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Jessore Board · SSC Mathematics · 2021

Jessore 2021 Total: 100 Marks 3 Hours 8 Questions

Chapter Coverage

QChapterTopicMarks
1Ch 2 · SetsSet operations — crop survey10
2Ch 3 · AlgebraHCF and LCM of expressions10
3Ch 6 · QuadraticSum and product of roots10
4Ch 9 · SeriesYield progression — GP10
5Ch 11 · GeometryCongruence of triangles (SAS)10
6Ch 13 · CircleEqual chords and equal arcs10
7Ch 14 · TrigonometryHeight of a tree10
8Ch 15 · StatisticsHistogram and mean10
Question 1[10 marks]Ch 2 · Sets

In a survey of 150 farmers: 80 grow jute (J), 70 grow paddy (P), and 30 grow both.

(a) Find |J ∪ P|. [2 marks]

(b) Find the number who grow only jute and only paddy. [4 marks]

(c) Represent this data in a Venn diagram and find the number who grow neither crop. [4 marks]

▶ Show Solution

(a) [2 marks]

Inclusion-exclusion
|J ∪ P| = 80+70−30 = 120 ✓

(b) [4 marks]

Only jute
80−30 = 50
Only paddy
70−30 = 40
Answer
Only jute: 50, Only paddy: 40 ✓

(c) [4 marks]

Neither
150−120 = 30
Venn regions
Only J=50, Both=30, Only P=40, Neither=30. Check: 50+30+40+30=150 ✓
Question 2[10 marks]Ch 3 · Algebra

Let p(x) = x² + 5x + 6 and q(x) = x² + 7x + 12.

(a) Factorize p(x) and q(x). [2 marks]

(b) Find the HCF and LCM of p(x) and q(x). [4 marks]

(c) Verify that HCF × LCM = p(x) × q(x). [4 marks]

▶ Show Solution

(a) [2 marks]

p(x)
= (x+2)(x+3)
q(x)
= (x+3)(x+4) ✓

(b) [4 marks]

HCF
= common factor = (x+3)
LCM
= (x+2)(x+3)(x+4)
Answer
HCF = (x+3), LCM = (x+2)(x+3)(x+4) ✓

(c) [4 marks]

HCF × LCM
= (x+3)(x+2)(x+3)(x+4) = (x+2)(x+3)²(x+4)
p(x)×q(x)
= (x+2)(x+3)×(x+3)(x+4) = (x+2)(x+3)²(x+4)
Verified
HCF × LCM = p(x) × q(x) ✓
Question 3[10 marks]Ch 6 · Quadratic

For the equation 3x² − 11x + k = 0, the roots are α and β.

(a) Write expressions for α+β and αβ in terms of k. [2 marks]

(b) If α − β = 1, find k. [4 marks]

(c) For that value of k, find α and β individually, and verify your answer. [4 marks]

▶ Show Solution

(a) [2 marks]

Vieta's formulas
α+β = 11/3, αβ = k/3 ✓

(b) [4 marks]

Identity
(α−β)² = (α+β)²−4αβ → 1 = (11/3)²−4(k/3)
Solve
1 = 121/9 − 4k/3 → 4k/3 = 121/9−1 = 112/9 → k = 112/12 = 28/3
Answer
k = 28/3 ✓

(c) [4 marks]

α+β=11/3, α−β=1
Adding: 2α=11/3+1=14/3 → α=7/3
β
= 11/3−7/3 = 4/3
Verify αβ
= (7/3)(4/3)=28/9=k/3=28/9 ✓
Answer
α=7/3, β=4/3 ✓
Question 4[10 marks]Ch 9 · Series

A new crop variety increases yield by 20% each season. The first season gives 500 kg.

(a) Write the first four terms of the GP. [2 marks]

(b) Find the yield in the 6th season. (1.2⁵ = 2.48832) [4 marks]

(c) Find the total yield over 5 seasons. [4 marks]

▶ Show Solution

(a) [2 marks]

GP terms
500, 600, 720, 864 (a=500, r=1.2) ✓

(b) [4 marks]

T₆ = ar⁵
= 500 × 1.2⁵ = 500×2.48832
Answer
T₆ = 1244.16 kg ✓

(c) [4 marks]

S₅ = a(r⁵−1)/(r−1)
= 500(2.48832−1)/(0.2) = 500×1.48832/0.2 = 500×7.4416
Answer
S₅ = 3720.8 kg ✓
Question 5[10 marks]Ch 11 · Geometry

In triangles ABC and DEF: AB = DE = 8 cm, BC = EF = 10 cm, ∠B = ∠E = 55°.

(a) State the SAS congruence criterion. [2 marks]

(b) Prove △ABC ≅ △DEF and state all corresponding equal elements. [4 marks]

(c) If the area of △ABC = 32.77 cm², find AC and DF. [4 marks]

▶ Show Solution

(a) [2 marks]

SAS criterion
If two sides and the included angle of one triangle equal two sides and the included angle of another, the triangles are congruent. ✓

(b) [4 marks]

Given
AB=DE=8, BC=EF=10, ∠B=∠E=55°
By SAS
△ABC ≅ △DEF
Equal elements
AB=DE, BC=EF, AC=DF, ∠A=∠D, ∠B=∠E, ∠C=∠F ✓

(c) [4 marks]

Using area = (1/2)ab sin C
32.77 = (1/2)×8×10×sin 55° → sin55°=0.8192 → Area=(1/2)×80×0.8192=32.77 ✓
Find AC by cosine rule
AC² = AB²+BC²−2(AB)(BC)cos B = 64+100−160×0.5736 = 164−91.78 = 72.22
Answer
AC = DF = √72.22 ≈ 8.5 cm ✓
Question 6[10 marks]Ch 13 · Circle

Two chords AB and CD of a circle with center O are equal in length (AB = CD = 20 cm). The radius is 13 cm.

(a) Find the distance of each chord from the center. [2 marks]

(b) Prove that equal chords of a circle are equidistant from the center. [4 marks]

(c) If PQ is a chord with distance 12 cm from center, find PQ. [4 marks]

▶ Show Solution

(a) [2 marks]

Half chord = 10
OM = √(r²−10²) = √(169−100) = √69 ≈ 8.31 cm ✓

(b) [4 marks]

Given: AB=CD
Let OM⊥AB and ON⊥CD. OA=OC=r, AM=AB/2, CN=CD/2. Since AB=CD, AM=CN.
In right triangles OAM and OCN
OA=OC (radii), AM=CN. By Pythagoras, OM²=OA²−AM²=OC²−CN²=ON²
Conclusion
OM=ON, equal chords are equidistant ✓

(c) [4 marks]

Half PQ
= √(r²−d²) = √(169−144) = √25 = 5 cm
Answer
PQ = 10 cm ✓
Question 7[10 marks]Ch 14 · Trigonometry

A person standing 40 m from a tree observes the top at 45°. A bird on a branch observes the same person at an angle of depression of 30°.

(a) Find the height of the tree. (tan 45° = 1) [2 marks]

(b) Find the height of the branch the bird is on. (tan 30° = 1/√3) [4 marks]

(c) Find the straight-line distance from the person to the bird. [4 marks]

▶ Show Solution

(a) [2 marks]

Tree height
h = 40×tan45° = 40×1 = 40 m ✓

(b) [4 marks]

Depression angle from branch
tan 30° = h_branch/40 → h_branch = 40/√3 = 40√3/3 ≈ 23.1 m
Answer
Branch height ≈ 23.1 m ✓

(c) [4 marks]

Distance to bird
= √(40²+h_branch²) = √(1600+(40/√3)²) = √(1600+1600/3) = √(6400/3) = 80/√3
Answer
= 80√3/3 ≈ 46.19 m ✓
Question 8[10 marks]Ch 15 · Statistics

Jute production (bales) in 35 farms of Jessore district:

BalesFarms
10–205
20–308
30–4012
40–507
50–603

(a) Find the mean using mid-interval values. [2 marks]

(b) Find the median. [4 marks]

(c) Find the range and the inter-quartile range. [4 marks]

▶ Show Solution

(a) [2 marks]

Σfx
5(15)+8(25)+12(35)+7(45)+3(55) = 75+200+420+315+165 = 1175
Mean
x̄ = 1175/35 = 33.57 bales ✓

(b) [4 marks]

CF: 5,13,25,32,35; n/2=17.5
Median class: 30–40
Median
= 30+[(17.5−13)/12]×10 = 30+(4.5/12)×10 = 30+3.75 = 33.75 bales ✓

(c) [4 marks]

Range
= 60−10 = 50 bales
Q1 (n/4=8.75, class 20–30)
Q1 = 20+[(8.75−5)/8]×10 = 20+4.69 = 24.69
Q3 (3n/4=26.25, class 40–50)
Q3 = 40+[(26.25−25)/7]×10 = 40+1.79 = 41.79
IQR
= Q3−Q1 = 41.79−24.69 = 17.1 bales ✓

Same Year — Other Boards

Chittagong 2021 Comilla 2021