Class 10th : MCQs 1
Class 10 Maths Practice Set : āĻĻāĻļāĻŽ āĻļā§ā§°ā§āĻŖā§ā§° āĻāĻŖāĻŋāĻ¤ āĻ āĻ¨ā§āĻļā§āĻ˛āĻ¨ā§ āĻā§āĻ
Section A : MCQs : (1 × 5 = 5 marks)
Q1. If the HCF of 306 and 657 is 9, find their LCM. :: 306 āĻā§°ā§ 657 ā§° HCF āĻ¯āĻĻāĻŋ 9 āĻšāĻ¯āĻŧ, āĻ¤ā§āĻ¨ā§āĻ¤ā§ LCM āĻāĻŋāĻŽāĻžāĻ¨ ?
Options: (a) 22338 (b) 20412 (c) 19836 (d) 21024
Ans: (a) 22338
Working / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
HCF × LCM = Product of numbers
LCM = (306 × 657) ÷ 9 = 22338
Q2. The zero of the linear polynomial 3x – 12 is :: 3x – 12 ā§° āĻļā§āĻ¨ā§āĻ¯ āĻŽāĻžāĻ¨ āĻā§āĻ¨āĻā§ ?
Options: (a) 2 (b) 4 (c) –4 (d) 6
Answer: (b) 4
Working / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
3x – 12 = 0
⇒ 3x = 12
⇒ x = 4
Section B : Very Short Answer
Q3. Find the value of k for which x² + kx + 9 = 0 has equal roots. x² + kx + 9 = 0 ā§° āĻ¸āĻŽāĻžāĻ¨ āĻŽā§āĻ˛ āĻš’āĻŦāĻ˛ā§ k ā§° āĻŽāĻžāĻ¨ āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻāĻāĨ¤
Options: (a) 6 (b) –6 (c) ±6 (d) 9
Ans: (c) ±6
Working / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
For equal roots,
b² − 4ac = 0
k² − 4(1)(9) = 0
k² − 36 = 0
⇒ k = ±6
Section B : Short Answer Questions
Q4. Find the LCM of 4, 5 and 220.
4, 5 āĻā§°ā§ 220 ā§° LCM āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻāĻāĨ¤
Ans / āĻāĻ¤ā§āĻ¤ā§°: 220
Working / āĻŦā§āĻ¯āĻžāĻā§āĻ¯āĻž:
4 = 2²
5 = 5
220 = 2² × 5 × 11
LCM = 2² × 5 × 11 = 220
Q5. Simplify: āĻ¸ā§°āĻ˛ā§āĻā§°āĻŖ āĻā§°āĻ: 102.5 − 45.08 − 57.42
Ans / āĻāĻ¤ā§āĻ¤ā§°: 0
Working / āĻŦā§āĻ¯āĻžāĻā§āĻ¯āĻž:
102.5 − 45.08 = 57.42
57.42 − 57.42 = 0
Q6. Find the smallest number which leaves remainder 5 when divided by 15 and 20. 15 āĻā§°ā§ 20 ā§°ā§ āĻāĻžāĻ āĻĻāĻŋāĻ˛ā§ 5 āĻ āĻŦāĻļāĻŋāĻˇā§āĻ āĻĨāĻāĻž āĻ¸ā§°ā§ āĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Ans / āĻāĻ¤ā§āĻ¤ā§°: 65
Working / āĻŦā§āĻ¯āĻžāĻā§āĻ¯āĻž:
LCM of 15 and 20 = 60
Required number = 60 + 5 = 65
Q7. Write the factors of 48. 48 ā§° āĻā§āĻŖāĻ¨ā§āĻ¯āĻŧāĻāĻ¸āĻŽā§āĻš āĻ˛āĻŋāĻāĻžāĨ¤
Ans / āĻāĻ¤ā§āĻ¤ā§°: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Q8. A wire bent from a rectangle of perimeter 124 cm to form a square. Find the side of the square. 124 cm āĻĒā§°āĻŋāĻ¸ā§āĻŽāĻž āĻĨāĻāĻž āĻāĻ¯āĻŧāĻ¤āĻā§āĻˇā§āĻ¤ā§ā§°ā§° āĻ¤āĻžā§°ā§ āĻŦā§°ā§āĻ āĻŦāĻ¨āĻžāĻ˛ā§, āĻŦā§°ā§āĻā§° āĻŦāĻžāĻšā§ āĻāĻŋāĻŽāĻžāĻ¨ ?
Ans / āĻāĻ¤ā§āĻ¤ā§°: 31 cm
Working / āĻŦā§āĻ¯āĻžāĻā§āĻ¯āĻž:
Perimeter of square = 124
Side = 124 ÷ 4 = 31 cm
Q9. Find the area of a square whose side is 31 cm. āĻŦāĻžāĻšā§ 31 cm āĻĨāĻāĻž āĻŦā§°ā§āĻā§° āĻā§āĻˇā§āĻ¤ā§ā§°āĻĢāĻ˛ āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻāĻāĨ¤
Ans / āĻāĻ¤ā§āĻ¤ā§°: 961 cm²
Working / āĻŦā§āĻ¯āĻžāĻā§āĻ¯āĻž:
Area = Side × Side
= 31 × 31
= 961 cm²
Q10: At 12 noon temperature is 12°C, decreasing 4°C per hour. Find temperature at 8 PM.
āĻĻā§āĻĒā§° 12 āĻŦāĻāĻžāĻ¤ āĻ¤āĻžāĻĒ = 12°C, āĻĒā§ā§°āĻ¤āĻŋāĻāĻŖā§āĻāĻž 4°C āĻāĻŽāĻŋāĻā§āĨ¤ ā§°āĻžāĻ¤āĻŋ 8 āĻŦāĻāĻžā§° āĻ¤āĻžāĻĒāĻŽāĻžāĻ¨ āĻŦāĻŋāĻāĻžā§°āĻžāĨ¤
Options: None
Ans / āĻāĻ¤ā§āĻ¤ā§°: −20°C
Explanation / āĻŦā§āĻ¯āĻžāĻā§āĻ¯āĻž:
Hours from 12 noon to 8 PM = 8 hours
ā§§ā§¨ āĻŦāĻāĻžā§° āĻĒā§°āĻž āĻ¸āĻ¨ā§āĻ§āĻŋāĻ¯āĻŧāĻž ā§Ž āĻŦāĻāĻžāĻ˛ā§ āĻ¸āĻŽā§ = ā§Ž āĻāĻŖā§āĻāĻž
Decrease per hour = 4°C
āĻĒā§ā§°āĻ¤āĻŋ āĻāĻŖā§āĻāĻžāĻ¤ āĻšā§ā§°āĻžāĻ¸ = ā§Ē°C
Total decrease = 8 × 4 = 32°C
āĻŽā§āĻ āĻšā§ā§°āĻžāĻ¸ = ā§Ž × ā§Ē = ā§Šā§¨°C
Temperature at 8 PM = 12 − 32 = −20°C
āĻ¸āĻ¨ā§āĻ§āĻŋāĻ¯āĻŧāĻž ā§Ž āĻŦāĻāĻžā§° āĻ¤āĻžāĻĒāĻŽāĻžāĻ¤ā§āĻ°āĻž = ā§§ā§¨ − ā§Šā§¨ = −ā§¨ā§Ļ°C
An: −20°C