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 / āĻāϤā§āϤ⧰: 220
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Prime factors: 4 = 2², 5 = 5, 220 = 2² × 5 × 11
LCM = Highest powers of all primes = 2² × 5 × 11 = 220
Prime factor āĻšāĻŋāĻāĻžāĻĒā§: 4 = 2², 5 = 5, 220 = 2² × 5 × 11
LCM = āϏāĻāϞ⧠prime ā§° āϏ⧰ā§āĻŦā§āĻā§āĻ āĻāĻžāϤ = 2² × 5 × 11 = 220
Q23(b). Simplify: 102.5 − 45.08 − 57.42 / 102.5 − 45.08 − 57.42 āϏ⧰āϞā§āĻā§°āĻŖ
Options: None
Ans / āĻāϤā§āϤ⧰: 0
Explanation / āĻŦā§āϝāĻžāĻā§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 āĻā§°ā§ 20 ā§°ā§ āĻāĻžāĻ āĻĻāĻŋāϞ⧠5 āĻ
āĻŦāĻļāĻŋāώā§āĻ āĻĨāĻāĻž āϏ⧰⧠āϏāĻāĻā§āϝāĻžāĻā§ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 65
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
LCM of 15 and 20 = 60
Smallest number leaving remainder 5 = 60 + 5 = 65
15 āĻā§°ā§ 20 ā§° LCM = 60
5 āĻ
āĻŦāĻļāĻŋāώā§āĻā§° āϏā§āϤ⧠āϏ⧰⧠āϏāĻāĻā§āϝāĻž = 60 + 5 = 65
Q24(b). Factors of 48 / 48 ā§° factor
Options: None
Ans / āĻāϤā§āϤ⧰: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Factors are numbers that divide 48 exactly: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factor āĻŽāĻžāύ⧠48-āĻā§ āύāĻŋāĻā§āĻāϤāĻāĻžā§ąā§ āĻāĻžāĻ āĻā§°āĻž āϏāĻāĻā§āϝāĻž
Q25(a): Wire bent from rectangle of perimeter 124 cm to square. Find side of square.
124 cm perimeter āĻĨāĻāĻž āĻāϝāĻŧāϤāĻā§āώā§āϤā§ā§°ā§° āϤāĻžā§°ā§ āĻŦā§°ā§āĻ āĻŦāĻžāύāĻžāϞā§, āĻŦā§°ā§āĻā§° āĻŦāĻžāĻšā§ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 31 cm
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Perimeter of square = Perimeter of rectangle = 124
Side of square = 124 ÷ 4 = 31 cm
āĻŦā§°ā§āĻā§° perimeter = āĻāϝāĻŧāϤāĻā§āώā§āϤā§ā§°ā§° perimeter = 124
āĻŦāĻžāĻšā§ = 124 ÷ 4 = 31 cm
Q25(b): Find area of square with side 31 cm.
āĻŦāĻžāĻšā§ 31 cm āĻĨāĻāĻž āĻŦā§°ā§āĻāĻā§āώā§āϤā§ā§°ā§° āĻā§āώā§āϤāĻĢāϞ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: 961 cm²
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Area = Side × Side = 31 × 31 = 961 cm²
āĻā§āώā§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.
āĻĻā§āĻĒā§° 12 āĻŦāĻāĻžāϤ āϤāĻžāĻĒ = 12°C, āĻĒā§ā§°āϤāĻŋāĻāĻŖā§āĻāĻž 4°C āĻāĻŽāĻŋāĻā§āĨ¤ ā§°āĻžāϤāĻŋ 8 āĻŦāĻāĻžā§° āϤāĻžāĻĒāĻŽāĻžāύ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: None
Ans / āĻāϤā§āϤ⧰: −20°C
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
Hours from 12 noon to 8 PM = 8
Decrease = 8 × 4 = 32
Temperature = 12 − 32 = −20°C
12 āĻŦāĻāĻžāϤ 8 āĻāĻŖā§āĻāĻž → 8 × 4 = 32
āϤāĻžāĻĒ = 12 − 32 = −20°C