CBSE Class 8 Mathematics Periodic Test 2 Sample Paper 01 with Answers (Old NCERT Syllabus)

This Class 8 Maths Periodic Test 2 sample paper carries 40 marks and has 17 questions in five sections (A to E), and every question comes with a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising Algebraic Expressions and Identities, Mensuration and Exponents and Powers. The questions test adding and multiplying polynomials, surface area and volume of cubes, cuboids and cylinders, laws of exponents with negative powers, standard form, and a case study on covering a box with square sheets of paper.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Add the following polynomials:
2x²y² − 3xy + 4, 5 + 7xy − 3x²y², and 4x²y² + 10xy
Answer:
Write the sum and group like terms:
(2x²y² − 3x²y² + 4x²y²) + (−3xy + 7xy + 10xy) + (4 + 5)
= 3x²y² + 14xy + 9
So the sum is 3x²y² + 14xy + 9.
In simple words: Add the x²y² terms, the xy terms and the numbers separately.

Teacher's Note:
1. Like terms have exactly the same variables with the same powers.
2. Write the answer with the highest-degree term first.

Question 8. Find the side of a cube whose surface area is 600 cm².
Answer:
Let the side of the cube be a cm.
Surface area of a cube = 6a².
6a² = 600, so a² = 100 and a = 10.
So the side of the cube is 10 cm.
In simple words: A cube has 6 equal square faces, so one face is 600 ÷ 6 = 100 cm², and its side is 10 cm.

Teacher's Note:
1. Divide by 6 first, then take the square root.
2. Side length is never negative, so take only the positive square root.

Question 9. Simplify and write the answer in the exponential form: \( \left(2^{5} \div 2^{8}\right)^{5} \times 2^{-5} \)
Answer:
Inside the bracket: \( 2^{5} \div 2^{8} = 2^{5-8} = 2^{-3} \).
\( \left(2^{-3}\right)^{5} = 2^{-15} \).
\( 2^{-15} \times 2^{-5} = 2^{-15-5} = 2^{-20} \).
So the answer is \( 2^{-20} \), which is the same as \( \frac{1}{2^{20}} \).
In simple words: Subtract powers for division, multiply powers for a power of a power, and add powers for multiplication.

Teacher's Note:
1. Laws used: a^m ÷ aⁿ = a^m−n, (a^m)ⁿ = a^mn, a^m × aⁿ = a^m+n.
2. Be careful with signs when adding negative exponents.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Simplify: (a + b)(c − d) + (a − b)(c + d) + 2(ac + bd)
Answer:
(a + b)(c − d) = ac − ad + bc − bd.
(a − b)(c + d) = ac + ad − bc − bd.
2(ac + bd) = 2ac + 2bd.
Add all three: ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd.
The ad terms cancel, the bc terms cancel, and −bd − bd + 2bd = 0.
This leaves ac + ac + 2ac = 4ac.
In simple words: After multiplying out, everything cancels except the ac terms, which add up to 4ac.

Teacher's Note:
1. Multiply each bracket fully before collecting like terms.
2. Keep track of signs: −b × d = −bd and −b × c = −bc.

Question 11. Simplify: \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \)
Answer:
Write the composite numbers as products of primes: 10 = 2 × 5, 6 = 2 × 3 and 125 = 5³.
\( \frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^{3}}{5^{-7} \times 2^{-5} \times 3^{-5}} \)
Powers of 3: \( 3^{-5-(-5)} = 3^{0} = 1 \). Powers of 2: \( 2^{-5-(-5)} = 2^{0} = 1 \).
Powers of 5: \( 5^{-5+3-(-7)} = 5^{5} \).
So the value is \( 5^{5} \) = 3125.
In simple words: Break every number into primes, then add or subtract the powers of each prime.

Teacher's Note:
1. (2 × 5)⁻⁵ = 2⁻⁵ × 5⁻⁵: a power of a product splits over each factor.
2. Dividing by 5⁻⁷ is the same as multiplying by 5⁷.

Question 12. Simplify 7x²(3x − 9) + 3 and find its values for x = 4 and x = 6.
Answer:
7x²(3x − 9) = 7x² × 3x − 7x² × 9 = 21x³ − 63x².
So 7x²(3x − 9) + 3 = 21x³ − 63x² + 3.
For x = 4: 21 × 64 − 63 × 16 + 3 = 1344 − 1008 + 3 = 339.
For x = 6: 21 × 216 − 63 × 36 + 3 = 4536 − 2268 + 3 = 2271.
In simple words: Multiply 7x² into the bracket, then put x = 4 and x = 6 in the simplified expression.

Teacher's Note:
1. Check with the original form: for x = 4, 7 × 16 × (12 − 9) + 3 = 336 + 3 = 339.
2. Work out the powers (4³, 4², 6³, 6²) first, then multiply.

Question 13. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Answer:
Volume of the cuboid = 60 × 54 × 30 = 97200 cm³.
Volume of one small cube = 6 × 6 × 6 = 216 cm³.
Number of cubes = 97200 ÷ 216 = 450.
Check: 6 divides 60, 54 and 30 exactly (10 × 9 × 5 = 450 cubes along the three edges).
So 450 cubes can be placed in the cuboid.
In simple words: Divide the volume of the box by the volume of one cube: 97200 ÷ 216 = 450.

Teacher's Note:
1. Dividing volumes works here because each edge of the cuboid is an exact multiple of 6 cm.
2. Counting along the edges (10, 9 and 5 cubes) is a quick check.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Simplify:
(i) (x + y)(2x + y) + (x + 2y)(x − y)
(ii) (x + y)(x² − xy + y²)
Answer:
(i) (x + y)(2x + y) = 2x² + xy + 2xy + y² = 2x² + 3xy + y².
(x + 2y)(x − y) = x² − xy + 2xy − 2y² = x² + xy − 2y².
Sum = 2x² + 3xy + y² + x² + xy − 2y² = 3x² + 4xy − y².
(ii) (x + y)(x² − xy + y²) = x(x² − xy + y²) + y(x² − xy + y²)
= x³ − x²y + xy² + x²y − xy² + y³
The x²y terms cancel and the xy² terms cancel, so the result is x³ + y³.
In simple words: Multiply every term of the first bracket by every term of the second, then collect like terms.

Teacher's Note:
1. In (ii), the middle terms cancel in pairs, which is a useful pattern to remember.
2. Check (i) with x = 1, y = 1: 2 × 3 + 3 × 0 = 6 and 3 + 4 − 1 = 6.

Question 15. Express the following numbers in usual form.
(i) 3.02 × 10⁻⁶ (ii) 4.5 × 10⁴ (iii) 3 × 10⁻⁸ (iv) 1.0001 × 10⁹
Answer:
A negative power of 10 moves the decimal point to the left; a positive power moves it to the right.
(i) 3.02 × 10⁻⁶: move the point 6 places left: 0.00000302.
(ii) 4.5 × 10⁴: move the point 4 places right: 45000.
(iii) 3 × 10⁻⁸: move the point 8 places left: 0.00000003.
(iv) 1.0001 × 10⁹: move the point 9 places right: 1000100000.
In simple words: Positive power: number gets bigger (point moves right); negative power: number gets smaller (point moves left).

Teacher's Note:
1. Fill empty places with zeros while moving the decimal point.
2. Check (iii): there are 7 zeros after the decimal point before 3.

Question 16. If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Answer:
Let the edge of the cube be a. Then surface area S = 6a² and volume V = a³.
(i) New edge = 2a. New surface area = 6(2a)² = 6 × 4a² = 4 × 6a² = 4S.
So the surface area becomes 4 times the original.
(ii) New volume = (2a)³ = 8a³ = 8V.
So the volume becomes 8 times the original.
In simple words: Doubling the edge multiplies area by 2 × 2 = 4 and volume by 2 × 2 × 2 = 8.

Teacher's Note:
1. Area depends on length squared and volume on length cubed.
2. Strictly, 'becomes 4 times' means the increase is 3 times the original; say which one you mean.

Section E: Case Study Based Question (4 Marks)

Question 17. Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it (see the figure). She must know the exact quantity of paper to buy for this purpose. The box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively and she wants to use square sheets of paper of side 40 cm for the decoration.
Answer the following questions based on the above information:
(a) Find the total surface area of the box. (1)
(b) Find the area of each sheet of the paper. (1)
(c) Find the number of sheets required. (2)
Answer:
(a) Total surface area of a cuboid = 2(lb + bh + hl)
= 2(80 × 40 + 40 × 20 + 20 × 80) = 2(3200 + 800 + 1600) = 2 × 5600 = 11200 cm².
(b) Area of one square sheet = 40 × 40 = 1600 cm².
(c) Number of sheets = total surface area ÷ area of one sheet = 11200 ÷ 1600 = 7 sheets.
In simple words: The box needs 11200 cm² of paper and each sheet gives 1600 cm², so 7 sheets are needed.

Teacher's Note:
1. The formula uses lb, bh and hl: each pair of opposite faces is counted twice.
2. This assumes the whole box is covered, including the bottom, with no overlap or waste.