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Jessore Board · SSC Mathematics · 2025

Jessore 2025 Total: 100 Marks 3 Hours 8 Questions

Chapter Coverage

QChapterTopicMarks
1Ch 2 · SetsVenn diagram — agricultural survey10
2Ch 3 · AlgebraCrop yield formula — algebraic expressions10
3Ch 6 · EquationsFarmer profit optimization10
4Ch 9 · SeriesIrrigation schedule — AP10
5Ch 12 · TrianglesTriangle congruence and similarity10
6Ch 12 · AreaField division — area theorem10
7Ch 15 · StatisticsCrop yield grouped frequency — mean/SD10
8Ch 16 · MensurationAgricultural irrigation tank (cylinder)10
Question 1[10 marks]Ch 2 · Sets

In a village of 200 farming households, 120 grow rice (R), 90 grow wheat (W), and 40 grow both rice and wheat. Some households grow neither.

(a) Find |R ∪ W| using the inclusion-exclusion principle. [2 marks]

(b) How many households grow only rice? Only wheat? [4 marks]

(c) How many households grow neither? Express the data as a completed Venn diagram with all regions labeled. [4 marks]

▶ Show Solution

(a) [2 marks]

Inclusion-exclusion
|R ∪ W| = |R|+|W|−|R ∩ W| = 120+90−40 = 170 ✓

(b) [4 marks]

Only rice
|R only| = 120−40 = 80 households
Only wheat
|W only| = 90−40 = 50 households
Answer
Only rice: 80, Only wheat: 50 ✓

(c) [4 marks]

Neither
200 − |R ∪ W| = 200−170 = 30 households
Venn diagram regions
Only R = 80, Both = 40, Only W = 50, Neither = 30 (total = 200 ✓)
Answer
Neither = 30 households ✓
Question 2[10 marks]Ch 3 · Algebraic Expressions

A farmer models crop yield Y (in quintals) as Y = (a+b)² − (a−b)², where a = land area (bigha) and b = fertilizer units.

(a) Simplify Y = (a+b)² − (a−b)². [2 marks]

(b) If a = 5 bigha and b = 3 fertilizer units, find Y. [4 marks]

(c) For what ratio a:b does the yield Y equal 4 times the product ab? Verify. [4 marks]

▶ Show Solution

(a) [2 marks]

Expand
(a+b)²−(a−b)² = (a²+2ab+b²)−(a²−2ab+b²)
Simplify
Y = 4ab ✓

(b) [4 marks]

Substitute
Y = 4×5×3 = 60 quintals ✓

(c) [4 marks]

Condition
Y = 4ab, and we want Y = 4ab → this is always true for any ratio a:b
Correct interpretation
The yield already equals 4ab. For a specific yield of, say, 100 quintals: 4ab=100, so ab=25. One solution: a=5, b=5 (ratio 1:1)
Verification
Y = 4×5×5 = 100, and (5+5)²−(5−5)² = 100−0 = 100 ✓
Question 3[10 marks]Ch 6 · Equations

A farmer sells mangoes and jackfruits. 3 kg of mangoes and 2 kg of jackfruit sell for 440 taka. 5 kg of mangoes and 3 kg of jackfruit sell for 710 taka.

(a) Set up the simultaneous equations for price per kg of mango (m) and jackfruit (j). [2 marks]

(b) Solve the system. [4 marks]

(c) A second farmer wants to maximize revenue from 20 kg of mangoes and 10 kg of jackfruit. Calculate total revenue. [4 marks]

▶ Show Solution

(a) [2 marks]

Equation 1
3m + 2j = 440
Equation 2
5m + 3j = 710

(b) [4 marks]

Multiply Eq1×3 and Eq2×2
9m+6j=1320 and 10m+6j=1420
Subtract
m = 100 taka/kg
Substitute
3(100)+2j=440 → 2j=140 → j=70 taka/kg
Answer
Mango = 100 taka/kg, Jackfruit = 70 taka/kg ✓

(c) [4 marks]

Revenue
= 20×100 + 10×70 = 2000+700 = 2700 taka ✓
Question 4[10 marks]Ch 9 · Finite Series

An irrigation pump waters fields on a schedule. On the first day it pumps 300 litres, and each day increases by 50 litres (arithmetic progression).

(a) Find the amount pumped on day 8. [2 marks]

(b) Find the total water pumped in 3 weeks (21 days). [4 marks]

(c) The tank holds 20,000 litres. On which day will the cumulative total first exceed half the tank capacity? [4 marks]

▶ Show Solution

(a) [2 marks]

T₈ = a+(n−1)d
= 300 + 7×50 = 300+350 = 650 litres ✓

(b) [4 marks]

S₂₁
= 21/2 × [2(300)+20(50)] = 21/2 × [600+1000] = 21/2 × 1600
Answer
S₂₁ = 21×800 = 16,800 litres ✓

(c) [4 marks]

Half tank
= 10,000 litres
S_n > 10000
n/2[600+(n−1)50] > 10000 → n[600+50n−50] > 20000 → n(550+50n) > 20000
Simplify
50n²+550n−20000 > 0 → n²+11n−400 > 0
Solve n²+11n−400=0
n = (−11+√(121+1600))/2 = (−11+√1721)/2 ≈ (−11+41.48)/2 ≈ 15.24
Answer
Day 16 (S₁₆ = 16/2×[600+750] = 8×1350 = 10,800 > 10,000) ✓
Question 5[10 marks]Ch 12 · Triangles

In two triangles △ABC and △DEF: AB = 6 cm, BC = 8 cm, AC = 10 cm, DE = 9 cm, EF = 12 cm, DF = 15 cm.

(a) Show that △ABC ~ △DEF. [2 marks]

(b) Find the ratio of their perimeters. [4 marks]

(c) Find the ratio of their areas and calculate both areas. [4 marks]

▶ Show Solution

(a) [2 marks]

Check ratios
AB/DE = 6/9 = 2/3, BC/EF = 8/12 = 2/3, AC/DF = 10/15 = 2/3
Conclusion
All sides in ratio 2:3, so △ABC ~ △DEF (SSS similarity) ✓

(b) [4 marks]

Perimeter ABC
= 6+8+10 = 24 cm
Perimeter DEF
= 9+12+15 = 36 cm
Ratio
24:36 = 2:3 (same as side ratio) ✓

(c) [4 marks]

Area ratio
= (2/3)² = 4/9
Area of △ABC (right triangle: 6²+8²=100=10²)
= (1/2)×6×8 = 24 cm²
Area of △DEF
= 24 × (9/4) = 54 cm²
Answer
Area ratio = 4:9, Areas = 24 cm² and 54 cm² ✓
Question 6[10 marks]Ch 12 · Area Theorems

A trapezoidal paddy field ABCD has AB ∥ CD. AB = 24 m, CD = 16 m, and the perpendicular height is 10 m. E is the midpoint of BC.

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

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

(c) Prove that triangle ADE and the remaining region AECD have specific areas and find them. [4 marks]

▶ Show Solution

(a) [2 marks]

Trapezoid area
= (1/2)(AB+CD)×h = (1/2)(24+16)×10 = (1/2)×40×10 = 200 m² ✓

(b) [4 marks]

Set up coordinates
A=(0,0), B=(24,0), C=(20,10), D=(4,10) [one valid configuration]
E = midpoint of BC
E = ((24+20)/2, (0+10)/2) = (22, 5)
Area △ABE
= (1/2)|base×height|. Using shoelace: A(0,0), B(24,0), E(22,5): = (1/2)|0(0−5)+24(5−0)+22(0−0)| = (1/2)|120| = 60 m²
Answer
Area △ABE = 60 m² ✓

(c) [4 marks]

Area △ADE
= Total area − Area△ABE − Area△AECD... use: Area△ADE = Area trapezoid − Area△ABE − Area△ECD
Area △ECD
E=(22,5), C=(20,10), D=(4,10): = (1/2)|22(10−10)+20(10−5)+4(5−10)| = (1/2)|0+100−20| = 40 m²
Area △ADE
= 200−60−40 = 100 m²
Answer
Area △ADE = 100 m², Area region AECD = 140 m² ✓
Question 7[10 marks]Ch 15 · Statistics

The following table shows crop yield (quintals/bigha) for 50 farms in Jessore district:

Yield (q/bigha)Frequency
10–145
14–1812
18–2218
22–2610
26–305

(a) Find the modal class and the mode. [2 marks]

(b) Calculate the mean yield using the assumed mean method (assumed mean = 20). [4 marks]

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

▶ Show Solution

(a) [2 marks]

Modal class
Highest frequency = 18, in class 18–22
Mode (formula)
Mode = L + [(f₁−f₀)/(2f₁−f₀−f₂)] × h = 18 + [(18−12)/(36−12−10)] × 4 = 18 + (6/14)×4 = 18+1.71 ≈ 19.71 ✓

(b) [4 marks]

Midpoints and deviations d=x−20
x: 12,16,20,24,28; d: −8,−4,0,4,8; f: 5,12,18,10,5
Σfd
5(−8)+12(−4)+18(0)+10(4)+5(8) = −40−48+0+40+40 = −8
Mean
x̄ = 20 + (−8/50) = 20−0.16 = 19.84 q/bigha ✓

(c) [4 marks]

Σfd²
5(64)+12(16)+18(0)+10(16)+5(64) = 320+192+0+160+320 = 992
Variance
σ² = Σfd²/n − (Σfd/n)² = 992/50 − (−8/50)² = 19.84 − 0.0256 = 19.814
SD
σ = √19.814 ≈ 4.45 q/bigha ✓
Question 8[10 marks]Ch 16 · Mensuration

An agricultural irrigation tank is cylindrical with a hemispherical bottom. The cylinder has inner radius 2.1 m and height 4 m. The hemispherical base has the same radius. (π = 22/7)

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

(b) Find the total water capacity (cylinder + hemisphere). [4 marks]

(c) Find the total inner curved surface area of the tank (cylinder CSA + hemisphere CSA). [4 marks]

▶ Show Solution

(a) [2 marks]

V_cyl
= πr²h = (22/7)×(2.1)²×4 = (22/7)×4.41×4 = (22/7)×17.64
Calculate
= 22×2.52 = 55.44 m³ ✓

(b) [4 marks]

V_hemisphere
= (2/3)πr³ = (2/3)×(22/7)×(2.1)³ = (2/3)×(22/7)×9.261
Calculate
= (2/3)×(22×1.323) = (2/3)×29.106 = 19.404 m³
Total
55.44 + 19.404 = 74.844 m³ ≈ 74.84 m³ ✓

(c) [4 marks]

CSA cylinder
= 2πrh = 2×(22/7)×2.1×4 = 2×(22/7)×8.4 = 2×26.4 = 52.8 m²
CSA hemisphere
= 2πr² = 2×(22/7)×4.41 = 2×13.86 = 27.72 m²
Total
52.8+27.72 = 80.52 m² ✓

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