Monday, February 9, 2026

Class 8 SESSION ENDING EXAMINATION (2026) SAMPLE PAPER-1 WITH ANSWERS

SESSION ENDING EXAMINATION (2026)
SAMPLE PAPER-1

Portion

Part -1

Ch - 5 Number Play

Ch - 6 We distribute, yet things multiply

Ch - 7 Proprtional Reasoning - 1

Part -2

Ch - 1 Fractions in Disguise

CH2 PART 2 THE BAUDHĀYANA-PYTHAGORAS THEOREM

Ch - 3 Proportional Reasoning - 2

Ch - 7 Area

Class: VIII Subject: Mathematics

Time: 2 ½ Hrs. Max Marks: 60

General Instructions:

All questions are compulsory.
The question paper consists of 30 questions divided into five sections A, B, C ,D and E.
Section A contains 13MCQ questions and 2 assertion and reasoning questions of one mark each.
Section B contains 5 short answer questions of 2 marks each.
Section C contains 5 short answer questions of 3 marks each.
Section D contains 3 long answer questions of 5 marks each.
Section E 2 case based questions of each4 marks.

Section A – MCQ & Assertion-Reason

The digital root of 63855 is
a) 6 b) 9 c) 3 d) 12
If a number is divisible by 4 and 8 it must be divisible by
a) 8 b) 12 c) 18 d) 24
Distributive property of (a+b)² can be expressed as
a) (a+b)×(a+b) b) (a+b)×(a−b) c) a+b,a+b d) none
A Square of any odd number is always
a) even b) multiple of 4 c) one more than multiple of 8 d) a multiple of 8
Which ratio is proportional to 96 and 144
a) 2:5 b) 2:3 c) 3:2 d) 3:3
If 72 liters of milk sufficient for 288 people, milk required for 450 people
a) 96 b) 108 c) 112.5 d) 120
25% of 160 =
a) 80 b) 90 c) 40 d) 42
Esha scored 42 marks out of 50 on English test, Score as a percentage
a) 84% b) 90% c) 100% d) 52%
Is √2 value less than or greater than 1
a) <1 b) >1 c) =1 d) none
If the hypotenuse of an isosceles right triangle is √72, the other two sides
a) 6,6 b) 6,7 c) 8,9 d) 12,6
Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
a) 150 cm² b) 300 cm² c) 900 cm² d) 1500 cm²
Area of each triangle in the rectangle of length 7 cm and breadth 4 cm
a) 14 cm² b) 15 cm² c) 20 cm² d) none
Area of the parallelogram obtained by
a) base × height b) 1 × b c) 2(l+b) d) ½ × b × h

Assertion-Reason:
14. Assertion: (a+b)² − (a−b)² = 4a
Reason: On expansion, square terms cancel
15. Assertion: Area of trapezium = ½ × height × sum of parallel sides
Reason: Parallel sides of trapezium 12 cm and 16 cm and height 5 cm then area = 70 cm²

Section B (2 marks each)
16. Write 2 multiples of 36 between 45,000 and 47,000.
17. Draw the Venn diagram which captures the relationship between the multiples of 4,8 and 32.
18. Expand (a+b)(a+b) OR Expand (3+u)(v−3).
19. Nandini has 25 marbles of which 15 are white. What percentage of her marbles are white?
20. Find the diagonal of a square with side length 5 cm.

Section C (3 marks each)
21. The sum of four consecutive numbers is 34. What are these numbers?
OR The digital root of an 8-digit number is 5. What will be the digital root of number + 10?
22. Find the HCF of 72 and 96 OR Divide ₹4500 into two parts in the ratio 2:3.
23. Find the value and draw bar models: i) 7% of 10 kg ii) 62% of 360.
24. A right-angled triangle has short side 8 cm and hypotenuse 17 cm. Find the third side using Pythagoras theorem.
25. Find area of quadrilateral ABCD given AC=22 cm, BM=3 cm, DN=3 cm, BM ⟂ AC, DN ⟂ AC.

Section D (Long Answer – 4 marks each)
26. Compute using suitable identities: i) 397 × 403 ii) (2x−1)²
27. One acre of land costs ₹15,00,000. What is the cost of 2,400 sq ft of same land?
28. If (a,b,c) is a Pythagorean triple, then (ka,kb,kc) is also a triple for positive integer k. Verify with example.

Section E (Case-Based – 4 marks each)
29. Gangaram opened a book exhibition stall He aims to achieve a daily sales of atleast RS 5000 The sales on the first 2 days were Rs 2000 and Rs3500

i) What percentage of his target did he achieve?on day 1 and day2 (1m)

ii) In the next two days i.e day 3 and day4 ,he made Rs 5000 and 6000 respectively what percentage of his target are these values? ( 1m)

iii) What is the short of his target on day1 (2m)

iv)what percentage of the target was achieved on day 4?

30. Farmer’s field shapes – areas of quadrilaterals.

In a Village a farmer having the field in the form of following shapes the farmer son studied civil engineering .one day his father asked him to survey the areas of the fields the area of the following shapes will be ?

i)find the area of quadrilateral shape of field i) shown in figure? 1m

ii)what is the are a field ii) in figure? 1m

iii)what is the difference between area of fig iii) and iv) 2m

iv)what is the area of fig iii) in meter square

ANSWER KEY SAMPLE Paper 1 – Section A – All MCQs with Explanations

Question 1

The digital root of 63855 is
a) 6 b) 9 c) 3 d) 12

🔍 Step 1: Understand the Problem
Given number = 63855. Need to repeatedly sum digits until a single digit is obtained.

🧠 Step 2: Identify the Concept Used
Concept: Digital root (sum of digits until one digit remains).

✏️ Step 3: Solve Step by Step
Sum digits: 6 + 3 + 8 + 5 + 5 = 27
Sum again: 2 + 7 = 9

✅ Final Answer
👉 Option b) 9

🧠 Key Idea to Remember
Digital root = remainder when number divided by 9, except if remainder 0 → digital root is 9.

⚠️ Common Mistake to Avoid
Stopping at 27 thinking it’s final — must reduce to single digit.

🌍 Real-Life Application
Used in check digits, quick divisibility checks, puzzles.

📝 Practice Question
Digital root of 45678?

🎯 Exam Tip
If sum of digits is divisible by 9, digital root = 9.

Question 2

If a number is divisible by 4 and 8 it must be divisible by
a) 8 b) 12 c) 18 d) 24

🔍 Step 1: Understand the Problem
Divisible by 4 and 8 → find number that is always a divisor.

🧠 Step 2: Identify the Concept Used
Concept: LCM of divisors.

✏️ Step 3: Solve Step by Step
LCM of 4 and 8 = 8
If divisible by 8, automatically divisible by 4, but not necessarily by 12, 18, or 24. So must be divisible by 8.

✅ Final Answer
👉 Option a) 8

🧠 Key Idea to Remember
If N divisible by a and b, then N divisible by LCM(a,b).

⚠️ Common Mistake to Avoid
Choosing 24 (not necessary unless divisible by 3 too).

🌍 Real-Life Application
Planning cycles, scheduling events.

📝 Practice Question
If divisible by 6 and 9, must be divisible by __?

🎯 Exam Tip
"Must be divisible" means always true for any such number.

Question 3

Distributive property of (a+b)² can be expressed as
a) (a+b)×(a+b) b) (a+b)×(a−b) c) a+b,a+b d) none

🔍 Step 1: Understand the Problem
(a+b)² means (a+b) multiplied by itself.

🧠 Step 2: Identify the Concept Used
Concept: Algebraic identity/expansion.

✏️ Step 3: Solve Step by Step
(a+b)² = (a+b)(a+b) → matches option a.

✅ Final Answer
👉 Option a) (a+b)×(a+b)

🧠 Key Idea to Remember
Distributive property: a(b+c) = ab + ac; here (a+b)(a+b) uses distribution.

⚠️ Common Mistake to Avoid
Confusing with (a+b)(a-b) = a² – b².

🌍 Real-Life Application
Area of a square of side (a+b), tiling in construction.

📝 Practice Question
Expand (x+3)² using distributive property.

🎯 Exam Tip
Memorize identities: (a+b)² = a² + 2ab + b².

Question 4

A Square of any odd number is always
a) even b) multiple of 4 c) one more than multiple of 8 d) a multiple of 8

🔍 Step 1: Understand the Problem
Let odd number = 2n+1, find property of (2n+1)².

🧠 Step 2: Identify the Concept Used
Concept: Number theory — odd squares.

✏️ Step 3: Solve Step by Step
(2n+1)² = 4n² + 4n + 1 = 4n(n+1) + 1
n(n+1) is even → 4 × even = multiple of 8
So 4n(n+1) is multiple of 8
Thus 4n(n+1) + 1 = one more than multiple of 8.

✅ Final Answer
👉 Option c) one more than multiple of 8

🧠 Key Idea to Remember
Odd² = 1 mod 8.

⚠️ Common Mistake to Avoid
Saying odd square is even (clearly false).

🌍 Real-Life Application
Cryptography, error detection codes.

📝 Practice Question
Check for 7², 11².

🎯 Exam Tip
Test with example: 5² = 25 = 3×8 + 1.

Question 5

Which ratio is proportional to 96 and 144
a) 2:5 b) 2:3 c) 3:2 d) 3:3

🔍 Step 1: Understand the Problem
Proportional means 96:144 simplifies to which ratio.

🧠 Step 2: Identify the Concept Used
Concept: Ratio in simplest form.

✏️ Step 3: Solve Step by Step
96 : 144
Divide both by 48 → 2 : 3

✅ Final Answer
👉 Option b) 2:3

🧠 Key Idea to Remember
Simplify ratio by dividing by HCF.

⚠️ Common Mistake to Avoid
Writing 144:96 instead of 96:144 as given order.

🌍 Real-Life Application
Mixing ingredients, scaling drawings.

📝 Practice Question
Simplify ratio 150 : 225.

🎯 Exam Tip
Check if options match after simplification.

Question 6

If 72 liters of milk sufficient for 288 people, milk required for 450 people
a) 96 b) 108 c) 112.5 d) 120

🔍 Step 1: Understand the Problem
Given milk for 288 people = 72 L. Direct proportion: more people → more milk.

🧠 Step 2: Identify the Concept Used
Concept: Unitary method / direct proportion.

✏️ Step 3: Solve Step by Step
Milk per person = 72 / 288 = 0.25 L
For 450 people = 0.25 × 450 = 112.5 L

✅ Final Answer
👉 Option c) 112.5

🧠 Key Idea to Remember
Find one unit first, then multiply.

⚠️ Common Mistake to Avoid
Inversely relating people and milk.

🌍 Real-Life Application
Catering, event planning.

📝 Practice Question
If 50 L for 200 people, how much for 350 people?

🎯 Exam Tip
Keep track of units (Liters).

Question 7

25% of 160 =
a) 80 b) 80 c) 40 d) 42

🔍 Step 1: Understand the Problem
Find 25% (one-fourth) of 160.

🧠 Step 2: Identify the Concept Used
Concept: Percentage to fraction.

✏️ Step 3: Solve Step by Step
25% = 1/4
1/4 × 160 = 40

✅ Final Answer
👉 Option c) 40

🧠 Key Idea to Remember
25% = 1/4, 50% = 1/2.

⚠️ Common Mistake to Avoid
Calculating 0.25 × 160 incorrectly.

🌍 Real-Life Application
Discounts, sales tax.

📝 Practice Question
Find 30% of 200.

🎯 Exam Tip
For 25%, just divide by 4.

Question 8

Esha scored 42 marks out of 50 on English test, Score as a percentage
a) 84% b) 90% c) 100% d) 52%

🔍 Step 1: Understand the Problem
Marks = 42, Total = 50, find percentage.

🧠 Step 2: Identify the Concept Used
Concept: Percentage = (Part/Total)×100.

✏️ Step 3: Solve Step by Step
(42/50)×100 = 84%

✅ Final Answer
👉 Option a) 84%

🧠 Key Idea to Remember
Percentage compares out of 100.

⚠️ Common Mistake to Avoid
Dividing total by part.

🌍 Real-Life Application
Test scores, performance metrics.

📝 Practice Question
If 36 out of 60, what percentage?

🎯 Exam Tip
Double-check multiplication.

Question 9

Is √2 value less than or greater than 1
a) <1 b) >1 c) =1 d) none

🔍 Step 1: Understand the Problem
Find if √2 is less than, greater than, or equal to 1.

🧠 Step 2: Identify the Concept Used
Concept: Square roots of numbers between 1 and 4.

✏️ Step 3: Solve Step by Step
√1 = 1, √4 = 2
Since 1 < 2 < 4, √1 < √2 < √4 → 1 < √2 < 2
Thus √2 > 1

✅ Final Answer
👉 Option b) >1

🧠 Key Idea to Remember
If x > 1, then √x > 1.

⚠️ Common Mistake to Avoid
Thinking √2 ≈ 1.41 is less than 1.

🌍 Real-Life Application
Diagonal length of unit square is √2 > 1.

📝 Practice Question
Is √3 less than or greater than 1.5?

🎯 Exam Tip
Memorize √2 ≈ 1.414, √3 ≈ 1.732.

Question 10

If the hypotenuse of an isosceles right triangle is √72, the other two sides
a) 6,6 b) 6,7 c) 8,9 d) 12,6

🔍 Step 1: Understand the Problem
Isosceles right triangle → legs equal. Hypotenuse = √72.

🧠 Step 2: Identify the Concept Used
Concept: Pythagoras theorem: hypotenuse = leg × √2.

✏️ Step 3: Solve Step by Step
Let leg = a
Hypotenuse = a√2 = √72
a√2 = √72
a = √72 / √2 = √(72/2) = √36 = 6

✅ Final Answer
👉 Option a) 6,6

🧠 Key Idea to Remember
In isosceles right triangle, sides: a, a, a√2.

⚠️ Common Mistake to Avoid
Forgetting to simplify √72 = 6√2.

🌍 Real-Life Application
45° set square in engineering drawings.

📝 Practice Question
If hypotenuse = 10, find legs of isosceles right triangle.

🎯 Exam Tip
Direct formula: leg = hypotenuse ÷ √2.

Question 11

Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
a) 150 cm² b) 300 cm² c) 900 cm² d) 1500 cm²

🔍 Step 1: Understand the Problem
Diagonals d1 = 20 cm, d2 = 15 cm. Find area.

🧠 Step 2: Identify the Concept Used
Concept: Area of rhombus = ½ × d1 × d2.

✏️ Step 3: Solve Step by Step
Area = ½ × 20 × 15 = 10 × 15 = 150 cm²

✅ Final Answer
👉 Option a) 150 cm²

🧠 Key Idea to Remember
Rhombus area = half product of diagonals.

⚠️ Common Mistake to Avoid
Multiplying diagonals without halving.

🌍 Real-Life Application
Kite making, diamond-shaped tiles.

📝 Practice Question
Rhombus diagonals 12 cm and 16 cm, find area.

🎯 Exam Tip
Units: cm × cm = cm².

Question 12

Area of each triangle in the rectangle of length 7 cm and breadth 4 cm
a) 14 cm² b) 15 cm² c) 20 cm² d) none

🔍 Step 1: Understand the Problem
Rectangle length 7 cm, breadth 4 cm. "Each triangle" likely means when rectangle divided by a diagonal → two congruent right triangles.

🧠 Step 2: Identify the Concept Used
Concept: Area of triangle = half area of rectangle when split by diagonal.

✏️ Step 3: Solve Step by Step
Area of rectangle = 7 × 4 = 28 cm²
Each triangle area = 28 ÷ 2 = 14 cm²

✅ Final Answer
👉 Option a) 14 cm²

🧠 Key Idea to Remember
Diagonal divides rectangle into two equal triangles.

⚠️ Common Mistake to Avoid
Calculating perimeter instead of area.

🌍 Real-Life Application
Cutting rectangular cloth diagonally.

📝 Practice Question
Rectangle 10 cm × 5 cm, find triangle area when cut diagonally.

🎯 Exam Tip
Area of each triangle = ½ × length × breadth.

Question 13

Area of the parallelogram obtained by
a) base × height b) 1 × b c) 2(l+b) d) ½ × b × h

🔍 Step 1: Understand the Problem
Identify correct area formula for parallelogram.

🧠 Step 2: Identify the Concept Used
Concept: Area of parallelogram = base × height.

✏️ Step 3: Solve Step by Step
Standard formula: base × height.

✅ Final Answer
👉 Option a) base × height

🧠 Key Idea to Remember
Parallelogram area = bh, triangle area = ½ bh.

⚠️ Common Mistake to Avoid
Using ½ bh (that's for triangle).

🌍 Real-Life Application
Tiling with parallelogram-shaped tiles.

📝 Practice Question
Parallelogram base 8 cm, height 5 cm, find area.

🎯 Exam Tip
Height must be perpendicular to base.

Question 14 (Assertion-Reason)

Assertion: (a+b)² − (a−b)² = 4ab
Reason: On expansion, square terms cancel

🔍 Step 1: Understand the Problem
Check if assertion is true and if reason correctly explains it.

🧠 Step 2: Identify the Concept Used
Concept: Algebraic expansion and simplification.

✏️ Step 3: Solve Step by Step
(a+b)² = a² + 2ab + b²
(a−b)² = a² − 2ab + b²
Subtract: (a+b)² − (a−b)² = (a²+2ab+b²) − (a²−2ab+b²) = 4ab
Square terms (a² and b²) cancel indeed.
Assertion true, reason correct and explains it.

✅ Final Answer
👉 Both Assertion and Reason are true, Reason explains Assertion.

🧠 Key Idea to Remember
(a+b)² − (a−b)² = 4ab is a useful identity.

⚠️ Common Mistake to Avoid
Miscalculating signs during subtraction.

🌍 Real-Life Application
Difference of areas of two squares.

📝 Practice Question
Find (x+3)² − (x−3)².

🎯 Exam Tip
Memorize identity to save time.

Question 15 (Assertion-Reason)

Assertion: Area of trapezium = ½ × height × sum of parallel sides
Reason: Parallel sides of trapezium 12 cm and 16 cm and height 5 cm then area = 70 cm²

🔍 Step 1: Understand the Problem
Check if area formula is correct and if example in reason is right.

🧠 Step 2: Identify the Concept Used
Concept: Trapezium area formula.

✏️ Step 3: Solve Step by Step
Assertion: Correct formula = ½ × h × (a + b)
Reason: ½ × 5 × (12 + 16) = ½ × 5 × 28 = 70 cm² → correct
But reason is just an example, not an explanation of assertion.

✅ Final Answer
👉 Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

🧠 Key Idea to Remember
Example verifying formula doesn't explain why formula is true.

⚠️ Common Mistake to Avoid
Thinking example is explanation.

🌍 Real-Life Application
Area of tabletop, plot of land in trapezoidal shape.

📝 Practice Question
Trapezium parallel sides 10 m, 6 m, height 4 m, find area.

🎯 Exam Tip
In assertion-reason, check if reason logically justifies assertion.

SAMPLE Paper 1 – Section B & C – All Questions with Explanations

Section B (5 × 2 marks)

Question 16

Write 2 multiples of 36 between 45,000 and 47,000.

🔍 Step 1: Understand the Problem
Find two numbers divisible by 36 in the range 45,000 to 47,000.

🧠 Step 2: Identify the Concept Used
Concept: Multiples = numbers obtained by multiplying 36 with integers.

✏️ Step 3: Solve Step by Step
Divide 45,000 by 36: 45,000 ÷ 36 = 1,250 exactly (since 36 × 1250 = 45,000)
Next multiples:
45,000 + 36 = 45,036
45,036 + 36 = 45,072
Check: 45,072 + 36 = 45,108, etc. up to < 47,000

✅ Final Answer
👉 Two multiples: 45,036 and 45,072 (many possible answers)

🧠 Key Idea to Remember
Multiples of 36 = numbers divisible by both 4 and 9.

⚠️ Common Mistake to Avoid
Not checking divisibility rules for 36 (divisible by 4 and 9).

🌍 Real-Life Application
Packaging items in groups of 36.

📝 Practice Question
Find two multiples of 24 between 10,000 and 10,100.

🎯 Exam Tip
Start from lower bound, add the number to get next multiples.

Q17: Draw Venn diagram for multiples of 4, 5, and 32.
✅ Explanation:

Multiples of 32 are subset of multiples of 4 (since 32 = 4×8).
Multiples of 5 are separate, but LCM(4,5)=20 and LCM(32,5)=160 give overlaps.
Draw three circles: “Multiples of 4” large, contains “Multiples of 32” inside, “Multiples of 5” overlaps both.

Question 18 (Section B - 2 marks)

Expand (a+b)(a+b) OR Expand (3+u)(v−3).

🔍 Step 1: Understand the Problem
Option 1: (a+b)², Option 2: binomial multiplication.

🧠 Step 2: Identify the Concept Used
Concept: Distributive law / FOIL method.

✏️ Step 3: Solve Step by Step
Option 1: (a+b)(a+b) = a² + ab + ba + b² = a² + 2ab + b²
Option 2: (3+u)(v−3) = 3v − 9 + uv − 3u = uv + 3v − 3u − 9

✅ Final Answer
👉 For (a+b)(a+b): a² + 2ab + b²
👉 For (3+u)(v−3): uv + 3v − 3u − 9

🧠 Key Idea to Remember
(a+b)(a+b) = (a+b)² = a² + 2ab + b².

⚠️ Common Mistake to Avoid
Forgetting middle term 2ab in (a+b)².

🌍 Real-Life Application
Area expansion in algebra tiles.

📝 Practice Question
Expand (x+5)(x+5).

🎯 Exam Tip
Use identity to save time.

Question 19 (Section B - 2 marks)

Nandini has 25 marbles of which 15 are white. What percentage of her marbles are white?

🔍 Step 1: Understand the Problem
Total = 25, White = 15, find percentage.

🧠 Step 2: Identify the Concept Used
Concept: Percentage = (Part/Total) × 100.

✏️ Step 3: Solve Step by Step
(15/25) × 100 = (3/5) × 100 = 60%

✅ Final Answer
👉 60%

🧠 Key Idea to Remember
Fraction to percentage: multiply by 100.

⚠️ Common Mistake to Avoid
Calculating 15/100 instead of 15/25.

🌍 Real-Life Application
Statistics, survey results.

📝 Practice Question
Out of 40 students, 24 are girls. What percentage are girls?

🎯 Exam Tip
Simplify fraction first: 15/25 = 3/5.

Question 20 (Section B - 2 marks)

Find the diagonal of a square with side length 5 cm.

🔍 Step 1: Understand the Problem
Square side = 5 cm, find diagonal.

🧠 Step 2: Identify the Concept Used
Concept: Pythagoras: diagonal = side × √2.

✏️ Step 3: Solve Step by Step
Diagonal = 5 × √2 = 5√2 cm ≈ 7.07 cm

✅ Final Answer
👉 5√2 cm (exact answer)

🧠 Key Idea to Remember
Diagonal of square = side√2.

⚠️ Common Mistake to Avoid
Using formula for rectangle diagonal (√(l²+b²)) unnecessarily.

🌍 Real-Life Application
Finding screen size (TV diagonal).

📝 Practice Question
Square side 10 cm, find diagonal.

🎯 Exam Tip
Leave in √ form unless decimal asked.

Question 21 (Section C - 3 marks)

The sum of four consecutive numbers is 34. What are these numbers?
OR
The digital root of an 8-digit number is 5. What will be the digital root of number + 10?

🔍 Step 1: Understand the Problem
Two parts: choose one.
Part 1: Find 4 consecutive numbers summing to 34.
Part 2: Digital root changes when adding 10.

🧠 Step 2: Identify the Concept Used
Concept: Algebra for consecutive numbers; Digital root property.

✏️ Step 3: Solve Step by Step
Option 1: Let numbers be n, n+1, n+2, n+3
Sum: 4n + 6 = 34 → 4n = 28 → n = 7
Numbers: 7, 8, 9, 10

Option 2: Digital root 5 means remainder 5 when divided by 9.
Add 10 → remainder increases by 10 mod 9 = 1
New remainder = 5 + 1 = 6 → digital root = 6
(If remainder 0, digital root 9)

✅ Final Answer
👉 Option 1: 7, 8, 9, 10
👉 Option 2: 6

🧠 Key Idea to Remember
Consecutive numbers: n, n+1, n+2,...
Digital root = mod 9 (except 0 → 9).

⚠️ Common Mistake to Avoid
Forgetting to check sum in consecutive numbers.

🌍 Real-Life Application
Number puzzles; Checksums.

📝 Practice Question
Sum of 3 consecutive even numbers is 48. Find them.

🎯 Exam Tip
For digital root, adding 9 doesn't change it.

Question 22 (Section C - 3 marks)

Find the HCF of 72 and 96
OR
Divide ₹4500 into two parts in the ratio 2:3.

🔍 Step 1: Understand the Problem
Two independent questions.

🧠 Step 2: Identify the Concept Used
Concept: HCF by prime factorization; Ratio division.

✏️ Step 3: Solve Step by Step
Option 1: HCF of 72 and 96
72 = 2³ × 3²
96 = 2⁵ × 3
HCF = 2³ × 3 = 24

Option 2: Ratio 2:3 → total parts = 5
First part = (2/5)×4500 = 1800
Second part = (3/5)×4500 = 2700

✅ Final Answer
👉 Option 1: HCF = 24
👉 Option 2: ₹1800 and ₹2700

🧠 Key Idea to Remember
HCF = product of common prime factors with lowest power.
Ratio division: divide total in given proportion.

⚠️ Common Mistake to Avoid
In ratio, not summing parts first.

🌍 Real-Life Application
Splitting inheritance; Cutting rods to equal pieces.

📝 Practice Question
HCF of 48 and 84; or divide ₹3600 in ratio 1:2:3.

🎯 Exam Tip
For HCF, use division method: 96 ÷ 72 = 1 rem 24, 72 ÷ 24 = 3 rem 0 → HCF = 24.

Question 23 (Section C - 3 marks)

Find the value and draw bar models: i) 7% of 10 kg ii) 62% of 360.

🔍 Step 1: Understand the Problem
Calculate percentages and represent with bar models.

🧠 Step 2: Identify the Concept Used
Concept: Percentage as fraction; visual bar model.

✏️ Step 3: Solve Step by Step
i) 7% of 10 kg = (7/100)×10 = 0.7 kg
Bar: 10 kg bar divided into 100 equal parts, 7 parts shaded.

ii) 62% of 360 = (62/100)×360 = 0.62×360 = 223.2
Bar: 360 bar divided into 100 parts, 62 parts shaded.

✅ Final Answer
👉 i) 0.7 kg ii) 223.2
Bar models: show total divided into 100 parts, shade given percentage.

🧠 Key Idea to Remember
% = per hundred; bar model helps visualize fractions.

⚠️ Common Mistake to Avoid
Calculating 62% as 0.62 but forgetting to multiply properly.

🌍 Real-Life Application
Discount calculations; nutrition percentages.

📝 Practice Question
Find 15% of 80 kg with bar model.

🎯 Exam Tip
For bar model, simple rectangle with divisions works.

Bar model: For (i) 100% = 10 kg, 7% small bar.

Question 24 (Section C - 3 marks)
A right-angled triangle has short side 8 cm and hypotenuse 17 cm. Find the third side using Pythagoras theorem.
🔍 Step 1: Understand the Problem
Given: side a = 8 cm, hypotenuse c = 17 cm, find side b.
🧠 Step 2: Identify the Concept Used
Concept: Pythagoras: a² + b² = c².
✏️ Step 3: Solve Step by Step
8² + b² = 17²
64 + b² = 289
b² = 289 − 64 = 225
b = √225 = 15 cm
✅ Final Answer
👉 15 cm
🧠 Key Idea to Remember
In right triangle, hypotenuse² = sum of squares of other sides.
⚠️ Common Mistake to Avoid
Subtracting incorrectly: 289 − 64 = 225 (not 225).
🌍 Real-Life Application
Construction, finding ladder length.
📝 Practice Question
Triangle sides 6 cm, hypotenuse 10 cm, find other side.
🎯 Exam Tip
Check if sides form Pythagorean triple (8,15,17 is a triple).
Question 25 (Section C - 3 marks)
Find area of quadrilateral ABCD given AC=22 cm, BM=3 cm, DN=3 cm, BM ⟂ AC, DN ⟂ AC.
🔍 Step 1: Understand the Problem
Quadrilateral ABCD with diagonal AC=22 cm, perpendiculars from B and D to AC are both 3 cm.
🧠 Step 2: Identify the Concept Used
Concept: Area = sum of areas of two triangles sharing base AC.
✏️ Step 3: Solve Step by Step
Area ΔABC = ½ × base AC × height BM = ½ × 22 × 3 = 33 cm²
Area ΔADC = ½ × base AC × height DN = ½ × 22 × 3 = 33 cm²
Total area = 33 + 33 = 66 cm²
✅ Final Answer
👉 66 cm²
🧠 Key Idea to Remember
Area of quadrilateral with perpendiculars to diagonal = ½ × diagonal × (sum of perpendiculars).
⚠️ Common Mistake to Avoid
Adding perpendiculars first: 3+3=6, then ½×22×6 = 66 cm².
🌍 Real-Life Application
Land area calculation with diagonal measurements.
📝 Practice Question
Quadrilateral with diagonal 30 m, perpendiculars 4 m and 5 m, find area.
🎯 Exam Tip
Formula: Area = ½ × d × (h₁ + h₂).
Question 26 (Section D - 4 marks)
Compute using suitable identities: i) 397 × 403 ii) (2x−1)²
🔍 Step 1: Understand the Problem
Use algebraic identities for quick calculation.
🧠 Step 2: Identify the Concept Used
Concept: (a−b)(a+b) = a² − b²; (a−b)² = a² − 2ab + b².
✏️ Step 3: Solve Step by Step
i) 397 × 403 = (400−3)(400+3) = 400² − 3² = 160000 − 9 = 159991
ii) (2x−1)² = (2x)² − 2(2x)(1) + 1² = 4x² − 4x + 1
✅ Final Answer
👉 i) 159991 ii) 4x² − 4x + 1
🧠 Key Idea to Remember
Recognize numbers near round numbers to use identities.
⚠️ Common Mistake to Avoid
In (2x−1)², forgetting to square coefficient: (2x)² = 4x² not 2x².
🌍 Real-Life Application
Mental math shortcuts; algebraic expansions.
📝 Practice Question
Use identity for 498 × 502; expand (3y−2)².
🎯 Exam Tip
For (a−b)², remember middle term is −2ab.
Question 27 (Section D - 4 marks)
One acre of land costs ₹15,00,000. What is the cost of 2,400 sq ft of same land?
*(Note: 1 acre = 43,560 sq ft)*
🔍 Step 1: Understand the Problem
Given cost per acre, find cost for 2,400 sq ft. Need conversion.
🧠 Step 2: Identify the Concept Used
Concept: Unitary method with unit conversion.
✏️ Step 3: Solve Step by Step
1 acre = 43,560 sq ft costs ₹15,00,000
Cost per sq ft = 15,00,000 ÷ 43,560 ≈ ₹34.44
Cost for 2,400 sq ft = 34.44 × 2400 = ₹82,656
✅ Final Answer
👉 Approximately ₹82,656 (exact: 15,00,000/43,560 × 2400)
🧠 Key Idea to Remember
Convert to common units before finding unitary rate.
⚠️ Common Mistake to Avoid
Forgetting to convert acre to sq ft.
🌍 Real-Life Application
Real estate pricing.
📝 Practice Question
If 1 hectare (10,000 m²) costs ₹25,00,000, find cost of 1,500 m².
🎯 Exam Tip
Keep more decimal places in intermediate steps for accuracy.
Question 28 (Section D - 4 marks)
If (a,b,c) is a Pythagorean triple, then (ka,kb,kc) is also a triple for positive integer k. Verify with example.
🔍 Step 1: Understand the Problem
Verify property: scaling Pythagorean triple gives another triple.
🧠 Step 2: Identify the Concept Used
Concept: Pythagorean theorem: a² + b² = c².
✏️ Step 3: Solve Step by Step
Take example triple: (3,4,5)
Check: 3²+4²=9+16=25=5² ✓
Multiply by k=2: (6,8,10)
Check: 6²+8²=36+64=100=10² ✓
General proof: (ka)²+(kb)² = k²a²+k²b² = k²(a²+b²) = k²c² = (kc)²
✅ Final Answer
👉 Verified with example (3,4,5) scaled by 2 gives (6,8,10) which is also Pythagorean triple.
🧠 Key Idea to Remember
Scaling a right triangle by k gives similar triangle with sides multiplied by k.
⚠️ Common Mistake to Avoid
Not checking both original and scaled satisfy a²+b²=c².
🌍 Real-Life Application
Scale models maintain proportions.
📝 Practice Question
Verify with triple (5,12,13) scaled by 3.
🎯 Exam Tip
Show both calculation and general proof for full marks.
Question 29 (Case-Based - 4 marks)
Gangaram opened a book exhibition stall. He aims for daily sales ≥ ₹5000. Sales: Day1=₹2000, Day2=₹3500, Day3=₹5000, Day4=₹6000.
i) What percentage of his target did he achieve on day 1 and day 2?
Day1: (2000/5000)×100 = 40%
Day2: (3500/5000)×100 = 70%
ii) In day 3 and day 4, what percentage of target?
Day3: (5000/5000)×100 = 100%
Day4: (6000/5000)×100 = 120%
iii) What is the shortfall from target on day 1?
Target = 5000, Actual = 2000
Shortfall = 5000 − 2000 = ₹3000
OR iv) What percentage of target achieved on day 4?
Already calculated: 120%
✅ Final Answers
i) Day1: 40%, Day2: 70%
ii) Day3: 100%, Day4: 120%
iii) ₹3000
OR iv) 120%
🧠 Key Idea to Remember
Percentage achievement = (Actual/Target)×100.
⚠️ Common Mistake to Avoid
For shortfall: subtract actual from target.
🌍 Real-Life Application
Business sales tracking.
📝 Practice Question
Target ₹8000, actual ₹6000, find percentage and shortfall.
🎯 Exam Tip
Read case carefully: "shortfall" means how much less than target.
Question 30 (Case-Based - 4 marks)
Farmer's field shapes – areas of quadrilaterals.
(i) first figure h=4 cm base = 7 cm (ii) h = 3 cm base = 5cm (iii) base 5 cm, h = 4.8 cm (iv) base 2 cm , h = 4.4 cm its a parallelogramStep 1 – Identify shape:
Figure (i) is a parallelogram (based on description).
Step 2 – Formula used:
$Area of parallelogram = base \times height$
Step 3 – Substitution:
$Area = 7 cm \times 4 cm = 28 {cm}^{2}$
✅ Final Answer (i):
Area of field (i) $= 28 {cm}^{2}$ .
Part (ii)
Step 1 – Shape:
Figure (ii) is a parallelogram.
Step 2 – Formula:
$Area of parallelogram = base \times height$
Step 3 – Calculation:
$Area = 5 cm \times 3 cm = 15 {cm}^{2}$
✅ Final Answer (ii):
Area of field (ii) $= 15 {cm}^{2}$ .
step 1 – Area of figure (iii): (Parallelogram)
Base = 5 cm, Height = 4.8 cm
$Area = base \times height = 5 cm \times 4.8 cm = 24 {cm}^{2}$
Step 2 – Area of figure (iv): (Parallelogram)
Base = 2 cm, Height = 4.4 cm
$Area = 2 cm \times 4.4 cm = 8.8 {cm}^{2}$
Step 3 – Difference:
$Difference = 24 {cm}^{2} - 8.8 {cm}^{2} = 15.2 {cm}^{2}$
✅ Final Answer (iii):
Difference between area of figure (iii) and (iv) $= 15.2 {cm}^{2}$ .
OR Part (iv)
Step 1 – Area of figure (iii) already calculated:
$Area = 24 {cm}^{2}$
Step 2 – Convert to square metres:
$1 m^{2} = 10, 000 {cm}^{2}$ ${Area in m}^{2} = \frac{24}{10, 000} = 0.0024 m^{2}$
✅ Final Answer (iv):
Area of figure (iii) $= 0.0024 m^{2}$ .
🧠 Key Idea to Remember
1. All figures are parallelograms → Same formula: Area = base × height
2. Unit conversion remains: $1 m^{2} = 10, 000 {cm}^{2}$
3. Difference = Larger area − Smaller area
⚠️ Common Mistake to Avoid
- Using different formulas for different figures (but here all same)
- Confusing base with side length
- Not paying attention to "perpendicular height"
📝 Practice Question
Four parallelograms have dimensions:
A: base = 6 cm, height = 4 cm
B: base = 8 cm, height = 3 cm
C: base = 5 cm, height = 5 cm
D: base = 7 cm, height = 2 cm
Find:
(a) Area of each
(b) Difference between largest and smallest area
🌍 Real-Life Application
Agricultural land measurement.
🎯 Exam Tip
Label which figure is which shape clearly in answer.

Sunday, February 8, 2026

Learning Mathematics Through Real-Life Applications

Mathematics Subject Enrichment Activity – Class 8

Learning Mathematics Through Real-Life Applications

Introduction

Mathematics is more than a school subject; it is a powerful tool that helps us understand and solve real-life problems logically and systematically. For Class 8 students, learning mathematics should not be limited to memorizing formulas or solving textbook problems. This is where Mathematics Subject Enrichment Activities play an important role.

Subject enrichment activities are designed to make learning interactive, practical, and meaningful. These activities help students apply mathematical concepts to real-world situations, improve thinking skills, and develop confidence in problem-solving. This post explains the importance of mathematics enrichment activities for Class 8, along with examples and learning outcomes.

What is a Subject Enrichment Activity in Mathematics?

A subject enrichment activity is a learning task that goes beyond routine exercises. It focuses on understanding concepts deeply and applying them in practical situations rather than rote learning.

In mathematics, enrichment activities help students:

Connect mathematical ideas with daily life
Improve logical and analytical thinking
Develop problem-solving strategies
Learn mathematics with interest and curiosity

Such activities support activity-based and competency-based learning, which is essential in modern education.

Objectives of Mathematics Subject Enrichment Activities (Class 8)

The main objectives of mathematics enrichment activities are:

To strengthen conceptual understanding
To encourage independent and creative thinking
To develop reasoning and analytical skills
To reduce fear and anxiety related to mathematics
To promote learning through observation and practice

These objectives help students build a strong foundation for higher-level mathematics.

Sample Mathematics Subject Enrichment Activities – Class 8

Activity 1: Data Collection and Representation

Topic: Data Handling

Activity Description:
Students are asked to collect data related to daily life, such as:

Daily temperature for one week
Favourite sports among classmates
Time spent on different activities in a day

Steps to Perform:

Collect data accurately
Organize the data in a table
Represent the data using a bar graph or pie chart
Interpret the information shown in the graph

Learning Outcome:
Students learn how data is collected, organized, represented, and interpreted, which improves their understanding of statistics used in real life.

Activity 2: Profit and Loss in Daily Life

Topic: Comparing Quantities

Activity Description:
Suppose a student buys a book for ₹120 and sells it for ₹150.

Solution:

Cost Price (CP) = ₹120
Selling Price (SP) = ₹150
Profit = SP − CP = ₹30

Profit Percentage:

$\frac{30}{120} \times 100 = 25\%$

Learning Outcome:
Students understand how profit and loss calculations are applied in everyday buying and selling activities.

Activity 3: Estimating Area Using Graph Paper

Topic: Mensuration

Activity Description:
Students draw an irregular shape on graph paper and estimate its area.

Steps:

Count the complete squares inside the shape
Combine partial squares to make full squares
Estimate the total area

Learning Outcome:
This activity helps students understand area estimation and improves spatial visualization skills.

Activity 4: Linear Graphs from Real-Life Situations

Topic: Linear Equations in One Variable

Activity Description:
If one pencil costs ₹5, find the cost of different numbers of pencils and represent it on a graph.

Number of Pencils	Total Cost (₹)
1	5
2	10
3	15
4	20

Plot these points on a graph.

Learning Outcome:
Students learn how linear relationships represent real-life situations and understand the concept of linear graphs.

Real-Life Applications of Mathematics (Class 8)

Mathematics is used in many aspects of daily life, such as:

Shopping: Calculating discounts, profit, and loss
Travel: Estimating speed, distance, and time
Construction: Measuring area, perimeter, and volume
Banking: Calculating simple interest and percentages

By relating mathematical concepts to real-life situations, students understand the usefulness of mathematics beyond examinations.

Importance of Mathematics Subject Enrichment Activities

Mathematics enrichment activities help students:

Develop logical thinking and reasoning
Improve accuracy and concentration
Gain confidence in solving problems
Understand concepts clearly instead of memorizing formulas
Build interest and curiosity in mathematics

These activities also prepare students for competitive exams and higher studies.

Role of Teachers in Enrichment Activities

Teachers play an important role in guiding students during enrichment activities. They:

Encourage questioning and exploration
Provide real-life examples
Help students correct mistakes
Motivate students to use different problem-solving methods

A supportive classroom environment makes learning mathematics enjoyable and effective.

Assessment Through Subject Enrichment Activities

Enrichment activities are often part of internal assessment. Students may be assessed based on:

Concept clarity
Step-by-step explanation
Neat presentation
Logical reasoning

Regular participation improves overall academic performance.

Tips for Students

Understand the concept before starting the activity
Show steps clearly and neatly
Relate activities to daily life
Practice regularly to improve confidence

Conclusion

Mathematics subject enrichment activities for Class 8 play a vital role in developing understanding, confidence, and problem-solving skills. These activities move students beyond rote learning and help them apply mathematical concepts in real-life situations.

Regular practice of enrichment activities strengthens mathematical foundations and makes learning enjoyable and meaningful.

Frequently Asked Questions (FAQ)

Q1. Are subject enrichment activities compulsory?
Yes, they are usually part of internal assessment and help in better understanding.

Q2. Do enrichment activities improve exam performance?
Yes, they improve concept clarity, which helps students solve exam problems confidently.

Q3. Can students design their own enrichment activities?
Yes, creating activities enhances creativity and logical thinking.

Friday, February 6, 2026

Mathematics Subject Enrichment Activity Chapter: Area Theme: Transforming Shapes without Changing Area (Śulba-Sūtras)

📘 Mathematics Subject Enrichment Activity

Chapter: Area

Theme: Transforming Shapes without Changing Area (Śulba-Sūtras)

🔷 Topic

Area Preservation through Geometric Dissection and Rearrangement

🔷 Aim

To understand that area remains the same when a figure is cut and rearranged into another shape, and to explore ancient Indian (Śulba-Sūtras) and Euclidean methods for transforming geometric figures.

🔷 Materials Required

Geometry box (ruler, compass, protractor)
Colour papers / chart paper
Scissors and glue
Pencil & eraser
Colour pencils

🔷 Learning Outcomes

Students will be able to:

Apply area formulas correctly
Perform geometric dissections
Appreciate the contribution of Śulba-Sūtras
Develop visual–spatial reasoning

🔷 Activity 1

Convert a Trapezium into a Rectangle of Equal Area

Procedure

Draw trapezium ABCD.
Extend non-parallel sides to form congruent triangles.
Cut triangles ∆AHI and ∆DGI, ∆BEJ and ∆CFJ.
Rearrange pieces to form rectangle EFGH.

Observation

Cut triangles are congruent.
Rearranged figure is a rectangle.
Area remains unchanged.

Conclusion

Dissection preserves area even though shape changes.

🔷 Activity 1(b)

Construct a Trapezium of Area 144 cm²

Solution (Example):
Take parallel sides = 18 cm, 6 cm
Height = 12 cm

\text{Area} = \frac{1}{2}(18+6)\times12 = 144 \text{ cm}^2

🔷 Activity 2

Convert an Isosceles Trapezium into a Rectangle

Method:
Cut off two congruent triangular ends and shift them inward to form a rectangle.

🔷 Activity 3

Rectangle → Rhombus (Equal Area)

Idea:
Cut along diagonals and rearrange to form a rhombus with same base × height.

🔷 Activity 4

Rhombus → Rectangle (Śulba-Sūtras Method)

Observation:
Height of rhombus becomes breadth of rectangle → area preserved.

🔷 Activity 5

Rectangle → Isosceles Triangle (Śulba-Sūtras)

Method:
Cut rectangle along diagonal and join halves to form triangle.

🔷 Activity 6

Area Comparison

(a) Equilateral Triangle vs Square (same side length a)

Triangle area = $\frac{\sqrt{3}}{4}a^2$
Square area = $a^2$

✅ Square has greater area

(b) Two equilateral triangles vs square

2\times \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{2}a^2 < a^2

✅ Square still has greater area

🔷 Activity 7

Triangle → Rectangle (Śulba-Sūtras)

Method:
Duplicate triangle, join base-to-base → rectangle.

🔷 Activity 8

Isosceles Triangle → Rectangle (Simpler Method)

Hint Used:
∆ADB and ∆ADC become two halves of a rectangle.

🔷 Activity 9

Rectangle of Twice the Area of a Triangle

Method 1:
Duplicate triangle → rectangle

Method 2:
Extend height or base proportionally

🔷 Activity 10

Quadrilateral of Half the Area of Another Quadrilateral

Method:
Divide quadrilateral by diagonal → take one triangle → rearrange.

🔷 Observations (Common)

Shape may change, area does not
Dissection relies on congruency
Visual reasoning is essential

🔷 Reflections (Student Writes)

I learnt that area depends on base and height, not shape.
Ancient mathematicians used logical geometric methods.
Dissection made geometry easier to understand.

🔷 HOTS (Higher-Order Thinking)

Can every polygon be converted into a rectangle of equal area?
Why are triangles most useful for dissection?
How is this concept used in land measurement?

🌟 Mathematical Heritage Link

Śulba-Sūtras (800–500 BCE) show that Indian mathematicians understood area conservation centuries before modern geometry.

1️⃣ Trapezium → Rectangle (Equal Area)

Labels to show on diagram (for students):

Trapezium ABCD
Height h
Parallel sides AB, CD
Cut triangles: ∆AHI ≅ ∆DGI, ∆BEJ ≅ ∆CFJ
Rearranged rectangle EFGH
Same height h

📝 Key Note for Students:

Pieces are rearranged, not resized → area remains same

2️⃣ Isosceles Trapezium → Rectangle

Highlight:

Equal non-parallel sides
Two congruent triangles cut and shifted

3️⃣ Rectangle → Rhombus (Śulba-Sūtras)

Label:

Same base
Same height
Area = base × height (unchanged)

4️⃣ Triangle → Rectangle (Śulba-Sūtras)

Student Hint on Diagram:

Two identical triangles → rectangle

🧠 PART B: Competency-Based Worksheet (with Answer Key)

🔹 Section A: Understanding

What happens to the area when a shape is dissected and rearranged?
Answer: Area remains unchanged.
Name the ancient Indian texts that discuss area transformation.
Answer: Śulba-Sūtras.

🔹 Section B: Application

A trapezium has parallel sides 12 cm and 8 cm, height 10 cm.
Find its area.
Answer:

$\frac{1}{2} (12 + 8) \times 10 = 100 {cm}^{2}$

If this trapezium is converted into a rectangle, what will be the area of the rectangle?
Answer: 100 cm²

🔹 Section C: Reasoning

Why are triangles commonly used in dissection methods?
Answer:
Because triangles are the simplest polygons and can easily form other shapes.
Explain why a rectangle formed from a triangle has the same area.
Answer:
No part is removed or added; only rearranged.

🔹 Section D: HOTS

Can a circle be converted into a rectangle using exact dissection? Why?
Answer:
No, because curved boundaries cannot be exactly rearranged into straight edges.
If the height of a triangle is doubled, what happens to its area?
Answer:
Area doubles.

🔹 Section E: Assertion–Reason

Assertion (A): Area depends on shape.
Reason (R): Rearrangement changes area.

Correct Answer: ❌ Both A and R are false.

📘 PART C: Teacher Rubric Page (Assessment)

Criteria	Excellent (4)	Good (3)	Satisfactory (2)	Needs Improvement (1)
Concept of Area	Clear & accurate	Mostly clear	Partial	Incorrect
Diagrams	Neat, labelled, colourful	Labelled	Rough	Missing
Mathematical Reasoning	Logical & clear	Mostly logical	Limited	Incorrect
Application of Śulba-Sūtras	Correct & explained	Correct	Partial	Not shown
Reflection	Deep insight	Relevant	Minimal	Missing

Total Marks: ____ / 20

🌟 Teacher Tip (Optional Page Note)

This activity integrates history of mathematics, geometry, and visual reasoning, aligning with NEP 2020 competency-based learning.

MATHEMATICS SUBJECT ENRICHMENT ACTIVITY Chapter: Proportional Reasoning – II Topic: Representation of Data Using Pie Chart

📘 MATHEMATICS SUBJECT ENRICHMENT ACTIVITY

Chapter: Proportional Reasoning – II

Topic: Representation of Data Using Pie Chart (Proportional Reasoning)

🎯 Aim

To collect data on favourite sports of Class 8 students and represent the data proportionally using a pie chart, and to identify:

The sport liked by maximum students
The sport liked by minimum students
Sports liked by equal number of students

🧰 Materials Required

Notebook
Pen / Pencil
Scale
Compass
Protractor
Colour pencils / sketch pens
Graph paper

📊 Data Collection

A survey was conducted among 40 classmates of Class 8 to find their favourite sport.

Collected Data Table

Sport	Number of Students
Cricket	14
Football	10
Badminton	6
Volleyball	6
Basketball	4
Total	40

🧮 Calculations (Proportional Reasoning)

Formula used:

\text{Angle of sector} = \frac{\text{Number of students}}{40} \times 360^\circ

Sport	Students	Fraction	Angle
Cricket	14	14/40	126°
Football	10	10/40	90°
Badminton	6	6/40	54°
Volleyball	6	6/40	54°
Basketball	4	4/40	36°

🎨 Colourful Pie Chart

(Students should draw this pie chart neatly using the above angles and colour each sector differently.)

🧑‍🏫 Procedure

Conduct a survey among 40 classmates.
Record the data in a table.
Convert the data into fractions of the total.
Calculate angles for each sport using proportional reasoning.
Draw a circle using a compass.
Mark angles with a protractor.
Colour each sector and label it clearly.

👀 Observations

Cricket occupies the largest sector.
Basketball occupies the smallest sector.
Badminton and Volleyball have equal-sized sectors.

📝 Conclusion

Maximum favourite sport: Cricket
Minimum favourite sport: Basketball
Same preference: Badminton and Volleyball
Pie charts help us understand proportional relationships visually.

💭 Reflection (Student Learning)

I learned how fractions and ratios are used in real-life data.
I understood the connection between angles and proportions.
Drawing a pie chart improved my data handling and reasoning skills.

🧠 Teacher’s Note (Assessment Link)

✔ Uses proportional reasoning
✔ Integrates data handling
✔ Encourages real-life application
✔ Supports visual learning

Competency-Based Questions (CBQs) perfectly aligned with
Class 8 – Ganita Prakash (Part 2)
Chapter: Proportional Reasoning – II
based on the Favourite Sports Pie Chart Activity.

🧠 Competency-Based Questions

(Data Handling & Proportional Reasoning)

Case Study

A survey was conducted among 40 Class 8 students to find their favourite sport.
The data collected is shown below:

Sport	Students
Cricket	14
Football	10
Badminton	6
Volleyball	6
Basketball	4

The data is represented using a pie chart.

🔹 Level 1: Understanding & Interpretation

1. What fraction of the students like Football?
a) 1/4
b) 1/5
c) 1/8
d) 3/10

2. Which sport is liked by the maximum number of students?

3. Name the sports which have equal representation in the pie chart.

🔹 Level 2: Application of Proportional Reasoning

4. If the total number of students is 40, what angle represents Cricket in the pie chart?

5. The angle representing Basketball is:
a) 36°
b) 54°
c) 72°
d) 90°

6. If 1 student represents 9°, verify whether the angle for Badminton is correct.

🔹 Level 3: Analysis & Reasoning

7. Compare the sectors of Cricket and Football.
Which is larger and why?

8. If 5 more students start liking Basketball, how will the angle of the Basketball sector change?

🔹 Level 4: Higher-Order Thinking (HOTS)

9. If the number of students increases to 60 but the ratio of preferences remains the same, find the new angle for Cricket.

10. Why is a pie chart more suitable than a bar graph for showing proportional data in this activity?

🔹 Assertion–Reason Type

11. Assertion (A): The total angle of all sectors in a pie chart is 360°.
Reason (R): A pie chart represents the whole data set as a circle.

a) Both A and R are true and R is the correct explanation of A
b) Both A and R are true but R is not the correct explanation of A
c) A is true but R is false
d) A is false but R is true

🔹 Real-Life Application

12. How can this method of data representation help a school decide which sports facilities to improve?

📝 Teacher Answer Key (Brief)

a)
Cricket
Badminton and Volleyball
126°
a)
6 × 9° = 54° ✔
Cricket, because it has more students
Angle increases proportionally
126° (ratio unchanged)
Shows part-to-whole relationship clearly
a)
Helps in decision-making using data

🎯 Competencies Assessed

✔ Data Interpretation
✔ Proportional Reasoning
✔ Mathematical Communication
✔ Critical Thinking
✔ Real-life Application

Monday, February 9, 2026

Class 8 SESSION ENDING EXAMINATION (2026) SAMPLE PAPER-1 WITH ANSWERS

SESSION ENDING EXAMINATION (2026) SAMPLE PAPER-1

Section A – MCQ & Assertion-Reason

ANSWER KEY SAMPLE Paper 1 – Section A – All MCQs with Explanations

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14 (Assertion-Reason)

Question 15 (Assertion-Reason)

SAMPLE Paper 1 – Section B & C – All Questions with Explanations

Section B (5 × 2 marks)

Question 16

Question 18 (Section B - 2 marks)

Question 19 (Section B - 2 marks)

Question 20 (Section B - 2 marks)

Question 21 (Section C - 3 marks)

Question 22 (Section C - 3 marks)

Question 23 (Section C - 3 marks)

Question 24 (Section C - 3 marks)

Question 25 (Section C - 3 marks)

Question 26 (Section D - 4 marks)

Question 27 (Section D - 4 marks)

Question 28 (Section D - 4 marks)

Question 29 (Case-Based - 4 marks)

Question 30 (Case-Based - 4 marks)

Part (ii)

OR Part (iv)

🧠 Key Idea to Remember

⚠️ Common Mistake to Avoid

📝 Practice Question

Sunday, February 8, 2026

Learning Mathematics Through Real-Life Applications

Mathematics Subject Enrichment Activity – Class 8

Learning Mathematics Through Real-Life Applications

Introduction

What is a Subject Enrichment Activity in Mathematics?

Objectives of Mathematics Subject Enrichment Activities (Class 8)

Sample Mathematics Subject Enrichment Activities – Class 8

Activity 1: Data Collection and Representation

Activity 2: Profit and Loss in Daily Life

Activity 3: Estimating Area Using Graph Paper

Activity 4: Linear Graphs from Real-Life Situations

Real-Life Applications of Mathematics (Class 8)

Importance of Mathematics Subject Enrichment Activities

Role of Teachers in Enrichment Activities

Assessment Through Subject Enrichment Activities

Tips for Students

Conclusion

Frequently Asked Questions (FAQ)

Friday, February 6, 2026

Mathematics Subject Enrichment Activity Chapter: Area Theme: Transforming Shapes without Changing Area (Śulba-Sūtras)

📘 Mathematics Subject Enrichment Activity

Chapter: Area

Theme: Transforming Shapes without Changing Area (Śulba-Sūtras)

🔷 Topic

🔷 Aim

🔷 Materials Required

🔷 Learning Outcomes

🔷 Activity 1

Convert a Trapezium into a Rectangle of Equal Area

Procedure

Observation

Conclusion

🔷 Activity 1(b)

Construct a Trapezium of Area 144 cm²

🔷 Activity 2

Convert an Isosceles Trapezium into a Rectangle

🔷 Activity 3

Rectangle → Rhombus (Equal Area)

SESSION ENDING EXAMINATION (2026)
SAMPLE PAPER-1