−20)+7(−10)+12(0)+6(10)+2(20) = −60−70+0+60+40 = −30
Mean
x̄ = 45+(−30/30) = 45−1 = 44 kg ✓

(c) [4 marks]

Cumulative freq
3,10,22,28,30; n/2=15
Median class
40–50 (cumulative first exceeds 15)
Median
= 40+[(15−10)/12]×10 = 40+(5/12)×10 = 40+4.17 = 44.17 kg ✓
Question 8[10 marks]Ch 16 · Mensuration

A rectangular storage room is 8 m long, 6 m wide, and 4 m high.

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

(b) Find the total surface area (including floor and ceiling). [4 marks]

(c) Find the length of the longest diagonal of the room. [4 marks]

▶ Show Solution

(a) [2 marks]

Volume
= 8×6×4 = 192 m³ ✓

(b) [4 marks]

Total SA
= 2(lb+bh+lh) = 2(48+24+32) = 2×104 = 208 m² ✓

(c) [4 marks]

Space diagonal
d = √(l²+b²+h²) = √(64+36+16) = √116
Answer
d = 2√29 ≈ 10.77 m ✓

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