CBSE Class 8 Mathematics Algebraic Expressions, Mensuration and Exponents Practice Paper 01 with Answers (Old NCERT Syllabus)

This Class 8 Maths practice paper covers Algebraic Expressions and Identities, Mensuration and Exponents and Powers. It carries 40 marks and has 17 questions in sections A to E, and every question has a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising these chapters. Students practise adding, subtracting and multiplying algebraic expressions, laws of exponents and usual form, and surface area and volume of cubes and cylinders.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm.
Answer:
Radius = 140 ÷ 2 = 70 cm = 0.7 m.
Volume = πr²h, so 1.54 = \( \frac{22}{7} \) × 0.7 × 0.7 × h = 1.54 × h.
h = 1.54 ÷ 1.54 = 1 m.
So the height is 1 m.
In simple words: Convert the radius to 0.7 m; the base area is 1.54 m², so a volume of 1.54 m³ needs a height of 1 m.

Teacher's Note:
1. Use the same unit for all lengths: the volume is in m³, so the radius must be in m.
2. \( \frac{22}{7} \) × 0.49 = 1.54.

Question 8. Find the value of m for which 5^m ÷ 5⁻³ = 5⁵
Answer:
Using \( a^{m} \div a^{n} = a^{m-n} \): \( 5^{m} \div 5^{-3} = 5^{m+3} \).
So \( 5^{m+3} = 5^{5} \), and m + 3 = 5.
m = 2.
So 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 means adding.
2. Check: 5² ÷ 5⁻³ = 5⁵.

Question 9. Subtract the following polynomials.
3xy + 5yz − 7xz + 1 from −4xy + 2yz − 2xz + 5xyz + 1
Answer:
(−4xy + 2yz − 2xz + 5xyz + 1) − (3xy + 5yz − 7xz + 1).
= −4xy + 2yz − 2xz + 5xyz + 1 − 3xy − 5yz + 7xz − 1.
Group like terms: (−4xy − 3xy) + (2yz − 5yz) + (−2xz + 7xz) + 5xyz + (1 − 1).
= −7xy − 3yz + 5xz + 5xyz.
So the answer is 5xyz − 7xy − 3yz + 5xz.
In simple words: Change every sign of the expression being subtracted, then add like terms.

Teacher's Note:
1. "Subtract A from B" means B − A.
2. Put the subtracted expression in brackets so every sign changes.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Simplify: \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \)
Answer:
Move negative powers across the fraction line: \( \frac{5^{7} \times 6^{5} \times 125}{3^{5} \times 10^{5}} \).
Write in prime factors: 6⁵ = 2⁵ × 3⁵, 10⁵ = 2⁵ × 5⁵ and 125 = 5³.
\( \frac{5^{7} \times 2^{5} \times 3^{5} \times 5^{3}}{3^{5} \times 2^{5} \times 5^{5}} = 5^{7+3-5} = 5^{5} \).
5⁵ = 3125.
So the value is 5⁵ = 3125.
In simple words: Turn every number into powers of 2, 3 and 5; the 2s and 3s cancel and 5⁵ = 3125 is left.

Teacher's Note:
1. A factor with a negative power in the numerator moves to the denominator with a positive power, and the other way round.
2. Breaking into prime powers makes cancelling easy.

Question 11. Simplify the following expressions:
(i) (x + y + z)(x + y − z)
(ii) x²(x − 3y²) − xy(y² − 2xy) − x(y³ − 5x²)
Answer:
(i) (x + y + z)(x + y − z) = x² + xy − xz + xy + y² − yz + xz + yz − z².
= x² + y² − z² + 2xy, so the answer is x² + y² + 2xy − z².
(Quick check: it is (x + y)² − z².)
(ii) x²(x − 3y²) = x³ − 3x²y²; −xy(y² − 2xy) = −xy³ + 2x²y²; −x(y³ − 5x²) = −xy³ + 5x³.
Add: x³ + 5x³ − 3x²y² + 2x²y² − xy³ − xy³ = 6x³ − x²y² − 2xy³.
In simple words: Multiply out every bracket, then collect like terms.

Teacher's Note:
1. In (i), take x + y as one term: (a + z)(a − z) = a² − z².
2. Watch the minus sign in front of each bracket in (ii).

Question 12. 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:
108 m³ = 108 × 1000 = 108000 litres.
Rate = 60 litres per minute = 60 × 60 = 3600 litres per hour.
Time = 108000 ÷ 3600 = 30 hours.
So it takes 30 hours to fill the reservoir.
In simple words: The reservoir holds 108000 litres and 3600 litres flow in each hour, so 30 hours.

Teacher's Note:
1. 1 m³ = 1000 litres.
2. Change the rate to litres per hour since the answer is wanted in hours.

Question 13. 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: ac + ac + 2ac + (−ad + ad) + (bc − bc) + (−bd − bd + 2bd) = 4ac.
So the expression simplifies to 4ac.
In simple words: Everything cancels except the ac terms, which add up to 4ac.

Teacher's Note:
1. Expand each product fully before collecting terms.
2. Check with a = b = c = d = 1: 0 + 0 + 4 = 4 = 4ac.

Section D: Long Answer Questions (4 Marks Each)

Question 14. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length 1 m.
Answer:
Radius r = 84 ÷ 2 = 42 cm; length h = 1 m = 100 cm.
Area levelled in one revolution = curved surface area = 2πrh = 2 × \( \frac{22}{7} \) × 42 × 100 = 26400 cm².
Area in 750 revolutions = 26400 × 750 = 19800000 cm².
1 m² = 10000 cm², so area = 19800000 ÷ 10000 = 1980 m².
So the area of the road is 1980 m².
In simple words: Each turn of the roller flattens 26400 cm²; 750 turns flatten 1980 m².

Teacher's Note:
1. One revolution of a roller covers its curved surface area once.
2. 1 m² = 100 cm × 100 cm = 10000 cm².

Question 15. Express the following numbers in usual form.
(i) 5.325 × 10⁻⁶   (ii) 5.45 × 10⁴   (iii) 7 × 10⁻⁸   (iv) 3.00201 × 10⁹
Answer:
(i) Move the decimal point 6 places to the left: 5.325 × 10⁻⁶ = 0.000005325.
(ii) Move the point 4 places to the right: 5.45 × 10⁴ = 54500.
(iii) Move the point 8 places to the left: 7 × 10⁻⁸ = 0.00000007.
(iv) Move the point 9 places to the right: 3.00201 × 10⁹ = 3002010000.
In simple words: A positive power of 10 moves the point to the right; a negative power moves it to the left.

Teacher's Note:
1. Fill in zeros where there are no digits left.
2. Check (iv): 3.00201 has 5 decimal digits, so 4 more zeros are needed for 9 places.

Question 16. 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: Open the bracket to get 21x³ − 63x² + 3, then put in x = 4 and x = 6.

Teacher's Note:
1. Use the distributive law a(b − c) = ab − ac.
2. Check with the original form: for x = 4, 7 × 16 × 3 + 3 = 339.

Section E: Case Study Based Question (4 Marks)

Question 17. Mohan purchased two cylinders for his Maths Project. He marked first cylinder as 'A' and second one as 'B' with marker pen. He wants to fill sand in two cylinders and also wants to wrap with orange paper around the curved surface of two cylinders as shown in the below figure. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm.

Answer the following questions based on the above information:
(a) Find the area of orange paper in Cylinder A.
(b) Find the area of orange paper in Cylinder B.
(c) Find the volume of sand in Cylinder A.
(d) Find the volume of sand in Cylinder B.
Answer:
Cylinder A: r = 3.5 cm, h = 14 cm. Cylinder B: r = 7 cm, h = 7 cm.
(a) Paper on A = 2πrh = 2 × \( \frac{22}{7} \) × 3.5 × 14 = 22 × 14 = 308 cm².
(b) Paper on B = 2 × \( \frac{22}{7} \) × 7 × 7 = 44 × 7 = 308 cm².
(c) Sand in A = πr²h = \( \frac{22}{7} \) × 3.5 × 3.5 × 14 = 22 × 3.5 × 7 = 539 cm³.
(d) Sand in B = \( \frac{22}{7} \) × 7 × 7 × 7 = 22 × 49 = 1078 cm³.
In simple words: Both cylinders need the same 308 cm² of paper, but B holds twice as much sand (1078 cm³) as A (539 cm³).

Teacher's Note:
1. Curved surface area depends on r × h, which is 49 for both, so the paper is equal.
2. Volume depends on r² × h, so doubling the radius matters more than doubling the height.