Jessore Board SSC Math 2023

Q	Chapter	Topic	Marks
1	Ch 1 · Real Numbers	Rational/irrational classification	10
2	Ch 3 · Algebra	Polynomial factorization	10
3	Ch 5 · Equations	Linear — market price problem	10
4	Ch 9 · Series	Geometric series — savings doubling	10
5	Ch 11 · Triangles	Mid-point theorem	10
6	Ch 13 · Circle	Tangent properties	10
7	Ch 15 · Statistics	Frequency polygon and cumulative frequency	10
8	Ch 16 · Mensuration	Sphere and composite solid	10

Question 1[10 marks]Ch 1 · Real Numbers

Classify and work with number types.

(a) Prove that √2 is irrational. [2 marks]

(b) Find a rational number between √3 and √5, and verify it is rational. [4 marks]

(c) Simplify: (√6+√3)(√6−√3) + (√5+√2)(√5−√2). [4 marks]

▶ Show Solution

(a) [2 marks]

Proof by contradiction

Suppose √2 = p/q (in lowest terms). Then 2 = p²/q² → p² = 2q² → p is even → p = 2k

Contradiction

Then 4k² = 2q² → q² = 2k² → q is even. But p,q both even contradicts lowest terms.

Conclusion

∴ √2 is irrational ✓

(b) [4 marks]

Values

√3 ≈ 1.732, √5 ≈ 2.236

Rational between

Try 2 (which is 2/1): 1.732 < 2 < 2.236 ✓

Verify

2 = 2/1 is rational, and √3 < 2 < √5 ✓

(c) [4 marks]

Difference of squares

(√6+√3)(√6−√3) = 6−3 = 3

Second term

(√5+√2)(√5−√2) = 5−2 = 3

Sum

3+3 = 6 ✓

Question 2[10 marks]Ch 3 · Algebra

Factorize the following expressions completely.

(a) Factorize: a² − b² − 2bc − c². [2 marks]

(b) Factorize: x⁴ − 5x² + 4. [4 marks]

(c) Factorize: 2x³ + 3x² − 11x − 6. [4 marks]

▶ Show Solution

(a) [2 marks]

Regroup

a² − (b²+2bc+c²) = a² − (b+c)²

Difference of squares

= (a+b+c)(a−b−c) ✓

(b) [4 marks]

Substitution

Let u = x²: u²−5u+4 = (u−1)(u−4) = (x²−1)(x²−4)

Further factor

= (x+1)(x−1)(x+2)(x−2) ✓

(c) [4 marks]

Try x = 2

2(8)+3(4)−11(2)−6 = 16+12−22−6 = 0 ✓

Factor out (x−2)

2x³+3x²−11x−6 = (x−2)(2x²+7x+3)

Factor 2x²+7x+3

= (2x+1)(x+3)

Answer

= (x−2)(2x+1)(x+3) ✓

Question 3[10 marks]Ch 5 · Equations

At a local market, onions cost x taka/kg and potatoes y taka/kg. A shopkeeper buys 4 kg onions + 6 kg potatoes for 760 taka. Another buys 7 kg onions + 4 kg potatoes for 850 taka.

(a) Set up the simultaneous equations. [2 marks]

(b) Solve for x and y. [4 marks]

(c) A family has 1500 taka to buy equal kg of each. How many kg of each can they buy? [4 marks]

▶ Show Solution

(a) [2 marks]

Equations

4x+6y=760 and 7x+4y=850

(b) [4 marks]

Multiply: Eq1×4, Eq2×6

16x+24y=3040 and 42x+24y=5100

Subtract

26x = 2060 → x = 2060/26 = 79.23...

Clean numbers: Multiply Eq1×2, Eq2×3

8x+12y=1520 and 21x+12y=2550; subtract: 13x=1030 → x=79.23

Try Eq1×7, Eq2×4

28x+42y=5320 and 28x+16y=3400; subtract: 26y=1920 → y=73.85

Answer

x ≈ 79.2 taka/kg (onion), y ≈ 73.8 taka/kg (potato) ✓

(c) [4 marks]

Equal kg k

k×x + k×y = 1500 → k(x+y) = 1500

x+y

≈ 79.2+73.8 = 153

k

= 1500/153 ≈ 9.8 kg

Answer

≈ 9.8 kg of each ✓

Question 4[10 marks]Ch 9 · Series

A savings plan doubles every 2 years. Initial deposit is 5000 taka.

(a) Write the GP showing amounts at year 0, 2, 4, 6. [2 marks]

(b) How much will be saved after 10 years? [4 marks]

(c) Find the total sum of all amounts over 10 years (5 terms). [4 marks]

▶ Show Solution

(a) [2 marks]

Sequence

5000, 10000, 20000, 40000 (a=5000, r=2) ✓

(b) [4 marks]

After 10 years = 5 periods

T₅ (0-indexed from T₁) = T at period 5 = 5000×2^(5−1) = 5000×16

Answer (year 10, 5th period)

Wait: year 0 → T₁=5000, year 2→T₂, year 4→T₃, year 6→T₄, year 8→T₅, year 10→T₆=5000×2⁵=160000

Answer

160,000 taka ✓

(c) [4 marks]

6 terms (years 0,2,4,6,8,10)

S₆ = a(r⁶−1)/(r−1) = 5000(64−1)/1 = 5000×63 = 315,000 taka ✓

Question 5[10 marks]Ch 11 · Triangles

In triangle ABC, M and N are midpoints of AB and AC respectively.

(a) State the midpoint theorem. [2 marks]

(b) If BC = 18 cm, find MN and the ratio MN:BC. [4 marks]

(c) Find the ratio of area of triangle AMN to area of trapezoid MNCB. [4 marks]

▶ Show Solution

(a) [2 marks]

Midpoint theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. ✓

(b) [4 marks]

MN

= (1/2)BC = (1/2)×18 = 9 cm

Ratio

MN:BC = 9:18 = 1:2 ✓

(c) [4 marks]

Similar triangles

△AMN ~ △ABC with ratio 1:2, so area ratio = 1:4

Area △AMN

= (1/4) Area △ABC

Area trapezoid MNCB

= Area △ABC − Area △AMN = (3/4) Area △ABC

Ratio

Area △AMN : Area MNCB = (1/4):(3/4) = 1:3 ✓

Question 6[10 marks]Ch 13 · Circle

PA and PB are tangents from external point P to a circle of center O and radius 6 cm. PA = 8 cm.

(a) Find OP. [2 marks]

(b) Prove PA = PB and find ∠APB. [4 marks]

(c) Find the area of quadrilateral PAOB. [4 marks]

▶ Show Solution

(a) [2 marks]

Right angle at tangent

OA ⊥ PA, so OP² = PA²+OA² = 64+36 = 100

Answer

OP = 10 cm ✓

(b) [4 marks]

Prove PA=PB

In △OAP and △OBP: OA=OB (radii), OP=OP (common), ∠OAP=∠OBP=90°. By RHS congruence, PA=PB ✓

∠OPA

cos(∠OPA) = PA/OP = 8/10 = 4/5 → ∠OPA = 36.87°

∠APB

∠APB = 2×36.87° ≈ 73.74° ✓

(c) [4 marks]

Area △OAP

= (1/2)×OA×PA = (1/2)×6×8 = 24 cm²

Total PAOB

= 2×24 = 48 cm² ✓

Question 7[10 marks]Ch 15 · Statistics

Wheat production (tonnes) in 40 farms of Jessore:

Production	Frequency
5–10	4
10–15	9
15–20	15
20–25	8
25–30	4

(a) Find the cumulative frequencies and median class. [2 marks]

(b) Calculate the median. [4 marks]

(c) Calculate the mean. [4 marks]

▶ Show Solution

(a) [2 marks]

Cumulative freq

4, 13, 28, 36, 40

Median class

n/2 = 20; cumulative freq first exceeds 20 at class 15–20

Answer

Median class = 15–20 ✓

(b) [4 marks]

Median formula

M = L + [(n/2 − F)/f] × h = 15 + [(20−13)/15] × 5

Calculate

= 15 + (7/15)×5 = 15 + 2.33 = 17.33 tonnes ✓

(c) [4 marks]

Midpoints × freq

7.5×4=30, 12.5×9=112.5, 17.5×15=262.5, 22.5×8=180, 27.5×4=110

Σfx

= 30+112.5+262.5+180+110 = 695

Mean

x̄ = 695/40 = 17.375 tonnes ✓

Question 8[10 marks]Ch 16 · Mensuration

A spherical water pot of radius 7 cm sits inside a cubical box of side 14 cm. (π = 22/7)

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

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

(c) Find the volume of the empty space in the box, and find the surface area of the sphere. [4 marks]

▶ Show Solution

(a) [2 marks]

V_sphere

= (4/3)πr³ = (4/3)×(22/7)×343 = (4/3)×22×49 = (4/3)×1078

Answer

V = 4312/3 ≈ 1437.3 cm³ ✓

(b) [4 marks]

V_cube

= 14³ = 2744 cm³ ✓

(c) [4 marks]

Empty space

= 2744 − 1437.3 = 1306.7 cm³ ✓

Surface area of sphere

= 4πr² = 4×(22/7)×49 = 4×22×7 = 616 cm² ✓

Jessore Board · SSC Mathematics · 2023

Chapter Coverage

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