Class 10th : MCQs
1. HCF of 24 and 36 / 24 āĻā§°ā§ 36 ā§° HCF
Q: Find the HCF of 24 and 36. 24 āĻā§°ā§ 36 ā§° āϏ⧰ā§āĻŦāĻžāϧāĻŋāĻ āϏāĻžāϧāĻžā§°āĻŖ āĻā§āĻŖāύā§ā§āĻ (HCF) āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: (a) 6 (b) 12 (c) 18 (d) 24
Ans / āĻāϤā§āϤ⧰: (b) 12
Explanation:
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Highest common factor = 12
āĻŦā§āϝāĻžāĻā§āϝāĻž:
24 ā§° āĻā§āĻŖāύā§ā§āĻ: 1, 2, 3, 4, 6, 8, 12, 24
36 ā§° āĻā§āĻŖāύā§ā§āĻ: 1, 2, 3, 4, 6, 9, 12, 18, 36
āϏ⧰ā§āĻŦāĻžāϧāĻŋāĻ āϏāĻžāϧāĻžā§°āĻŖ āĻā§āĻŖāύā§ā§āĻ = 12
Q2: Which of the following is a prime number ? āϤāϞ⧰ āĻā§āύāĻā§ āĻŽā§āϞāĻŋāĻ āϏāĻāĻā§āϝāĻž ?
Options: (a) 21 (b) 33 (c) 47 (d) 49
Ans / āĻāϤā§āϤ⧰: (c) 47
Explanation:
21 is divisible by 3, 33 by 3, and 49 by 7.
47 has only two factors (1 and 47), so it is prime.
āĻŦā§āϝāĻžāĻā§āϝāĻž:
21, 3 ā§°ā§ āĻŦāĻŋāĻāĻžāĻā§āϝ; 33, 3 ā§°ā§ āĻŦāĻŋāĻāĻžāĻā§āϝ; 49, 7 ā§°ā§ āĻŦāĻŋāĻāĻžāĻā§āϝāĨ¤
47 ā§° āĻā§ā§ąāϞ āĻĻā§āĻāĻž āĻā§āĻŖāύā§ā§āĻ (1 āĻā§°ā§ 47) āĻāĻā§, āϏā§ā§ā§ āĻ āĻŽā§āϞāĻŋāĻāĨ¤
Q3: Find the value of 3.45 × 100. 3.45 × 100 ā§° āĻŽāĻžāύ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: (a) 34.5 (b) 3.45 (c) 345 (d) 3450
Ans / āĻāϤā§āϤ⧰: (c) 345
Explanation: Multiplying by 100 shifts the decimal two places to the right.
āĻŦā§āϝāĻžāĻā§āϝāĻž: 100 ā§°ā§ āĻā§āĻŖ āĻā§°āĻŋāϞ⧠āĻĻāĻļāĻŽāĻŋāĻ āĻĻā§āĻāĻž āϏā§āĻĨāĻžāύ āϏā§āĻāĻĢāĻžāϞ⧠āϏ⧰āĻŋ āϝāĻžā§āĨ¤
Q4: A triangle with all sides equal is called wha? āϏāĻāϞ⧠āĻŦāĻžāĻšā§ āϏāĻŽāĻžāύ āĻĨāĻāĻž āϤā§ā§°āĻŋāĻā§āĻāĻ āĻāĻŋ āĻā§ā§ąāĻž āĻšā§ ?
Options: (a) Isosceles (b) Equilateral (c) Scalene (d) Right triangle
Ans / āĻāϤā§āϤ⧰: (b) Equilateral
Explanation: An equilateral triangle has three equal sides.
āĻŦā§āϝāĻžāĻā§āϝāĻž: āϏāĻŽāĻŦāĻžāĻšā§ āϤā§ā§°āĻŋāĻā§āĻā§° āϤāĻŋāύāĻŋāĻāĻāĻž āĻŦāĻžāĻšā§ āϏāĻŽāĻžāύ āĻšā§āĨ¤
Q5: Find the perimeter of a square with side 5 cm. 5 cm āĻŦāĻžāĻšā§ āĻĨāĻāĻž āĻāĻāĻž āĻŦā§°ā§āĻā§° āĻĒā§°āĻŋāϏā§āĻŽāĻž āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: (a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm
Ans / āĻāϤā§āϤ⧰: (c) 20 cm
Explanation: Perimeter = 4 × side = 4 × 5 = 20 cm
āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻĒā§°āĻŋāϏā§āĻŽāĻž = 4 × āĻŦāĻžāĻšā§ = 4 × 5 = 20 cm
Q6: What is the Roman numeral for 49 ? 49 ā§° Roman āϏāĻāĻā§āϝāĻž āĻāĻŋ ?
Options: (a) XLIX (b) IL (c) XXXXIX (d) LIX
Ans / āĻāϤā§āϤ⧰: (a) XLIX
Explanation:
40 = XL, 9 = IX
49 = XL + IX = XLIX
āĻŦā§āϝāĻžāĻā§āϝāĻž:
40 = XL, 9 = IX
49 = XL + IX = XLIX
Q7: Find the sum of 2/7 and 1/7. 2/7 āĻā§°ā§ 1/7 ā§° āϝā§āĻāĻĢāϞ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: (a) 1/7 (b) 2/7 (c) 3/7 (d) 4/7
Ans / āĻāϤā§āϤ⧰: (c) 3/7
Explanation: Same denominator → Add numerators: 2 + 1 = 3. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻšā§° āĻāĻā§ → āĻ āĻāĻ āĻĻā§āĻāĻž āϝā§āĻ āĻā§°āĻ: 2 + 1 = 3āĨ¤
Q8: Find 15% of 200. 200 ā§° 15% āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: (a) 15 (b) 25 (c) 30 (d) 35
Ans / āĻāϤā§āϤ⧰: (c) 30
Explanation:
15% = 15/100
(15/100) × 200 = 30
āĻŦā§āϝāĻžāĻā§āϝāĻž:
15% = 15/100
(15/100) × 200 = 30
Q9: How many lines of symmetry does an equilateral triangle have ? āϏāĻŽāĻŦāĻžāĻšā§ āϤā§ā§°āĻŋāĻā§āĻāϤ āĻāĻŋāĻŽāĻžāύāĻāĻž āϏāĻŽāĻŽāĻŋāϤāĻŋ ā§°ā§āĻāĻž āĻĨāĻžāĻā§ ?
Options: (a) 1 (b) 2 (c) 3 (d) 4
Ans / āĻāϤā§āϤ⧰: (c) 3
Explanation: Each vertex forms one symmetry line → Total 3. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻĒā§ā§°āϤāĻŋāĻā§ āĻļā§ā§°ā§āώ⧰ āĻĒā§°āĻž āĻāĻāĻž āϏāĻŽāĻŽāĻŋāϤāĻŋ ā§°ā§āĻāĻž āĻšā§ → āĻŽā§āĻ 3āĻāĻžāĨ¤
Q10: Find the value of |−9| + |−5|. |−9| + |−5| ā§° āĻŽāĻžāύ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: (a) 14 (b) -14 (c) 4 (d) -4
Ans / āĻāϤā§āϤ⧰: (a) 14
Explanation: Absolute value makes numbers positive. |−9| = 9, |−5| = 5 → 9 + 5 = 14
āĻŦā§āϝāĻžāĻā§āϝāĻž: Absolute value āĻ āĻāĻŖāĻžāϤā§āĻŽāĻ āϏāĻāĻā§āϝāĻžāĻ āϧāύāĻžāϤā§āĻŽāĻ āĻā§°ā§āĨ¤ |−9| = 9, |−5| = 5 → 9 + 5 = 14
Section B – Short Answer Questions
11. LCM of 12 and 18 / 12 āĻā§°ā§ 18 ā§° LCM
Q: Find the LCM of 12 and 18. 12 āĻā§°ā§ 18 ā§° LCM āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 36
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Prime factors:
12 = 2² × 3
18 = 2 × 3²
LCM = Product of highest powers of all prime factors = 2² × 3² = 36
Prime factor āϞ⧠LCM = āϏāĻāϞ⧠prime ā§° āϏ⧰ā§āĻŦā§āĻā§āĻ āĻāĻžāϤ⧰ āĻā§āĻŖāĻĢāϞ
LCM = 2² × 3² = 36
12. Arrange in ascending order / āĻā§āϰāĻŽāĻŦāϰā§āϧāĻŽāĻžāύāϤ āϏāĻžāĻāĻžāĻāĻ
Q: Arrange the numbers 0.03, 0.3, 0.33, 0.033 in ascending order.
0.03, 0.3, 0.33, 0.033 āϏāĻāĻā§āϝāĻžāĻŦā§ā§°āĻ āϏ⧰⧠āĻĒā§°āĻž āĻĄāĻžāĻā§°āϞ⧠āϏāĻžāĻāĻžāĻāĻāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 0.03, 0.033, 0.3, 0.33
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
As decimals: 0.03 → 0.030, 0.033 → 0.033, 0.3 → 0.300, 0.33 → 0.330
Ascending order: 0.030, 0.033, 0.300, 0.330 → 0.03, 0.033, 0.3, 0.33
Decimal āĻšāĻŋāĻāĻžāĻĒā§: 0.03 = 0.030, 0.033 = 0.033, 0.3 = 0.300, 0.33 = 0.330
āϏ⧰⧠āĻĒā§°āĻž āĻĄāĻžāĻā§°: 0.03, 0.033, 0.3, 0.33
13. Simplify: − [15 − 8/4 × (4 − 2)] / āϏ⧰āϞā§āĻā§°āĻŖ
Q: Simplify − [15 − 8/4 × (4 − 2)]
− [15 − 8/4 × (4 − 2)] āϏ⧰āϞ āĻā§°āĻāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: −11
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
1st: 4 − 2 = 2 → − [15 − 8/4 × 2]
2nd: 8 ÷ 4 = 2 → − [15 − 2 × 2]
3rd: 2 × 2 = 4 → − [15 − 4]
4th: 15 − 4 = 11 → −11
1st: 4 − 2 = 2 → − [15 − 8/4 × 2]
2nd: 8 ÷ 4 = 2 → − [15 − 2 × 2]
3rd: 2 × 2 = 4 → − [15 − 4]
4th: 15 − 4 = 11 → −11
14. Convert to improper fraction / āĻ āϏāĻžāϧāĻžā§°āĻŖ āĻāĻā§āύāĻžāĻāĻļāϞ⧠⧰ā§āĻĒāĻžāύā§āϤ⧰
Q: Convert 5 1/7 to improper fraction. 5 1/7 āĻ āĻ āϏāĻžāϧāĻžā§°āĻŖ āĻāĻā§āύāĻžāĻāĻļāϤ ā§°ā§āĻĒāĻžāύā§āϤ⧰ āĻā§°āĻāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 36/7
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Mixed fraction = Whole × Denominator + Numerator
5 × 7 + 1 = 36 → 36/7
āĻŽāĻŋāĻļā§ā§° āĻāĻā§āύāĻžāĻāĻļ = āĻĒā§ā§°ā§āĻŖāϏāĻāĻā§āϝāĻž × āĻšā§° + āĻ
āĻāĻ
5 × 7 + 1 = 36 → 36/7
15. Area of rectangle / āĻāϝāĻŧāϤāĻā§āώā§āϤā§ā§°ā§° āĻā§āώā§āϤāĻĢāϞ
Q. Find the area of a rectangle with length 8 cm and breadth 5 cm. āĻĻā§ā§°ā§āĻā§āϝ 8 cm āĻā§°ā§ āĻĒā§āϰāϏā§āĻĨ 5 cm āĻĨāĻāĻž āĻāϝāĻŧāϤāĻā§āώā§āϤā§ā§°ā§° āĻā§āώā§āϤāĻĢāϞ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 40 cm²
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž: Area = Length × Breadth = 8 × 5 = 40 cm². āĻā§āώā§āϤāĻĢāϞ = āĻĻā§ā§°ā§āĻā§āϝ × āĻĒā§ā§°āϏā§āĻĨ = 8 × 5 = 40 cm²
Q16. Solve: 2x + 7 = 15 / āϏāĻŽā§āĻā§°āĻŖ
Soln
Solve 2x + 7 = 15.
2x + 7 = 15 āϏāĻŽāĻžāϧāĻžāύ āĻā§°āĻāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: x = 4
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
2x + 7 = 15 → 2x = 15 − 7 → 2x = 8 → x = 4
2x + 7 = 15 → 2x = 15 − 7 → 2x = 8 → x = 4