Time: 2½ Hrs
Class: X MATHEMATICS Total Score: 80
| Q.No | POSSIBLE ANSWERS | SCORE |
|---|---|---|
| 1 | D. -5 | 1 |
| 2 | C. (4, 7) | 1 |
| 3 | B. 2√3 | 1 |
| 4 | C. 70 | 1 |
| 5 | D. 4 | 1 |
| 6 | C. (ii) and (iv) | 1 |
| 7 | C) Both the statements are true, Statement II is the reason of Statement I. | 1 |
| 8 | C) Both the statements are true, Statement II is the reason of Statement I. | 1 |
| 9 |
i) 12/20 = 3/5 ii) (12+x)/(20+x) = 2/3 36 + 3x = 40 + 2x x = 4 4 white beads |
3 |
| 10 |
i) x₅ = 19 ii) x₁₇ = x₁₁ + 6d = 27 + 2 × 4 = 35 iii) 12 × (x₁₇ + x₈) 12 × (35 + 23) 12 × 58 = 696 |
4 |
| 11 |
i) B (6, 1) D (2, 4) ii) Diagonal - 5 |
3 |
| 12 |
A (5, 7) B (3, -1) C (7, 5) |
3 |
| 13 |
A. i) (5, 0) (-5, 0) ii) No, √(32² + 22²) = √13 not equal to radius iii) (3, 4) OR B. i) A(2, 2√3) B(-2, -2√3) |
4 |
| ii) 8 | |
| i) | 5 |
| ii) 5 × 4 = 20 | |
| i) (30 × 31)/2 = 465 | 3 |
| ii) 2 × 465 + 1 × 30 = 960 | |
|
A. i) 4n + 2 ii) 2n² + 4n = 880 n² + 2n = 440 n² + 2n + 1 = 441 n = 20 B. OR i) x(30 - x) = 216 ii) 30x - x² = 216 x² - 30x = 216 x² - 30x + 15² = -216 + 15² (x - 15) = √9 = 3 x = 3 + 15 = 18 18 and 12 |
4 |
|
i) (x - 12)(x + 3) ii) x = 12 x = -3 |
4 |
|
i) ½x(x + 7) = 60 x² + 7x = 60 × 2 x² + 7x = 120 |
5 |
ii) x⁷ + 7x - 120 = 0
x = (-7 ± √(49 - x1 × 710))/(2 × 1)
= (-7 ± √529)/2
= (-7 ± 23)/2
= (-7 ± 23)/2, (-7-13)/2
= 16/2
= 8
19 A. Let O is the centre of circle. A, B are points where tangents meet and P is intersection of tangents.
BO = AO (radius)
PO (Common side)
PB = √(PO² - BO²)
= √(PO² - AO²) = PA
B. In the figure ABC is an equilateral triangle. Let O is the centre and D is the point where in radius touch the chord AB
∠ODA = 90° (radius tangent)
∠DAO = 30° (angle bisector)
If OD = r then AO = 2 × r
20 r² - d² = 4 × 3
r² - 2² = 12
r² = 12 + 4
r = 4
21 A.
i) sin 40 = OA/5
5 × sin 40 = OA
5 × 64 = OA
3.2 = OA
Cos 40 = OB/5
Cos 40 × 5 = OB
.76 × 5 = OB
3.8 = OB
ii) AC = 6.4
BC = 7.6
B. OR
i) d = 9/sin 60 = 9 + √3/2
= 9 × 2/√3 = 6√3
r = 3√3
22
i) ps = 4
ii) ½ × (4√3 + 4) × 9
= (8√3 + 8)cm²
23
tan 35 = h/(10+x)
(10 + x).7 = h
tan 60 = h/x
x tan 60 = h
x × 1.7 = h
(10+x).7 = 1.7x
7 + .7x = 1.7x
7 = 1.7x - .7
7 = 1x
Height of flag post
= 1.7 × 7 + 1.6
= 13.5 Meter
35 60 10 1.6
B. OR
ii) tan 40 = h/x
x tan 40 = h
x × .84 = h
tan 50 = h/(10-x)
(10 - x) tan 50 = h
(10 - x) 1.2 = h
12 - 1.2x = .84x
12 = 2.04x
5.88 = x
h = x × .84
= 5.88 × .84
= 4.9
Height of the electric post 4.9 + 1.5 = 6.4 meter
h
|
|
|
40 x
50
|
|
|
10
24 i) CR = 12-6 = 6
AR = 12-10 = 2
AB = 12-8 = 4
ii) Consider △APOR
RO = OP (Radius)
∠R = ∠P = 90 (Angles between radius and tangents)
∴ AP = AR (Tangent from same point)
∴ APOR is a rectangle with adjacent sides are equal.
∴ It's a Square
R
|
O
|
P
i) 60
ii) 80
iii) ∠P = 95
∠S = 110
∠Q = 85
∠R = 70
25
7 cm 2.5
26
4.5 6 7
27 i) √24
ii)