Class 8 SESSION ENDING EXAMINATION (2026) SAMPLE PAPER-4 WITH ANSWERS
SAMPLE Paper 4
SECTION A
Digital root of 34761097: a) 37 b) 10 c) 1 d) 91
True proportion: a) 4:7::12:21 b) 8:3::24:6 c) 24:8::9:3 d) 7:12::12:71
How many m² in 1 km²: a) 100 b) 1,000 c) 10,000 d) 1,000,000
Nandini: 25 marbles, 15 white → % white: a) 10% b) 15% c) 60% d) 25%
672 divisible by: a) 3&5 b) 2,3,4 c) 2&3 only d) 2&5
Algebraic expression for "sum of cube of two integers": a) 3+x b) 5+x³ c) x³+27 d) (x+1)³
Area of rhombus: a) base×height b) ½×base×height c) ½×product of diagonals d) ½×height×sum of parallel sides
Primitive Pythagorean triple: a) 2,3,4 b) 3,4,5 c) 9,12,15 d) 18,24,30
100% of m=215 → 10% of m = ? a) 10 b) 2.15 c) 21.5 d) 21
Square of odd number always 1 more than multiple of 8: a) always true b) sometimes c) never d) can't say
Ratio proportional to 54:81: a) 4:8 b) 12:18 c) 16:27 d) 20:25
p invested 4 years at 6% p.a. no compounding → total amount expression: a) p×6×4 b) p×0.6×4 c) p+(p×0.06×4) d) p×1.6×4
Diagonal of square side 5 cm: a) 10 cm b) 25 cm² c) 5√2 cm d) 2√5 cm
Assertion-Reason (14-15)
14. A: (a+b)²−(a−b)² always divisible by 4 R: Simplifies to 4ab
15. A: Price increases 20% then decreases 20% → same R: Increase & decrease by same % cancel
SECTION B (2 marks)
16. Middle of 5 consecutive even numbers=5p → express all in terms of p.
17. Find missing side in right triangle (diagram).
18. 104² using identity OR (3a−9b)(3a+9b)
19. Sum of 4 consecutive numbers=34 → find them.
20. Fill: i) 3 is ___% of 300 ii) ___ is 40% of 4 OR Shirt ₹300, 25% discount → Mohan pays?
SECTION C (3 marks)
21. Midday meal: 120 students→15 kg rice, 80 students→? kg OR Car: 48 L→432 km, 20 L→? km.
22. Side of rhombus diagonals 24 units & 70 units.
23. SP of 3 pencils = CP of 5 pencils → profit/loss %?
24. In parallelogram PQRS, find QN (diagram).
25. Smallest number remainder 2 when ÷3, remainder 4 when ÷5.
SECTION D (4 marks)
26. Pattern: i) Draw next figure ii) Basic units in Step 10 iii) Expression for Step y.
27. Mixture 40 kg sand:cement=3:1 → add cement to make ratio 5:2.
28. Lotus stem tip 1 unit above water, bends to touch water 3 units away → depth of lake OR Generate 2 scaled Pythagorean triples from (5,12,13) and 2 more using (n−1)²+(2n−1)=n².
SECTION E (Case-Based)
29. Three shops: A: Buy1 Get1, B: Buy2 Get1, C: Buy3 Get1. Price ₹100/item.
i) Effective price per item in A
ii) % discount each shop
iii) Arrange cheapest to costliest
30. Rectangular path around garden:
a) Name two rectangles used
b) Formula for path area
c) If garden 20m×12m, path width 2m → area of path
d) Does path area change if outer rectangle shifted?
SAMPLE Paper 4 – Complete Solutions with Explanations
Section A – MCQs with Explanations
Q1: Digital root of 34761097?
a) 37 b) 10 c) 1 d) 91
✅ Answer: c) 1
Explanation: Sum digits: 3+4+7+6+1+0+9+7=37 → 3+7=10 → 1+0=1.
Q2: Which proportion is true?
a) 4:7 :: 12:21 b) 8:3 :: 24:6 c) 24:8 :: 9:3 d) 7:12 :: 12:71
✅ Answer: a) 4:7 :: 12:21
Explanation: Cross-check: 4×21=84, 7×12=84 ✔.
Q3: How many m² in 1 km²?
a) 100 b) 1,000 c) 10,000 d) 10,00,000
✅ Answer: d) 10,00,000
Explanation: 1 km² = 1,000,000 m².
Q4: Nandini has 25 marbles, 15 white → % white?
a) 10% b) 15% c) 60% d) 25%
✅ Answer: c) 60%
Explanation: 15/25 = 0.6 = 60%.
Q5: 672 divisible by:
a) 3 and 5 b) 2,3, and 4 c) 2 and 3 only d) 2 and 5
✅ Answer: b) 2,3, and 4
Explanation: 672 even → divisible by 2. Sum digits=15 → divisible by 3. Last two digits 72 → divisible by 4.
Q6: Algebraic expression for “sum of cube of two integers”?
a) 3+x b) 5+x³ c) x³+27 d) (x+1)³
✅ Answer: None match exactly — but likely c) x³+27 (if integers are x and 3). But best generic: a³+b³. Not in options.
Q7: Area of rhombus is:
a) Base×height b) ½×base×height c) ½×product of diagonals d) ½×height×sum of parallel sides
✅ Answer: c) ½×product of diagonals
Explanation: Standard formula.
Q8: Which is primitive Baudhayana (Pythagorean) triplet?
a) 2,3,4 b) 3,4,5 c) 9,12,15 d) 18,24,30
✅ Answer: b) 3,4,5
Explanation: Primitive means coprime: 3,4,5 have HCF=1.
Q9: 100% of m = 215 → 10% of m = ?
a) 10 b) 2.15 c) 21.5 d) 21
✅ Answer: c) 21.5
Explanation: m=215, 10% = 21.5.
Q10: Square of any odd number is always 1 more than multiple of 8:
a) Always true b) sometimes true c) never true d) can’t say
✅ Answer: a) Always true
Explanation: Odd number = 2k+1 → (2k+1)² = 4k²+4k+1 = 4k(k+1)+1, k(k+1) even → 8m+1.
Q11: Ratio proportional to 54:81?
a) 4:8 b) 12:18 c) 16:27 d) 20:25
✅ Answer: b) 12:18
Explanation: 54:81 = 2:3, 12:18 = 2:3 ✔.
Q12: p invested for 4 years at 6% p.a., no compounding → total amount?
a) p×6×4 b) p×0.6×4 c) p+(p×0.06×4) d) p×1.6×4
✅ Answer: c) p+(p×0.06×4)
Explanation: Simple interest formula.
Q13: Diagonal of square side 5 cm:
a) 10 cm b) 25 sq cm c) 5√2 cm d) 2√5 cm
✅ Answer: c) 5√2 cm
Explanation: Diagonal = side√2.
Assertion-Reason
Q14:
A: (a+b)² − (a−b)² always divisible by 4.
R: Expression simplifies to 4ab.
✅ Answer: (a) Both true, R explains A
Explanation: (a+b)²−(a−b)² = 4ab, which is divisible by 4.
Q15:
A: Price increases 20% then decreases 20% → final price same.
R: Increase and decrease by same % cancel each other.
✅ Answer: (d) A false, R false
Explanation: If original=100, after +20% →120, after −20% →96 (not same). They don’t cancel multiplicatively.
Section B – Short Answers (2 marks)
Q16: Middle of 5 consecutive even numbers = 5p. Express others in terms of p.
✅ Answer: Let numbers: 5p−4, 5p−2, 5p, 5p+2, 5p+4.
Q17: Find missing side in right triangle. Diagram needed.
If sides given, use Pythagoras.
Q18: Compute 104² using identity.
✅ Answer: 104² = (100+4)² = 10000+800+16 = 10,816.
OR (3a−9b)(3a+9b) = 9a²−81b².
Q19: Sum of four consecutive numbers = 34 → numbers?
✅ Answer: Let n, n+1, n+2, n+3 → 4n+6=34 → n=7 → 7,8,9,10.
Q20(i): 3 is ___% of 300.
✅ Answer: (3/300)×100 = 1%.
(ii): ___ is 40% of 4.
✅ Answer: 0.4×4 = 1.6.
OR Shirt ₹300, 25% discount → Mohan pays 75% of 300 = ₹225.
Section C – Short Answers (3 marks)
Q21: Midday meal: 120 students → 15 kg rice. 80 students → ? kg rice.
✅ Answer: Proportion: 15/120 = x/80 → x = (15×80)/120 = 10 kg.
OR Car: 48 L → 432 km, 20 L → ? km.
Fuel efficiency = 432/48 = 9 km/L → 20 L → 180 km.
Q22: Side of rhombus with diagonals 24 units and 70 units.
✅ Answer: Side = √[(24/2)² + (70/2)²] = √(12²+35²) = √1369 = 37 units.
Q23: SP of 3 pencils = CP of 5 pencils → profit/loss %?
✅ Answer: Let CP per pencil = ₹1, CP of 5=₹5=SP of 3 → SP per pencil=5/3≈1.67.
Profit % = [(5/3 − 1)/1]×100 = (2/3)×100 ≈ 66.67%.
Q24: In parallelogram PQRS, find QN? Diagram needed. If heights given, use area equality.
Q25: Smallest number leaves remainder 2 when ÷3, remainder 4 when ÷5.
✅ Answer: Let N = 3a+2 = 5b+4. Test:
3a+2 = 5b+4 → 3a−5b=2.
Try b=1 → 3a=7 no. b=2 → 3a=12 → a=4 → N=14.
Check: 14÷3=4 rem 2, 14÷5=2 rem 4 ✔. Smallest is 14.
Section D – Long Answers (4 marks)
Q25 (Pattern):
(i) Draw next figure: likely growing pattern of squares/triangles.
(ii) Basic units in Step 10: if Step n has n²+something, plug n=10.
(iii) Expression for Step y: e.g., y²+2y.
Q26: Mixture 40 kg sand:cement = 3:1 → sand=30 kg, cement=10 kg. Want ratio 5:2 after adding cement.
Let x kg cement added. New cement = 10+x, sand=30.
Ratio 30/(10+x) = 5/2 → 60 = 50+5x → 5x=10 → x=2 kg.
Q27: Lotus stem tip 1 unit above water, bends to touch water 3 units away horizontally → depth?
✅ Answer: Let depth = d, stem length = d+1.
Pythagoras: (d+1)² = d² + 3² → d²+2d+1 = d²+9 → 2d=8 → d=4 units.
OR
(a) Scaled triple from (5,12,13): multiply by k=2 → (10,24,26), k=3 → (15,36,39).
(b) Using (n−1)²+(2n−1)=n²: Choose n=5 → 4²+9=25 → triple (4,3,5)?? Check: (n−1, 2n−1, n)?? Actually 2n−1 odd. For n=5: (4,9,√97) not integer triple. Maybe misprint.
Section E – Case-Based (4 marks)
Q28: Shop deals:
A: Buy 1 get 1 free → effective price per item = half.
B: Buy 2 get 1 free → effective price per item = (2×price)/3.
C: Buy 3 get 1 free → effective price per item = (3×price)/4.
Price = ₹100.
(i) Shop A effective price = ₹50.
(ii) % discount:
A: discount 50%.
B: pay 200 for 3 → per item 66.67 → discount 33.33%.
C: pay 300 for 4 → per item 75 → discount 25%.
(iii) Cheapest to costliest: A (50), B (~66.67), C (75).
Q30: Rectangular path around garden:
a) Two rectangles: ABCD (outer), EFGH (inner).
b) Area of path = Area(ABCD) − Area(EFGH).
c) If length=20 m, breadth=12 m, path width=2 m → outer: 24×16 m.
Area path = (24×16)−(20×12) = 384−240 = 144 m².
d) Area of path changes if outer rectangle shifts? Yes, if dimensions change.
