Class 10 Mathematics
Previous Board Examination – Full Solved Answers (Set 1)
The following solutions are written in clear student handwriting style, with proper steps as expected by CBSE examiners.
Section A – MCQs (Answers)
1. −3/2
2. 30°
3. 6
4. 5
5. 1
Section B – Very Short Answer
6. Find the zero of the polynomial x − 5.
Zero of polynomial = 5
7. Evaluate √144.
√144 = 12
8. Find the HCF of 15 and 25.
15 = 3 × 5
25 = 5 × 5
HCF = 5
9. Write the value of sin 90°.
sin 90° = 1
10. Find the distance between (1,2) and (1,5).
Distance = √[(1−1)² + (5−2)²]
= √[0 + 9]
= 3 units
Section C – Short Answer
11. Find the zeros of x² − 7x + 10.
x² − 7x + 10 = 0
(x − 5)(x − 2) = 0
x = 5, 2
12. Solve: 2x + 3 = 7.
2x = 7 − 3
2x = 4
x = 2
13. Find HCF of 36 and 84 using Euclid’s Division Algorithm.
84 = 36 × 2 + 12
36 = 12 × 3 + 0
HCF = 12
14. Prove: sin²A + cos²A = 1.
Using trigonometric identity,
sin²A + cos²A = 1
Hence proved.
15. Find the distance between (−2,3) and (4,3).
Distance = √[(4+2)² + (3−3)²]
= √[36]
= 6 units
16. Find the area of a circle of radius 7 cm.
Area = πr²
= 22/7 × 7 × 7
= 154 cm²
Section D – Long Answer
17. Solve x² − 5x + 6 = 0 by factorisation.
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or x = 3
18. Find the value of sin 30° / cos 60°.
sin 30° = 1/2
cos 60° = 1/2
Value = (1/2) ÷ (1/2) = 1
19. Find the midpoint of the line joining (2,4) and (6,8).
Midpoint = ((2+6)/2 , (4+8)/2)
= (4,6)
20. Find the area of a sector of radius 14 cm and angle 90°.
Area = (θ/360) × πr²
= (90/360) × 22/7 × 14 × 14
= 154 cm²
Section E – Case Study Based
21.
Length = 50 m, Breadth = 30 m
a) Perimeter = 2(l + b)
= 2(50 + 30) = 160 m
b) Area = l × b = 50 × 30 = 1500 m²
c) Fencing required = Perimeter = 160 m
Shaktimatha CBSE Learning
Class 10 Mathematics – Previous Board Exam Full Answers (Set 1)
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