Class 10th : MCQs 1


Section B – Short Answer Questions


Q. 23(a). LCM of 4, 5, 220 / 4, 5, 220 ā§° LCM


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 220


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Prime factors: 4 = 2², 5 = 5, 220 = 2² × 5 × 11
LCM = Highest powers of all primes = 2² × 5 × 11 = 220


Prime factor āĻšāĻŋāϚāĻžāĻĒ⧇: 4 = 2², 5 = 5, 220 = 2² × 5 × 11
LCM = āϏāĻ•āϞ⧋ prime ā§° āĻ¸ā§°ā§āĻŦā§‹āĻšā§āϚ āϘāĻžāϤ = 2² × 5 × 11 = 220


Q23(b). Simplify: 102.5 − 45.08 − 57.42 / 102.5 − 45.08 − 57.42 āϏ⧰āϞ⧀āϕ⧰āĻŖ


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 0


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
1st: 102.5 − 45.08 = 57.42
2nd: 57.42 − 57.42 = 0


āϧāĻžāĻĒ 1: 102.5 − 45.08 = 57.42
āϧāĻžāĻĒ 2: 57.42 − 57.42 = 0


Q24(a): Find the smallest number which leaves remainder 5 when divided by 15 and 20.
15 āφ⧰⧁ 20 ⧰⧇ āĻ­āĻžāĻ— āĻĻāĻŋāϞ⧇ 5 āĻ…āĻŦāĻļāĻŋāĻˇā§āϟ āĻĨāĻ•āĻž āϏ⧰⧁ āϏāĻ‚āĻ–ā§āϝāĻžāĻŸā§‹ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 65


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
LCM of 15 and 20 = 60
Smallest number leaving remainder 5 = 60 + 5 = 65


15 āφ⧰⧁ 20 ā§° LCM = 60
5 āĻ…āĻŦāĻļāĻŋāĻˇā§āϟ⧰ āϏ⧈āϤ⧇ āϏ⧰⧁ āϏāĻ‚āĻ–ā§āϝāĻž = 60 + 5 = 65


Q24(b). Factors of 48 / 48 ā§° factor


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Factors are numbers that divide 48 exactly: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factor āĻŽāĻžāύ⧇ 48-āϕ⧇ āύāĻŋāϖ⧁āρāϤāĻ­āĻžā§ąā§‡ āĻ­āĻžāĻ— āϕ⧰āĻž āϏāĻ‚āĻ–ā§āϝāĻž


Q25(a): Wire bent from rectangle of perimeter 124 cm to square. Find side of square.
124 cm perimeter āĻĨāĻ•āĻž āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§°ā§° āϤāĻžā§°ā§‡ āĻŦā§°ā§āĻ— āĻŦāĻžāύāĻžāϞ⧇, āĻŦā§°ā§āĻ—ā§° āĻŦāĻžāĻšā§ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 31 cm


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Perimeter of square = Perimeter of rectangle = 124
Side of square = 124 ÷ 4 = 31 cm


āĻŦā§°ā§āĻ—ā§° perimeter = āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§°ā§° perimeter = 124
āĻŦāĻžāĻšā§ = 124 ÷ 4 = 31 cm


Q25(b): Find area of square with side 31 cm.
āĻŦāĻžāĻšā§ 31 cm āĻĨāĻ•āĻž āĻŦā§°ā§āĻ—āĻ•ā§āώ⧇āĻ¤ā§ā§°ā§° āĻ•ā§āώ⧇āϤāĻĢāϞ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: 961 cm²


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Area = Side × Side = 31 × 31 = 961 cm²


āĻ•ā§āώ⧇āϤāĻĢāϞ = āĻŦāĻžāĻšā§ × āĻŦāĻžāĻšā§ = 31 × 31 = 961 cm²


Q26(a): At 12 noon temperature is 12°C, decreasing 4°C per hour. Find temperature at 8 PM.
āĻĻ⧁āĻĒā§° 12 āĻŦāϜāĻžāϤ āϤāĻžāĻĒ = 12°C, āĻĒā§ā§°āϤāĻŋāϘāĻŖā§āϟāĻž 4°C āĻ•āĻŽāĻŋāϛ⧇āĨ¤ ā§°āĻžāϤāĻŋ 8 āĻŦāϜāĻžā§° āϤāĻžāĻĒāĻŽāĻžāύ āĻŦāĻŋāϚāĻžā§°āĻžāĨ¤


Options: None


Ans / āωāĻ¤ā§āϤ⧰: −20°C


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Hours from 12 noon to 8 PM = 8
Decrease = 8 × 4 = 32
Temperature = 12 − 32 = −20°C


12 āĻŦāϜāĻžāϤ 8 āϘāĻŖā§āϟāĻž → 8 × 4 = 32
āϤāĻžāĻĒ = 12 − 32 = −20°C