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CLASS VIII • MATHEMATICS PRACTICE TEST 01 (2026–27)

New NCERT Curriculum: Ganita Prakash (Part-1, Chapters 1 to 7)

⏱️ DURATION: 1½ Hours

🎯 MAXIMUM MARKS: 40

📋 BOOK: Ganita Prakash Part-1

General Instructions:

All questions are compulsory.
This question paper contains 20 questions divided into five Sections: A, B, C, D, and E.
Section A contains 10 Multiple Choice Questions of 1 mark each (Q1 to Q10 including 2 Assertion-Reason questions).
Section B contains 4 Short Answer questions of 2 marks each (Q11 to Q14).
Section C contains 3 Short Answer questions of 3 marks each (Q15 to Q17).
Section D contains 1 Long Answer question of 5 marks (Q18).
Section E contains 2 Case Study-based questions of 4 marks each (Q19 & Q20).
Use of calculators is strictly prohibited.

1. In Queen Ratnamanjuri's 100-locker puzzle, how many lockers remain open at the end? [1]

(a) 10

(b) 50

(d) 1

✓ Solution (1 Mark): Only square numbers have an odd number of factors and thus remain open after an odd number of toggles. Between 1 and 100, the square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 (Total = 10). (Correct Option: a)

2. The value of 2¹⁰⁰ ÷ 2²⁵ expressed as a power of 2 is: [1]

(a) 2⁴

(b) 2⁷⁵

(d) 2⁵⁰

✓ Solution (1 Mark): Using the exponent division law: nᵃ ÷ nᵇ = nᵃ⁻ᵇ.
2¹⁰⁰ ÷ 2²⁵ = 2¹⁰⁰⁻²⁵ = 2⁷⁵. (Correct Option: b)

3. The ancient Mesopotamian (Babylonian) number system was based on which number? [1]

(a) Base 10

(b) Base 20

(d) Base 12

✓ Solution (1 Mark): Mesopotamians developed a positional system with landmark powers of 60 (Base-60 sexagesimal system), which is still used today in measuring hours, minutes, and seconds. (Correct Option: c)

4. A quadrilateral whose diagonals are equal in length and bisect each other at 90° must be a: [1]

(a) Rectangle

(b) Rhombus

(d) Trapezium

✓ Solution (1 Mark): If diagonals are equal and bisect each other, it is a rectangle. If they bisect at right angles (90°), it is a rhombus. A figure satisfying both conditions is a Square. (Correct Option: c)

5. What is the digital root of the number (9a + 36b + 13), where a and b are any integers? [1]

(a) 1

(b) 4

(d) 5

✓ Solution (1 Mark): 9a + 36b + 13 = 9(a + 4b + 1) + 4.
Multiples of 9 have a digital root of 9. Adding 4 gives digital root = 9 + 4 = 13 ⇒ 1 + 3 = 4. (Correct Option: b)

6. Given that 125² = 15625, the value of 126² can be quickly found using which expression? [1]

(a) 15625 + 126

(b) 15625 + 251

(d) 15625 + 26²

✓ Solution (1 Mark): (n + 1)² = n² + 2n + 1. Here n = 125.
126² = 125² + 2(125) + 1 = 15625 + 251 = 15876. (Correct Option: b)

7. A 40 kg mixture contains sand and cement in the ratio 3 : 1. How much cement must be added to make the ratio 5 : 2? [1]

(a) 2 kg

(b) 4 kg

(d) 6 kg

✓ Solution (1 Mark): Sand = (3/4) × 40 = 30 kg; Initial cement = (1/4) × 40 = 10 kg.
In new ratio (5 : 2), sand remains 30 kg. New cement = (2/5) × 30 = 12 kg.
Cement to add = 12 - 10 = 2 kg. (Correct Option: a)

8. The smallest number expressible as the sum of two cubes in two different ways (1³ + 12³ = 9³ + 10³) is: [1]

(a) 1331

(b) 1729

(d) 4104

✓ Solution (1 Mark): 1729 is the celebrated Hardy-Ramanujan Taxicab number: 1³ + 12³ = 1 + 1728 = 1729, and 9³ + 10³ = 729 + 1000 = 1729. (Correct Option: b)

9. Assertion (A): In a parallelogram, opposite angles are equal and adjacent angles add up to 180°.
Reason (R): Opposite sides of a parallelogram are parallel, and interior angles on the same side of a transversal are supplementary. [1]

(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

(d) A is false but R is true

✓ Solution (1 Mark): Both statements are true, and transversal properties of parallel lines explain why adjacent angles sum to 180°. (Correct Option: a)

10. Assertion (A): If a number is divisible by 9, then the sum of its digits is divisible by 9.
Reason (R): Every number can be written as (multiples of 9) + (sum of its digits). [1]

(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

(d) A is false but R is true

✓ Solution (1 Mark): Place values 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1 directly prove that the remainder on dividing by 9 is the sum of digits. Both A and R are true and R explains A. (Correct Option: a)

11. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Also find the square root of the product. [2]

✓ Solution (2 Marks): Prime factorisation of 9408: 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7 = (2&sup6;) × 3 × (7²).
The prime factor 3 is unpaired. Hence, 9408 must be multiplied by 3.
Product = 9408 × 3 = 28224.
Square root = √28224 = 2³ × 3 × 7 = 8 × 21 = 168.

12. Express the following large numbers in scientific standard form (x × 10ʸ):
(a) Mass of the Earth: 59,76,00,00,00,00,00,00,00,00,00,00,000 kg
(b) Distance between the Sun and Earth: 1,49,60,00,00,000 m [2]

✓ Solution (2 Marks): (a) Mass of the Earth = 5.976 × 10²⁴ kg.
(b) Distance between Sun and Earth = 1.496 × 10¹¹ m.

13. In rectangle PQRS, diagonals PR and QS intersect at O. If ∠QOR = 110°, calculate the measures of ∠OPS and ∠OQR. [2]

✓ Solution (2 Marks): In ΔOQR, OQ = OR (diagonals of rectangle are equal and bisect each other).
∠OQR + ∠ORQ + 110° = 180° ⇒ 2(∠OQR) = 70° ⇒ ∠OQR = 35°.
Since vertically opposite angles are equal, ∠POS = ∠QOR = 110°.
In isosceles ΔPOS, OP = OS ⇒ ∠OPS = (180° - 110°) / 2 = 35°.

14. Anagh mixes 600 mL of orange juice with 900 mL of apple juice to prepare a fruit drink.
(a) Write the ratio of orange juice to apple juice in simplest form.
(b) How much orange juice is needed to prepare 3000 mL of this drink with the same taste? [2]

✓ Solution (2 Marks): (a) Ratio = 600 : 900 = 2 : 3 (dividing by HCF 300).
(b) Total ratio parts = 2 + 3 = 5.
Orange juice needed = (2 / 5) × 3000 mL = 1200 mL (1.2 litres).

15. (a) Expand the product using the distributive property: (4y + 7)(y + 11z - 3)
(b) Using the algebraic identity (a + b)(a - b) = a² - b², evaluate: 397 × 403 [3]

✓ Solution (3 Marks): (a) (4y + 7)(y + 11z - 3) = 4y(y + 11z - 3) + 7(y + 11z - 3)
= 4y² + 44yz - 12y + 7y + 77z - 21 = 4y² + 44yz + 77z - 5y - 21. [1.5 Marks]
(b) 397 × 403 = (400 - 3)(400 + 3) = 400² - 3² = 160000 - 9 = 159991. [1.5 Marks]

16. (a) If the 6-digit number 48a23b is divisible by 18, find two possible pairs of values for (a, b).
(b) Solve the cryptarithm where each letter is a unique digit: JK × 6 = KKK [3]

✓ Solution (3 Marks): (a) Divisible by 18 means divisible by 2 and 9. Thus b is even: {0, 2, 4, 6, 8}.
Sum of digits = 4 + 8 + a + 2 + 3 + b = 17 + a + b (must be multiple of 9).
If b = 0 ⇒ 17 + a = 18 ⇒ a = 1. Pair: (1, 0).
If b = 2 ⇒ 19 + a = 27 ⇒ a = 8. Pair: (8, 2). Other valid pairs: (6, 4), (4, 6), (2, 8). [1.5 Marks]
(b) KKK = 111 × K. So JK × 6 = 111 × K ⇒ JK = (37 × K) / 2.
Since JK is a 2-digit integer, K must be even. For K = 4, JK = 37 × 2 = 74.
Hence J = 7, K = 4. Verification: 74 × 6 = 444. [1.5 Marks]

17. In a rhombus GAME, the diagonals GE and AM intersect at point O.
(a) Using triangle congruence, prove that the diagonals are perpendicular (intersect at 90°).
(b) If diagonal GE = 6 cm and diagonal AM = 8 cm, calculate the length of each side of the rhombus. [3]

✓ Solution (3 Marks): (a) In ΔGEO and ΔMEO: GE = ME (rhombus sides are equal), OE = OE (common side), OG = OM (diagonals bisect each other).
By SSS congruence, ΔGEO ≅ ΔMEO ⇒ ∠GOE = ∠MOE.
Since they form a linear pair: ∠GOE + ∠MOE = 180° ⇒ 2(∠GOE) = 180° ⇒ ∠GOE = 90°. [1.5 Marks]
(b) Diagonals bisect at 90°: OE = 6/2 = 3 cm, OA = 8/2 = 4 cm.
In right-angled ΔAOE, by Pythagoras theorem:
Side AE = √(3² + 4²) = √(9 + 16) = √25 = 5 cm. [1.5 Marks]

18. An architect is designing an eco-park in Dhauli. The layout features two identical square plots, each of area g² sq. ft., reserved for green lawn cover. Surrounding both squares is a uniform tiled walking pathway of width w ft. on all sides and between them. [5]

(a) Write the algebraic expression for the total length and breadth of the rectangular park in terms of g and w. [1 Mark]
(b) Show algebraically that the area of the walking path that needs to be tiled is 8w(g + w) sq. ft. [2 Marks]
(c) If g = 20 ft and w = 5 ft, calculate the numerical area of the tiled walking path, and find the ratio of the green lawn area to the tiled walking path area in simplest form. [2 Marks]

✓ Solution (5 Marks): (a) Length of park = w + g + 2w + g + w = 2g + 4w ft. Breadth of park = w + g + w = g + 2w ft. [1 Mark]
(b) Total Area of park = (2g + 4w)(g + 2w) = 2g² + 4gw + 4gw + 8w² = 2g² + 8gw + 8w².
Green lawn area = 2 × g² = 2g².
Tiled Walking Area = Total Area - Green Area = (2g² + 8gw + 8w²) - 2g² = 8gw + 8w² = 8w(g + w) sq. ft. [2 Marks]
(c) For g = 20 ft, w = 5 ft:
Tiled Area = 8(5)(20 + 5) = 40 × 25 = 1000 sq. ft.
Green Lawn Area = 2 × (20)² = 2 × 400 = 800 sq. ft.
Ratio (Green Area : Tiled Area) = 800 : 1000 = 4 : 5. [2 Marks]

19. Case Study 1 (Chapter 1): Wonders of Squares and Hardy-Ramanujan Numbers [4]

During a math club activity on ancient number patterns, students explored how every square number n² can be represented as the sum of the first n consecutive odd natural numbers. In the same discussion, they learned about taxicab numbers, named after Srinivasa Ramanujan who famously showed that 1729 is the smallest number expressible as the sum of two cubes in two different ways (1³ + 12³ = 9³ + 10³).

(a) Express 121 as the sum of consecutive odd natural numbers starting from 1. [1 Mark]

(b) The next taxicab number after 1729 is 4104. Show that 4104 can be expressed as the sum of two cubes in two different ways using pairs from (2, 16) and (9, 15). [1 Mark]

(d) Can a perfect cube end with exactly two zeros (00)? Justify your answer mathematically. [1 Mark]

✓ Solution (4 Marks): (a) 121 = 11² = Sum of the first 11 consecutive odd numbers:
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. [1 Mark]
(b) Pair 1: 2³ + 16³ = 8 + 4096 = 4104.
Pair 2: 9³ + 15³ = 729 + 3375 = 4104. [1 Mark]
(c) 10648 = (2 × 2 × 2) × (11 × 11 × 11) = 2³ × 11³.
Cube root = ³√10648 = 2 × 11 = 22. [1 Mark]
(d) No. If a number ends in k zeros, its cube must end in 3k zeros (a multiple of 3 zeros such as 3, 6, 9 zeros). Therefore, a perfect cube can never end with exactly two zeros. [1 Mark]

20. Case Study 2 (Chapter 6): The Calendar Diagonal Mystery & Algebraic Expansion [4]

While exploring calendar patterns, students selected any 2 × 2 block of dates. For dates 4, 5 on top and 11, 12 below, the diagonal products are 4 × 12 = 48 and 5 × 11 = 55, giving a difference of 55 - 48 = 7. They modeled this algebraically using the general 2 × 2 square: top row [a, a+1] and bottom row [a+7, a+8]. In algebra, the distributive law also allows geometric visualization of identities like (a + b)², (a - b)², and (a + b)(a - b).

(a) Prove algebraically that for any 2 × 2 calendar square, the difference between the diagonal products (a + 1)(a + 7) and a(a + 8) is always equal to 7. [1 Mark]

(b) Using the identity (a + b)(a - b) = a² - b², evaluate the product 98 × 102 without manual column multiplication. [1 Mark]

(c) A square playground has a side length of (2x + 3) metres. Expand (2x + 3)² using the identity (a + b)² = a² + 2ab + b². [1 Mark]

(d) If (a - b) = 5 and ab = 14, find the value of (a² + b²) using the identity (a - b)² = a² + b² - 2ab. [1 Mark]

✓ Solution (4 Marks): (a) Product 1 = (a + 1)(a + 7) = a² + 7a + a + 7 = a² + 8a + 7.
Product 2 = a(a + 8) = a² + 8a.
Difference = (a² + 8a + 7) - (a² + 8a) = 7 (proven constant for all dates). [1 Mark]
(b) 98 × 102 = (100 - 2)(100 + 2) = 100² - 2² = 10000 - 4 = 9996. [1 Mark]
(c) (2x + 3)² = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9. [1 Mark]
(d) (a - b)² = a² + b² - 2ab
⇒ 5² = (a² + b²) - 2(14)
⇒ 25 = (a² + b²) - 28 ⇒ a² + b² = 25 + 28 = 53. [1 Mark]

Turn Your Kindness Into Knowledge.

Sample Paper 1 Class8 Mathematics Part 1 (chapter1 to 7)

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CLASS VIII • MATHEMATICS PRACTICE TEST 01 (2026–27)