CBSE Class 8 Mathematics Periodic Test 2 Sample Paper 02 with Answers (Old NCERT Syllabus)

This Class 8 Maths Periodic Test 2 sample paper is a 40-mark test with 19 questions in five sections (A to E), each solved step by step. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it useful for state-board students and for revising Algebraic Expressions and Identities, Mensuration, Exponents and Powers and Direct and Inverse Proportions. It checks multiplying monomials and binomials, areas of a rhombus and a trapezium, volume of a cuboid, solving exponent equations, standard form, direct and inverse variation word problems, and a kaleidoscope case study on the curved surface area and volume of a cylinder.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Teacher's Note:
1. Finding the unit cost first (unitary method) keeps the numbers small.
2. In direct variation the ratio x/y stays the same.

Question 11. Multiply (i) (x − 4) and (2x + 3) (ii) (x − y) and (3x + 5y)
Answer:
(i) (x − 4)(2x + 3) = x(2x + 3) − 4(2x + 3) = 2x² + 3x − 8x − 12 = 2x² − 5x − 12.
(ii) (x − y)(3x + 5y) = x(3x + 5y) − y(3x + 5y) = 3x² + 5xy − 3xy − 5y² = 3x² + 2xy − 5y².
In simple words: Multiply each term of the first bracket by the whole second bracket, then combine like terms.

Teacher's Note:
1. xy and yx are like terms, so 5xy − 3yx = 2xy.
2. Check (i) with x = 1: (−3)(5) = −15 and 2 − 5 − 12 = −15.

Question 12. A school has 8 periods a day, each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Answer:
Total teaching time is fixed, so more periods means shorter periods: this is inverse variation.
Let each period be x minutes. Then 8 × 45 = 9 × x.
x = 360 ÷ 9 = 40.
So each period would be 40 minutes.
In simple words: The total time is 8 × 45 = 360 minutes; shared among 9 periods, each gets 40 minutes.

Teacher's Note:
1. In inverse variation, the product of the two quantities stays the same.
2. Check: 9 × 40 = 360 = 8 × 45.

Question 13. Find the value of m for which 5^m ÷ 5⁻³ = 5⁵.
Answer:
Using a^m ÷ aⁿ = a^m−n: 5^{m − (−3)} = 5^{m + 3}.
So 5^{m + 3} = 5⁵.
The bases are equal, so m + 3 = 5, giving m = 2.
In simple words: Dividing by 5⁻³ adds 3 to the power, so m + 3 = 5 and m = 2.

Teacher's Note:
1. Subtracting a negative exponent is the same as adding.
2. Check: 5² ÷ 5⁻³ = 5² × 5³ = 5⁵.

Section C: Short Answer Questions (3 Marks Each)

Question 14. Simplify 3x(4x − 5) + 3 and find its values for (i) x = 3 (ii) x = \( \frac{1}{2} \).
Answer:
3x(4x − 5) + 3 = 12x² − 15x + 3.
(i) x = 3: 12 × 9 − 15 × 3 + 3 = 108 − 45 + 3 = 66.
(ii) x = \( \frac{1}{2} \): 12 × \( \frac{1}{4} \) − 15 × \( \frac{1}{2} \) + 3 = 3 − \( \frac{15}{2} \) + 3 = 6 − \( \frac{15}{2} \) = \( \frac{-3}{2} \).
In simple words: Expand first to 12x² − 15x + 3, then substitute each value of x.

Teacher's Note:
1. Square the fraction before multiplying: (½)² = ¼.
2. Write 6 as 12/2 to subtract 15/2.

Question 15. Simplify: \( \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \) (t ≠ 0)
Answer:
Move the negative powers from the denominator to the numerator: \( \frac{25 \times 5^{3} \times t^{-4} \times t^{8}}{10} \).
\( t^{-4} \times t^{8} = t^{4} \) and 25 × 5³ = 25 × 125 = 3125.
\( \frac{3125\, t^{4}}{10} = \frac{625\, t^{4}}{2} \).
So the answer is \( \frac{625\, t^{4}}{2} \).
In simple words: Bring the negative powers up from the bottom, add the powers of t, and cancel 3125/10 to 625/2.

Teacher's Note:
1. 1/a⁻ⁿ = aⁿ, so 5⁻³ in the denominator becomes 5³ in the numerator.
2. Reduce the number part fully: both 3125 and 10 are divisible by 5.

Question 16. If y varies inversely as x, and if y = 4 when x = 72, find the following:
(a) the constant of variation
(b) the value of y when x = 9.
Answer:
For inverse variation, xy = k (constant).
(a) k = 72 × 4 = 288.
(b) When x = 9, y = \( \frac{k}{x} \) = \( \frac{288}{9} \) = 32.
In simple words: The product x × y is always 288, so when x is 9, y must be 32.

Teacher's Note:
1. x became 8 times smaller (72 to 9), so y became 8 times larger (4 to 32).
2. Always find k first in a variation problem.

Section D: Long Answer Questions (5 Marks Each)

Question 17. Simplify: (i) (x + y)(x² − xy + y²) (2)
(ii) (a + b + c)(a + b − c) (3)
Answer:
(i) (x + y)(x² − xy + y²) = x(x² − xy + y²) + y(x² − xy + y²)
= x³ − x²y + xy² + x²y − xy² + y³
= x³ + (−x²y + x²y) + (xy² − xy²) + y³ = x³ + y³.
(ii) (a + b + c)(a + b − c) = a(a + b − c) + b(a + b − c) + c(a + b − c)
= a² + ab − ac + ab + b² − bc + ac + bc − c²
= a² + 2ab + b² − c² (the ac and bc terms cancel).
So the answer is a² + 2ab + b² − c².
In simple words: Multiply each term of the first bracket by the second bracket; the middle terms cancel in pairs.

Teacher's Note:
1. In (ii) you can also use (p + c)(p − c) = p² − c² with p = a + b: (a + b)² − c².
2. Collect like terms carefully; ab and ba are the same term.

Question 18. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Answer:
1 m³ = 1000 litres, so volume of the reservoir = 108 × 1000 = 108000 litres.
Water poured in 1 hour = 60 × 60 = 3600 litres.
Time taken = 108000 ÷ 3600 = 30 hours.
So the reservoir fills in 30 hours.
In simple words: The tank holds 108000 litres and fills at 3600 litres every hour, so it takes 30 hours.

Teacher's Note:
1. Change everything to the same units (litres and hours) before dividing.
2. Remember 1 m³ = 1000 L and 1 hour = 60 minutes.

Section E: Case Study Based Question (4 Marks)

Question 19. Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope (see the figure). She wanted to make a kaleidoscope of length 25 cm with a 7 cm diameter. You may take π = \( \frac{22}{7} \).
Answer the following questions based on the above information:
(i) Find the radius of the kaleidoscope. (1)
(ii) Find the area of chart paper required. (1)
(iii) Find the cost of chart paper if the rate of chart paper is Rs 2 per 100 cm². (1)
(iv) Find the volume of the kaleidoscope. (1)
Answer:
(i) Radius = diameter ÷ 2 = 7 ÷ 2 = 3.5 cm.
(ii) Chart paper needed = curved surface area = 2πrh = 2 × \( \frac{22}{7} \) × 3.5 × 25 = 550 cm².
(iii) Cost = \( \frac{550}{100} \) × Rs 2 = 5.5 × 2 = Rs 11.
(iv) Volume = πr²h = \( \frac{22}{7} \) × 3.5 × 3.5 × 25 = 962.5 cm³.
In simple words: Halve the diameter for the radius, use 2πrh for the paper, and πr²h for the space inside.

Teacher's Note:
1. Only the curved surface is covered, so the two circular ends are not included in (ii).
2. 22/7 × 3.5 = 11 is a handy shortcut in both (ii) and (iv).