Class 10th : MCQs


1. HCF of 24 and 36 / 24 āφ⧰⧁ 36 ā§° HCF


Q: Find the HCF of 24 and 36. 24 āφ⧰⧁ 36 ā§° āĻ¸ā§°ā§āĻŦāĻžāϧāĻŋāĻ• āϏāĻžāϧāĻžā§°āĻŖ āϗ⧁āĻŖāĻ¨ā§€ā§ŸāĻ• (HCF) āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: (a) 6 (b) 12 (c) 18 (d) 24


Ans / āωāĻ¤ā§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§āϝāĻžāĻ–ā§āϝāĻž:
24 ā§° āϗ⧁āĻŖāĻ¨ā§€ā§ŸāĻ•: 1, 2, 3, 4, 6, 8, 12, 24
36 ā§° āϗ⧁āĻŖāĻ¨ā§€ā§ŸāĻ•: 1, 2, 3, 4, 6, 9, 12, 18, 36
āĻ¸ā§°ā§āĻŦāĻžāϧāĻŋāĻ• āϏāĻžāϧāĻžā§°āĻŖ āϗ⧁āĻŖāĻ¨ā§€ā§ŸāĻ• = 12


Q2: Which of the following is a prime number ? āϤāϞ⧰ āϕ⧋āύāĻŸā§‹ āĻŽā§ŒāϞāĻŋāĻ• āϏāĻ‚āĻ–ā§āϝāĻž ?


Options: (a) 21 (b) 33 (c) 47 (d) 49


Ans / āωāĻ¤ā§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§āϝāĻžāĻ–ā§āϝāĻž:
21, 3 ⧰⧇ āĻŦāĻŋāĻ­āĻžāĻœā§āϝ; 33, 3 ⧰⧇ āĻŦāĻŋāĻ­āĻžāĻœā§āϝ; 49, 7 ⧰⧇ āĻŦāĻŋāĻ­āĻžāĻœā§āϝāĨ¤
47 ā§° āĻ•ā§‡ā§ąāϞ āĻĻ⧁āϟāĻž āϗ⧁āĻŖāĻ¨ā§€ā§ŸāĻ• (1 āφ⧰⧁ 47) āφāϛ⧇, āĻ¸ā§‡ā§Ÿā§‡ āχ āĻŽā§ŒāϞāĻŋāĻ•āĨ¤


Q3: Find the value of 3.45 × 100. 3.45 × 100 ā§° āĻŽāĻžāύ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: (a) 34.5 (b) 3.45 (c) 345 (d) 3450


Ans / āωāĻ¤ā§āϤ⧰: (c) 345


Explanation: Multiplying by 100 shifts the decimal two places to the right.


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: 100 ⧰⧇ āϗ⧁āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĻāĻļāĻŽāĻŋāĻ• āĻĻ⧁āϟāĻž āĻ¸ā§āĻĨāĻžāύ āϏ⧋āρāĻĢāĻžāϞ⧇ āϏ⧰āĻŋ āϝāĻžā§ŸāĨ¤


Q4: A triangle with all sides equal is called wha? āϏāĻ•āϞ⧋ āĻŦāĻžāĻšā§ āϏāĻŽāĻžāύ āĻĨāĻ•āĻž āĻ¤ā§ā§°āĻŋāϭ⧁āϜāĻ• āĻ•āĻŋ āĻ•ā§‹ā§ąāĻž āĻšā§Ÿ ?


Options: (a) Isosceles (b) Equilateral (c) Scalene (d) Right triangle


Ans / āωāĻ¤ā§āϤ⧰: (b) Equilateral


Explanation: An equilateral triangle has three equal sides.


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āϏāĻŽāĻŦāĻžāĻšā§ āĻ¤ā§ā§°āĻŋāϭ⧁āϜ⧰ āϤāĻŋāύāĻŋāĻ“āϟāĻž āĻŦāĻžāĻšā§ āϏāĻŽāĻžāύ āĻšā§ŸāĨ¤


Q5: Find the perimeter of a square with side 5 cm. 5 cm āĻŦāĻžāĻšā§ āĻĨāĻ•āĻž āĻāϟāĻž āĻŦā§°ā§āĻ—ā§° āĻĒā§°āĻŋāϏ⧀āĻŽāĻž āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: (a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm


Ans / āωāĻ¤ā§āϤ⧰: (c) 20 cm


Explanation: Perimeter = 4 × side = 4 × 5 = 20 cm


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻĒā§°āĻŋāϏ⧀āĻŽāĻž = 4 × āĻŦāĻžāĻšā§ = 4 × 5 = 20 cm


Q6: What is the Roman numeral for 49 ? 49 ā§° Roman āϏāĻ‚āĻ–ā§āϝāĻž āĻ•āĻŋ ?


Options: (a) XLIX (b) IL (c) XXXXIX (d) LIX


Ans / āωāĻ¤ā§āϤ⧰: (a) XLIX


Explanation:
40 = XL, 9 = IX
49 = XL + IX = XLIX


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
40 = XL, 9 = IX
49 = XL + IX = XLIX


Q7: Find the sum of 2/7 and 1/7. 2/7 āφ⧰⧁ 1/7 ā§° āϝ⧋āĻ—āĻĢāϞ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: (a) 1/7 (b) 2/7 (c) 3/7 (d) 4/7


Ans / āωāĻ¤ā§āϤ⧰: (c) 3/7


Explanation: Same denominator → Add numerators: 2 + 1 = 3. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻšā§° āĻāϕ⧇ → āĻ…āĻ‚āĻ• āĻĻ⧁āϟāĻž āϝ⧋āĻ— āϕ⧰āĻ•: 2 + 1 = 3āĨ¤


Q8: Find 15% of 200. 200 ā§° 15% āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: (a) 15 (b) 25 (c) 30 (d) 35


Ans / āωāĻ¤ā§āϤ⧰: (c) 30


Explanation:


15% = 15/100 


(15/100) × 200 = 30


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
15% = 15/100
(15/100) × 200 = 30


Q9: How many lines of symmetry does an equilateral triangle have ? āϏāĻŽāĻŦāĻžāĻšā§ āĻ¤ā§ā§°āĻŋāϭ⧁āϜāϤ āĻ•āĻŋāĻŽāĻžāύāϟāĻž āϏāĻŽāĻŽāĻŋāϤāĻŋ ⧰⧇āĻ–āĻž āĻĨāĻžāϕ⧇ ?


Options: (a) 1 (b) 2 (c) 3 (d) 4


Ans / āωāĻ¤ā§āϤ⧰: (c) 3


Explanation: Each vertex forms one symmetry line → Total 3. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻĒā§ā§°āϤāĻŋāĻŸā§‹ āĻļā§€ā§°ā§āώ⧰ āĻĒā§°āĻž āĻāϟāĻž āϏāĻŽāĻŽāĻŋāϤāĻŋ ⧰⧇āĻ–āĻž āĻšā§Ÿ → āĻŽā§āĻ  3āϟāĻžāĨ¤


Q10: Find the value of |−9| + |−5|. |−9| + |−5| ā§° āĻŽāĻžāύ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: (a) 14 (b) -14 (c) 4 (d) -4


Ans / āωāĻ¤ā§āϤ⧰: (a) 14


Explanation: Absolute value makes numbers positive. |−9| = 9, |−5| = 5 → 9 + 5 = 14


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: Absolute value āĻ āĻ‹āĻŖāĻžāĻ¤ā§āĻŽāĻ• āϏāĻ‚āĻ–ā§āϝāĻžāĻ• āϧāύāĻžāĻ¤ā§āĻŽāĻ• āϕ⧰⧇āĨ¤ |−9| = 9, |−5| = 5 → 9 + 5 = 14


Section B – Short Answer Questions


11. LCM of 12 and 18 / 12 āφ⧰⧁ 18 ā§° LCM


Q: Find the LCM of 12 and 18. 12 āφ⧰⧁ 18 ā§° LCM āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 36


Explanation / āĻŦā§āϝāĻžāĻ–ā§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 = āϏāĻ•āϞ⧋ prime ā§° āĻ¸ā§°ā§āĻŦā§‹āĻšā§āϚ āϘāĻžāϤ⧰ āϗ⧁āĻŖāĻĢāϞ
LCM = 2² × 3² = 36


12. Arrange in ascending order / āĻ•ā§āϰāĻŽāĻŦāĻ°ā§āϧāĻŽāĻžāύāϤ āϏāĻžāϜāĻžāĻ“āĻ•


Q: Arrange the numbers 0.03, 0.3, 0.33, 0.033 in ascending order.
0.03, 0.3, 0.33, 0.033 āϏāĻ‚āĻ–ā§āϝāĻžāĻŦā§‹ā§°āĻ• āϏ⧰⧁ āĻĒā§°āĻž āĻĄāĻžāϙ⧰āϞ⧈ āϏāĻžāϜāĻžāĻ“āĻ•āĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 0.03, 0.033, 0.3, 0.33


Explanation / āĻŦā§āϝāĻžāĻ–ā§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, 0.3300.03, 0.033, 0.3, 0.33


Decimal āĻšāĻŋāϚāĻžāĻĒ⧇: 0.03 = 0.030, 0.033 = 0.033, 0.3 = 0.300, 0.33 = 0.330
āϏ⧰⧁ āĻĒā§°āĻž āĻĄāĻžāϙ⧰: 0.03, 0.033, 0.3, 0.33


13. Simplify: − [15 − 8/4 × (4 − 2)] / āϏ⧰āϞ⧀āϕ⧰āĻŖ


Q: Simplify − [15 − 8/4 × (4 − 2)]
− [15 − 8/4 × (4 − 2)] āϏ⧰āϞ āϕ⧰āĻ•āĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: −11


Explanation / āĻŦā§āϝāĻžāĻ–ā§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 āĻ­āĻ—ā§āύāĻžāĻ‚āĻļāϞ⧈ ā§°ā§‚āĻĒāĻžāĻ¨ā§āϤ⧰


Q: Convert 5 1/7 to improper fraction. 5 1/7 āĻ• āĻ…āϏāĻžāϧāĻžā§°āĻŖ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļāϤ ā§°ā§‚āĻĒāĻžāĻ¨ā§āϤ⧰ āϕ⧰āĻ•āĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 36/7


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Mixed fraction = Whole × Denominator + Numerator
5 × 7 + 1 = 36 → 36/7


āĻŽāĻŋāĻļā§ā§° āĻ­āĻ—ā§āύāĻžāĻ‚āĻļ = āĻĒā§‚ā§°ā§āĻŖāϏāĻ‚āĻ–ā§āϝāĻž × āĻšā§° + āĻ…āĻ‚āĻ•
5 × 7 + 1 = 36 → 36/7


15. Area of rectangle / āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§°ā§° āĻ•ā§āώ⧇āϤāĻĢāϞ


Q. Find the area of a rectangle with length 8 cm and breadth 5 cm. āĻĻā§ˆā§°ā§āĻ˜ā§āϝ 8 cm āφ⧰⧁ āĻĒā§āϰāĻ¸ā§āĻĨ 5 cm āĻĨāĻ•āĻž āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§°ā§° āĻ•ā§āώ⧇āϤāĻĢāϞ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 40 cm²


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: Area = Length × Breadth = 8 × 5 = 40 cm². āĻ•ā§āώ⧇āϤāĻĢāϞ = āĻĻā§ˆā§°ā§āĻ˜ā§āϝ × āĻĒā§ā§°āĻ¸ā§āĻĨ = 8 × 5 = 40 cm²


Q16. Solve: 2x + 7 = 15 / āϏāĻŽā§€āϕ⧰āĻŖ


Soln


Solve 2x + 7 = 15.
2x + 7 = 15 āϏāĻŽāĻžāϧāĻžāύ āϕ⧰āĻ•āĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: x = 4


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
2x + 7 = 15 → 2x = 15 − 7 → 2x = 8 → x = 4


2x + 7 = 15 → 2x = 15 − 7 → 2x = 8 → x = 4