Karnataka SSLC Maths Model Question Paper 3 Kannada Medium

ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80

I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)

Question 1.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತ 16 : 25 ಆದರೆ ಅವುಗಳ ಬಾಹುಗಳ ಅನುಪಾತ
(A) 3 : 4
(B) 4 : 5
(C) 5 : 6
(D) 6 : 7

Question 2.
7 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತದಲ್ಲಿ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ
(A) 25.67 cm²
(B) 35.32 cm²
(C) 15.25 cm²
(D) 77 cm²

Question 3.
ಮೂಲಬಿಂದು ಮತ್ತು (6, 8) ಬಿಂದುವಿನ ನಡುವಿನ ದೂರ
(A) 6 ಮೂ.ಮಾ.
(B) 8 ಮೂ.ಮಾ.
(C) 10 ಮೂ.ಮಾ.
(D) 14 ಮೂ.ಮಾ.

Question 4.
(15, 20) ರ ಮ.ಸಾ.ಅ. 5 ಆದರೆ ಅವುಗಳ ಲ.ಸಾ.ಅ.
(A) 15
(B) 20
(C) 40
(D) 60

Question 5.
\(\frac { { sin18 }^{ 0 } }{ { cos72 }^{ 0 } }\) ಯ ಬೆಲೆ
(A) 1
(B) 0
(C) -1
(D) 2

Question 6.
x² – 25 = 0 ಈ ಸಮೀಕರಣದ ಮೂಲಗಳು
(A) (+5, -5)
(B) (+5, +5)
(C) (-5, -5)
(D) (25, -25)

Question 7.
ಒಂದು ನಿರ್ದಿಷ್ಟ ದಿನದಲ್ಲಿ ಮಳೆ ಬೀಳುವ ಸಂಭವನೀಯತೆಯು 0.64 ಆಗಿದೆ. ಅದೇ ದಿನ ಮಳೆ ಬೀಳದಿರುವ ಸಂಭವನೀಯತೆ.
(A) -0.64
(B) 64
(C) 0.36
(D) -0.36

Question 8.
ಶಂಕುವಿನ ಭಿನ್ನಕದ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ

II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)

Question 9.
ಪೈಥಾಗೊರಸ್ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,

Question 10.
ಒಂದು ಸಂಖ್ಯೆಯನ್ನು 14 ರಿಂದ ಭಾಗಿಸಿದಾಗ ಶೇಷ 5 ಆದರೆ 7ರಿಂದ ಅದೇ ಸಂಖ್ಯೆಯನ್ನು ಭಾಗಿಸಿದಾಗ ದೊರೆಯುವ ಶೇಷವೆಷ್ಟು?

Question 11.
α ಮತ್ತು β ಗಳು ax² + bx + c ಎಂಬ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳಾದರೆ αβ ದ ಬೆಲೆಯೇನು?

Question 12.
‘O’ ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ PA ಮತ್ತು PB ಗಳು ಬಾಹ್ಯ ಬಿಂದು P ನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಮತ್ತು ∠APB = 60° ಹಾಗು AP = 8 cm ಆದಾಗ AB ಜ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 13.
(1 + sin²90°)² ನ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 14.
12, 16, 20, 24, 28 ರ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)

Question 15.
a = 5, d = 3, a_n = 50 ಆದರೆ S_n ಕಂಡುಹಿಡಿಯಿರಿ.

Question 16.
ಚಿತ್ರದಲ್ಲಿ ∆ABCಯ ಎತ್ತರಗಳಾದ AD ಮತ್ತು CE ಗಳು ಪರಸ್ಪರ P ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ∆AEP ~ ∆CDP ಎಂದು ಸಾಧಿಸಿ.

ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ 0ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.

Question 17.

Question 18.
5x – 3y = 11, -10x + 6y = -22 ಈ ಸಮೀಕರಣಗಳು ಸ್ಥಿರವಾಗಿವೆಯೆ? ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೆ? ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 19.
ಚಿತ್ರದಲ್ಲಿ ಕೇಂದ್ರ O ಇರುವ ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 7 cm ಮತ್ತು 14 cm ಇವೆ. ∠AOC = 40° ಆದರೆ ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 20.
4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕಗಳ ನಡುವಿನ ಕೋನ 70° ಇರುವಂತೆ ಒಂದು ಜೊತೆ ಸ್ಪರ್ಶಕಗಳನ್ನು ಎಳೆಯಿರಿ.

Question 21.
AB ವ್ಯಾಸವಾಗಿರುವ ವೃತ್ತದ ಕೇಂದ್ರ (2, -3) ಮತ್ತು B ಯು (1, 4) ಆದರೆ A ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 22.
(8, 1), (k, -4), (2, -5) ಎಂಬ ಬಿಂದುಗಳು ಸರಳರೇಖಾಗತವಾಗಿದ್ದರೆ k ಯ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗು ಗುಣಲಬ್ದಗಳು ಕ್ರಮವಾಗಿ 4 ಮತ್ತು 1 ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
2x² + 3x – 5 = 0 ಈ ವರ್ಗ ಸಮೀಕರಣವನ್ನು ಸೂತ್ರದಿಂದ ಬಿಡಿಸಿ,

Question 25.
20 m ಎತ್ತರದ ಕಟ್ಟಡವೊಂದರ ಮೇಲೆ ಸ್ಥಾಪಿಸಲಾದ ಪ್ರಸರಣೆಯ ಗೋಪುರವೊಂದರ ಮೇಲುದಿ ಮತ್ತು ಪಾದಗಳನ್ನು ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ನೋಡಿದಾಗ ಉನ್ನತ ಕೋನಗಳು ಕ್ರಮವಾಗಿ 60° ಮತ್ತು 45° ಇದೆ. ಪ್ರಸರಣೆಯ ಗೋಪುರದ ಎತ್ತರ ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
50√3 m ಎತ್ತರದಲ್ಲಿರುವ ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ನೆಲದ ಮೇಲಿರುವ ಒಂದು ಕಾರನ್ನು ನೋಡಿದಾಗ ಉಂಟಾದ ಅವನತ ಕೋನವು 60° ಆಗಿರುತ್ತದೆ. ಹಾಗಾದರೆ ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 27.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಮಧ್ಯಾಂಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 28.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. (i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ (ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 29.
ಸಮನಾದ ತ್ರಿಜ್ಯವುಳ್ಳ ಒಂದು ಶಂಕುವನ್ನು ಒಂದು ಅರ್ಧಗೋಳಾಕೃತಿಯ ಮೇಲೆ ಜೋಡಿಸಿ ಒಂದು ಆಟಿಕೆಯನ್ನು ಮಾಡಲಾಗಿದೆ. ಶಂಕುವಿನ ಭಾಗದ ವ್ಯಾಸವು 6 cm ಮತ್ತು 4 cm ಎತ್ತರ ಇದ್ದರೆ ಈ ಘನವಸ್ತುವಿನ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಶಂಕುವಿನ ಭಿನ್ನಕದ ಓರೆ ಎತ್ತರ 10 cm, ಅದರ ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 8 cm ಮತ್ತು 6 cm ಆಗಿದೆ. ಆ ಭಿನ್ನಕದ ವಕ್ರಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.

IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)

Question 31.
ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮನಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು 5 cm ಮತ್ತು 3 cm ಆಗಿವೆ. ಚಿಕ್ಕ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಿಸುವಂತೆ ದೊಡ್ಡ ವೃತ್ತದ ಹ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 32.
BC = 7 cm, ∠A = 45°, ∠B = 105° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು, ∆ABCಯ ಅನುರೂಪ ಬಾಹುಗಳ \(\frac { 3 }{ 4 }\) ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,

Question 33.
3 ವರ್ಷಗಳ ಹಿಂದಿನ ರೆಹಮಾನನ ವಯಸ್ಸು ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವ್ಯತ್ಯಮಗಳ ಮೊತ್ತ \(\frac { 1 }{ 3 }\) ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ರೈಲು 360km ದೂರವನ್ನು ಏಕರೂಪ ಜವದೊಂದಿಗೆ ಕ್ರಮಿಸುತ್ತದೆ. ಅದರ ಜವವು 5km/hr ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಅಷ್ಟೇ ದೂರವನ್ನು ಕ್ರಮಿಸಲು ಅದು 1 ಗಂಟೆ ಕಡಿಮೆ ತೆಗೆದುಕೊಳ್ಳುತ್ತಿತ್ತು. ರೈಲಿನ ಜವವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.
(cos A + sec A)² + (sin A + cosec A)² = 7 + tan²A + cot²A ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
A, B ಮತ್ತು C ಗಳು ∆ABCಯ ಒಳಕೋನಗಳಾದರೆ sin(\(\frac { B+C }{ 2 }\)) = cos(\(\frac { A }{ 2 }\)) ಎಂದು ಸಾಧಿಸಿ.

Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ,

(ಅಧಿಕ ಇರುವ ವಿಧಾನ)

Question 36.
4u² – 8u ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿದು ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆನೋಡಿ.
ಅಥವಾ
p(x) = x⁵ – 4x³ + x² + 3x + 1 ನ್ನು g(x) = x³ – 3x + 1 ರಿಂದ ಭಾಗಿಸಿ g(x), p(x) ನ ಅಪವರ್ತನವನ್ನು ಪರೀಕ್ಷಿಸಿ.

V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)

Question 37.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ 4ನೇ ಮತ್ತು 8ನೇ ಪದಗಳ ಮೊತ್ತವು 24ಹಾಗು ಅದೇ ಶ್ರೇಢಿಯ 6ನೇ ಮತ್ತು 10ನೇ ಪದಗಳ ಮೊತ್ತವು 44 ಆದರೆ ಶ್ರೇಢಿಯ ಮೊದಲ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಸುರುಳಿಯನ್ನು ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲಾಗಿದೆ ಅವುಗಳ ಕೇಂದ್ರಗಳು ಪರ್ಯಾಯವಾಗಿ A & B ನಲ್ಲಿದ್ದು A ಕೇಂದ್ರದಿಂದ ಆರಂಭವಾಗಿ ತ್ರಿಜ್ಯಗಳು 0.5 cm, 1 cm, 1.5 cm, 2 cm …….. ಹೀಗೆ ಇದೆ. ಈ ರೀತಿಯ 13 ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲ್ಪಟ್ಟ ಒಟ್ಟು ಉದ್ದ ಏನು?

Question 38.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 39.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ, 2x – y = 5 ಮತ್ತು x + 3y = 6.

Question 40.
ಒಂದು ಅರ್ಧಗೋಳಾಕಾರದ ಬಟ್ಟಲಿನ ಒಳಜ್ಯವು 18 cm ಇದ್ದು ಅದರ ತುಂಬ ಹಣ್ಣಿನ ರಸವಿದೆ. ಈ ರಸವನ್ನು 3 cm ತ್ರಿಜ್ಯವಿರುವ ಮತ್ತು 9 cm ಎತ್ತರವಿರುವ ಸಿಲಿಂಡರಿನಾಕೃತಿಯ ಬಾಟಲಿಗಳಿಗೆ ತುಂಬಬೇಕು. ಬಟ್ಟಲನ್ನು ಖಾಲಿ ಮಾಡಲು ಎಷ್ಟು ಬಾಟಲಿಗಳ ಅವಶ್ಯಕತೆ ಇದೆ?

Solutions

I.
Solution 1.
(B) 4 : 5

Solution 2.
(A) 25.67 cm²

Solution 3.
(C) 10 ಮೂ.ಮಾ.

Solution 4.
(D) 60

Solution 5.
(A) 1

Solution 6.
(A) (+5, -5)

Solution 7.
(C) 0.36

Solution 8.
(B) π (r₁ + r₂) l

II.
Solution 9.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ಮೇಲಿನ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮ.

Solution 10.
ಶೇಷ – 5

Solution 11.
αβ = \(\frac { c }{ a }\)

Solution 12.

∆APB ನಲ್ಲಿ PA = PB = 8cm
∠PAB = ∠PBA = 60°
∠APB = 60°
∆APB ಒಂದು ಸಮಬಾಹು ತ್ರಿಭುಜ
AB = 8 cm

Solution 13.
(1 + sin²90°)² = (1 + 1)² = 4

Solution 14.

III.
Solution 15.

Solution 16.

ಸಾಧನೀಯ: ∆AEP ~ ∆CDP
∆AEP & ∆CDP ಗಳಲ್ಲಿ
∠AEP = ∠CDP = 90°
∠APE = ∠CPD (ಶೃಂಗಾಭಿಮುಖ ಕೋನಗಳು)
∆AEP ~ ∆CDP
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ O ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.

Solution 17.

Solution 18.

Solution 19.
R = 14 cm, r = 7 cm, θ = 40°
ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣ

Solution 20.
ತ್ರಿಜ್ಯ = 4cm
ಕೇಂದ್ರ ಕೋನ = 180° – 70° = 110°

Solution 21.

Solution 22.

Solution 23.

Solution 24.

Solution 25.

ಗೋಪುರದ ಎತ್ತರ = AB = x ಆಗಿರಲಿ
ಕಟ್ಟಡದ ಎತ್ತರ = BC = 20 m
tan 45° = \(\frac { BC }{ CD }\)
1 = \(\frac { 20 }{ CD }\)
CD = 20 m
tan 60° = \(\frac { AC }{ CD }\)
⇒ √3 = \(\frac { x+20 }{ 20 }\)
⇒ 20√3 = x + 20
⇒ x + 20 = 20√3
⇒ x = 20(√3 – 1)
ಗೋಪುರದ ಎತ್ತರ = 20(√3 – 1) m

Solution 26.

ಕಟ್ಟಡ ಎತ್ತರ = AB = 50√3 m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ BC ಆಗಿರಲಿ
tan 60° = \(\frac { AB }{ BC }\)
⇒ √3 = \(\frac { 50\surd 3 }{ BC }\)
⇒ BC = 50m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ = 50m

Solution 27.

Solution 28.
ಫಲಿತಗಳ ಸಂಖ್ಯೆ = n(S) = 6 {1, 2, 3, 4, 5, 6}
(i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ = A = {2, 3, 5}
n(A) = 3
ಸಂಭವನೀಯ = P(A) = \(\frac { n(A) }{ n(S) }\) = \(\frac { 3 }{ 6 }\)
(ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆ = B = {2, 3, 4}
n(B) = 3
ಸಂಭವನೀಯತೆ = P(B) = \(\frac { n(A) }{ n(S) }\) = \(\frac { 3 }{ 6 }\)

Solution 29.

Solution 30.
3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ ಎಂದು ಊಹಿಸೋಣ.

ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ = ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
ನಮ್ಮ ಊಹೆ 3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂಬುದು ತಪ್ಪು
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ

IV.
Solution 31.

ದತ್ತ: O ವೃತ್ತಕೇಂದ್ರ PA ಮತ್ತು PB ಗಳ ಬಾಹ್ಯಬಿಂದು P ನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು.
ಸಾಧನೀಯ: PA = PB
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔPOA & ΔPOB ಗಳಲ್ಲಿ
OA = OB
∠PAO = ∠PBO = 90°
OP = OP
ΔPOA = ΔPOB
PA = PB
ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು ತ್ರಿಜ್ಯ ಮತ್ತು ಸ್ಪರ್ಶಕದ ನಡುವಿನ ಕೋನ ಉಭಯಸಾಮಾನ್ಯ ಲಂ.ವಿ.ಬಾ. ಜ್ವಸಿದ್ಧಾಂತ ಸರ್ವಸಮ ತ್ರಿಭುಜದ ಅನುರೂಪ ಭಾಗಗಳು.
ಅಥವಾ

ΔPOB ನಲ್ಲಿ
PB² = OB² – OP²
⇒ PB² = 5² – 3² = 25 – 9 = 16
⇒ PB² = 16
⇒ PB = 4
AB = AP + PB
⇒ AB = 4 + 4 (∴ AP = PB)
⇒ AB = 8 cm ಜ್ಯಾದ ಉದ್ದ 8 cm

Solution 32.
∠A = 45°, ∠B = 105°, BC = 7 cm
∠C = 180° – 150° = 30°

ರಚಿಸಬೇಕಾದ ತ್ರಿಭುಜ A’BC’

Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು x ಆಗಿರಲಿ
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ಅವನ ವಯಸ್ಸು = x – 3 ವರ್ಷ
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸ್ಸು = x + 5 ವರ್ಷ

⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
⇒ x (x – 7) + 3 (x – 7) = 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 ಅಥವಾ x + 3 = 0
⇒ x = 7 ಅಥವಾ x = -3
ರೆಹಮಾನನ ಈಗಿನ ವಯಸು 7 ವರ್ಷ
ಅಥವಾ
ರೈಲಿನ ಜನ x km/hr, ಆಗಿರಲಿ
ದೂರ = d = 360 km/hr

Solution 34.

Solution 35.

Solution 36.

V.
Solution 37.

Solution 38.

ದತ್ತ: ΔABC ನಲ್ಲಿ ∠BAC = 90°
ಸಾಧನೀಯ: BC² = AB² + AC²
ರಚನೆ: AD ⊥ BC ಗೆ ಎಳೆಯಿರಿ,
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔBAC & ΔBDA ಗಳಲ್ಲಿ
∠ABC = ∠ABD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠BDA = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔBDA ಕೊ .ಕೋ .ನಿ.ಗು.
\(\frac { BA }{ BD }\) = \(\frac { BC }{ BA }\)
AB² = BD.BC …….. (1)
ΔBAC & ΔADC ಗಳಲ್ಲಿ
∠ACB = ∠ACD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠ADC = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔADC ಕೋ .ಕೋ .ನಿ.ಗು.
\(\frac { AC }{ DC }\) = \(\frac { BC }{ AC }\)
AC² = DC × BC ……. (2)
AB² + AC² = BD.BC + DC.BC
(1) + (2) ರಿಂದ
BC(BD + DC) = BC × BC
AB² + AC² = BC²

Solution 39.

Solution 40.

Karnataka SSLC Maths Model Question Papers

Karnataka SSLC Maths Model Question Paper 2 Kannada Medium

ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80

I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)

Question 1.
ಚಿತ್ರದಲ್ಲಿ QE = 7.2 cm, PF = 1.8 cm, FR = 5.4cm ಆದರೆ PE ಯು

(A) 2 cm
(B) 2.4 cm
(C) 2.8 cm
(D) 3.2 cm

Question 2.
‘r’ ವೃತ್ತ ತ್ರಿಜ್ಯ ಹಾಗೂ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ

Question 3.
P (-2, -1), ರಲ್ಲಿ ಲಂಬದೂರವು
(A) -2
(B) -1
(C) 1
(D) 2

Question 4.
‘0’ ಮೂಲಬಿಂದು ಮತ್ತು P (4, 3) ರ ನಡುವಿನ ದೂರ (ಏಕಮಾನಗಳಲ್ಲಿ) OP = …….
(A) 5
(B) 4
(C) 3
(D) 2

Question 5.
P (x) = 5x – 10 ರ ಶೂನ್ಯತೆ.
(A) 2
(B) -2
(C) 5
(D) -5

Question 6.
x + \(\frac { 2 }{ x }\) = 3 ಸಮೀಕರಣದ ಆದರ್ಶರೂಪ
(A) x² + 2x – 3 = 0
(B) x² + 3x + 2 = 0
(C) x² – 3x + 2 = 0
(D) x² – 2x + 3 = 0

Question 7.
ಎರಡು ನಾಣ್ಯಗಳನ್ನು ಏಕಕಾಲಕ್ಕೆ ಚಿಮ್ಮಿದಾದ, ಶಿರ H ಮತ್ತು ಪ್ರಚ್ಛ T ಇರುವಂತೆ ಸಾಧ್ಯ ಫಲಿತಗಳು
(A) {T, H, H, T}
(B) {TT, HH, HT, TH}
(C) {T, H}
(D) {TT, HH}

Question 8.
ಸಿಲಿಂಡರ್‌ನ ಎತ್ತರ ‘h’ ಮತ್ತು ಪಾದದ ತ್ರಿಜ್ಯ ‘r’ ಆದಾಗ ಸಿಲಿಂಡರ್‌ನ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ
(A) 2π (r + h)
(B) 2πr (r + h)
(C) 2πrh
(D) \(\frac { 2\pi r }{ h }\)

II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)

Question 9.
ಒಂದು 5m ಎತ್ತರದ ಏಣಿಯನ್ನು ನೆಲದ ಮೇಲಿನಿಂದ 4 m ಎತ್ತರದ ಕಿಟಕಿಯನ್ನು ತಲುಪುವಂತೆ ಗೋಡೆಗೆ ಒರಗಿಸಿದೆ. ಗೋಡೆಯ ಪಾದದಿಂದ ಏಣಿಯ ಪಾದಕ್ಕಿರುವ ದೂರ ಲೆಕ್ಕಿಸಿ.

Question 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?

Question 11.
‘ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ’ ವನ್ನು ಬರೆಯಿರಿ.

Question 12.
ಯೂಕ್ಲಿಡ್‌ನ ಭಾಗಾಕಾರ ಅನುಪ್ರವೇಯವನ್ನು ತಿಳಿಸಿ.

Question 13.
ಘನಪದೋಕ್ತಿಯು ಹೊಂದಿರಬಹುದಾದ ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?

Question 14.
ಎರಡು ಸಂಖ್ಯೆಗಳ ಮೊತ್ತ 27 ಮತ್ತು ಗುಣಲಬ್ಧ 182 ಆದರೆ ಆ ಸಂಖ್ಯೆಗಳು ಯಾವುವು?

III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)

Question 15.
ಮೊತ್ತ ಕಂಡುಹಿಡಿಯಿರಿ: 34 + 32 + 30 + …………… + 10,

Question 16.
ಒಂದು ಏಣಿಯ ಪಾದವು ನೆಲದ ಮೇಲೆ ಗೋಡೆಯಿಂದ 2.5m ದೂರದಲ್ಲಿ ಹಾಗೂ ಅದರ ತುದಿಯು ನೆಲದ ಮೇಲಿಂದ 6 m ಎತ್ತರದಲ್ಲಿರುವ ಕಿಟಕಿಯನ್ನು ಮುಟ್ಟುವಂತೆ ಏಣಿಯನ್ನು ಗೋಡೆಗೆ ಒರಗಿಸಿ ಇಡಲಾಗಿದೆ. ಏಣಿಯ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 17.
ಒಂದು ಎರವಲು ಗ್ರಂಥಾಲಯದಲ್ಲಿ ಮೊದಲ 3 ದಿನಕ್ಕೆ ಒಂದು ನಿಗದಿತ ಶುಲ್ಕವಿರುತ್ತದೆ. ಆ ದಿನದ ಪ್ರತಿಯೊಂದೂ ದಿನಕ್ಕೂ ಒಂದು ಹೆಚ್ಚುವರಿ ಶುಲ್ಕವಿರುತ್ತದೆ. ಪುಸ್ತಕವನ್ನು 7 ದಿನ ತನ್ನಲ್ಲಿ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸರಿತಾ ₹ 27 ನ್ನು ಪಾವತಿಸಿದರೆ, ಪುಸ್ತಕವನ್ನು 5 ದಿನ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸೂಸಿ ₹ 21 ನ್ನು ಪಾವತಿಸಿದಳು. ನಿಗದಿತ ಶುಲ್ಕ ಮತ್ತು ಪ್ರತಿಯೊಂದು ಹೆಚ್ಚುವರಿ ದಿನದ ಶುಲ್ಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 18.
ನೀರಿನ ಒಳಭಾಗದಲ್ಲಿರುವ ಬಂಡೆಗಳ ಬಗ್ಗೆ ಎಚ್ಚರಿಸಲು ಒಂದು ದೀಪಸ್ಥಂಭವು 80° ಕೋನವಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದಲ್ಲಿ 16.5 km ದೂರಕ್ಕೆ ಕೆಂಪು ಬೆಳಕನ್ನು ಹರಡುತ್ತದೆ. ಹಡಗುಗಳನ್ನು ಎಚ್ಚರಿಸುವ ಈ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಚಿತ್ರದಲ್ಲಿ ತೋರಿಸಿರುವಂತೆ 4 cm ಬಾಹುವುಳ್ಳ ಒಂದು ಚೌಕದ ಪ್ರತೀ ಮೂಲೆಯಲ್ಲಿ 1 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತ ಚತುರ್ಥಕವನ್ನು ಮತ್ತು 2 cm ವ್ಯಾಸವಿರುವ ಒಂದು ವೃತ್ತವನ್ನು ಕತ್ತರಿಸಿದೆ. ಚೌಕದ ಉಳಿದ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 19.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1\(\frac { 1 }{ 2 }\) ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,

Question 20.
(4, -3) ಮತ್ತು (8, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡವನ್ನು ಆಂತರಿಕವಾಗಿ 3 : 1 ಅನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುವ ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 21.
ಶೃಂಗಗಳು ಈ ಕೆಳಗಿನಂತಿರುವ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ. (5, -1), (3, -5), (5, 2)

Question 22.
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.

Question 23.
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ವರ್ಗಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
ಮೂಲಗಳ ಸ್ವಭಾವವನ್ನು ವಿವೇಚಿಸಿ, ವಾಸ್ತವ ಮೂಲಗಳಿದ್ದಲ್ಲಿ, ಅವುಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ: 2x² – 3x + 5 = 0

Question 25.
1.6m ಎತ್ತರದ ಪ್ರತಿಮೆಯೊಂದನ್ನು ಒಂದು ಪೀಠದ ಮೇಲ್ಬಾಗದಲ್ಲಿ ಇರಿಸಲಾಗಿದೆ. ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ಪ್ರತಿಮೆಯ ಮೇಲಿನ ಉನ್ನತ ಕೋನವು 60° ಮತ್ತು ಅದೇ ಬಿಂದುವಿನಿಂದ ಪೀಠದ ಮೇಲ್ತುದಿಯ ಉನ್ನತ ಕೋನವು 45° ಇದ್ದರೆ, ಪೀಠದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
10 ಪಂದ್ಯಗಳಲ್ಲಿ ಒಬ್ಬ ಬೌಲರ್‌ನು ಪಡೆದ ವಿಕೆಟ್‌ಗಳ ಸಂಖ್ಯೆಯು ಈ ಕೆಳಗಿನಂತಿದೆ.

Question 27.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. 2 ಮತ್ತು 6ರ ನಡುವೆ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 28.
ಒಂದು ಪಾತ್ರೆಯ ಆಕಾರವು ಟೊಳ್ಳಾದ ಸಿಲಿಂಡರಿನ ಒಂದು ಪಾದದ ಮೇಲೆ ಟೊಳ್ಳಾದ ಅರ್ಧಗೋಳಾಕೃತಿಯನ್ನು ಕೂಡಿಸಿ ಮಾಡಿದೆ. ಅರ್ಧಗೋಳದ ವ್ಯಾಸವು 14 cm ಮತ್ತು ಪಾತ್ರೆಯ ಒಟ್ಟು ಎತ್ತರ 13 cm ಇದೆ. ಈ ಪಾತ್ರೆಯ ಒಳಮೇ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಘನ ಸಿಲಿಂಡರಿನ ಎತ್ತರ 2.4 m ಮತ್ತು ವ್ಯಾಸ 1.4 m ಇದೆ. ಇದರಿಂದ ಒಂದೇ ಎತ್ತರ ಮತ್ತು ಒಂದೇ ವ್ಯಾಸವನ್ನು ಹೊಂದಿರುವ ಶಂಕುವಿನಾಕಾರದ ಹಳ್ಳವನ್ನು ಕೊರೆದು ಟೊಳ್ಳಾಗಿಸಿದೆ. ನೂತನ ಘನದ ಒಟ್ಟು ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಅತ್ಯಂತ ಸಮೀಪದ ಬೆಲೆಗೆ cm² ನಲ್ಲಿ ಕಂಡುಹಿಡಿಯಿರಿ.

Question 29.
\(\frac { 3 }{ 2 }\) x + \(\frac { 5 }{ 3 }\) y = 7: 9x – 10y = 14 ಈ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳಲ್ಲಿ \(\frac { { a }_{ 1 } }{ { a }_{ 2 } }\), \(\frac { { b }_{ 1 } }{ { b }_{ 2 } }\) ಮತ್ತು \(\frac { { c }_{ 1 } }{ { c }_{ 2 } }\) ಅನುಪಾತಗಳನ್ನು ಹೋಲಿಸುವ ಮೂಲಕ ಜೋಡಿಗಳು ಸ್ಥಿರವಾಗಿವೆಯೇ ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೇ ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
150 cm ಎತ್ತರವಿರುವ ಒಬ್ಬ ವ್ಯಕ್ತಿಯು ತನ್ನ ನೆರಳಿನ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಅದು ಅವನ ಪಾದದಿಂದ 150√3 cm ದೂರದಲ್ಲಿರುವುದು ಕಂಡುಬರುತ್ತದೆ. ಹಾಗಾದರೆ ಅವನ ನೋಟದಲ್ಲಿ ಉಂಟಾದ ಅವನತ ಕೋನವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)

Question 31.
‘ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಸಮ ಎಂದು ಸಾಧಿಸಿ

Question 32.
BC = 6 cm, AB = 5cm ಮತ್ತು ∠ABC = 60° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ತ್ರಿಭುಜ ABC ಯ ಅನುರೂಪ ಬಾಹುಗಳ \(\frac { 3 }{ 4 }\) ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ.

Question 33.
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ರೆಹಮಾನನ ವಯಸ್ಸು (ವರ್ಷಗಳಲ್ಲಿ) ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವೃಶ್ಯಮಗಳ ಮೊತ್ತ \(\frac { 1 }{ 3 }\) ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಮೂರು ಕ್ರಮಾನುಗತ ಸಂಖ್ಯೆಗಳಲ್ಲಿ, ಮೊದಲನೆಯ ವರ್ಗ ಮತ್ತು ಉಳಿದೆರಡು ಸಂಖ್ಯೆಗಳ ಗುಣಲಬ್ದಗಳ ಮೊತ್ತ 154 ಆಗಿದೆ. ಹಾಗಾದರೆ ಆ ಮೂರು ಸಂಖ್ಯೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.
ಈ ಕೆಳಗೆ ವ್ಯಾಖ್ಯಾನಿಸಲ್ಪಟ್ಟ ಹೇಳಿಕೆಗಳ ಕೋನಗಳು ಲಘಕೋನಗಳು. ಈ ಕೆಳಗಿನ ಸಮೀಕರಣಗಳನ್ನು ಸಾಧಿಸಿ.

ಅಥವಾ

Question 35.
ಒಂದು ಕಾರ್ಖಾನೆಯ 50 ಕೆಲಸಗಾರರ ದೈನಂದಿನ ಆದಾಯವನ್ನು ಕೆಳಗಿನ ವಿತರಣೆಯು ನೀಡುತ್ತಿದೆ.

ಮೇಲಿನ ವಿತರಣೆಯನ್ನು ಕಡಿಮೆ ಇರುವ ವಿಧಾನದ’ ಸಂಚಿತ ಆವೃತ್ತಿ ವಿತರಣೆಯಾಗಿ ಬದಲಾಯಿಸಿ ಮತ್ತು ಅದರ ಓಜೀವ್ ಎಳೆಯಿರಿ.

Question 36.
ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯಿಂದ ಭಾಗಿಸಿ ಹಾಗೂ ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯು ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯ ಅಪವರ್ತನವಾಗಿದೆಯೇ ಎಂದು ಪರೀಕ್ಷಿಸಿ.
t² – 3 ; 2t⁴ + 3t³ – 2t² – 9t – 12
ಅಥವಾ
ಬಹುಪದೋಕ್ತಿ p(x) ನ್ನು ಬಹುಪದೋಕ್ತಿ g(x) ನಿಂದ ಭಾಗಿಸಿ, ಭಾಗಲಬ್ಧ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
p(x) = x⁴ – 3x + 4x + 5
g(x) = x² + 1 – x

V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)

Question 37.
ಎರಡು ತ್ರಿಭುಜಗಳಲ್ಲಿ ಒಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳು ಮತ್ತೊಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳೊಡನೆ ಸಮಾನುಪಾತ ಹೊಂದಿದ್ದರೆ, ಅವುಗಳ ಅನುರೂಪ ಕೋನಗಳು ಸಮವಾಗಿರುತ್ತವೆ ಮತ್ತು ಅದರಿಂದಾಗಿ ಆ ಎರಡು ತ್ರಿಭುಜಗಳು ಸಮರೂಪಿಗಳಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.

Question 38.
ನಕ್ಷೆಯ ಮೂಲಕ ಬಿಡಿಸಿ: y = 2x + 1; x = 2y – 5

Question 39.
(4, -1) ಮತ್ತು (-2, -3) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಭಾಜಕ ಬಿಂದುಗಳ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 40.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ ಪದದ ವರ್ಗವು ಅದರ 8ನೇ ಪದಕ್ಕೆ ಸಮವಾಗಿದೆ ಹಾಗೂ 8ನೇ ಪದವು ನಾಲ್ಕನೇ ಪದಕ್ಕಿಂತ 24 ಹೆಚ್ಚಾಗಿದೆ. ಹಾಗಾದರೆ ಶ್ರೇಢಿಯ ಪದಗಳನ್ನು ಬರೆಯಿರಿ.

Solutions

I.
Solution 1.
(B) 2.4 cm

Solution 2.
(C) \(\frac { { \pi r }^{ 2 } }{ 6 }\)

Solution 3.
(B) -1

Solution 4.
(A) 5

Solution 5.
(A) 2

Solution 6.
(C) x² – 3x + 2 = 0

Solution 7.
(B) {TT, HH, HT, TH}

Solution 8.
(C) 2πrh

II.
Solution 9.
ತ್ರಿವಳಿ: 5, 4, 3
ಏಣಿಯ ಪಾದದಿಂದ ಗೋಡೆಯ
ಪಾದಕ್ಕಿರುವ ದೂರ = 3 ಮೀ.

Solution 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ = 1

Solution 11.
(x₁, y₁) ಮತ್ತು (x₂, y₂) ಬಿಂದುಗಳಿಗೆ ಸಂಬಂಧಿಸಿದಂತೆ : ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ (m₁ : m₂ ಅನುಪಾತದಲ್ಲಿ)

Solution 12.
ಭಾಜ್ಯ = ಭಾಜಕ × ಭಾಗಲಬ್ದ + ಶೇಷ
a = b × q + r

Solution 13.
ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳು = 3

Solution 14.
x + y = 27
13 + 14 = 27
xy = 182
13 × 14 = 182
ಮೊದಲ ಸಂಖ್ಯೆ = 13, ಎರಡನೆಯ ಸಂಖ್ಯೆ = 14

Solution 15.
a₁ = 38, a₁₆ = 73, a₃₁ = ?
a₁₆ = a + 15d
73 = 38 + 15d
73 = 38 + 15d
d = 3
a₃₁ = a₁ + 30d = 38 + 30(3) = 38 + 90 = 128
a₃₁ =128

Solution 16.

AB ಯು ಏಣಿ, CA ಯು ಗೋಡೆ ಮತ್ತು ಕಿಟಕಿಯಾಗಿರಲಿ.
BC = 2.5m ಮತ್ತು CA = 6m
AB² = BC² + CA² = (2.5)² + (6)² = 42.25
AB = 6.5
ಏಣಿಯ ಉದ್ದ ಆಗಿದೆ.

Solution 17.
ಮೊದಲ 3 ದಿನದ ಒಂದು ಶುಲ್ಕ = x
ಪ್ರತೀ ದಿನದ ಹೆಚ್ಚುವರಿ ಶುಲ್ಕ = y
x + 4y = 27
x + 2y = 21
2y = 6
y = 3
x + 2y = 21
x + 2(3) = 21
x = 21 – 6
x = 15
ಮೂರು ದಿನಗಳ ಶುಲ್ಕ = x = ₹ 15
ಹೆಚ್ಚುವರಿ ಪ್ರತೀ ದಿನದ ಶುಲ್ಕ = y = ₹ 13

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ.
√5 = \(\frac { p }{ q }\) p, q ∈ I
p ಮತ್ತು q ಗಳು 1 ನ್ನು ಹೊರತುಪಡಿಸಿ ಬೇರೆ ಸಾಮಾನ್ಯ ಅಪವರ್ತನ ಹೊಂದಿದ್ದರೆ, ಸಾಮಾನ್ಯ ಅಪವರ್ತನದಿಂದ ಭಾಗಿಸಬಹುದು, ಆದ್ದರಿಂದ p ಮತ್ತು Q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳೆಂದು ಭಾವಿಸೋಣ
p = q√5
p² = 5q² (ಎರಡು ಕಡೆ ವರ್ಗ ಮಾಡಿದಾಗ)
5, p² ನ್ನು ಭಾಗಿಸುತ್ತದೆ…… (1)
5, p ಮತ್ತು q ಗಳ ಸಾಮಾನ್ಯ ಅಪವರ್ತನವಾಗಿದೆ.
ಏಕೆಂದರೆ, 5, q ನ್ನೂ ಸಹ ಭಾಗಿಸುತ್ತದೆ.
ಅಂದರೆ p = 5r
5q² = 5²r²
q = 5r² ………(2)
p ಮತ್ತು q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳು ಎಂಬ ಸತ್ಯಸಂಗತಿಗೆ ವಿರುದ್ಧವಾಗಿದೆ.
√5 ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆಯಾಗಿದೆ.

Solution 23.
ಶೂನ್ಯಗಳ ಮೊತ್ತ = α + β = 1
ಶೂನ್ಯಗಳ ಗುಣಲಬ್ದ = αβ = 1
ವರ್ಗಬಹುಪದೋಕ್ತಿ: x² – (α + β) x + (αβ)
ಯಲ್ಲಿ ಆದೇಶಿಸಿದರೆ x² – 1x + 1
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Solution 24.
2x² – 3x + 5 = 0
ax² + bx + c = 0,
a = 2, b = -3, c = 5
ಶೋಧಕ: b² – 4ac = 9 – 4(2)(5) = 9 – 40 = -31 < 0
ಮೂಲಗಳು ಸಂಮಿಶ್ರ ಸಂಖ್ಯೆಗಳಾಗಿವೆ.

Solution 25.

Solution 26.
ಆವೃತ್ತಿ ವಿತರಣಾ ಪಟ್ಟಿ

ಬೌಲರ್ ಗರಿಷ್ಠ ಪಂದ್ಯಗಳಾದ 3 ರಲ್ಲಿ ಪಡೆದ ವಿಕೆಟ್‌ಗಳ ಸಂಖ್ಯೆ 2 ಆಗಿದೆ. ಆದ್ದರಿಂದ ದತ್ತಾಂಶಗಳ ಬಹುಲಕ 2,

Solution 27.
ಫಲಿತಗಣ S = {1, 2, 3, 4, 5, 6}
n(S) = 6
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆಗಳು A = {3, 4, 5}
n(A) = 3
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆ ಪಡೆಯುವ ಸಂಭವನೀಯತೆ
P(A) = \(\frac { n(A) }{ n(S) }\) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

Solution 28.

Solution 29.

Solution 30.

IV.
Solution 31.

ವೃತ್ತದ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’
ದತ್ತ: O ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ OP ತ್ರಿಜ್ಯ ಮತ್ತು XY ಸ್ಪರ್ಶಕ ಎಳೆದಿದೆ.
ಸಾಧನೀಯ: OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತೋಂದು ಬಿಂದು Q ಗುರುತಿಸಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: Q ಬಿಂದುವು ವೃತ್ತದ ಹೊರಭಾಗದಲ್ಲಿರಬೇಕು.
Q ಬಿಂದುವು ವೃತ್ತದ ಒಳಭಾಗದಲ್ಲಿದ್ದರೆ, XY ವೃತ್ತಕ್ಕೆ ಛೇದಕವಾಗುತ್ತದೆಯೇ ಹೊರತು ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವಾಗುವುದಿಲ್ಲ. ಆದ್ದರಿಂದ OQ ಇದು ವೃತ್ತದ ತ್ರಿಜ್ಯ OP ಗಿಂತ ಉದ್ದವಾಗಿದೆ. ಅಂದರೆ OQ > OP.
P ಬಿಂದುವನ್ನು ಹೊರತುಪಡಿಸಿ, XY ಮೇಲಿನ ಎಲ್ಲಾ ಬಿಂದುಗಳಿಗೂ ಇದು ಅನ್ವಯಿಸುವುದರಿಂದ, O ಬಿಂದುವಿನಿಂದ XYನ ಮೇಲಿನ ಇತರೆ ಬಿಂದುಗಳಿಗಿರುವ ದೂರಕ್ಕಿಂತ OP ಯು ಕನಿಷ್ಟ ಉದ್ದ ಹೊಂದಿದೆ
OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ಅಥವಾ

ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮ.
ದತ್ತ: 0ವೃತ್ತಕೇಂದ್ರ, P ಬಾಹ್ಯಬಿಂದು, PQ & PR ಗಳು ಬಾಹ್ಯಬಿಂದು Pನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು
ಸಾಧನೀಯ: PQ = PR
ರಚನೆ: OP, OR, OP ಸೇರಿಸಿ
ಸಾಧನೆ: ΔOQP ಮತ್ತು ΔORP ಗಳಲ್ಲಿ
OQ = OR (ತ್ರಿಜ್ಯಗಳು)
∠OQP = ∠ORP = 90°
OP = OP (ಉಭಯಸಾಮಾನ್ಯ)
ΔOQP = ΔORP
PQ = PR

Solution 32.

Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು = x ಆಗಿರಲಿ
3 ವರ್ಷಗಳ ಹಿಂದೆ, ಅವನ ವಯಸ್ಸು = x – 3
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸು = x + 5

Solution 34.

Solution 35.

Solution 36.

Solution 37.

Solution 38.

Solution 39.

Solution 40.

Karnataka SSLC Maths Model Question Papers

Karnataka State Syllabus SSLC Maths Model Question Paper 1 with Answers in Kannada

Karnataka State Syllabus SSLC Maths Model Question Paper 4 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117, then the value of m is
a) 4
b) 3
c) 11
d) 2
Answer:
Solution:
The H.C.F. of 65 and 117 = 65m- 117
Now 13 = 65m- 117
65m 117+13
65rn = 130
m = 130/65 = 2
∴ The value of m = 2
∴ HCF(65, 117)
= 13

Question 2.
If sinx = sin 60° cos 30° – cos 60 sin 30!l, then the value of x is
a) 0°
b) 30°
c) 45°
d) 60°
Answer:
b) 30°
Solution:
Sinx = sin 60°cos30 – cos 60 sin 30

Question 3.
The angle between the radius of a circle and the tangent drawn at the point of contact is
a) 0°
b) 60°
c) 90°
d) 30°
Answer:
c) 90°

Question 4.
The T.S.A. of a cuboid of dimension, l = 30cm, b = 20cm, c = 10cm, is ______
a) 600cm²
b) 60cm²
c) 6000cm²
d) 2200 cm²
Answer:

Question 5.
Which of the following is a polynomial
a) x² – 5x + 3√x
b) x^1/2 + x^1/2 – x +1
c) \(\sqrt{x}-\frac{1}{\sqrt{x}}\)
d) x² – 4x + √2
Answer:
x² – 4x + √2
Solution:
x² – 4x + √2 because variable has no negative or decimal powers.

Question 6.
The value of p if x, 2x + p and 3x + 6 are in A.P.
a) p = 3
b) p = 2
c) p = 1
d) p = 0
Answer:
Solution:
x, 2x + p, 3x + 6 are inAP a, A, b are A.P.

2(2x + p) = 2(2x + 3)
2x + p = 2x + 3
p = 2x + 3 – 2k
P = 3

Question 7.
In triangle PQR, The value of y is
a) 4√3
b) 6√3
c) 5√3
d) √3
Answer:
b) 6√3
Solution:

QS² = PS x SR
QS² = \(\sqrt{9 \times 12}\)
QS = \(\sqrt{9 \times 4 \times 3}\)
QS = 3 x 2√3
QS = 6√3

Question 8.
When 2 unbiased coins are tossed at ‘ a time, the probability of getting 2 heads is _____
a) 1/4
b) 1/2
c) 1
d) 0
Answer:
a) 1/4
Solution:
S = {(H, H) (H, T) (T, H) (T, T)}
n(s) = 4
An event of getting 2 heads = A={(H, H)}
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{1}{4}\)

II. Answer the following Questions : ( 1 x 8 = 8 )

Question 9.
If the product of zeros of polynomial f(y) = ay³ – 6y² + 11y – 6 is 4 then find the value of ‘a’.
Answer:
α β γ = \(-\frac{\mathrm{d}}{\mathrm{a}}\)
4 = \(\frac{-(-6)}{a}\)
4a = 6
a = \(\frac{6}{4}\)
a = \(\frac{3}{2}\)

Question 10.
What is the value of C, if ax² + bx + c = 0 has equal roots?
Answer:
If ax² + bx + c = 0 has equal roots then,
b² – 4ac = 0
b² = 4ac
4ac = b²
c = \(\frac{b^{2}}{4 a}\)

Question 11.
Find the second term if sum of the ‘n’ term of an AP is 2n² + 1.
Answer:
Sn = 2n² + 1
S₁ = 2(1)² + l= 2xl + l= 2+ l= 3
∴ S₁ = T₁ = a = 3
S₂ = 2(2)² + 1 =8+ 1 =9
a + T₂ = 9
T₂ = 9 – 3 = 6

Question 12.
State converse of Pythagoras Theorem.
Answer:
“The square on the greatest side of a triangle is equal to the sum of the squares on the other two sides, then the other two sides contian the right angle”.

Question 13.
What is the form of \(\frac{p}{q}\) (p,q ∈ z q ≠ 0) form of 0.57 ?
Answer:
\(0.5 \overline{7}=\frac{57-5}{90}=\frac{52}{90}=\frac{26}{45}\)

Question 14.
If sinθ = \(\frac{1}{3}\) then find the value of (cot²θ + 2)
Answer:
AC² = AB² + BC²
BC² = AC² – AB²
= 3² – 1²
= 9 – 1 = 8
BC = √8 = 2√2
cotθ = \(\frac{2 \sqrt{2}}{1}\) = 2√2
2cot²θ + 2 = 2(2√2)² + 2
= 2 x 8 + 2
= 16 + 2
= 18

Question 15.
In sin(A + B) = \(\frac{\sqrt{3}}{2}\) and cos (A – B) = 1, 0< A + B < 90°, A ≥ B
Answer:
sin(A + B) = \(\frac{\sqrt{3}}{2}\) = sin60°
A + B = 60° …. (1)
cos (A – B) = 1 = cos0°
A – B = 0 ….(2)
Solve (1) and (2)
A + B = 60
A – B = 0
2A = 60 or A = 30°
A + B = 60
30 + B = 60
B = 60 – 30° = 30°
∴ A = 30°, B = 30°

Question 16.
The surface area of a sphere is same as the C.S.A. of a right circular cylinder whose height and diameter are 4 cm each. Find the radius of the sphere.
Soln:
4πr² = 2πrh
4R² = dh .
4R²= 4 x 4
R² = \(\frac{4 \times 4}{4}\) = 4
R = √4 = 2cm
A = 30°, B = 30°
∴ Radius of the sphere = 2cm.

III. Answer the following : ( 2 x 8 = 16 )

Question 17.
By Euclid’s division lemma, show that thte square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answer:
Let x be any positive integer. Then it is of the form 3q, 3q+1 or 3q + 2.
Now, (3q)² = 9q² = 3(3q²) = 3m where m = 3q²
(3q + 1)² = 9q² + 6q+ 1 = 3(3q² + 2q) + 1
= 3m + 1, where m = 3q² + 2q and (3q+2)² = 9q² + 12q + 4
= 3(3q² + 4q + 1) +1
=3m +1, where m = 3q² + 4q + 1
Hence, it is proved.

Question 18.
Solve : \(\frac{x+y}{x y}\) = 2 and \(\frac{x-y}{x y}\) = 6
Answer:
x + y = 2xy
x – y = 6xy
2x = 8xy
2 = 8y

Question 19.
Solve : y² – (√3 + 1)y + √3 = 0
Answer:
y² – √3y – y + √3 = 0
y(y – √3) – 1(y – √3) = 0
(y – √3)(y – 1) = 0
y – √3 = 0, y – 1 = 0
y = √3 and y = 1

Question 20.
Show that the points (3, 2) (-2, -3) and (2, 3) are collinear Or non – collinear.
Answer:
Let A= (3, 2) B = (-2, -3) C = (2, 3)

AC ≠ AB + BC
∴ They are non – collinear.

Question 21.
In the given fig ∆DGH ~ ∆DEF, DH = 8cm, DF = 12cm, DG = (3x – l)cm and DE = (4x +. 2)cm, Find the lengths of DG and DE.
Answer:
∆DGH ~ ∆DEF
\(\frac{\mathrm{DG}}{\mathrm{DE}}=\frac{\mathrm{DH}}{\mathrm{DF}}\)
\(\frac{3 x-1}{4 x+2}=\frac{8}{12}\)

12 (3x – 1) = 8(4x + 2)
36x – 12 = 32x + 16
36x – 32x = 16 + 12
4x = 28
x = 28/4
x = 7
∴ DG = 3x – 1 = 3 x 7 – 1 = 21 – 1 = 20
DE = 4x + 2 = 4 x 7 + 2 = 28 + 2 = 30

OR

D is a point on the side BC of A ABC such that ∠ADC = ∠BAC. Prove that
\(\frac{\mathbf{C A}}{\mathbf{C D}}=\frac{\mathbf{C B}}{\mathbf{C A}}\)
Answer:
In ∆ ABC and ∆ DAC,
∠ADC = ∠BAC and
∠C = ∠C
∆ABC ~ ∆DAC

Question 22.
A card is drawn at random from a box containing 21 cards numbered 1 to 21. Find the probability that the card drawn is
a) Prime number b) Divisible by 3.
Answer:
n(s) = 21
A= {2,3,5,7,11,13,17,19}
n(A) = 8
P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{21}\)

b) n(S) = 21
B= {3,6, 9,12,15,18}
n(B) = 6
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{6}{21}=\frac{2}{7}\)

Question 23.
Draw a circle of radius 3cms. Construct a pair of tangents to it, from a point 5cm away from the circle.
Answer:
r = 3cm d = 3 + 5 = 8cm

Question 24.
Express sinA and sec A in terms of cot A.
Answer:
W.K.T cosec²A = 1 + cot²A

OR

If tanθ + sinθ = m and tan θ – sin θ= n, S.T m² – n² = 4\(\sqrt{\mathbf{m} \mathbf{n}}\)
Answer:
L.H.S = m² – n² = (m +n) (m – n)
= {(tan θ + sin θ ) + (tan θ – sin θ )x(tan θ +sin θ) – (tan θ – sin θ)}
= {(2tanθ) x (2sinθ) = 4tanθsinθ

= 4\(\sqrt{\mathbf{m} \mathbf{n}}\)
= RHS.

IV. Answer the following : ( 3 x 9 = 27 )

Question 25.
The sum of the numerator and denominator of a fraction is 24. If 4 is subtracted from the numerator and 5 from its denominator, then it reduces to 1/4. Find the fraction.
Answer:
Let the fraction be \(\frac{\mathrm{x}}{\mathrm{y}}\)
sum of its numerator and denominator = 24.
x + y = 24 ……..(1)
If 4 is subtracted from the numerator and
5 from its denominator we get = 1/4
⇒ \(\frac{x-4}{y-5}=\frac{1}{4}\)
4(x – 4) = 1(y – 5)
4x – 16 = y – 5
4x – y = -5 + 16
4x – y = 11 ……..(2)
From (1) and (2)

Consider,
x + y = 24
7 + y = 24
y = 24 – 7
y = 17
∴ The fraction is \(\frac{x}{y}=\frac{7}{17}\)

OR

Two women and five men can together finish an embroidary work in 4 days. While three women and 6 men can finish in 3 days. Find the time taken by one women alone and also that taken by one man alone.
Answer:
Let 1 woman can finish the embroidary in x days and 1 man can finish the embroidary in y days.
1 woman’s 1 day’s work = \(\frac{1}{x}\)
1 man’s 1 day’s work = \(\frac{1}{y}\)

Let \(\frac{1}{x}\) = u, \(\frac{1}{y}\) = v
2u + 5v = \(\frac{1}{4}\)
3u + 6v = \(\frac{1}{3}\)
solve (1) and (2)
(8u + 20v = 1) x 9
(9u+ 18v= 1) x 8

v = \(\frac{1}{36}\)
Put v = \(\frac{1}{36}\)…….(1)
2u + 5v = \(\frac{1}{4}\)
2u + 5\(\frac{1}{36}\) = \(\frac{1}{4}\)

Thus, 1 woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.

Question 26.
If two zeroes of the polynomial p(x) = x⁴ – 6x³ – 26x² + 138x – 35 are 2 – √3 and 2 + √3. Find the other two zeroes.
Answer:
The two zeroes of p(x) are,
(2 – √3) and (2 + √3)
⇒ [x – (2 – √3] and [x – [2 + √3] are the factors p(x).
⇒ [ x – 2 + √3] [x – 2 – √3]
(x – 2)² – (√3)²
x² – 4x + 4 – 3
⇒ x² – 4x + 1 is a factor of p(x)
Divide p(x) by x² – 4x + 1 to get quotient

Consider the quotient, x² – 2x – 35 = 0
x² – 7x + 5x – 35 = 0
x(x – 7) + 5(x – 7) = 0
(x – 7) (x + 5) = 0
x-7 = 0 orx + 5 = 0
x = 7 or x = -5
∴ The other two zeroes are 7, -5.

Question 27.
A two digit number is such that the product of its digits is 18. when 63 is subtracted from the number, the digits interchange their places. Find the number.
Answer:
Let the two digits be x and y.
⇒ The two digit number is = 10x + y
The product of three digits = xy = 18
When 63 is subtracted from the number, the digits interchange their places.
10x + y – 63 = 10y + x
10x – x – 10y + y = 63
9x – 9y = 63
9(x – y) = 63
x – y = 63/9 = 7
x – y = 7
x = 7 + y
Consider xy = 18 and substitute x = (y + 7) in it.
y(y + 7) = is . .
y² + 7y – 18 = 0
y² + 9y – 2y – 18 = 0
y (y + 9) – 2(y + 9) = 0 –
(y + 9) (y – 2) = 0 .
y + 9 = 0, y -2 = 0
y = -9 or y = 2
xy= 18
x(2) = 18
x = 18/2 = 9
x = 2
But negative digit is not considered.
∴ The two digit number is
= 10x + y
= 10(9) + 2 = 92

OR

A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km / hr from its usual speed. Find its usual speed.
Answer:
Let the usual speed of the plane be x km / hr. Time taken to cover 1500 km with the usual speed.
\(\frac{1500}{x}\) hrs
Time taken to cover 1500 km with the speed of (x + 250) km / hr = \(\frac{1500}{x+250}\)

2(1500 x 250) = x2 + 250x
x² + 250x-750000 = 0
x² + 1000x – 750x – 750000 = 0
x(x + 1000) – 750(x + 1000) = 0
(x+ 1000)(x-750) = 0
x + 1000 = 0 or x – 750 = 0
x = -1000 or x= 750 ,
x = 750
Hence, the usual speed of the plane = 750 km/hr.

Question 28.
If the co – ordinates of the mid points of ∆ ABC are D(1, 2) E(0, -1) and (2, -1). Find the respective co – ordinates of ∆ ABC.
Answer:

D is the midpoint of BC.

x₁ + x₂ = 4 and y₁ + y₂ = -2 ………(1)
E is the mid point of BC

x₂ + x₃ = 0 and y₂ + y₃ = -2 …….(2)
F is the mid point of BC

x₁ + x₃ = 4 and y₂ + y₃ = -2 ….(3)
Solve (1) and (3)

x₂ = 0
x₂ = 0
0 + x₃ = 0
x₃ = 0
x₁ + x₃ = 4
x₁ + 0 = 4
x₁ = 4 – 0 = 4
x₁ = 4

y₂ = -i
y₂ + y₃ = -2
-1 + y₃ = -2
y₃ = -2 +1
y₃ = -1
y₁ + y₃ = -2
y₁ + (-1) = -2
y₁ -1 = -2
y₁ = -2 + 1 = -1
y₁ = -1
A(x₁,y₁) = A(4,-1)
B (x₂ y₂) = B(0, -1)
C(x₃, y₃) = C(0,-1)

OR

Find the length of the median through the vertex A(5,1) drawn to the triangle ABC where other two vertices are B(1, 5) andiC(-3, -1)
Answer:

Question 29.
Prove that the tangents drawn from an external point are equal.
Answer:

Data : ‘O is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that : ∠APB + ∠AOB = 180°
∠OAP + ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (Common side)
OA = OB (Radii of the same circle)
According to RHS postulate
∆APO = ∆BPO
∠ OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° +90° =180°
⇒ Opposite angles of OAPB quadrilateral are supplementary.
OAPB is a cyclic quadrilateral
⇒ ∠ APB + ∠ AOB = 180°

Question 30.
If a chord of circle of radius 10cm subtend an angle of 60° at the centre of the circle. Find the area of the corresponding segment of the circle.
(Take π =3.14, √3 = 1.7)
Answer:
O is the centre of the circle having radius 10 cm.
∠ AOB = 60° is angle subtended at the centre O of the circle by a chord AB.
Draw BL ⊥ OA
In right angled ∆ OLB,


Area of segment ∆ APB
= The area of sector OAPB – Area of ∆ OAB

OR

Find the area of the shaded region where PQRS is a square of side lOcms and semicircles are drawn with each side of square as diameter.
Answer:
The four shaded regions are marked as I, II, III, IV.
All the four regions have equal areas due to § symmetry.
Area of I + Area of III
= Area of square ABCD – Sum of the areas of two semicircles of each of radius 5cm.

{10 x 10 – 2 x \(\frac{1}{2}\) π x 5²}cm²
= {100 – 25 x 3.14} cm²
= {100 – 78.5} cm²
= 21.5 cm²
Similarly Area of II + Area of IV = 21.5 cm2. Total Area of shaded region = 21.5 + 21.5
= 43 cm²

Question 31.
The following table gives the production yield per hectare of paddy of 50 farms of a village.

Answer:

Question 32.
Find the mean of the following frequency distribution.

Answer:

A = 50, h = 20, N = 100 and Σ f_iu_i = 15
Mean = A + h\(\left\{\frac{1}{\mathrm{N}} \sum \mathrm{f}_{i} \mathrm{u}_{\mathrm{i}}\right\}\)
Mean = 50 + 20 x \(\frac{15}{100}\)
Mean = 50 + 3 = 53

Question 33.
Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose Sides are 2/3 of the corresponding sides of it.
Answer:

V. Answer the following : ( 4 x 4 = 16 )

Question 34.
Solve the pair of equations graphically.
4x – 3y + 4 = 0
4x + 3y – 20 = 0
Answer:
4x – 3y + 4 = 0 ….. (1)
4x + 4 = 3y
y = \(\frac{4 x+4}{3}\)

4x + 3y – 20 = 0
3y = -4x + 20
y = \(\frac{-4 x+20}{3}\)

Question 35.
How many terms of the series 93 + 90 + 87 + ……….. amounts to 975. Find also the last term.
Answer:
93 + 90 + 87 + …………
a = 93, d = 90 – 93 = -3, S_n = 975, n = ?,
T = ?
S = \(\frac{n}{2}\)[2a + (n – 1)d]
975 = \(\frac{n}{2}\)[2(93) + (n-1)(-3)]
975 x 2 = n[186 – 3n + 3]
1950 = n [189 – 3n]
1950 = n[189 – 3n]
1950 = 189n – 3n2
(3n²-189n+ 1950-0) H-3
n² – 63n + 650 = 0
n² – 50n-13n +650 = 0
n(n – 50) – 13(n – 50) = 0
(n – 50) (n -13) = 0
n – 50 = 0 or n – 13 = 0
n= 50 or n= 13
When n = 50, the last term
T_n = a + (n – 1)d
= 93 + (49) (-3)
= 93 – 147
T₅₀ =-54
When n = 13, the last term
T_n = a + (n – 1)d ”
=93 + (13-1) (-3)
= 93 – 36
T₁₃ = 57

OR

If m times the mth term of an A.P is equal to n times its n,h term, show that (m + n)th term is zero.
Soln: Let T₁ = a and c.d = d
Given : m times m,h term = n times nthterm
=> mT_m = nT_n
m[a + (m – 1)d] = n[a + (n – 1)d]
m[a +(m – l)d] – n[a+(n -l)d] = 0
m[a + md – d] – n[a + nd – d] = 0
ma + m²d – md – na – n²d + nd = 0
a(m – n) + d(m – n) (m + n) – d(m – n) = 0
(m – n) {a + d(m +n) – d} = 0 . (m – n) {a + (m + n – l)d} = 0
m-n = 0, a + (m + n-l)d = 0
m= n T_m+n , = 0
But m ≠ n

Question 36.
A tower is 50m high. Its shadow is x mtrs shorter when the suns altitude is 45° than when it is 30°. Find the value of x.
Answer:

Let PQ be the tower and let PA and PB be its shadows when the altitudes of the sum are 45° and 30° respectively.
∠ PAQ = 45°, ∠ PBQ = 30°
∠ BPQ = 90° and PQ = 50m ,
Let AB = x metres
From right angled ∆ APQ, we have
\(\frac{\mathrm{AP}}{PQ}\) = cot 45°=1
\(\frac{\mathrm{AP}}{50}\) = 1 => AP = 50m
From right angled ∆ BPQ,
\(\frac{\mathrm{BP}}{\mathrm{PQ}}\) = cot30° = √3
⇒ \(\frac{x+50}{50}\) = √3
x = 50(√3 – 1)m
= 50 (1.73 – l)m
= 50 x 0.732
= 36.6 m

Question 37.
Prove that “In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other sides.”
Answer:

Verify
BC = 8cm
BC’= \(\frac{5}{3} \times 8=\frac{40}{3}\)
AB =6cms
A’B= \(\frac{5}{3} \times 6\)
=10cms.

VI. Answer the following : ( 5 x 1 = 5 )

Question 38.
The height of cone is 20m. A small cone is cut off from it at its top by the plane parallel to the base. If the volume of small cone 1/1000 th of the volume of given cone, at what height above the base the section is made.
Answer:
In A OPB and A OQC A OPB ~aOQC OP _ PB OQ ~ QC

Karnataka State Syllabus SSLC English Previous Year Question Paper March 2018 (2nd Language)

Time: 3 hours
Maximum Marks: 80

I. Four alternatives are given for each of the following questions / incomplete statements.
Only one of them is correct or most appropriate. Choose the correct alternative and write the complete answer along with its letter of the alphabet. [3]

Question 1.
Father did not like Swami
(A) joining the cricket club
(B) sleeping in the office room
(C) sleeping beside his granny
(D) biting a burglar’s leg.

Question 2.
Roma Talreja felt that she could never repay Baleshwar because he
(A) informed Dinesh about the accident
(B) requested all the motorists to save her
(C) risked his life to save her
(D) pleaded the passengers to save her.

Question 3.
The concept “immersion in science doesn’t go with practical sense” suggests
(A) foolishness of scientists
(B) absentmindedness of scientists
(C) scientists do not bother about their research results
(D) scientists seldom work together.

II. Answer the following questions in 2 – 3 sentences each: [4×2=8]

Question 4.
How did the tempo truck driver help Baleshwar?

Question 5.
What was the intention of Smita’s going to the concert?

Question 6.
The artists visiting Anant was unusual. Justify.

Question 7.
Despite low income, why do some people choose to become scientists?

III. Read the following extracts and answer the questions that follow: [9]

Question 8.
“Why don’t you join the police when you are grown up ?”
(a) Who gave this suggestion?
(b) Why did he give this suggestion?
(c) What did the listener actually want to become in his later life?

Question 9.
“The trees in the orchard are not mine.”
(a) Who is the speaker here?
(b) According to the speaker, who else do the trees belong to?
(c) Why does he feel that the trees don’t belong to him ?

Question 10.
“Who could have dreamt that one born to a Mahar family would one day become not only a Law
Minister but a Law-maker ?”
(a) Who is referred to as ‘Law Minister’ here?
(b) Who chose him to be the ‘Law Minister’?
(c) Why is he recognised as the Law-maker?

IV. Answer the following question in about 8-10 sentences: [4]

Question 11.
How did the crew members express their discontent with Columbus?
OR
Explain the incident that forced Satish to stay back at home. How did it affect his formal education?

Poetry

V. Choose the most appropriate one and write the complete answer with the letter of alphabet in the answer-book: [1]

Question 12.
In the line from the poem, ‘You say you own me’, the terms ‘you’ and ‘me’ refer to
(A) children and the farmer
(B) man and the land
(C) soldier and the land
(D) buyer and seller of the land.

VI. Answer the following questions in 2 – 3 sentences each: [4]

Question 13.
How does the poet praise India being developed in industry and technology?

Question 14.
The speaker in ‘Off to Outer Space Tomorrow Morning’ is uncertain of his return. How does he express this?

VII. Read the given extract and answer the questions that follow: [3]

Question 15.
“Isn’t God upon the ocean Just the same as on the land ?”
(a) Who is the speaker here?
(b) When was this said?
(c) What was the result of these words?

VIII.
Question 16.
Quote from memory: [4]
“It is…………………………………………………..
………………………………………………himself
And ……………………………………………………
………………………………………………………justice.”
OR
“Then let…………………………………………….
…………………………………………….. destroy:
Whilst………………………………………………..
……………………………………………………… boy.”

IX. Answer the following question in 8 to 10 sentences: [4]

Question 17.
Describe in your own words how the poet’s grandma was a genius and how she upheld ‘her right to reside in a tree’.
OR
The physical appearance of the Jazz player does not match when he plays on his saxophone. Explain.

Supplementary Reading

X. Answer the following questions in 2 – 3 sentences each: [4]

Question 18.
Why were Babu and Manju a bit disappointed with the way the students were marching?
OR
How did the lofty Himalayas inspire Dicky Dolma?

Question 19.
How did Hema Aziz teach her children the message that “one’s duty is the most important thing in life”?
OR
What changes came over Wangjia as the Bird of Happiness caressed him?

Vocabulary

XI.
Question 20.
Which one of the following words has one syllable? [1]
Father, Cheque, Office, Travel.

Question 21.
Combine the word in Column-A with its collocative word in Column-B: [1]

Question 22.
Fill in the blanks choosing the appropriate words given in brackets: [1]

The_______ ( not, knot) of the rope was so strong that it could________(knot, not) be removed by us.

Question 23.
Give one word for ‘very eager for knowledge’. [1]

SECTION – B

Reading

XII. Read the following passage and answer the questions that follow: [4]

Once there was a slave. His name was Andrades. His master was very cruel. He treated him badly. So, the slave was sick of him and ran away into a forest. There he saw a lion crying with pain. The lion sat before him and held out its paw. There was a big thorn in it. The slave drew it out and dressed the wound. They became friends.

After a few days, the slave was caught. He was ordered to be thrown before a hungry lion. Many people came to see the act of cruelty.

A lion was caught for this purpose. It was kept hungry for many days. On the fixed day, the lion was let loose on the slave. It recognised him. It began to lick his feet. ‘

All the people were surprised. The slave told the whole story. The master was pleased and set Andrades free.

Question 24.
What made Andrades go to the forest?

Question 25.
How did Andrades help the lion?

Question 26.
What act of cruelty did the people go to watch?

Question 27.
Why didn’t the lion kill Andrades?

Question 28.
Rearrange the jumbled words into a meaningful sentence: [1]
garden / flowers / in / the / pluck / don’t

Question 29.
Look at the two lines. Write a sentence using ‘longer than’ based on what you see: [1]

XIII. Read the following conversation and answer the questions that follow:

News Reporter: Congratulations Rashmi! Your feat is laudable. By the way, did you expect first rank in the CET this year?
Rashmi: No. Actually I expected to be one in top ten ranks.
News reporter: How could you achieve this?
Rashmi: I worked hard from the beginning. My teachers, my college library, my parents were of great help to me.

Question 30.
Why does the news reporter congratulate Rashmi? [1]

Question 31.
What was the secret of Rashmi’s achievement? [1]

Writing

XIII.
Question 32.
Given below is the profile of Rabindranath Tagore. Write a paragraph using the clues [4]

OR
Write a paragraph, using the clues given below. You may add some more points if necessary :
Twelve fools start on a journey — cross a river — count themselves — one man lost—a passer-by offers help — gives each a blow on the back — they count twelve strokes — satisfied — moral

Question 33.
Imagine that you are Divya / Dinesh, residing at ‘Sai Nivas’, #74,1 Cross, Vivek Nagar, Hiriyur.
Write a letter to your cousin inviting him/her to come to your house to spend summer vacation. [4]
OR
Write a letter to the Chief Officer, Town Municipal Council, Hiriyur about the garbage dumped in your locality and request him/her to do the needful.

Editing

Question 34.
The following paragraph has two errors. Edit the paragraph and rewrite it in the answer-book. Clues are given. [2]
Whales are normally gentle. Many whales exibit strong family ties. The young ones remain with their parents for up to fifteen years or more. How wonderful it is.
Clues:
(a) Spelling mistake to be corrected
(b) Use correct punctuation mark.

GRAMMAR ( Language Use )

XIV.
Question 35.
Read the given conversation and choose the language function for the underlined sentence: [1]
Visitor: Excuse me, can you help me in locating the book, ‘War And Peace’?
Librarian: Sure. Go to the sixth row and the book is on the third shelf.
Visitor: Thank you very much.
(A) Advice
(B) Giving directions
(C) Order
(D) Request.

Question 36.
Read the conversation and rewrite into a reported form:
Ravi: Good morning daddy.
Father: Good morning Ravi. Go to the railway station at 4 P.M. and bring your grandpa.
Ravi: I will daddy, don’t worry.
Ravi and father greeted each other.
Father asked Ravi…………………………….
……………………………………………………………
Ravi replied………………………………………
……………………………………………………………

Question 37.
Fill in the blanks with appropriate ‘article’ and ‘preposition’: [1]
There are_______ number of newspapers_______ Kannada and English.

Question 38.
Gopi broke the glass,_______? [1]
The question tag to be used for the above statement is
(A) did he
(B) was he
(C) wasn’t he
(D) didn’t he.

Question 39.
Fill in the blanks with appropriate tense forms of the verbs given in brackets: [2]
Maya is a lively woman. She_______ (love) visiting beaches. Last year, she_______ (have + visit) seven beautiful beaches of Indian islands.

Question 40.
Fill in the blanks using the suitable linking words given in brackets: [1]
The flight in which I travelled was delayed_______ the sky was very cloudy._______, my parents
were much worried about my safety, (but, so, because, that)

Question 41.
Read the conversation and fill in the blanks with correct ‘if clause’ choosing from the given alternatives: [1]
Neighbour: Hi Ranjith, Did you get a seat in the college you desired?
Ranjith: No uncle, my application was rejected due to my low scores.
Neighbour: I am sorry to hear this. If you had worked harder, you_______ got a seat in the college.
(A) will have
(B) shall have
(C) would have
(D) should have.

Question 42.
Read the conversation and identify the infinitive: [1]
Cook: What shall I cook for dinner tonight, Sir?
Master: Prepare some special items. I am expecting two guests to come for dinner.
Cook: Alright Sir, what about Pulao and Halwa?
Master: Okay, also make some fruit salad.
Cook: Yes Sir.
(a) Prepare
(b) Expecting
(C) To come
(d) Make

Reference Skill

XV.
Question 43.
Arrange the following words in the order in which they are found in dictionary: [2]
adverb, advocate, advise, advice.

Question 44.
Write it in the normal way: [1]
my sis gt 80% in maths test

Question 45.
Which source of information would you refer to know about global warming? Choose the correct source: [1]
Thesaurus, Encyclopaedia, Atlas, Dictionary.

Answers

Answer 1:
(C) Sleeping beside his granny.

Answer 2:
(C) he risked his life to save her.

Answer 3:
(C) Scientists do not bother about their research results.

Answer 4:
The tempo truck driver stepped out of his vehicle. He was a middle-aged man. He responded to Baleshware’s request and helped Baleshwar lay Roma down in the back of his truck and helped him to take Roma to hospital nearby.

Answer 5:
The intension of smitha’s going to the concert was to request Pandith Ravishankar to come to their house and play sitar for her brother Ananth suffering from cancer.

Answer 6:
The artists who visited anant were Pandit Ravishankar a great Performer of Sitar instrument and Ustad Alla Rakha well – known Tabla artist.

Answer 7:
Science is an exciting adventure where major reqard comes from the discovery itself. What they get instead of money is freedom, camaradesire and independence. The honours and promotions will depend on themselves and feeling of uncovering one of nature’s mysteries.

Answer 8:
(a) The Inspector gave this suggestion.
(b) Because swami helped police in catching the notorious house-breakes of the district.
(c) The listener made his mind to be an engine driver, a railway guard or a bus conductor later on life.

Answer 9:
(a) Don Anselmo is the speaker here.
(b) The trees belong to the descendants of RIO en Medio.
(c) Because since he took the posession of that house, he planted a tree for the children of Rio – en Medio.

Answer 10:
(a) B.R. Ambedkar is referred to as ‘Law Minister’ here.
(b) Jawaharlal Nehru Chose him to be the law minister.
(c) He is recognised as the law-makes because of his skills in te field of law and legislation and his vision of social justice.

Answer 11:
The crew members were expressing them discontent by singing the song again and again. They were feeling like bats trying to fly by day.
They were not ready to be in the hands of Columbus who was a man of quick temper. Columbus was always trying to control them. He was suspecious about whatever they talk. The crew members were unhappy about being in voyage for many days. They have become impatient and thinking that whether is god dumb and not listening to their prayer. They were fed up with the dictatorial attitutde of columbus. They were drinking and singing the song for their relaxation. But columbus was Irritated hearing to it several times. The crew member and sea men became so much dissatisfied that they were not obeying the order of columbus.

OR
These had been an accident when satish went living with his father and brother Inder. They were crossing a rickety bridge over some rapids. The boys stood and looked down at swisling water below them. Just then Satish slipped and losing his balance. He fell into the rapids, when he regained consciousness. His legs were in a plaster cast and his head lust badly. Though his legs healed, they remained weak, requiring several operations on them. He suffered frequently from bows of fever and infections, especially of the ear. He was gesting terrible headaches and them feel as of everything is going dark and silent.

This situation made satish to take leave again and again from school. His frequent absence made the school to say that they couldn’t keep him in the school and suggested Satish’s father to put his son to the school for deaf and dumb. His father made unsuccessful expedition to other schools. Satish become very moody and couldn’t talk freely with anyone.

Answer 12:
(b) Man and the land.

Answer 13:
The poet says that he can 5 mg. about the dam and the lake. We appreciates the steel molls, ship building yard and praises the hard work of men behind that further, talking about technology he sings about atomic age and ironmen came in their wake.

Answer 14:
The poet expresses the uncertainly of his returning by saying that we can take a last look at him by passing the helmet to him from as plastic hook. His name can be crossed out from the telephone book because he will be off to outer space tomorrow morning.

Answer 15:
(a) The seamen are the speakers here.
(b) This was said when they were on voyage.
(c) The seamen were feeling relaxation after singing these words.

Answer 16:
“It is enthroned in the heart of kings:
It is an attribute to god himself;
and certainly power doth then show likes god’s.
When mercy seasons justice.

OR
Then let not what
I cannot have my cheer of mind destroy;
Whrist thus I sing, I a making.
Although a poor blind boy.

Answer 17:
The poet’s grandmother is a genius. She had the gift of being happies on a tree than on a left from childhood. She could climb trees, spreading or high. She would be up their branches in a trice when she was sixty two. Once she climbed a tree and couldn’t come down. After the rescue, she had to take a week’s rest. But after she felt stronger. She called poet’s father and told him to build a house on a tree top. After the construction of that house, she moved up to her tree- house and thepoet climb to her room everyday with glasses and tray. The poet climb to her room everyday with glasses and tray. The poet and granny were sitting there and drinking sherry. This way poet’s grandmother uphold “her right to reside in a tree”.

OR
Jazz player’s physical appearance is not matching with his talent when he plays on saxophone. He is a pathetic figure but at the same time commanding artist. He is silent on speech, ugly in physical appearance but eloguent on music. He has wrinkled old face, faded-blue old shirt dark with sweat, old necktre drops loosely about the warn, holding his stagging stomach, run-down shoes work papers on them. In betwen all these he lofts his saxophone supported from his neck by a wire coat hunger and preaches the blaag Gospel of Jazz with words of screaming notes and chords and fires in his music high and high and makes people listen.

Answer 18:
The students walked in complete silence. There were no slogans, no shouts, just the shuffle of feet, the drop of rain and a low murmur from the watching crowd.

OR
Dicky Dolma was fascinated by the grandeur of the snow-clad peaks of the Himalayas from child hood every morning, these beautiful mountain peak she used to see first.

Answer 19:
Hema Aziz often left hanif and his brother alone and travelled with the performance wing of the information and broadcasting ministry. This made her kids hanif and his brother responsible and they were getting up ready for without her having to tell them. Her going to work gave them the message that one’s duty is the most important thing in life.

OR
The birds of happiness casessed wangjia with its wings and sang for him. His eyeballs flew back to their sockets and now he saw much more brightly than before. All his wounds were healed and he was stronger than eve.

Answer 20:
Cheque.

Answer 21:
Rusting Waves.

Answer 22:
The knot of the rope was so strong that it could not be removed by us.

Answer 23:
Earnest

Answer 24:
Andrade’s master was very cruel and treated him badly.

Answer 25:
Andrades drew out the big thorn in the paw of the iron and dressed the wound.

Answer 26:
The throwing of slave before hungry lion.

Answer 27:
The lion didn’t kill andrakes because it recognised him and began to kill his feet.

Answer 28:
Don’t pluck the flowers in the garden.

Anser 29:
B is longer than A

Answer 30:
News suporter congratulated rashmi for securing first rank on the CET that year.

Answer 31:
The hardwork from the beginning was the secret of Rashmi’s achievement.

Answer 32:
Rabindranath Tagore was known as Gurudeva in Bengal. He wa bom on 7th May 1861. His parents were Debendranath Tagore and Sarala Devi. Rabindranath Tagore was a writer, painter and singer. His notable works are Gitanjali, Jana Gana Mana, Rabindra Sangeet. Amar Sonar Bangla etc. He got the Nobel prize for literature in 1913 for his work ‘Gitanjali’. He died on 7th August, 1941.

OR
Once twelve fools started on a journey they had to cross a river they decided to confirm that all of them crossed the river. So one of them started counting themselves. But everytime only 11 members were there and one man lost. They asked the help of a passes by. That passes by gave each a blow on the back and these fools counted 12 strokes. They were satisfied that all 12 were there. They decided to cross the river and reached their destination.
Moral: We should never be foolish.

Answer 33:
From:
Divya / Dinesh
“Sai Nivas’, #741 Cross,
Vivek Nagar, Hiriyur.
Date:………..
Dear Madhu,

I am fine here. I hope you are also fine. Our summer vacation begins from next week. I am inviting you to come to my house to spend summer vacation with me. Our village is a beautiful place. There are historical spots in our village. We can enjoy the beauty of nature. Please, come to my house and we will have an enjoyful summer holidays.

Your lovely Sister/Brother
Sincerly
(Divya / Dinesh)

OR

From:
ABC
Date:……..
Respected Sir/Madam,

Subject: Dumping of Garbage in our locality.

Hereby bringing to your notice, the garbage dumped in our locality has become a big problem. It led to the mosquitoes in the surrounding areas and people of our locality are suffering from fever that is very dangerous. We requested many times to the concerned department but action was not yet taken kindly, consider the seriousness of the situation, and do the needful as early as possible.

Thanking you,

Your’s Sincerely,
ABC

Answer 34:
Whales are normally gentle. Many whales exhibit strong family ties. The young ones remain with their parents for up to fifteen years or more. How wonderful it is!.

Answer 35:
(B) Giving directions.

Answer 36:
Father asked ravi to go to the railway station at 4 pm. to bring his grandpa. Ravi replied he would and not to worry.

Answer 37:
There are a number of newspapers in kannada and english.

Answer 38:
(D) didn’t he?

Answer 39:
Maya is a lively woman she loves visiting beaches. She had visited seven beautiful beaches of Indian islands.

Answer 40:
The flight in which I travelled was delayed because the sky was very cloudy. So my parents were much worried about my safety.

Answer 41:
(C) Would have

Answer 42:
(C) to come

Answer 43:
adverb, advice, advise, advocate.

Answer 44:
My sister got 80% in Maths test.

Answer 45:
Encyclopaedia.