Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

Karnataka SSLC Maths Model Question Paper 4 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.
ಕೆಳಗಿನ ಸಂಖ್ಯೆಗಳ ಗಣಗಳಲ್ಲಿ ಸಮರೂಪ ತ್ರಿಭುಜಗಳಾಗುವಂತಹ ಜೋಡಿಯು
(A) (3, 4, 6) (9, 12, 24)
(B) (3, 4, 6) (9, 12, 18)
(C) (2,4, 6) (2,3, 14)
(D) (5, 10, 15) (10, 30, 45)

Question 2.
ಚತುರ್ಥಕದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
(A) πr2
(B) \(\frac { 1 }{ 2 }\) πr2
(C) \(\frac { 1 }{ 3 }\) πr2
(D) \(\frac { 1 }{ 4 }\) πr2

Question 3.
(2, 3) ಮತ್ತು (4, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಮಧ್ಯಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳು
(A) (3, 4)
(B) (4, 5)
(C) (5, 6)
(D) (6, 7)

Question 4.
ಧನ ಪೂಣಾಂಕ ‘a’ ಅನ್ನು ಪೂಣಾಂಕ ‘b’ ನಿಂದ ಭಾಗಿಸಿದಾಗ ಅನನ್ಯ ಪೂರ್ಣಾಂಕಗಳಾದ ‘q’ ಮತ್ತು ‘r’ ಗಳು ಇರುತ್ತವೆ. ಇದಕ್ಕೆ ಸರಿಹೊಂದುವ ಹೇಳಿಕೆ.
(A) b = a × q + r
(B) q = a × b + r
(C) a = b × q – r
(D) a = b × q + r

Question 5.
tan260° ಯ ಬೆಲೆ
(A) \(\frac { 1 }{ 3 }\)
(B) \(\frac { 1 }{ \surd 3 }\)
(C) 3
(D) √3

Question 6.
m ನ ಯಾವ ಧನಾತ್ಮಕ ಬೆಲೆಗೆ 3x2 + ka + 3 = 0 ಸಮೀಕರಣದ ಮೂಲಗಳು ಸಮವಾಗಿರುತ್ತವೆ.
(A) 2
(B) 3
(C) 5
(D) 6

Question 7.
ಒಂದು ಪ್ರಯೋಗದ ಎಲ್ಲಾ ಪ್ರಾಥಮಿಕ ಘಟನೆಗಳ ಸಂಭವನೀಯತೆಗಳ ಮೊತ್ತವು
(A) 0
(B) 1
(C) 2
(D) 3

Question 8.
ಒಂದು ಸಿಲಿಂಡರ್‌ನ ಪಾದದ ವಿಸ್ತೀರ್ಣ 24 cm2 ಮತ್ತು ಎತ್ತರ 10 cm ಆದರೆ ಸಿಲಿಂಡರ್‌ನ ಘನಫಲ
(A) 24 cm3
(B) 48 cm3
(C) 240 cm3
(D) 480 cm3

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

Question 9.
ಥೇಲ್ಸ್‌ನ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,

Question 10.
135 ಮತ್ತು 225 ರ ಮ.ಸಾ.ಅ. ವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 11.
ಬಹುಪದೋಕ್ತಿ x2 – 3x +5 ರ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 12.
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕದ ಉದ್ದವು 8 cm ಮತ್ತು ವೃತ್ತಕೇಂದ್ರ ಹಾಗು ಬಾಹ್ಯಬಿಂದುವಿನ ದೂರ 10 cm ಆದರೆ ವೃತ್ತದ ತ್ರಿಜ್ಯವೆಷ್ಟು?

Question 13.
sin A = \(\frac { 3 }{ 4 }\) ಆದರೆ cosec A ನ ಬೆಲೆಯೇನು?

Question 14.
ಹಂತ ವಿಚಲನಾ ವಿಧಾನದಿಂದ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರವನ್ನು ಬರೆಯಿರಿ.

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

Question 15.
ವಾರ್ಷಿಕ ಸಂಬಳ ₹ 5000 ಮತ್ತು ಪ್ರತಿವರ್ಷಕ್ಕೆ ಹೆಚ್ಚುವರಿ ಬತ್ಯ ₹ 200 ಇರುವ ಕೆಲಸಕ್ಕೆ ಸುಬ್ಬರಾವ್ 1995 ರಲ್ಲಿ ಸೇರಿದರು. ಯಾವ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ?

Question 16.
ಚಿತ್ರದಲ್ಲಿ LM || CB ಮತ್ತು LN || CD ಆದರೆ \(\frac { AM }{ AB }\) = \(\frac { AN }{ AD }\) ಎಂದು ಸಾಧಿಸಿ.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium Q16

Question 17.
ಈ ಜೋಡಿ ಸಮೀಕರಣಗಳನ್ನು ವರ್ಜಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ, 3x + 4y = 10 & 2x – 2y = 2.

Question 18.
ಕೆಳಗಿನ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳ ಜೋಡಿಗಳು ಪ್ರತಿನಿಧಿಸುವ ಸರಳರೇಖೆಗಳು ಒಂದು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆಯೇ? ಸಮಾಂತರವಾಗಿವೆಯೆ? ಅಥವಾ ಐಕ್ಯಗೊಂಡಿವೆಯೆ? ಕಂಡುಹಿಡಿಯಿರಿ
9x + 3y + 12 = 0; 18x + 6y + 24 = 0.

Question 19.
ಪರಧಿಯು 22 cm ಇರುವ ಒಂದು ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 20.
4 cm ಮತ್ತು 6 cm ತ್ರಿಜ್ಯಗಳಿರುವ ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳಿವೆ. 6 cm ತ್ರಿಜ್ಯದ ವೃತ್ತದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ 4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವನ್ನು ರಚಿಸಿ.

Question 21.
(2, -5) ಮತ್ತು (-2, 9) ರಿಂದ ಸಮಾನ ದೂರದಲ್ಲಿರುವ x-ಅಕ್ಷದ ಮೇಲಿನ ಬಿಂದುವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 22.
ಶೃಂಗಬಿಂದುಗಳು (1, -3), (4, 1) ಮತ್ತು (2, 3) ಆಗಿರುವ ತ್ರಿಭುಜದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ \(\frac { -1 }{ 4 }\) ಮತ್ತು ಗುಣಲಬ್ಧ \(\frac { 1 }{ 4 }\) ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
x2 + 6x – 7 = 0 ಸಮೀಕರಣವನ್ನು ವರ್ಗಪೂರ್ಣಗೊಳಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ.

Question 25.
ಗೋಪುರದ ಪಾದದಿಂದ 30m ದೂರದ ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ, ಗೋಪುರದ ತುದಿಯನ್ನು ನೋಡಿದಾಗ ಉಂಟಾಗುವ ಉನ್ನತ ಕೋನವು 30° ಆದರೆ, ಗೋಪುರದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ಹಾಗೂ ಕೆಳಗಿನಿಂದ ಬೆಟ್ಟದ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಉಂಟಾದ ಉನ್ನತ ಕೋನವು 45° ಮತ್ತು 60° ಆಗಿದೆ. ಕಟ್ಟಡದ ಎತ್ತರ 24 m ಆದರೆ ಬೆಟ್ಟದ ಎತ್ತರವೇನು?

Question 27.
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium Q27
ಅಥವಾ
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಬಹುಲಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium Q27.1

Question 28.
ಒಂದು ಚೀಲದಲ್ಲಿ 3 ಕೆಂಪು ಚೆಂಡುಗಳು ಮತ್ತು 5 ಕಪ್ಪು ಚೆಂಡುಗಳಿವೆ. ಚೀಲದಿಂದ ಯಾದೃಚ್ಛಿಕವಾಗಿ ಒಂದು ಚೆಂಡನ್ನು ತೆಗೆಯಲಾಗಿದೆ. ತೆಗೆದ ಚಂಡು ಕೆಂಪು ಆಗಿರುವ ಸಂಭವನೀಯತೆ ಎಷ್ಟು?

Question 29.
64 cm3 ಘನಫಲವನ್ನು ಹೊಂದಿರುವ 2 ವರ್ಗ ಘನಗಳ ಮುಖಗಳನ್ನು ಸೇರಿಸಿ ಒಂದು ಆಯತ ಘನಾಕೃತಿ ಮಾಡಿದೆ. ಈ ಘನಾಕೃತಿಯ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
12 ಮತ್ತು 15 ರ ಲ.ಸಾ.ಅ. ಮತ್ತು ಮ.ಸಾ.ಅ. ವನ್ನು ಅವಿಭಾಜ್ಯ ಅಪವರ್ತನ ವಿಧಾನದಿಂದ 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
ಒಂದು ಸಮಾಂತರ ಚತುರ್ಭುಜದಲ್ಲಿ ವೃತ್ತವು ಅಂತಸ್ಥವಾದಾಗ ಸಮಾಂತರ ಚತುರ್ಭುಜವು ವಜ್ರಾಕೃತಿಯಾಗುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 32.
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 33.
ಎರಡು ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 290 ಆದರೆ ಆ ಪೂಣಾರ್ಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಆಯತಾಕಾರದ ಹೊಲದ ಕರ್ಣವು ಅದರ ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 60 m ಹೆಚ್ಚಾಗಿದೆ. ಅದರ ದೊಡ್ಡ ಬಾಹುವು ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 30 m ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಆ ಹೊಲದ ಬಾಹುಗಳ ಉದ್ದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium Q34
ಅಥವಾ
sec A (1 – sin A) (sec A + tan A) = 1 ಎಂದು ಸಾಧಿಸಿ.

Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ, (ಕಡಿಮೆ ವಿಧಾನ)
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium Q35

Question 36.
3x2 – x – 4 ಎಂಬ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆ ನೋಡಿ.
ಅಥವಾ
x4 – 3x2 + 4x + 5 ನ್ನು x2 + 1 – x ದಿಂದ ಭಾಗಿಸಿ ಭಾಗಲಬ್ದ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

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

Question 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳ ಮೊತ್ತವು 15 ಮತ್ತು ಅವುಗಳ ಅಂತ್ಯಪದಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 58 ಆಗಿದೆ. ಶ್ರೇಢಿಯ ಆ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ 5 ಪದಗಳ ಮೊತ್ತವು ಮುಂದಿನ 5 ಪದಗಳ ಮೊತ್ತದ ನಾಲ್ಕನೇ ಒಂದು ಭಾಗದಷ್ಟಿದೆ. ಮೊದಲ ಪದ 2 ಆದರೆ, a20 = -112 ಎಂದು ಸಾಧಿಸಿ ಮತ್ತು S20 ನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 38.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತವು ಅವುಗಳ ಅನುರೂಪ ಬಾಹುಗಳ ವರ್ಗಗಳ ಅನುಪಾತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 39.
ಒಂದು ಬಾವಿಯ ವ್ಯಾಸ 3 m ಮತ್ತು ಆಳ 14 m ಇರುವಂತೆ ತೋಡಿದೆ. ಭೂಮಿಯಿಂದ ತೆಗೆದ ಮಣ್ಣನ್ನು ಬಾವಿಯ ಸುತ್ತಲು ಸಮವಾಗಿ ಹರಡಿ 4 m ಅಗಲವಿರುವ ವೃತ್ತಾಕಾರದ ಕಟ್ಟೆಯನ್ನು ಕಟ್ಟಿದೆ. ಕಟ್ಟೆಯ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 40.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ.
2x + y = 8
x + 2y = 7

Solutions

I.
Solution 1.
(B) (3, 4, 6) (19, 12, 18)

Solution 2.
(D) \(\frac { 1 }{ 4 }\) πr2

Solution 3.
(A) (3, 4)

Solution 4.
(D) a = b × q + r

Solution 5.
(C) 3

Solution 6.
(D) 6

Solution 7.
(B) 1

Solution 8.
(C) 240 cm3

II.
Solution 9.
ತ್ರಿಭುಜದ ಎರಡು ಬಾಹುಗಳನ್ನು ಎರಡು ವಿಭಿನ್ನ ಬಿಂದುಗಳಲ್ಲಿ ಛೇದಿಸುವಂತೆ ಒಂದು ಬಾಹುವಿಗೆ ಸಮಾಂತರವಾಗಿ ಎಳೆದ ಸರಳರೇಖೆಯು ಉಳಿದೆರಡು ಬಾಹುಗಳನ್ನು ಸಮಾನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುತ್ತದೆ.

Solution 10.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S10

Solution 11.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S11

Solution 12.
OA2 = OP2 – AP2 =102 – 82
⇒ OA2 = 100 – 64
⇒ OA2 = 100 – 64
⇒ OA = √36
⇒ OA = 6 cm
ವೃತ್ತದ ತ್ರಿಜ್ಯ = 6 cm

Solution 13.
cosec A = \(\frac { 4 }{ 3 }\)

Solution 14.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S14

III.
Solution 15.
5000, 5200, 5400, ……… 7000.
a = 5000, d = 200 an = 7000, n = ?
an = a + (n – 1) d
⇒ 7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ 2000 = 200 (n – 1)
⇒ (n – 1) = 10
⇒ n = 10 + 1
⇒ n = 11
11 ನೇ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ.

Solution 16.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S16

Solution 17.
3x + 4y = 10 …. (1)
2x – 2y = 2 …… (2)
ಸಮೀಕರಣ 2ನ್ನು 2ರಿಂದ ಗುಣಿಸಿದಾಗ
3x + 4y = 10
4x – 4y = 4
…………………
7x = 14
x = 2
3x + 4y = 10
3(2) + 4y = 10
4y = 10 – 6
y = 1
∴ x = 2, y = 1

Solution 18.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S18

Solution 19.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S19
ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣ = \(\frac { 77 }{ 4 }\) cm2

Solution 20.
R = 6 cm, r = 4 cm
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S21

Solution 22.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S22

Solution 23.
ಬಹುಪದೋಕ್ತಿ = ax2+ bx + c ಆಗಿರಲಿ
ಶೂನ್ಯತೆಗಳು α ಮತ್ತು β ಆಗಿರಲಿ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S23

Solution 24.
x2 + 6x – 7 = 0
⇒ x2 + 6x = 7
⇒ x2 + 6x + 32 – 32 = 7
⇒ (x + 3)2 – 9 = 7
⇒ (x + 3)2 = 16
⇒ x + 3 = ±√16
⇒ x + 3 = ±4
⇒ x = ±4 – 3
⇒ x = +4 – 3 ಅಥವಾ x = -4 – 3
⇒ x = 1 ಅಥವಾ x = -7

Solution 25.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S25

Solution 26.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S26
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S26.1

Solution 27.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S27

Solution 28.
ಚೀಲದಲ್ಲಿರುವ ಒಟ್ಟು ಚೆಂಡುಗಳು = n(S) = 3 + 5 = 8
ಕೆಂಪು ಚೆಂಡುಗಳು ಸಂಖ್ಯೆ = 3 = n(A)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = \(\frac { n(A) }{ n(S) }\)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = \(\frac { 3 }{ 8 }\)

Solution 29.
ವರ್ಗಗಳ ಘನಫಲ = 64cm3
ಬಾಹುವಿನ ಅಳತೆ = \(\sqrt [ 3 ]{ 64 }\) = 4 cm
ಆಯತ ಘನದ ಅಗಲ b = 4 cm
ಆಯತ ಘನದ ಉದ್ದ l = 8 cm (4 + 4)
ಆಯತ ಘನದ ಎತ್ತರ h = 4 cm
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 160 cm2
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 160 cm2
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Solution 30.
12 = 2 × 2 × 5
15 = 3 × 5
ಮ.ಸಾ.ಅ. = 5
ಲ.ಸಾ.ಅ. = 2 × 2 × 3 × 5
ಲ.ಸಾ.ಅ. = 60

IV.
Solution 31.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S31
ದತ್ತ: ‘O’ ವೃತ್ತಕೇಂದ್ರ, XY ಸ್ಪರ್ಶಕ, Pಸ್ಪರ್ಶಬಿಂದು
ಸಾಧನೀಯ: OP ⊥ XY
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತ್ತೊಂದು ಬಿಂದು Q ಆಗಿರಲಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: OQ ವೃತ್ತವನ್ನು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸಿದೆ.
OP = OR (ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು)
OQ = OR + RQ
OQ > OR
OQ > OP (OP = OR)
OP ಯು O ನಿಂದ ಸ್ಪರ್ಶಕಕ್ಕೆ ಕನಿಷ್ಟ ದೂರವಾಗಿದೆ
OP ⊥ XY
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S31.1
ಸಾಧನೀಯ: ABCD ಒಂದು ವಜ್ರಾಕೃತಿ. (AB = BC = CD = AD)
ಸಾಧನೆ: AB = CD ಮತ್ತು AD = BC …… (1)
AP = AS, BP = BQ, CQ = CR, DS = DR
(ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು)
AB + CD = AP + PB + DR + CR
⇒ AB + CD = AS + BQ + DS + CQ
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ AB + AB = AD + AD [AB = CD, AD = BC]
⇒ 2AB = 2AD
⇒ AB = AD ….. (2)
∴ AB = BC = CD = AD (ಸಮೀಕರಣ (1) & (2) ರಿಂದ)

Solution 32.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S32

Solution 33.
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು x & x + 2
x2 + (x + 2)2 = 290
⇒ x2 + x2 + 4x + 4 – 290 = 0
⇒ 2x2 + 4x – 286 = 0
⇒ x2 + 2x – 143 = 0
⇒ x2 + 13x – 11x – 143 = 0
⇒ x (x + 13) – 11(x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x + 13 = 0 ಅಥವಾ x – 11 = 0
⇒ x = -13 ಅಥವಾ x = 11
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು 11 & 13
[∴ x = 11 & x + 2 = 11 + 2 = 13]
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S33
ಚಿಕ್ಕ ಬಾಹು x ಆಗಿರಲಿ
AC2 = AB2 + BC2
⇒ (x + 60)2 = x2 + (x + 30)2
⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60x
⇒ x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
⇒ -x2 + 2700 + 60x = 0
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x (x – 90) + 30 (x – 90) = 0
⇒ x – 90 = 0 ಅಥವಾ x + 30 = 0
⇒ x = 90 ಅಥವಾ x = -30
ಚಿಕ್ಕ ಬಾಹುವಿನ ಉದ್ದ = x = 90 m
ದೊಡ್ಡ ಬಾಹುವಿನ ಉದ್ದ = x + 30 = 90 + 30 = 120 m

Solution 34.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S34
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S34.1

Solution 35.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S35

Solution 36.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S36
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S36.1

V.
Solution 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳು
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
(a – d)2 + (a + d)2 = 58
⇒ a2 + d2 – 2ad + a2 + d2 + 2ad = 58
⇒ 2a2 + 2d2 = 58
⇒ 2(a2 + d2) = 58
⇒ a2 + d2 = 29
⇒ 52 + d2 = 29
⇒ d2 = 29 – 25
⇒ d2 = 4
⇒ d = ± 2
⇒ d = 2 ಅಥವಾ d = -2
ಮೂರು ಪದಗಳು
∴ a – d = 5 – 2 = 3
∴ a = 5
∴ a + d = 5 + 2 = 7
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S37
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S37.1

Solution 38.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S38
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S38.1

Solution 39.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S39
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S39.1

Solution 40.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S40
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S40.1
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium S40.2

Karnataka SSLC Maths Model Question Papers

Karnataka SSLC Maths Model Question Paper 3 Kannada Medium

Karnataka SSLC Maths Model Question Paper 3 Kannada Medium

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
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q8

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, an = 50 ಆದರೆ Sn ಕಂಡುಹಿಡಿಯಿರಿ.

Question 16.
ಚಿತ್ರದಲ್ಲಿ ∆ABCಯ ಎತ್ತರಗಳಾದ AD ಮತ್ತು CE ಗಳು ಪರಸ್ಪರ P ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ∆AEP ~ ∆CDP ಎಂದು ಸಾಧಿಸಿ.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q16
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ 0ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.

Question 17.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q17

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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q19

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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q27

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,
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q35
(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?
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium Q37

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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S12
∆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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S14

III.
Solution 15.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S15

Solution 16.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S16
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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S17

Solution 18.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S18

Solution 19.
R = 14 cm, r = 7 cm, θ = 40°
ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣ
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S19

Solution 20.
ತ್ರಿಜ್ಯ = 4cm
ಕೇಂದ್ರ ಕೋನ = 180° – 70° = 110°
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S21

Solution 22.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S22

Solution 23.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S23

Solution 24.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S24

Solution 25.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S25
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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S26
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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S27

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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S29
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S29.1

Solution 30.
3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ ಎಂದು ಊಹಿಸೋಣ.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S30
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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S31
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
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S31.1
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°
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S32
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
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S33
⇒ 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
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S33.1

Solution 34.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S34

Solution 35.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S35

Solution 36.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S36
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S36.1

V.
Solution 37.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S37
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S37.1

Solution 38.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S38
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.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S39
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S39.1

Solution 40.
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium S40

Karnataka SSLC Maths Model Question Papers

Karnataka SSLC Maths Model Question Paper 2 Kannada Medium

Karnataka SSLC Maths Model Question Paper 2 Kannada Medium

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
Karnataka SSLC Maths Model Question Paper 2 Q1
(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
Karnataka SSLC Maths Model Question Paper 2 Q2

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.
Karnataka SSLC Maths Model Question Paper 2 Q18

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.
Karnataka SSLC Maths Model Question Paper 2 Q20

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.
Karnataka SSLC Maths Model Question Paper 2 Q34
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 2 Q34.1

Question 35.
ಒಂದು ಕಾರ್ಖಾನೆಯ 50 ಕೆಲಸಗಾರರ ದೈನಂದಿನ ಆದಾಯವನ್ನು ಕೆಳಗಿನ ವಿತರಣೆಯು ನೀಡುತ್ತಿದೆ.
Karnataka SSLC Maths Model Question Paper 2 Q35
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)
Karnataka SSLC Maths Model Question Paper 2 S11

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.
Karnataka SSLC Maths Model Question Paper 2 S16
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.
Karnataka SSLC Maths Model Question Paper 2 S18

Solution 19.
Karnataka SSLC Maths Model Question Paper 2 S19

Solution 20.
Karnataka SSLC Maths Model Question Paper 2 S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 2 S21

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.
Karnataka SSLC Maths Model Question Paper 2 S25
Karnataka SSLC Maths Model Question Paper 2 S25.1

Solution 26.
ಆವೃತ್ತಿ ವಿತರಣಾ ಪಟ್ಟಿ
Karnataka SSLC Maths Model Question Paper 2 S26
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.
Karnataka SSLC Maths Model Question Paper 2 S28

Solution 29.
Karnataka SSLC Maths Model Question Paper 2 S29
Karnataka SSLC Maths Model Question Paper 2 S29.1

Solution 30.
Karnataka SSLC Maths Model Question Paper 2 S30

IV.
Solution 31.
Karnataka SSLC Maths Model Question Paper 2 S31
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
Karnataka SSLC Maths Model Question Paper 2 S31.1
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.
Karnataka SSLC Maths Model Question Paper 2 S32

Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು = x ಆಗಿರಲಿ
3 ವರ್ಷಗಳ ಹಿಂದೆ, ಅವನ ವಯಸ್ಸು = x – 3
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸು = x + 5
Karnataka SSLC Maths Model Question Paper 2 S33

Solution 34.
Karnataka SSLC Maths Model Question Paper 2 S34
Karnataka SSLC Maths Model Question Paper 2 S34.1

Solution 35.
Karnataka SSLC Maths Model Question Paper 2 S35

Solution 36.
Karnataka SSLC Maths Model Question Paper 2 S36

Solution 37.
Karnataka SSLC Maths Model Question Paper 2 S37

Solution 38.
Karnataka SSLC Maths Model Question Paper 2 S38
Karnataka SSLC Maths Model Question Paper 2 S38.1

Solution 39.
Karnataka SSLC Maths Model Question Paper 2 S39
Karnataka SSLC Maths Model Question Paper 2 S39.1

Solution 40.
Karnataka SSLC Maths Model Question Paper 2 S40

Karnataka SSLC Maths Model Question Papers

Karnataka SSLC Maths Model Question Paper 1 with Answer in Kannada

Students can Download Karnataka SSLC Maths Model Question Paper 1 with Answers in Kannada, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

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

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Karnataka SSLC Maths Model Question Paper 1 with Answers

Students can Download Karnataka SSLC Maths Model Question Paper 1 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Maths Model Question Paper 1 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.
The pair of linear equations 3a + 4b = k, 9a + 12b = 6 have infinitely many solutions when,
a) K = -2
b) K = 3
c) K = 2
d) K = -3
Answer:
c) K = 2

Question 2.
The HCF and LCM of 12, 15 and 21 are ________ and ________ respectively
a) 3,1
b) 420,3
c) 3, 420
d) 420,1
Answer:
c) 3,420
Solution:
Factors of 12 = 2 × 2 × 3
= 3 × 5
= 7 × 3

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 3.
The value of \(\frac{\sin 18}{\cos 72}\) is ________
a) 1 / 2
b) 1
c) 0
d) √3 / 2
Answer:
b) 1
Solution:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 1

Question 4.
The distance between the points p(2, 3) and Q(4, 1) is
a) 3√6
b) 2√2
c) 2√4
d) 3√9
Answer:
b) 2√2
Solution:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 2

Question 5.
In the given figure^ABC, DE||BC. If DE=5cm, BC=8cm and AD=3.5 cm, then the length of AB is
a) 5.6cm
b) 4.8cm
c) 5.2cm
d) 6.4 cm
Answer:
a) 5.6cm
Solution:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 3

Question 6.
6. The 10th term of an A.P : 2, 7, 12,
a) 1
b) 47
c) 9
d) 17
Answer:
b) 47
Solution:
a = 2, d= 7 – 2 = 5, ri= 10
an = a + (n – 1)d
= 2 + (10-1) (5)
= 2 + 45 = 47
∴ 10th term of an AP is 47.

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 7.
A box consists, of 4 red, 5 black and 6 white balls. One ball is drawn out at random, find the probability that the ball drawn is black.
a) 1/15
b) 1
c) 1/4
d) 1/3
Answer:
d) 1/3
Solution:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 4

Question 8.
In the given fig, the angle of elevation θ is
a) 30°
b) 45°
c) 90°
d) 60°
Answer:
b) 45°
Solution:
Tan θ  = \(\frac{A B}{B C}=\frac{1}{1}\)
Tan θ = Tan 45°
θ = 45°
Karnataka SSLC Maths Model Question Paper 1 with Answers - 5

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

Question 9.
7 × 11 × 13 + 13 is a composite number. Why?
Answer:
7 × 11 × 13 + 13 = 1014 is a number divisible by some other numbers other than itself and by one.
∴ 1014 is a composite number.

Question 10.
If one of the roots of the quadratic equation 6x2 – x – 2 = 0 is 2/3. Find the other.
Answer:
Let a and p are the roots.
Karnataka SSLC Maths Model Question Paper 1 with Answers - 6

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 11.
Find the Value of cos90° + tan 45° Soln: cos90 + tam45 = 0+1 = 1
Answer:
cos 90 + tan 45 = 0 + 1 = 1

Question 12.
If F(2, P) is the mid point of the Sine segment joining the points A(6, -5) and B(-2, 11). Find the value of P.
Answer:
Since P is the mid point of AB,
Co – ordinates of P are
Karnataka SSLC Maths Model Question Paper 1 with Answers - 7
Karnataka SSLC Maths Model Question Paper 1 with Answers - 8
Co – ordinates of P are (2, 3)

Question 13.
If the probability of winning a game is 5/11, what is the probability of losing it.
Answer:
PE(E) + P(Ē) = 1
\(\frac{5}{11}\) + P(Ē) = 1
P(Ē) = 1 – \(\frac{5}{11}\) = 1 – \(\frac{5}{11}\)
∴ P(Ē) = \(\frac{6}{11}\)

Question 14.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then the what is the measure of ∠POA = ?
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 9
In ∆ POA , ∠OAP = 90°
∠APO = y = 40
∠POA = 90 – 40 = 50°
∴ ∠POA = 50°

Question 15.
Write the formula to find the volume of a hemisphere.
Answer:
Volume of hemisphere = 2/3 πr3

Question 16.
What is the Area of the shaded region?
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 10
Area of the shaded region = Area of square – Area of circle
= (14)2 \(\frac{22}{7}\) × 7 × 7
= 196 – 154
= 42 sq.cms .

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

Question 17.
P.T 5 – √3 is irrational.
Answer:
Let 5 – √3 be rational no. in the form of p/q
where p and q are integers and q ≠ 0.
5 – √3 = p/q
5 = \(\frac{p}{q}+\sqrt{3}\)
⇒ 5 is rational number, \(\frac{p}{q}+\sqrt{3}\) is irrational which leads to the contradiction.
Hence our assumption is wrong.
5 – √3 is not a rational number ,
⇒ 5 – √3 is an irrational number.

Question 18.
In the figure DE || AC and DF || AE. BF BE
Prnvp that = \(\frac{\mathbf{B F}}{\mathbf{F E}}=\frac{\mathbf{B E}}{\mathbf{E C}}\)
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 11
By Thales theorem,
In ∆ ABC, DE || AC
Karnataka SSLC Maths Model Question Paper 1 with Answers - 12

Karnataka SSLC Maths Model Question Paper 1 with Answers - 14
OR

In fig LM || CB and LN || CD Prove that \(\frac{A M}{A B}=\frac{A N}{A D}\)
Karnataka SSLC Maths Model Question Paper 1 with Answers - 13
Answer:
In ∆ ABC
LM || CB
According to Thales theorem,
\(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AL}}{\mathrm{AC}}\) ….. (1)
In ∆ ADC LN || CD
According to corollary of Thales
\(\frac{A N}{A D}=\frac{A L}{A C}\) ……. (2)
From (1) and (2)
\(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Question 19.
On comparing the ratios \(\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}}, \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}\) and \(\frac{\mathbf{c}_{1}}{\mathbf{c}_{2}}\) find out whether the lines representing the following pair of linear equations interesect at a point, are parallel or coincident
5x – 4y + 8 = 0
7x + 6y – 9 = 0
5x – 4y + 8 = 0 is the line a1x + b1y + C1 = 0
7x + 6y – 9 = 0 is the line a2x + b2y +c2 =0
a1 = 5, b1 = -4, c1 = 8
a2 = 7, b2 = 6, c2 = -9
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) The two lines are intersecting lines.

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 20.
Draw a circle of radius 4cm and construct a pair of tangents to the circle from a point 8cm away from the centre.
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 15

Question 21.
Find a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively.
Answer:
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = -b/a = -3
αβ = c/a = 2
Equation is x2 – ( α + β) x + αβ
x2 – (-3)x + 2
x2 + 3x + 2

Question 22.
A kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that is no slack in the string.
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 16

OR

Two solid right circular cones have same height and the radii of bases are r1 and r2. Both the cones are metled and recasted to form a cylinder of same height. Show that the radius of the base of the cylinder is \(\sqrt{\frac{\mathbf{r}_{1}^{2}+\mathbf{r}_{2}^{2}}{3}}\)
Answer:
Volume of the cylinder = Sum of the volumes 2 cones
Karnataka SSLC Maths Model Question Paper 1 with Answers - 17

Question 23.
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Answer:
Distance covered by the wheel in one revolution = Distance moved / Number of revolutions
= \(\frac{11}{5000}\) cm
= \(\frac{11}{5000}\) × 1000 × 100cm
= 220cm
∴ Circumference of the wheel = 220 cm
2πr = 220cm
= 2 × \(\frac{22}{7}\) × r
= 220
=3- r = 35 cm
∴ Diameter = 2r = 2 × 35 = 70cm
Hence, the diameter of the wheel = 70cm.

Question 24.
A conical vessel whose internal radius is 5cm and height 24cm is full of water. The water is emptied into a cylindrical vessel with internal radius lOcms. Find the height to which the water rises.
Answer:
r1 = radius o the conical vessel = 5cm
h1 = height of the conical vessel = 24cm
r2 = radius of the cylindrical vessel = 10cm.
∴ Volume of water in conical vessel = Volume of water in cylindrical vessel.
Karnataka SSLC Maths Model Question Paper 1 with Answers - 18
∴ Height of the water in the cylindrical vessel = 2cm.

III. Answer the following : ( 3 × 9 = 27 )

Question 25.
If 2 is added to numerator and denominator to a fraction, it becomes 9/10 If 3 is subtracted from numerator and denominator, the fraction becomes 4/5 Find the function.
Answer:
Let the fraction be x / y
where x < y
If 2 is added to Nr and Dr, we get 9/10
∴ \(\frac{x+2}{y+2}=\frac{9}{10}\)
10(x + 2) = 9(y + 2)
10x + 20 = 9y + 18
10x – 9y + 20 – 18 = 0
10x – 9y + 2 = 0
l0x – 9y = -2 ..(1)
If 3 is subtracted from same fraction, we
\(\frac{x-3}{y-3}=\frac{4}{5}\)
5(x – 3) = 4(y – 3)
5x – 15 = 4y – 12
5x – 4y = -12 + 15
5x – 4 = 3….. (2)
Solve (1) and (2)
(10x – 9y = -2) x 5
(5x – 4y = 3) x 10
Karnataka SSLC Maths Model Question Paper 1 with Answers - 19
Substitute the value of y in (1)
10x – 9y = -2
10x – 9(8) = -2
10x – 72 = -2
10x = -2 + 72
10x = 70
x = 7
∴ Fraction = \(\frac{x}{y}=\frac{7}{8}\)

OR

A girl is twice as old as her sister. Four years, hence the product of their ages (in years) will be 160. Find their
present ages.
Answer:
Let Girls present age be x years.
Sister’s age = 2x years.
After 4 years, their respective ages will be (x + 4) years and (2x + 4) years.
(x + 4) (2x + 4) =160
2x2 + 4x+ 8x + 16= 160
2x2 + 12x+ 16= 360
x2 + 6x + 8 = 80
x2 + 6x + 8 – 80 = 0
x2 + 6x – 72 = 0
x2+ 12x – 6x – 72 = 0
x(x + 12) – 6(x + 12) = 0
(x + 12) (x – 6) = 0
x + 12 = 0
x = -12 or x – 6 = 0, x = 6.
Girls present age = 6 years.
Sister’s age = 12 years.

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 26.
The difference of squares of two natural numbers is 84. The square of the larger number is 25 times the smaller number. Find the numbers.
Answer:
Let the two natural numbers be x and y such that x > y.
x2 – y2 = 84 ……..(1)
x2 = 25y …….(2)
By eq.(1) and (2)
25y – y2 = 84
y2 – 25y + 84 = 0
y2 – 21y – 4y + 84 = 0
y (y – 21) – 4(y – 21) = 0
(y – 21) (y – 4) = 0
y – 21 = 0,y = 4
Substitute the value of y in (2)
x2 = 25 x 21
x = \(\sqrt{25 \times 21}=5 \sqrt{21}\)
x2 = 25y
x2 = 25 × 4= 100
x = √10o = 1o
∴ Two numbers are 10 and 4.

Question 27.
Find the value of K if the points A(2, 3) B (4, k) and C(6, -3) are collinear.
Answer:
Since the given points are collinear, the area of the triangle formed by them must be O.
\(\frac{1}{2}\)[2(k + 3) + 4(-3 -3) + 6(3 – k)] = 0 1
\(\frac{1}{2}\)[2k + 6 + 4(-6) + 6(3 – k)] = 0
\(\frac{1}{2}\)[2k + 6 – 24 + 18 – 6k] = 0
\(\frac{1}{2}\)(-4k) = 0
-2k = 0 or k = 0.

OR

Find the area of a triangle whose vertices are (1, -1) (-4, 6) and (-3, -5)
Answer:
Area of the triangle =
= \(\frac{1}{2}\)[x1(y2 – y3) + x2 (y3 – y2) + x3 (y1 – y2)]
x1 = 1 x2 = -4 x3 = -3
y1 = -1 y2 = 6 y3 = -5
Area = \(\frac{1}{2}\)[1(6 + 5) + (-4) (-5 + 1) + (-3)(-l- 6)]
= \(\frac{1}{2}\) [11 + (-4(-4) +(-3) (-7)]
= \(\frac{1}{2}\) (11 + 16 + 21)
= \(\frac{1}{2}\) (48) = 24 .
∴ Area of the triangle = 24 square units.

Question 28.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Karnataka SSLC Maths Model Question Paper 1 with Answers - 20
Answer:
Given
centred ‘O’ at the point of contact ‘P’.
To Prove that : OP ⊥ AB
Construction : Take any point Q, other than P, on the tangent AB, Join OQ, OQ cuts the circle at R.
Proof : “Of all the line segments joining the point ‘O’ to the tangent AB, perpendicular OP is the shortest distance from O to AB.
∴ OP = OR [ Radii of the same circle]
OQ = OR + RQ .
OQ > OR
OQ > OP (∵ OP = OR)
OP > OQ
∴ ‘OP’ is the shortest distance from point O to AB.
OP ⊥ AB

Question 29.
How many spherical bullets can be made out of a solid cube of lead whose edge measures 44cm, each bullet being 4cm in diameter.
Answer:
d = 4cm, r = 2cm, a = 44cm.
Volume of cube = a3 = 443cm3.
Karnataka SSLC Maths Model Question Paper 1 with Answers - 21

OR

A circus tent is cylinderica! upto a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of th conial part is 53m, find the total canvas used in making the tent.
Answer:
d= 105m, r = 105/2 = 52.5m, l = 53m,
h = 3m.
Total canvas used = C.S.A of cylinder + C.S.A. of cone
Karnataka SSLC Maths Model Question Paper 1 with Answers - 22
Karnataka SSLC Maths Model Question Paper 1 with Answers - 23

Question 30.
Solve the following system of equations graphically.
x + 3y = 6
2x – 3y = 12
Answer:
x + 3y = 6
3y = 6- x
y = \(\frac{6-x}{3}\)
Karnataka SSLC Maths Model Question Paper 1 with Answers - 24
2x – 3y = 12
-3y = 12 – 2x
3y = 2x – 12
y = \(\frac{2 x-12}{3}\)

Karnataka SSLC Maths Model Question Paper 1 with Answers - 25
Karnataka SSLC Maths Model Question Paper 1 with Answers - 26

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 31.
Draw less than Ogive for the given data:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 27
Answer:

C.I F C.f
20- 30 5 5
30-40 4 9
40-50 3 12
50-60 8 20
60-70 7 27

Karnataka SSLC Maths Model Question Paper 1 with Answers - 28

Question 32.
How many terms of the series 54, 51, 48…. be taken. So that their sum is 513? Find the last term.
Answer:
n = ? a = 54, d = 51 – 54 = -3, Sn = 513. S„=
Sn\(\frac{\mathrm{n}}{2}\) [2a + (n-1)d]
513 = \(\frac{\mathrm{n}}{2}\) [2(54) + (n – 1)(-3)]
513 = \(\frac{\mathrm{n}}{2}\) (l08-3n + 3)
n(111 -3n)= 1026
111n – 3n2 – 1026 = 0
-3n2 + 111n- 1026 = 0
Divide by -3.
n2-37n + 342 = 0
n2 – 19n – 18n + 342 = 0
n(n – 19) – 18(n -19) = 0
(n – 19) (n – 18) = 0
∴ n = 19 or n = 18
T19 = a + 18d
= 54-54 =54-51
T19 = 0
OR
T18 = a + 17d
= 54 + 18(-3)
= 54 + 17(-3)
T18 = 3
∴ Sum of 18termsas well as that of 19 terms is 513.

Question 33.
The sum of two numbers is 15. If the sum of their reciprocals is 3/10. Find the numbers.
Answer:
Let the numbes be x and (15 – x)
By data,
Karnataka SSLC Maths Model Question Paper 1 with Answers - 29
45x – 3x2 = 150
3x2 – 45x + 150 = 0
Divide by 3 .
x2 – 45x + 150=0
Divide by 3
x2 – 15x + 50 = 0
x2 – 10x – 5x + 50 = 0
x(x – 10) – 5(x – 10) = 0
(x – 10) (x – 5) = 0
x -10 = 0
x = 10 OR
x – 5 = 0
x = 5
Hence the two numbers are 10 and 5.

OR

Seven years ago Keshav’s age was five times the square of Raghav’s age. Three years hence Raghav’s age will be two fifth of Keshav’s age. Find their present ages.
Answer:
Seven years ago, let Raghav’s age be x years. Then, seven years ago Keshav’s age was 5x2 years.
Raghav’s present age = (x + 7) years.
Keshav’s present age = (5x2 + 7) years.
Three years hence, we have R
aghav’s age = (x + 7 + 3) years = (x + 10) years.
Keshav’s age = (5x2 + 7 + 3) years = (5x2 +10) years.
It is given that three years hence Raghav’s 2
age will be 2/5 of Keshav’s age.
x + 10 = \(\frac{2}{5}\) (5x2 +10)
x + 10 = 2x2 + 4
2x2 + 4-x – 10 = 0
2x2 – 6 – x = 0
2x2 – 4x + 3x – 6 = 0
2x(x – 2) + 3(x – 2) = 0 ,
(x – 2) (2x + 3) =0
x – 2 = 0, 2x + 3 = 0
x = 2, x = -3/2
Raghav’s present age = (2 + 7) years = 9 years.
Keshav’s present age = (5 x 22 + 7) = 27 years.

Karnataka SSLC Maths Model Question Paper 1 with Answers

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

Question 34.
The first and the last term of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Soln: Let a be the first term and d be the common difference.
Answer:
Let 1 be its last term. Then a = 17,1 = an =350 and d = 9
1 = an = 350 ”
a + (n – 1)d = 350
9(n – 1) = 350 – 17 = 333
n – 1 = 333/9 = 38
n = 37 + 1 = 38
Putting a = 17,1 = 350 and n = 38
Sn = \(\frac{n}{2}\) (a + 1)
S38 = \(\frac{38}{2}\) (17 + 350) .
= 19 × 367
= 6973
Hence, there are 38 terms in A.P having their sum as 6973.

OR

Find the sum of first 51 terms of an A.P whose 2nd term and 3rd term are 14 and 18 respectively.
Answer:
Let a be the first term and d be the common difference of given A.P. Then
a2= 14 and a3 = 18 a + d = 14 and a + 2d = 18
Solve
Karnataka SSLC Maths Model Question Paper 1 with Answers - 30
d= 4
a + d= 14
a + 4 = 14
a = 14 – 4 = 10
a = 10
Putting a = 10, d = 4 nd n = 51
Sn = \(\frac{n}{2}\) [2a + (n-l)d]
S5l= \(\frac{51}{2}\)[2×10 + (51 – 1) × 4]
= \(\frac{51}{2}\)[20 + 50 × 4]
= \(\frac{51}{2}\)(20 + 200)
= \(\frac{51}{2}\) × 220
= 51 × 110= 5610

Question 35.
Find the mean for the following data
Karnataka SSLC Maths Model Question Paper 1 with Answers - 31
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 32

Karnataka SSLC Maths Model Question Paper 1 with Answers

Question 36.
Prove the Identity
\(\frac{\tan ^{2} \theta}{\sin ^{2} \theta-1}+\frac{\csc ^{2} \theta}{\sec ^{2} \theta-\csc ^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}\)
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 33
Karnataka SSLC Maths Model Question Paper 1 with Answers - 34

Question 37.
Construct a triangle with sides 5cm. 6cm and 7cm and then another triangle
whose sides are 7/5 of the corresponding sides of the first triangle.
Answer:
Karnataka SSLC Maths Model Question Paper 1 with Answers - 35

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

Question 38.
Areas of similar triangles are proportional to the squares on the corresponding sides.
Answer:
Areas of similar triangles are proportional to the squares on the corresponding sides.
Karnataka SSLC Maths Model Question Paper 1 with Answers - 36
Data: ∆ ABC ∆DEF
∠A = ∠D
∠B = ∠F
∠C = ∠F
\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)
Karnataka SSLC Maths Model Question Paper 1 with Answers - 37

Karnataka SSLC Science Model Question Paper 2 With Answers

Students can Download Karnataka SSLC Science Model Question Paper 2 with Answers, Karnataka SSLC Science Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Science Model Question Paper 2 With Answers

Time: 3 Hours
Max Marks: 80

I. Four alternatives are provided for each of the following questions or incomplete statements. Choose the most appropriate alternative and write with its alphabet. ( 8 × 1 = 8 )

Question 1.
Part of the flower that develops into fruit and part the seed that develops into root respectively are
A) Ovary and plumule
B) Plumule and radicle
C) Ovary and radicle
D) Ovary and ovule
Answer:
C) Ovary and radicle

Question 2.
The chemical equation that represent neutralization reaction among the following is
A) BaCl2 + H2SO4 → BaSO4 + 2HCl
B) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
C) 2NaOH + H2SO4 → NaSO4 + 2H2O
D) AgNO3 + HCl → AgCl + HNO3
Answer:
C) 2NaOH + H2SO4 → NaSO4 + 2H2O

Question 3.
The action of bile can be termed as :
A) Esterification
B) Hydrogenation
C) Oxidation
D) Emulsification
Answer:
D) Emulsification

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 4.
Which of the following are environment friendly practices ?
A) Carrying cloth – bags to put purchases while shopping.
B) Switching of unnecessary lights & fans
C) Walking to school instead of getting your mother to drop you on her scooter
D) All the above
Answer:
D) All the above

Question 5.
Bauxite is an ore of which metal
A) Iron
B) Aluminium
C) Copper
D) Tin
Answer:
B) Aluminium

Question 6.
The frequency of power supply used in India
A) 70 HZ
B) 60 HZ
C) 50 HZ
D) 30 HZ
Answer:
C) 50 HZ

Question 7.
On passing excess, of CO2 gas in an aqueous solution of calcium carbonate milkness of the solution
A) Persists
B) Fades
C) Deepens
D) Disappears
Answer:
B) Fades

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 8.
Which of the following is a plant hormone
A) Thyroxin
B) Cytokinin
C) Insulin
D) Oestrogen
Answer:
B) Cytokinin

II. Answer the following questions ( 8 × 1 = 8 )

Question 9.
State ohm’s law.
Answer:
Under similar physical conditions and at constant temperature the current flowing thought a wire is directly proportional to the potential difference applied across its ends.

Question 10.
Why does a ray of light bend when it travels from one medium into another?
Answer:
Due to changes in velocity of light in the medium and to reduce the time to travel the same.

Question 11.
What id mutation.
Answer:
ways to protect their DNA from changes inspite of those mechanisms, however changes in the DNA occasionally do occur. Any changes in the DNA sequence is called a mutation.

Question 12.
Why does raw bread taste sweeter on mastication ?
Answer:
Salivary amylase which converts starch into sugars.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 13.
Why does a copper vessel develop with green coating in rainy season ?
Answer:
Oxygen, carbon dioxide and water vapour of the air attack copper metal forming green coloured basic copper carbonate. The formula is CuCO3, CU(OH)2

Question 14.
Will Current flow more easily through a thick wire or thin wire of the same material when connected to the same source ? Why ?
Answer:
The current will flow more easily through thick wire because the resistance of a conductor is inversely proportional to area of cross section. If thickness of the wire less is resistances and hence more easily the current flows.

Question 15.
Why does distilled water not conduct electricity whereas rain water does ?
Answer:
Distilled water cannot conduct electricity because it does not contain ions while rainwater conducts electricity as it contains ions due to presence of dissolved salts in it.

Question 16.
What is reflex arc ?
Answer:
The structural and functional unit that carries out reflex action is called a reflex arc.

III. Answer the following questions ( 8 × 2 = 16 )

Question 17.
An electric bulb of 200 Ω draws a current of 1 ampere. Calculate the power of the bulb. The potential difference at is ends and the energy in kwh consumed in burning it for 5h.
Answer:
Power of the bulb
Power of the bulb P = I2R= (1)2 x 200
P = 200 W
Energy consumed by bulb in 5h in burning
= Power x time
= 200 x 5
= 1000 wh = 1 kwh.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 18.
What are structural isomers ? Name the first member of alkanes that shows structural isomerism.
Answer:
Compounds with identical molecular formula but different structures are called structural isomers.
Butane or C4H10

Question 19.
Draw the circuit symbols of the following
i) Variable resistance
ii) Wire crossing without joining.
Answer:
Karnataka SSLC Science Model Question Paper 2 With Answers - 1

Question 20.
Describe how hydro-energy can be converted into electrical energy.
Answer:

  • High rise dams are constructed on the river to obstruct the flow of water to collect it at suitable height. The stored water has a lot of potential energy.
  • The water from a suitable height is allowed to fall on the blades of a turbine located at the bottom of a dam through a pipe.
  • Kinetic energy of flowing water rotates the turbine helping the armature coil of generator to rotate rapidly in the magnetic field. Thus hydroelectricity is generated.

OR

Explain how the tidal energy is harnessed and write one limitation of the use of tidal energy.
Answer:
Tidal energy is harnessed by constructing a dam near the shore. During the high tidal water flows into the dam during the low tides later flows out this flowing water rotates the turbine present at the opening of the dam and produces electricity.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 21.
Draw a diagram of excretory system in human beings label the following parts.
i) Urethra ii) Left renal vein
Answer:
Karnataka SSLC Science Model Question Paper 2 With Answers - 2

Question 22.
Draw the ray diagram showing the position of the object and image to get real the inverted image, whose size is same as the object using convex lens.
Answer:
Karnataka SSLC Science Model Question Paper 2 With Answers - 3

Question 23.
Why must we conserve our forests ? List any two causes for deforestation to take place.
Answer:
We must conserve our forests as they are of great value the reasons for conserving forests are.

  • Forests help in protection of land and retaining sub soil water.
  • Forests check floods and maintain ecosystem.
  • Therefore forests must be conserved for economic and social growth.

Two causes of deforestation taking place are.

  • For industrial needs.
  • For development projects like building of road or dams.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 24.
Give reasons :
i) “Alloys of iron-are more useful when compared to pure iron”
Answer:

  • Pure iron is very soft
  • Stretches, when easily when hot
  • Alloys are hard
  • The properties of iron can be changed if it is mixed with other substances.

ii) Aluminium oxide is called amphoteric oxide
Answer:
Aluminium oxide (Al2O3) reacts with both acids as well as bases to produce salt and water.

OR

i) Metals are regarded as electro positive elements
ii) Articles of aluminium do not corrode even though aluminium is an active metal.
Answer:
i) Metals have a tendency to lose one or more electrons to form positive ions, therefore metals are called electro positive elements.
ii) Aluminium being a reactive metal readily combines with oxygen to give a protective layer of Al2O3. As a result further corrosion stops.

IV. Answer the following questions. ( 9 × 3 = 27 )

Question 25.
a) Which disease is caused in human beings due to depletion of ozone layers in the atmorphere.
Answer:
Skin cancer is caused in human beings due to the depletion of ozone layer in the atmosphere.

b) What step is being taken to limit the damage to the ozone layer.
Answer:

  • Judicious use of aerosol spray propellants such as fluro carbons and chlorofluro- carbons which cause depeletion or hole in ozone layer.
  • Control over large scale nuclear explosions and limited use of supersonic planes.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 26.
a) State the modern periodic law.
Answer:
“Properties of elements are the periodic function of their atomic number”

b) How does the electronic configuration of atom of an element relate to its position in the modren periodic table? Explain with one example.
Answer:
The position of elements depend upon number of valence electrons which depend upon electronic configuration those elements which have same valence electrons occupy same group. Those elements which have one valence electrons belong to group 1 elements, which have two valence electrons belong to group 2. Period numbers is equal to the number of shell.

Example : atomic number of sodium (Na) is 11, so electronic configuration will be 2, 8, 1. Sodium has one valence electron in valence shell so it belong to group 1 as sodium has three shells, so it belong to 3rd period.

OR

a) How does the electronic configuration of atom relates to its position in the modern periodic table.
Answer:
In the periodic table elements are placed according to their atomic number. If an element has only one shell in its electronic configuration it is placed in the first period, if the element has two shells then it is placed in the second period and so on. Vertical column in the periodic table are called groups there are eighteen groups and in a group all the elements have same number of valence electron.

b) Would you place the two isotopes of chlorine Cl – 35 and Cl – 37, in different slots because of their different atomic masses are in the same slot because their chemical properties are the same. Justify your answer.
Answer:
No. Cl – 35 and Cl – 37 will be placed in same slot. The modem periodic table is based on atomic numbers not atomic mass since, both Cl – 35 and Cl – 37 have same atomic numbers hence they are placed in same slot.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 27.
What is meant by electric current? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current? Give justification for your answer.
Answer:
A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flows through any section of the conductor is 1 second (charge on electron 1.6 x 1019 Coulomb)

Electric current : The amount of charge Q flowing through a particular area of cross section in unit time ‘t’ is called electric current, ie., Electrict current, I = Q/t
SI unit of electric current is ampere

One ampere of current is that current which flow when one coulomb of electric charge flowing through a particular area of cross section of the conductor is one second, i.e, 1A = 1 CS-1

The direction of conventional current is A to B, ie., opposite to the direction of flow of electrons. In a metal, flow of electrons carrying negative charge constitutes the current. Direction of flow of electrons given the direction of electronic current by convention, the direction of flow of positive charge is taken as the direction of conventional current charge
= q = ne
For q = I Coulomb
Karnataka SSLC Science Model Question Paper 2 With Answers - 4

OR

a) Write two points of difference between electric energy and electric power
b) Out of 60 W and 40 W lamps which one has higher electrical resistance when in use.
c) What is the commercial unit of electric energy? Convert it into joules. Electric energy
Answer:
The work done or energy supplied by the source in maintaining the flow of electric current is called electrical energy. It appears in the from of heat given by
H = VIt = \(\frac{v^{2} t}{R}\) = I2RT

  • It is equal to be product of power and time E = P × t
  • It SI unit is joule (J)
    1J= 1W × 1S

Electric power:
The time rate at which electronic energy is consumed or dissptied by an electrical device is called electric power and is given by
P = VI = \(\frac{v^{2}}{R}\) = I2R
It is equal to the rate of doing work by an energy source
P = \(\frac{\mathrm{W}}{\mathrm{t}}\)
Its SI unit is watt(W)
1W = 1 JS-1

b) For the same applied voltage P & \(\frac{1}{\mathrm{R}}\)
i,e less the same power of electrical device higher it is electrical resistance.

c) Kilowatt hour – commercial unit of electrical energy.
1 kwh = 1000 wh = 1000 J/S × 3600 see
= 3600000 J = 3.6 × 106J

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 28.
Draw the diagram of the arrangement of apparatus showing the reaction of zinc granules with dilute sulphuric acid and testing hydrogen gas by burning and label the soap bubbles filled with hydrogen.
i) Soap solution ii) Delivery tube
Answer:
Karnataka SSLC Science Model Question Paper 2 With Answers - 5

Question 29.
What is Tyndall effect ? What is its cause? Explain two phenomenon observed in daily life which are based on Tyndall effect.
Answer:
Tyndal effect : When a beam of light is passed through a colloidal solution placed in a dark room the path of beam becomes illuminated when observed through a microscope placed perpendicular to the path of light. This effect is called Tyndall effect.

Causes of tyndal effect : The size of the collidal partical is relatively larger then the solute particals of a true solution. The collidal particles first absorb energy from the incident light and then a part of energy from their surface. Thus tyndall effect is due to scattering of light by the colloidal particals and the colloidal particles are seen to be moving as points of lights moving againist a dark background.

Daily life examples :

  • When sunlight passes though a canopy of a dense forest the tiny water droplets in the mist scatter light and become visible.
  • When a fine beam of sunlight enters a smoke filled room through a small hole the smoke particals become visible due to the scattering of light.

OR

a) What is myopia ? State the two causes of myopia
Answer:
Myopic eye can see objects at short distance inability of the eye in viewing long distant objects the image falls before retina causes of myopia.

  • Elongation of eye ball
  • Excessive curvature in cornea.

b) Why is the normal eye unable to focus on an object placed whith in 10 cm from the eye?
Answer:
A normal eye is unable to clearly see the objects placed closer then 25 cm because the ciliary muscles of eyes are unable to contract beyond a certain limit.

When the object is placed at a distance less or more than 25 cm from the eye than the object appears blurred and induces strain in the eyes.

Question 30.
Draw a diagram of human brain label the parts.
i) Cranium ii) Pituitary gland
Answer:
Karnataka SSLC Science Model Question Paper 2 With Answers - 6

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 31.
a) State in brief the function of the following organs in the human female reproductive system. Ovary’, fallopian tube, uterus
Answer:
Function of ovary :

  • Production of female gamete.
  • Production of female hormone

Function of fallopian tube :

  • Site of fertilization
  • Transfer of female garnets from ovary

Function of uterus:

  • Implantation of zygote fertilizes the egg.
  • Nourishment to the developing embryo

b) Why mensuration take place ?
Answer:
Menstruation : It is the periodic breakdown of uterine lining and it is removed along with blood & mucous in human female uterine lining is required to nourish the embryo that is formed if fertilization takes place. In absence of fertilization the lining is not required and hence is shed in the from of menstruation.

Question 32.
a) Name two metals which can evolve hydrogen gas by reacting with very dilute nitric acid.
b) Explain the formation of MgO by transfer of Electrons.
Answer:
a) Magnesium and Manganese
b) Formation loses two electrons , and forms
Magnesium ions to complete its octet.
Mg → Mg2+ + 2e-
oxygen gains that two electrons to become O2-
Mg2+ + O2 → MgO
Karnataka SSLC Science Model Question Paper 2 With Answers - 7

Question 33.
Define genetics what is the contribution of Mendel in the field of genetics ?
Answer:
The branch of biology that deals with the study of heredity and variation is called genetics Gregor Johann Mendel was the first person to carry out experiment regarding the heredity of certain characters from one generation to another in a scientific manner. He worked mainly on the garden pea plant.

His observations regarding the occurance of contrasting characters in various generations of garden pea led him to interpret that these are controlled by units which he called factors. These factors are today know as genes. He is also know as the ‘father of genetics’.

OR

a) How can we trace evolutionary relationships ?
Answer:
Evolutionary relationships can be traced by studying fossils, by studying homologous and analogous organs, by comparing the embryo or of different animals and by comparing the DNA’s of different species.

b) Define variation in relation to a species. Why is variation beneficial to the species.
Answer:
Variation are differences that occur between the organisms of same species in spite of the same basic features.
Variation in species promotes survival of an organism in the changing environment by increasing the adaptability.

V. Answer the following questions. ( 4 × 4 = 16 )

Question 34.
a) Explain why the planets do not twinkle ?
Answer:
Planets are much closer to the earth and are seen as extended sources, so a planet may be considered as a collection of a large number of point size light sources, although light coming from individual point sized sources flicks but the total amount of light entering our eye from all the individual point sized sources average out to be constant thereby planets appear equally bright and there is no twinkling of planets.

b) What is the colour of the clear sky during day time? Give the reason for it.
Answer:
Clear sky appears blue when sunlight passes through the atmosphere having the molecules of air and other fine particles whose size is smaller then the wavelength of visible light. These molecules and particular scatter the blue colour more strongly then the other colours of spectrum. As the wave length of blue colour is more, this scattered blue light enters our eye, so, the colour of sky appears blue to us during day time.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 35.
What is an esterification reaction ?
Answer:
When carbonic acid reacts with alcohol in the presence of cone. H2SO4, pleasant fruity smelling compound is formed this type of reaction is called esterification reaction.
Karnataka SSLC Science Model Question Paper 2 With Answers - 8

b) Write a chemical equation in each case to represent -the following types of chemical reaction of organic compounds :
i) Oxidation reaction
ii) Addition reaction
iii) Substitution reaction.
Answer:
i) Oxidation reaction :
Karnataka SSLC Science Model Question Paper 2 With Answers - 9
ii) Addition reaction :
Karnataka SSLC Science Model Question Paper 2 With Answers - 10
iii) Substitution reaction :
Karnataka SSLC Science Model Question Paper 2 With Answers - 10

OR

a) What is meant by saponification? Give an explanation.
Answer:
It is the reaction that forms soap when an ester reacts with water in the presence of base a salt of carboxylic acid and an alcohol are produced. Such a reaction is called saponification.
For example r When ethyl ethanote is heated with a solution of sodium hydroxide sodium ethanote and ethanol are produced.
CH3COO C2H5 + NaOH → CH3COONa + C2H5OH

b) Draw the structure of the following compound and identify functional group present in them, i) Butanoic acid ii) Bromopropane.
Answer:
i) Function group is carboxylic acid
Karnataka SSLC Science Model Question Paper 2 With Answers - 12

ii) Function group is Bromine
Karnataka SSLC Science Model Question Paper 2 With Answers - 13

Question 36.
How are fats digested in our bodies ? Where does this process take place ? Explain.
Answer:
Digestion of fats takes place in the small intestine the fats are digested by the digestive enzymes. The fats are present in the from of large globle in the small intestine. Bile juice secreted by the liver is produced in the intestine along with pancreatic juice. The bile salts present in the bile juice emulsify the large globles of fats.

So by emulsification large globles breakdown into fine globules to provide large surface area to act upon by the enzymes. Enzyme present in the pancratic juice cause breakdown of emulsified fats. Glands present in the wall of small intestine secretes intestinal juice which contains lipase enzyme that converts fats into fatty acids and glycerol.

Karnataka SSLC Science Model Question Paper 2 With Answers

Question 37.
a) What do you mean by a precipitation reaction? Explain by giving example.
b) Explain the following in terms of gain or loss of oxygen with examples.
Answer:
a) Any reaction in with an insoluble solid is produced is called a precipitate reaction. For example:
Karnataka SSLC Science Model Question Paper 2 With Answers - 14

b) Oxidation : It is defined as process which involves gain of oxygen
For example :
Karnataka SSLC Science Model Question Paper 2 With Answers - 15
Reduction : It is defined al process which involves loss of oxygen.
For example :
Karnataka SSLC Science Model Question Paper 2 With Answers - 16

Karnataka SSLC Science Model Question Paper 2 With Answers

VI. Answer the following question. ( 1 × 5 = 5 )

Question 38.
What are magnetic field lines ? How is the direction of a magnetic field at point determined? Mention two important properties of the magnetic field lines.
Answer:
The space surrounding a magnet in which magnetic force is exerted is called a magnetic filed. Magnetic field lines are the lines that are drawn at every point indicating the direction in which a north pole would move if placed at that point. They are determined by placing an imaginary hypothetical north pole at that point and finding the direction in which it would move due to the magnetic filed at that point of a compass needle gets deflected when placed near a magnet due to the magnetic force exerted by the magnet on it.

Some important properties of magnetic field lines are :

  • The tangent drawn at any point on the field line indicates the direction in which a north pole would move if placed at that point
  • The relative strength of the field is proportional to the degree of closeness of the lines. The more clustered they are the stronger the field in the region.
  • The magnetic field lines never intersect. This is because a pole can move only in zone direction and if the lines intersect they would have to move in two direction simultaneously which is impossible.

Karnataka SSLC Science Model Question Paper 3 With Answers

Students can Download Karnataka SSLC Science Model Question Paper 3 with Answers, Karnataka SSLC Science Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Science Model Question Paper 3 With Answers

Time: 3 Hours
Max Marks: 80

I. Four alternatives are provided for each of the following, questions or incomplete statements. Choose the most appropriate alternative and write with its alphabet. ( 8 × 1 = 8 )

Question 1.
In the following sketch of stomatal apparatus the parts I, II and III where labeled differently by four students, the correct labeling out of the following is
Karnataka SSLC Science Model Question Paper 3 With Answers - 1
A) I Guard cell, II stomatal pore, III chloroplast.
B) I Stomatal pore, II guard cell, III chloroplast
C) I chloroplast, II stomatal pore, III Guard cells
D) I Guard cell, II chloroplast, III stomatal pores.
Answer:
A) I Guard cell, II stomatal pore, III chloroplast.

Question 2.
The sulphide ore among the following is
A) Haematite
B) Bauxite
C) Argentite
D) Zinc blende.
Answer:
D) Zinc blende.

Question 3.
Which of the following it most suitable for the core of electromagnets
A) Air
B) Soft iron
C) Steel
D) Cu-Ni alloy
Answer:
B) Soft iron

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 4.
Decomposers of an ecosystem consists of
A) Organisms which make organic compound out of inorganic compounds
B) Organisms which use radiant energy to produce biomass
C) Certain blue green algae and algae
D) Certain fungi and bacteria.
Answer:
D) Certain fungi and bacteria.

Question 5.
Phototropism is caused to differential distribution of
A) Abscisic asid
B) Gibberellin
C) Cytokinin
D) Auxin
Answer:
D) Auxin

Question 6.
Most of the sources of energy we use represent stored solar energy which of the following is not ultimately derived from the sun’s energy
A) Geothermal energy
B) Wind energy
C) Nuclear energy
D) Bio mass
Answer:
C) Nuclear energy

Question 7.
Which of the following represent N2 molecules
Karnataka SSLC Science Model Question Paper 3 With Answers - 2
Answer:
Karnataka SSLC Science Model Question Paper 3 With Answers - 3

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 8.
Galvanisation of iron means coating iron with
A) Zinc
B) Nickle
C) Chromium
D) Tin
Answer:
A) Zinc

II. Answer the following questions ( 8 × 1 = 8 )

Question 9.
What is the principle of a electric motor ?
Answer:
An electric motor is based on the principle that when a current is passed through conductor placed a uniformed magnetic field it experiences a force which maker the conductor rotate.

Question 10.
Why does an aqueous solution of an acid conduct electricity ?
Answer:
The presence of hydrogen (H+) or hydronium (H3O+) ions in aqueous solution of an acid are responsible for conducting electricity.

Question 11.
Write a mathematical expression for Joule’s law of heating. Name on device which work on this principle.
Answer:
H = I2Rt
An electric heater works on Joule’s law of heating

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 12.
How is the increase in demand for energy affecting our environment adversely ?
Answer:
The increase in demand for energy affects our environment adversely due to the increase pollutants like 2, CO, SO4 etc. are released into the atmosphere which lead to greenhouse affect.

Question 13.
If a person is working on a treadmill in a gymnasium will it effect his rate of breathing ? How ?
Answer:
Yes, it will affect his rate of breathing. It will become fast to supply more oxygen to meet the increased demand of energy.

Question 14.
When is the force experienced by current carrying conductor placed in a magnetic field largest ?
Answer:
The force experienced by current carrying conductor placed in a magnetic field is largest when the current carrying conductor is placed in a direction perpendicular to that of magnetic field.

Question 15.
What change will you observe it you test soap with Litmus paper (red and blue)
Answer:
Soap is basic in nature and hence turns red litmus blue.

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 16.
Define Rancidity ?
The oxidation of oils or fats in food resulting in a bad smell and bad taste is called rancidity.

III. Answer the following questions. ( 8 × 2 = 16 )

Question 17.
Define the term functional group identify the functional group present in the following.
Karnataka SSLC Science Model Question Paper 3 With Answers - 4
It is a atom or group of atom or reactive part of compound which largely determine the chemical properties of compound
i) Methanol: Alcohol (-OH)
ii) Ethanoic acid : caboxylic acid (-COOH)

OR

What are saturated and unsaturated hydrocarbon? Write the structure of the simplest hydrocarbon ?
Answer:

  • The hydrocarbon in which valency of carbon is saturated hydrocarbon.
  • The valencies of all the atoms are satisfied by double or triple bonds between them are called unsaturated hydrocarbon.
  • The simplest hydrocarbon is methane.
    Karnataka SSLC Science Model Question Paper 3 With Answers - 5

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 18.
Draw the diagram of an electric circuit in which the resistors R1,R2, and R3, are connected in parallel including an ammeter and a voltmeter and mark the direction of the current.
Answer:
Karnataka SSLC Science Model Question Paper 3 With Answers - 6

Question 19.
Name the part of the eye where image is formed by the eye lens. What is the nature of the image formed ? How is their sent to the brain ?
Answer:
The image of the object formed by the eye lens is at the retina of the eye. The image formed on the retina is real and inverted the image is sent to the brain with the help of the optic nerve.

Question 20.
How are the alveoli designed to maximise the exchange of gases ?
Answer:
There are millions of thin walled air sacs called alveoli in the lungs. The presence of these alveoli provides a large area for exchange of gases the availability of large surface area maximises the exchange of gases for example if all the alveoli from two human beings are unfolded they would give an area of about 80 square metres.

OR

How is the amount of urine produced in human regulated ?
Answer:
The amount of urine produced is regulated by reabsorption of water and some of the dissolved subtances into the blood through blood capillaries surrounded by the tabules of nephrons. The amount of urine produced depends upon how much excess water is present in the body and how much of dissolved salts are to be excreted.

Question 21.
Draw a diagram of testing the conductivity of a salt solution ?
Answer:
Karnataka SSLC Science Model Question Paper 3 With Answers - 7

Question 22.
Distinguish between renewable and non-renewable sources of energy.
Answer:
Renewable source:
The sources of energy which can never be finished and are continuously produced in nature are known as renewable sources of energy.
Ex: wind, the sun, biogas, hydropower etc.

Non-renewable sources:
The sources of energy which are exahaustable and take lots of time to be formed again are known as non-renewable source of energy.
For ex: coal, natural gas petroleum etc.

Question 23.
Draw the diagram showing longitudinal section of a flower and lable the parts where pollination takes place.
Answer:
Karnataka SSLC Science Model Question Paper 3 With Answers - 8

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 24.
Describe two examples of different oxidations of ethanol. Name the products obtained in each case
Answer:
i) When ethanol is heated with copper at 573 k, Ethanol if is formed.
Karnataka SSLC Science Model Question Paper 3 With Answers - 9
ii) When Ethanol is oxidised with alkaline
potassium permanganate solution, ethanoic acid is formed.
Karnataka SSLC Science Model Question Paper 3 With Answers - 10

IV. Answer the following questions. ( 9 × 3 = 27 )

Question 25.
The atomic number of Na and Mg is 11 and 12 respectively and they belong to the same period.
a) Which one would have smaller atomic size.
b) Which one would be more electro positive.
c) To which group would each one belong
Answer:
a) Mg has smaller atomic size due to high nuclear charge (Na → 2, 8, 1; Mg → 2, 8, 2)
b) Na would be more electro positive because it can lose the valance electron more easily.
c) Na belongs to group 1 and Mg belong to group 2 as Na contains 1 and Mg contains 2 valence electron.

OR

How did Dobereiner classify the elements? Give one example of what were the limitations of this classification.
Answer:
He classified in group of three such that atomic No. of middle elements is average of first and third element.
Eg : Li(7), Na(23), K(39).

Limitations :

  • Dobereiner could find only three triads, i,e, total of 9 elements only.
  • However the total number of elements were more than that of those encompassed in Dobereiner’s triad.
  • Thus Dobereiner could not classify most of the elements known at that time.

Question 26.
What is meant by sustainable management ? The environmentalist are insisting upon “Sustainable natural resource management” State its four advantages.
Answer:
Sustainable management is the management of natural resources which requires a long term perspective so that it lasts for generation to come and should not be exploited for short term gains.
Its four advantages are :

  1. Resources last for a longer duration
  2. It provides steady economic growth
  3. It helps in ecological conservation
  4. It reduces pollution.

Question 27.
What are marginal rays ? A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm, the distance of the object from the lens is 16 cm find the position size and nature of the image formed using the lens formula.
Answer:
The ray which stricks the mirror surface near the periphery are called marginal rays for convex lens,
f = +24 cm, u= 16 cm
using lens formula
Karnataka SSLC Science Model Question Paper 3 With Answers - 11
The image is formed at a distance of 48 cm from the optical focus of the lens on the same side of the object. This is indicated by the negative sign. Size of image is three times the size of object i.e., 12 cm.
Nature of image: Positive sign in the image height indicates that image is virtual and erect.

OR

Define the power of lens. What is the meaning of “the power of lens is 1 dioptre” if the power of a lens -2.0D, then what type of lens is that ? when an object is kept at infinity from this type of lens, what is the size of the image formed.
Answer:
The ability of a lens to converge or diverge the ray of light after refraction is called power (p) of the lens. It is defined as the reciprocal of the focal length.
i.e, P = \(\frac{1}{f}\)
SI unit of power of a lens is diopter. A lens of focal length 100 cm has a power of 1 dioptre i,e., 1 dioptre = 1 m
If the power of a lens is 2 D then the type of lens is concave lens.
The size of the image in this lens is highly diminished point sized.

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 28.
Draw the diagram showing the sectional view of the human heart. Label the following parts.
i) Aorta
ii) Chamber of the heart that receives deoxygenated blood
Answer:
Karnataka SSLC Science Model Question Paper 3 With Answers - 12

Question 29.
a) Why do ionic compounds have high melting points ?
b) Write the equation for the reaction of
i) Iron with steam
ii) Calcium and potassium with water.
Answer:
In the formation of ionic compounds, positive ions and negative ions precipitate. These are closely packed and the ionic compounds exist as crystalline solids. They have strong inter – ionic process of attraction and have high metling and boiling points.
i) 3Fe + 4H2O → FC2O4 + H2
ii) Ca + 2H2O → Ca(OH)2 + H2
2K + 2H2O → 2KOH + H2

OR

a) What is thermite reaction ?
b) Q is an element which is one in copper, zinc, aluminium and iron, its properties are as follows.
i) Q2 Q3 is found in nature but not affected by water.
ii) Two chlorides with formula QCl2 and QCl3 are formed by the metal. Identify the metal
Answer:
a) The displacement reaction which is. highly exothermic and the metal produced is in molten state is called thermite reaction.
i) Since the oxide of the metal is Q2Q3 its valence is 3. therefore out of the mentioned metal only aluminium and iron can be the metal with properties mentioned above.
ii) QCl2 and QCl2 must be iron as it shows variable valency of two and three. FeCl2 and FeCl3 are the chlorides.

Question 30.
Draw the ray diagram showing the image formation by a convex lens, when the object is kept beyond 2F, with the help of the diagram mention the nature of the image formed.
Answer:
Karnataka SSLC Science Model Question Paper 3 With Answers - 13

Question 31.
What is the function of receptors in our body? think of situations where receptors do not work properly what problemsare likely to arise ?
Answer:
Receptors are specialised called tissues, organ game and nerve ending which are able to pick up specific stimuli e.g gustatory receptors help in detecting taste and olfactory receptors help in sensing smell receptors provide sensory input about external and internal environment. Without them an animal will not be able to observe handle and taste food.

It will, not be aware of an approaching enemy. The animal may not be able to correct its positions and will fall down repeatedly. Therefore animal will not be able to perform the normal activities connected with the defective receptors.

OR

a) Name the hormone secreted by an endocrine gland during emergency.
b) Compare and contrast nervous and hormonal mechanism for control and co – ordination in animals ?
Answer:
a) Adrenalin hormone is secreted by the adrenal gland. It helps to regulate heartbeat, blood pressure, metabolism in the times of stress or emergency to cope up with the situation.

b) Nervous system:

  • It consists of nerve impules between PNS, CNS and brain.
  • The axone and dendrite transmit through a coordinated effort.
  • The flow of information is rapid and the response is quick,
  •  Nerve impulses are not specific in their action.
  • Effects are short level.

Hormonal system:

  • It consists of endocrine system which secrete hormones directly into blood.
  • The information is transmitted or transported through blood.
  • The information travels slowly and the response is slow.
  • Each hormone has specific action
  • It has prolonged effects.

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 32.
Write the chemical equation of the reaction in which the following changes have taken place with an example of each.
Answer:
i) Change in colour
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag.
The solution will become blue in colour and shining silver metal will be deposited

ii) Change in temperature.
NaOH + HCl → NaCl + H2O + Heat.
The temperature will increase because heat is evolved.

iii) Formation of precipitate.
Pb(NO3)2(aq) +2KI(aq) → PbI2(s) +2KNO3(aq)
Yellow PPt
Yellow precipitate of Phi, will he formed.

Question 33.
State the principle of an electric generator. Write any two differences between electric motors and electric generators.
Answer:
An electric generator works on the principle of electromagnetic induction when, a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the direction of the field. The magnetic field lines passing through the coil changes and induce potential differences. Hence a current is set up in it.
Electric motor:

  • It converts electrical energy into mechanical energy,
  • It works on the principle of magnetic field of electric current.

Electric generator:

  • It converts mechanical energy into electrical energy.
  • It works on the principle of electromagnetic induction.

V. Answer the following questions. ( 4 × 4 = 16 )

Question 34.
An electric geyser rated 1500 W, 250 V is connected to a 250 V line mains. Calculate the electric current draw n by it, energy consumed by it in 50 hours and energy consumed if each unit coast Rs. 6.00.
Answer:
Given P = 1500 WV = 250 V
i) Current drawn I = \(\frac{P}{V}=\frac{1500}{250}\) = 6A

ii) Electric energy consumed by gyser in
50 hours = Power × time
= 1500 × 50 = 75,000 wh
= 75 kwh
Since 1 kwh = 1 unit
75 kwh = 75 units.
∴Energy consumed = 75 units

iii) Cost of one unit = 6.00 Cost
75 units = 75 × 6.00
Rs. 450.00

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 35.
i) Write the difference between homologous organs and analogous organs.
ii) Write the difference between the sex chromosomes of man and sex chromosomes of woman.
iii) Sex of a child is determined by the father how ?
Answer:
i) Difference between homologous analogous.
Homologous organs:

  • Organs of different organisms have common origin.
  • They have similar structure and perform different function.
  • Ex. Forelimbs of frog and forelimbs of bird.

Analogous organs:

  • Organs of different organisms have different origin they have different structure and perform similar function.
  • Wings of bird and wings of bat.

ii) Woman have a perfect pair of sex chromosomes both called X. Man has a normal sized chromosome X and another short sized chromosome Y.

iii) A child who inherits X chromosome from her father will be a girl and child who inherits Y chromosome from his father will be a boy both the girl and the boy inherit only X chromosomes from the mother therefore sex of a child is determined by the father.

OR

a) What is meant by the term speciation. List four factors which could lead to speciation.
b) Name the two laws of inheritance postulated by Mendel.
Answer:
a) Speciation is the evolution of reproduction isolation among once interbreeding population factors which can lead to speciation are:

  • Genetic drift : over generations, genetic drift may accumulate which lead to speciation.
  • Natural selection : Natural selection which may give rise to speciation.
  • Severe DNA change
  • A variation may occur which does not allow sexual act between two groups.

b) i) The law of regeneration
ii) The law of independence assortment.

Question 36.
Why does the sun appear reddish early in the morning. Why does the sky appear dark instead of blue to an astronaut.
Answer:
During sunrise the light rays coming from the sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes in this journey. The shorter waves of light are scattered out and only longer wavelengths are able to reach our eyes. Since blue colour has a shorter wave length and red colour has longer wave length the red colour is able to reach our eye after the atmospheric scattering of light, therefore the sun appears reddish early in the morning.

The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight as the sunlight reaches the eyes of the astronauts. Without scattering the sky appears black to them.

Karnataka SSLC Science Model Question Paper 3 With Answers

Question 37.
a) How is plaster of paris prepared ? What reaction takes place when it sets to a hard mass.
Answer:
Plaster of paris is prepared by heating gypsum to a temperature of 100°C. The following reaction takes place :
Karnataka SSLC Science Model Question Paper 3 With Answers - 14
On mixing with water it retracts to a hard mass to form gypsum again i.e., the reverse reaction takes place.

b) Explain the general properties of bases
Answer:

  • Bases possess a bitter taste
  • Bases are slippery to touch.
  • Bases turn red litmus blue
  • Bases turn phenolpthalein solution pink
  • Strong base like caustic soda (NaOH) caustic potash (KOH) are corrosive in nature and produce burning sensation on the skin
  • Solution of bases in water conduct electricity.
  • Bases react with acid, neutralise them.
  • Producing salt and water.

Karnataka SSLC Science Model Question Paper 3 With Answers

VI. Answer the following question. ( 1 × 5 = 5 )

Question 38.
a) What are the advantages of sexual reproduction over asexual reproduction.
b) List any four reasons for adopting contraceptive method
c) If a woman is using copper T, will it help in protecting her from sexually transmitted diseases. Why ?
Answer:
a) Advantages of sexual reproduction.

  • In sexual reproduction more variations are produced thus it ensures survival of species in a pollination.
  • The new formed individual has characteristics of both the parents.
  • Variations are more variable in sexual mode then in asexual one. This is because in asexual reproduction, DNA has to function inside the inherited cellular apparatus.

b) Four reasons for adopting contraceptive methods are

  • To increase the gap between two children.
  • To prevent unwanted pregnancy.
  • To prevent transmission of STD’s
  • To control populations growth.

c) If a woman is using copper – T it will not help in protecting her from sexually transmitted diseases. Copper – T prevents only implantation in the uterus.

Karnataka SSLC Science Model Question Paper 4 With Answers

Students can Download Karnataka SSLC Science Model Question Paper 4 with Answers, Karnataka SSLC Science Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Science Model Question Paper 4 With Answers

Time: 3 Hours
Max Marks: 80

I. Four alternatives are provided for each of the following questions or incomplete statements. Choose the most appropriate alternative and write with its alphabet. ( 8 × 1 = 8 )

Question 1.
The organ which perform different function but have the same basic structure are called.
A) Vestigal organ
B) Analogous organs
C) Homologous organs
D) Analytic organs
Answer:
C) Homologous organs

Question 2.
Identify the correct statement among the following with respect to plant hormones.
A) Cytoldnin promotes wilting of leaves.
B) Auxin inhibits stem elongation
C) Abscisic acid inhibits growth of plants
D) Gibberillin promotes falling of leaves.
Answer:
C) Abscisic acid inhibits growth of plants

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 3.
The watershed management
A) Increases drought and floods
B) Increases production and income of the watershed community
c) Decreases the biodiversity of the downstream reservoirs.
d) increases deforestation
Answer:
B) Increases production and income of the watershed community.

Question 4.
pH of ammonium chloride (NH4Cl) or copper sulphate (CuSO4) solution be
A) 7
B) >7
C) <7
D) 0
Answer:
C) <7

Question 5.
Which of the following has a triple bond.
A)C2H6
B) C3H4
C)C3H8
D) C3H6
Answer:
B) C3H4

Question 6.
Characters transmitted from parents to offspring are present in
A) Cytoplasm
B) Ribosomers
C) Golgi complex
D) Genes
Answer:
D) Genes

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 7.
What is that instrument which can detect the presence of electric current in a circuit.
a) Galvanometer
b) Motor
c) Generator
d) None of above
Answer:
a) Galvanometer

Question 8.
Aluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same
i) Good thermal conductivity
ii) Good electrical conductivity
iii) Ductility
iv) High melting point.
A) (i) and(ii)
B) (i) and (iii)
C) (ii) and (iii)
D) (i) and (iv)
Answer:
D) (i) and (iv)

II. Answer the following questions ( 8 × 1 = 8 )

Question 9.
Meat is easier to digest as compared to grass. Why ?
Answer:
It is easier to digest meat because our digestive juices contain enzymes which can easily digest meat but our body does not digest cellulose which is main component of grass.

Question 10.
Define accommodation of an eye ?
Answer:
The ability of the eye to focus both near and distant objects by adjusting the focal length is called accommodation of the eye.

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 11.
What happens if fallopian tube is blocked ?
Answer:
If fallopian tube is blocked the egg will not be able to reach the uterus in such case fertilization will not take place.

Question 12.
Convex mirror is commonly used as rear view mirror in vehicles. Why ?
Answer:

  • They always give an erect diminished image.
  • Also they have a wider field of view as they are curved outwards.

Question 13.
What is meant by the statement that the rating of fuse in a circuit is 5A.
Answer:
It means maximum current of 5A can pass through the fuse without melting it.

Question 14.
List any two causes for the failure of sustained availability of ground water
Answer:

  1. Loss of vegetation cover
  2. Pollution from industrial effluents

Question 15.
What is the role of the seminal vesicles and the prostate gland.
Answer:
The secretion from seminal vesicles and prostate glands lubricate the sperms and provide a fluid medium for easy transport the secretions also provide nutrient in the from of fructose calcium and some enzymes.

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 16.
What happens when magnesium ribbon burns in air ?
Answer:
When magnesium ribbon bums in air it combines with the oxygen to from magnesium oxide.
2Mg(s) + O2(g) → 2Mgo(s)

III. Answer the following questions ( 8 × 2 = 16 )

Question 17.
An ammeter is always connected in series across a circuit element what happens when it is connected in parallel with a circuit element.
Answer:
An ammeter is law resistance device when it is connected in parallel the resistance of the circuit reduces consider blu therefore a large current flows through the circuit by virtue of circuit element may damage.

OR

a) Define electric power express it in terms of potential difference V and resistance R.
b) What is meant by saying that the potential difference between two points is IV?
Answer:
a) Electric power : it is the rate of doing work by an energy source or the rate at which the electrical energy is dissipated or consumed per unit time in the electric circuit is called electric power.
Karnataka SSLC Science Model Question Paper 4 With Answers - 1
b) If 1 I of work is required to move a charge of amount 1C. from one point to another then it is said that the potential difference between the two points is 1V.

Question 18.
Draw a diagram of an excretory system in human beings.
Answer:
Karnataka SSLC Science Model Question Paper 4 With Answers - 2

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 19.
Explain the cause of shoots of the plants bending towards light.
Answer:
Stems are positively phototropic and bend towards the direction of light the movement is due to occurrence of more auxin on the darken side and lesser auxin on the illuminated side as a result there in more growth on the darker side which causes the stem to bend towards light.

Question 20.
Draw the diagram of arrangement of apparatus used to show the reaction of zinc granules with dilute sulphuric acid and testing hydrogen gas by burning. Label the following parts.
1) Soap solution ii) Delivery tube
Answer:
Karnataka SSLC Science Model Question Paper 4 With Answers - 3
Question 21.
Variations that confer an advantage to an individual organism only will survive in population. Justify.
Answer:
Variations is the difference in the characters or traits among the individuals of a species sexual reproduction of organisms produces variation. The variations produced in organisms during successive generations gets accumulated in the organism the significance of variations shows up only if it continues to be inherited by the offspring for several generation.

Question 22.
List the three kinds of blood vessels of human circulatory system and write their functions.
Answer:

  1. Arteries : They carry blood away from the heart to various organs of the body.
  2. Veins : They collect the blood from different organ and bring it back to the heart.
  3. Capillaries : Exchanges of material . between the blood and surrounding cells take place across the thin walls of capillaries.

OR

Why is necessary to separate oxygenated and deoxygenated blood by mammals and birds?
Answer:
Mammals and bird are warm blooded animal they constantly use energy to maintain their body temperature they have higher energy needs and so they require more oxygen to produce energy. Thus it is important that their oxygenated blood does not get mixed up with deoxygenated blood.

Question 23.
Draw the diagram of an electric generator and label the following parts.
i) Armature ii) Rings
Answer:
Karnataka SSLC Science Model Question Paper 4 With Answers - 4

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 24.
What is the function of receptors in our body ?
Answer:
Think of a situation where receptors do not work properly what problems are likely to arise.
Function of receptors:

  • They sense the external stimuli such as heat or pain
  • They also trigger an impulse in the sensory neuron which sends message to the spinal cord.

When the receptors are damaged the external stimuli transferring signal to the brain are not felt. For example in the case of damaged receptors if we accidentally touch any hot objects we cannot feel the pain or heat. Receptors cannot precive the external stimuli of heat and pain.

IV. Answer the following questions ( 9 × 3 = 27 )

Question 25.
Mention why is it not possible to make use of solar cells to meet all our energy needs ?
Answer:
State three reasons to support your answer, also mention three uses of solar cell.

  • It is not possible to make use of solar cells to melt all our energy needs because.
  • Limited availability of special grade semi conducting materials such as silicon and germanium.
  • Solar cell have lower efficiency as they depend entirely on intensity of solar radiation.
  • The process of manufacturing of solar cells is very expensive silver used for inter connection of cell in the panel further add to the cost.

Uses of solar cell:

  • They provide electric power to satellites and space probes.
  • They provide electric power to offshore drilling plants from light houses.
  • The T.V relay stations located in remate area use solar panels to get electric power for wireless transmission.

OR

Compare and contrast fossil fuel and the sun as direct sources of energy ?
Answer:
Fossil fuel are energy sources such as coal and petroleum obtained from underneath of earth’s crust they are directly available to human being for use hence fossil fuels are the direct source of energy these are limited in amount these are non renewable sources of energy because three cannot be replenished in nature fossil fuel-take million of years for their formation.

If the present fossil fuel of the earth gets exhausted its formations will take several years, fossil fulls are also very costly.
On the other hand solar energy is a renewable and direct source of energy and will do so for the next five billion years, solar energy is available free of cost to all in unlimited amount replenishes in the sun itself.

Question 26.
Draw the ray diagram showing the image formation by a convex lens, when the object is at principal focus fl. with the help of the diagram mention the nature of the image formed.
Answer:
Karnataka SSLC Science Model Question Paper 4 With Answers - 5
The nature of the image is real and inverted.

Question 27.
Explain the breakdown of glucose in aerobic respiration and anaerobic respiration.
Answer:
i) In the presence of oxygen : When breakdown of glucose is carried out in the presence of oxygen in a cell, it is called as aerobic respiration. Glucose is converted into a 3 carbon molecule called pyruvate which further breakdown in the presence of oxygen to form carbon dioxide and water. Energy is released in this process.
Karnataka SSLC Science Model Question Paper 4 With Answers - 6

ii) In the absence of oxygen : When breakdown of glucose is carried out in the absence of oxygen in a cell. It is called as anaerobic respiration. This process is called as fermentation in microbes. Ethyl alcohol or lactic acid is produced by the breakdown of pyruvate
Karnataka SSLC Science Model Question Paper 4 With Answers - 7

OR

Explain the process of transportation of substances in phloem.
Answer:
Trans-location, i.e., transportation of food , in plants takes place through phloem. Sieve tubes with the help of adjacent companion cells in the phloem move the soluble food in the form of sucrose, amino acids etc in both upward and downward directions. Live cells of phloem actively take up material like sucrose using energy from ATP. This increases the osmotic pressure of the tissue causing water to move into it.

This pressure moves the material in the phloem to tissues which have less pressure. According to the needs of the plants food from leaves to storage organs of roots, fruits and seeds and the growing regions or from stored parts such as roots or stem to the buds (in Spring) which needs energy to grow is transported.

Question 28.
a) Why do stars twinkle
Answer:
Stars twinkle due to atmospheric refraction of starlight as the stars are very far away. They behave as almost point sources of light as on account of atmospheric refraction. The path rays of light coming from the star goes on varying slightly the apparent position of the star fluctuates and the amount of starlight entering the eye flickers. So sometimes the star appears brighter and at some other time fainter, thus the stars twinkle.

b) State two properties of the image formed by the eye lens on the retina.
Answer:

  1. Image on the retina is real and inverted.
  2. Diminished in size.

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 29.
Draw a neat diagram of action of steam on a metal label the following parts,
i) Metal sample
ii) Delivery tube.
Answer:
Karnataka SSLC Science Model Question Paper 4 With Answers - 8

Question 30.
i) What is genetics ?
Answer:
Science which deals with the study of heridity and variations is called genetic.

ii) Give the common name of a plant on which Mendel performed its experiments.
Answer:
Pea plant.

iii) What for did Mendel use the term factors and what are these factors called now ?
Answer:
Mendel used the term factors for genes.

iv) What are genes ? Where are the genes located?
Answer:
Genes is the unit of inheritance it is a part of the chromosome which controls the appearance of a set of heredity character gender are located on the chromosomes.

OR

a) Define evalution? Describe the contribution of landmark.
Answer:
Evalution is referred to as the changes acquired by a species or a certain population of a species gradually over a long period of time these changes should be heritable contribution of landmark.

According to the theory of inheritance of acquired characters or Lamarkism, put forward by lamark. The use and disuse of an organ leads to acquiring change in the features of that organ.

these change are also inherited by the offsprings the favorable variations caused due to use and disue after a considerably long period of time result in evolution of a new species.

b) How do homologous organs provide evidence in support of evolution ?
Answer:
The presence of homologous organs indicates that all vertebrates have a common ancestry. Similarly all organs and system of the vertebrates show the fundamental similarities Which point towards common ancestry.

Question 31.
a) What are trophic levels ?
Answer:
Trophic levels are the feeding level in an ecosystem, the trophic level of living beings represent their placement in a food chain it also tell the order of consumption and energy transfer throughout the ecosystem.

b) What will happen if we kill all organisms in one tropic level.
Answer:
If we kill all the organisms in one trophic level the following effects will taken place

  • The population of organisms in previous tropic level will increase.
  • The organisms in next trophic level will not be able to get the food so they will migrate to some other ecosystem or die.
  • It will cause an ecological imbalance in the food chain.

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 32.
a) State the principle of an electric generator
Answer:
An electric generation is based on the principle of electromagnetic induction when a rectangular coil is rotated in a uniform magnetic field, an induced voltage is generated between the ends of the coil.

b) Write the difference between direct current and alternating current.
Answer:
Direct current:

  • Its magnitude is constant and flows in one direction only.
  • The frequency of DC is zero

Alternating current:

  • Its magnitude and direction reserves periodically.
  • The frequency of AC is finite.

Question 33.
a) What is a universal indicator ?
Answer:
A universal indicator is a mixture of a number of indicators which shows different colors at different pH values.

b) A milk man adds a very small amount of baking soda to fresh milk.
i) Why does he shift the pH of the fresh milk from 6 to slightly alkaline.
Answer:
Since the fresh milk is acidic it turns sour early in presence of baking soda, milk become alkaline and does not turn sour easily because of alkali does not allow the milk to become more acidic early.

ii) Why does this milk take a long time to set as curd.
Answer:
When the milk sets to curd the pH decreases i.e., it becomes more acidic the presence of alkali does not allow it to become more acidic easily hence it will take a long time to set as curd.

OR

a) Why does an aqueous solution of an acid conduct electricity.
Answer:
The presence of hydrogen (H+) or hydronium(H3O+) ions in the aqueous solution of an acid are responsible for conducting electricity.

b) How is plaster of paris prepared ? What reaction takes plaster place when its sets to a hard mass.
Answer:
Plaster of Paris is prepared by heating gypsum to a temperature of 100°C the following reaction takes place :
Karnataka SSLC Science Model Question Paper 4 With Answers - 9
On mixing with water it converts to a hard mass to form gypsum again i,e the reverse reaction takes place.

Karnataka SSLC Science Model Question Paper 4 With Answers

V. Answer the following questions. ( 4 × 4 = 16 )

Question 34.
When we say the resistors are in parallel ?
Write the expression for the resistors in parallel. How many 176 Ω resistors (in parallel) are required to carry on a 220 V line ?
Answer:
We can say that resistors are in parallel if same potential difference is applied across them.
For three resistors if we connect them in parallel the equivalent resistance is R
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
For x number of resistors of resistance 176 Ω the equivalent resistance of the resistors connected in parallel is given by ohm’s law as
V = IR
R = \(\frac{\mathrm{V}}{\mathrm{I}}\)
Where,supply voltage, V = 220 V
current I = 5A
Equivalent resistance of the combination = R given as
Karnataka SSLC Science Model Question Paper 4 With Answers - 10

Question 35.
a) List any three observation that determine that a chemical reaction has taken place. Also list three in formation that cannot be obtained about a chemical reaction merely by its chemical equation.
b) Balance the following chemical equations
i) Fe + H2O → Fe3O4 + H2
ii) CO2 + H2O → C6H12O6 + O2
Answer:
The three observations are

  1. Change of colour
  2. Change of temperature
  3. Evolution of gas

The three information are

  1. Atmospheric conditions
  2. catalyst iaoived
  3. Physical state of reactants and products

Balancing equations

  1. 3Fe+4H2O → Fe3O4 + 4H2
  2. 6CO2 + 6H2O → C6H12O6 + 6O2

OR

a) Explain the following in terms of gain or lose of oxygen with two examples each.
i) Oxidation ii) reduction.
Answer:
i) Oxidation : It is defined as process which involves gain of oxygen for example
Karnataka SSLC Science Model Question Paper 4 With Answers - 11

ii) Reduction: It is defined as the process which involves loss of oxygen for example
Karnataka SSLC Science Model Question Paper 4 With Answers - 12

b) Balance the following chemical equations.
i) HNO2+Ca(OH)2 → Ca(NO3)2 +H2O
ii) NaOH+ H2SO4 → Na2SO4+H2O
Answer:
i) 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O
ii) 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O (I)

Karnataka SSLC Science Model Question Paper 4 With Answers

Question 36.
a) State Mendeleev’s periodic law.
b) Did Mendeleev have gaps in his periodic table?
c) Any three limitations of Mendeleev’s classification.
d) Does electronic configurations of atoms change in a period with increase in atomic number?
Answer:
a) Mendeleev’s periodic law states that “the physical and chemical properties of all elements are the periodic function of their atomic masses”.
b) Gaps were left for undiscovered element in the Mendeleev’s periodic table.
c) i) Position of hydrogen was not justified
ii) Increasing order of atomic mass could not be maintained
iii) Isotopes have similar chemical properties but different atomic masses, they cannot be given separate places.
d) Number of shells remain the same, number of valence electrons goes on increasing from left to right in a period till octet is complete.
Karnataka SSLC Science Model Question Paper 4 With Answers - 13

Question 37.
a) State the two laws of reflection of light.
b) The refractive induces of four media A, B, C & D are given in the following table.
Karnataka SSLC Science Model Question Paper 4 With Answers - 14
If light travels from one medium to another in which case the change in speed will be
i) minimum ii) maximum
Answer:
a) Laws of reflection of light are :

  • The angle of incidence is equal to the angle of refection.
  • The incident ray the normal to the reflecting surface at the point of incidence and reflected ray from that point all lies in the same plane.

b)i) Minimum change is seen as light moves between 1.50 and 1.52 i,e B and
ii) Maximum change when light moves between 1.33 and 2.40 i,e A and D

Karnataka SSLC Science Model Question Paper 4 With Answers

VI. Answer the following question. ( 1 × 5 = 5 )

Question 38.
a) What is hydrogenation? What is its industrial application.
Answer:
Addition of hydrogen to unsaturated hydrocarbons in presence of a catalyst such as nickel or palladium to form saturated hydrocarbons is called hydrogenation For Example:
Karnataka SSLC Science Model Question Paper 4 With Answers - 15

Industrial application:
The process of hydrogenation is used in industry to convert vegetable oils to vanaspathi ghee
Karnataka SSLC Science Model Question Paper 4 With Answers - 16
The vegetable oil contains a number of unSaturated carbon chain having double bonds between carbon atoms when H2 is bubbled through vegetable oils in presence of nickel as catalyst at 437 K some of the double bonds add H2 to form saturated carbon chain. As a result of this partial hydrogenation, oils are converted into solid fats.

b) Give the names of the following
i) An aldehyde derived from ethane.
ii) Ketone derived from butane
iii) Compound obtained by the oxidation of ethanol by chromic anhydride.
Answer:
i) Ethanal (CH3CHO)
ii) Butanone (CH3CO CH2CH3)
iii) Ethanol(CH3CHO).

Karnataka SSLC Science Previous Year Question Paper March 2019

Students can Download Karnataka SSLC Science Previous Year Question Paper 2019 5 with Answers, Karnataka SSLC Science Previous Year Question Paper 2019s with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Science Previous Year Question Paper March 2019

Time: 3 Hours
Max Marks: 80

Question 1.
The change that occurs in the eye to see the distant objects clearly is
a. focal length of the eye lens decreases
b. curvature of the eye lens decreases
c. focal length of the eye lens increases
d. ciliary muscles of the eye contract.
Answer:
c. focal length of the eye lens increases

Question 2.
The functional groups present in propanol and propanal respectively are.
a. -OH and -CHO
b. -OH and – COOH
c. – COP and COOH
d. – CHO and – CO
Answer:
a. – OH and – CHO

Question 3.
The correct path of the movement of nerve impulses in the following diagram is
Karnataka SSLC Science Previous Year Question Paper March 2019 - 1
(A) Q → S → R → P
(B) P → Q → R → S
(C) S → R → Q → P
(D) P → R → S → Q
Answer:
(D) P → R → S → Q

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 4.
The resistance of a conductor is 27. If it is cut into three equal parts and connected in parallel, then its total resistance is
a. 6 Ω
b. 3 Ω
c. 9 Ω
d 27 Ω
Answer:
b. 3 Ω

Question 5.
The chemical equation that represents neutralization reaction among the following is
a. BaCl2 + H2SO4 → BaSO4 + 2HCl
b. MnO2 + 4 HCl → MnCl2 + 2H2O + Cl2
c. 2 NaOH + H2SO4 → NaSO4 + 2H2O
d AgNO3 + HCl → AgCl + HNO3
Answer:
c. 2 NaOH + H2SO4 → NaSO4 + 2H2O

Question 6.
By constructing Khadin check-dams in level terrains,
a. underground water level decreases
b. underground water level increases
c. vegetation in the nearby areas are destroyed due to excess moisture
d. underground water gets polluted
Answer:
b. underground water level increases

Question 7.
To obtain a diminished image of an object from a concave mirror, position of the object should be
(F = principal focus, C = centre of curvature, P – pole)
a. between C and F
b. beyond C
c. between P and F
d. at F
Answer:
b. beyond C

Question 8.
The electronic configuration of element x is 2, 8, 8, 1 and the electronic configuration of element Y is 2, 8, 7. Then the type of bond formed between these two elements is
a. covalent bond
b. hydrogen bond
c. metallic bond
d. ionic bond
Answer:
d. ionic bond

Question 9.
Part of the flower that develops into fruit and part of the seed that develops into root respectively are
(A) ovary and plumule
(B) plumule and radicle
(C) ovary and radicle
(D) ovary and ovule
Answer:
(C) ovary and radicle.

Question 10.
A pure dominant pea plant producing round – yellow seeds is crossed with pure recessive pea plant producing wrinkled – green seeds. The number of plants bearing round – green seeds in the F1 generation of Mendel’s experiment is
(A) 0
(B) 1
(C) 3
(D) 9
Answer:
(A) 0

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 11.
The functions of hormones are given in Column-A and the names of. the hormones are given in Column-B. Match them and write the answer along with its letters :

Column – A Column – B
(A) Prepares the body to deal with the situation (i)   Growth hormone
(B) Regulates metabolism for body growth (ii)    Testosterone
(C) Regulates blood sugar levels (iii)   Adrenaline
(D) Regulates the growth and development of the body (iv)    Progesterone
(v)   Insulin
(vi)   Thyroxine
(vii) Oestrogen

Answer:
(A) – (iii) Adrenaline
(B) – (vi) Thyroxine
(C) – (v) Insulin
(D) – (i) Growth hormone

Question 12.
Name the acid present in the stinging hairs of nettle leaves.
Answer:
Methanoic acid.

Question 13.
What are fossils ?
Answer:
The preserved traces of the living organisms are called fossils.

Question 14.
Convex mirror is commonly used as rear – view mirror in vehicles. Why?
Answer:

  • They always give an erect diminished image.
  • Also they have a wider field of view as they are curved outwards.

Question 15.
What is roasting in metallurgy?
Answer:
The sulphide ores are converted into oxides by heating strongly in the presence of excess air.

Question 16.
Observe the given figure. Name the eye defect indicated in the figure and also mention the lens used to correct this defect.
Karnataka SSLC Science Previous Year Question Paper March 2019 - 2
Answer:

  • Myopia
  • Concave lens

Question 17.
What is Tyndall effect?
Answer:
The phenomenon of scattering of light by the colloidal particles gives rise to tyndall effect.

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 18.
Under what condition lactic acid is produced in the muscle cells ?
Answer:
Lactic acid is produced when there is lack of oxygen in the muscle cells.

Question 19.
Draw the diagram of an electric circuit in which the resistors R1, R2 and R3 are connected in parallel including an ammeter and a voltmeter and mark the direction of the current.
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 3

Question 20.
Name the brown fumes liberated when lead nitrate is heated. Write the balanced chemical equation for this reaction.
Answer:
Nitrogen dioxide (NO2)
2 Pb(NO3)2 → 2PbO + 4NO2 + O2

Question 21.
Explain the process of trans location of food materials in plants;
Answer:

  • The transport of soluble products of photosynthesis is called trans location.
  • It occurs in the phloem.
  • The products of photosynthesis are transported from leaves to all parts of the plant body.
  • Translocation takes place in the sieve tubes with the help of companion cells.

OR

Explain the process of digestion in the small intestine of man.
Answer:
Digestion of food in small intestine :

  • Small intestine is the site of complete digestion of proteins, carbohydrates and fats.
  • Glands -present in the walls of small intestine secrete intestinal juice.
  • Enzymes in the intestinal juice convert proteins into amino acids, complex carbohydrates into glucose and fats into fatty acids and glycerol.
  • Digested food is absorbed by the villi present in the walls of intestine.

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 22.
Draw the diagram of a simple electric motor. Label the following parts :
i) Split rings
ii) Brushes
Answer:
i) Split rings
Karnataka SSLC Science Previous Year Question Paper March 2019 - 4
ii) Brushes
Karnataka SSLC Science Previous Year Question Paper March 2019 - 5

Question 23.
What are structural isomers? Name the first number of alkanes that shows structural isomerism.
Answer:
Compounds with identical molecular formula but different structures of called structural isomers.
Butane or C4H10

Question 24.
Draw the diagram showing the longitudinal section of a flower.
Label the following parts :
(i) Style (ii) Anther.
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 6

Question 25.
Draw the diagram of arrangement of apparatus used to show the reaction of zinc granules with dilute sulphuric acid and testing hydrogen gas by burning. Label the following parts.
i) Soap solution
ii) Delivery tube.
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 7

Question 26.
It is advantageous to connect electric devices in parallel instead of connecting them in series. Why?
Answer:

  • The appliances connected in series need currents of widely different values to operate properly.
  • In a series circuit, if one component fails, the circuit is broken and none of the components work.
  • But in a parallel circuit current divides through the electrical gadgets.
  • This is helpful particularly when each gadget had different resistance and requires different current to operate properly.

OR

According to Joule’s law of heating, mention the factors on which heat produced in a resistor depends. According to this law write the formula used to calculate the heat produced.
Answer:
Heat produced in a resistor is,

  • Directly proportional to the square of current for a given resistance
  • Directly proportional to resistance for a given current, and
  • Directly proportional to the time for which the current flows through the resistor
  • H – I2Rt

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 27.
List the disadvantages of using fossil fuels.
Answer:

  • Fossil fuels are formed from carbon and biomass which contains hydrogen, carbon, nitrogen and sulphur.
  • When these are burnt, the products are oxides of carbon, water, oxides of nitrogen and oxides of sulphur.
  • Oxides of nitrogen, oxides of sulphur and carbon monoxide are poisonous at high concentration. They may lead to acid rain.
  • Carbon dioxide is a greenhouse gas. When its concentration in the atmosphere increases continuously, leads to intence global warming.

OR

List the advantages of ‘reduce’ and ‘reuse’ to save environment. Advantages of reduce and reuse to save environment:
Answer:
Reduce :By the practice of ‘Reduce’, we can save

  • Electricity
  • Water
  • Food
  • Natural resources.

Reuse : By the practice of ‘Reuse’

  • Environment pollution can be controlled
  • Materials are available for immediate use
  • Energy can be saved
  • Use of raw materials can be minimised.

Question 28.
The focal length of a concave lens is 30 cm. At what distance should the object be placed from the lens so that it forms an image at 20 cm from the lens?
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 8

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 29.
Draw the diagram of the apparatus used in the electrolysis of water. Label the following parts.
i) Graphite rod
ii) Cathode
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 9
Question 30.
Growth of thread like structures along with the gradual spoilage of tomato can be observed when a cut tomato is kept aside for four days. Interpret the causes for this change.
Answer:

  • The thread like structures that grow on the tomato are hyphae of Rhizopus (Bread mould)
  • They have rod like structures called sporangia.
  • Sporangia contain spores, they are reproductive structures.
  • When spores come into contact with moist surface, they begin to grow. Therefore cut tomato gets spoiled gradually.

Question 31.
An “electric refrigerator rated 300 W is used for 8 hours a day. An electric iron box rated 750 W is used for 2 hours a day. Calculate the cost of using these appliances for 30 days, if the cost of 1 kWh is Rs. 3 /-
Answer:
The total energy consumed by the refrigerator in 30 days
= 400 × 8 × 30 = 96000 Wh = 96 kWh
The total energy consumed by the iron box in 30 days
=750 × 2 × 30 = 45000 Wh = 45 kWh
The total energy consumed by the refrigerator and iron box is
=96kWh + 45kWh = 141 kWh
The sum of bill amount for 141 kWh at rate of Rs. 3 per 1 kWh is
=141 × 3 = Rs. 423

Question 32.
There is no change in the colour of red litmus and blue litmus paper when introduced into an aqueous solution of sodium chloride. After passing direct current through the same solution, red litmus changes to blue colour. Which product is responsible for this change? Mention any two uses of this product.
Answer:
Sodium hydroxide / NaOH.

  • Degreasing metals
  • Soaps and detergents
  • Paper making
  • Artificial fibres.

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 33.
A food chain in a polluted aquatic ecosystem is given. Observe it and answer the following question.
Fresh water → Algae → Fishes → Birds.
(i) Which organisms are disturbed more due to biomagnification ? Why ?
Answer:

  • Birds are disturbed more due to biomagnification.
  • As the birds occupy the top most level in the given food chain, the maximum concentration of harmful chemicals causing biomagnification get accumulated in their body.

(ii) This ecosystem will be destroyed gradually due to biomagnification. Why?
Answer:

  • Biomagnification is the process of accumulation of non-degradable chemicals in the various trophic levels of food chains.
  • As the chemicals are non-degradable or cannot be washed, they cannot be removed from the organisms of the food chain. This leads to gradual destruction of the ecosystem.

OR

A student places a piece of cucumber, a glass piece, a banana peel and a plastic pen in a pit and closes it. What changes can be observed in these materials after a month ? Give scientific reason for these changes.
Answer:

  • Cucumber piece and banana peel are organic substances.
  • They are biodegradable substances, and are eco-friendly.
  • Glass piece and plastic pen are inorganic / synthetic substances.
  • They are non-biodegradable substances and cause soil pollution.

Question 34.
What is dispersion of light? Mention the colour that bends the least and the colour that bends the most when light undergoes dispersion through a prism.
Answer:
The splitting of light into its component colours is called dispersion.

  • The red colour bends the least.
  • The voilet colour bends the most.

OR

Mention any four phenomena that can be observed due to atmospheric refraction of light on the earth.
Answer:

  1. The sun is visible to us two minutes before the actual sun rise.
  2. The sun is visible to us two minutes after the actual sunset also.
  3. The apparent position of the star is slightly different from its actual position.
  4. Twinkling of star
  5. Formation of rainbow
  6. The apparent random wavering or flickering of objects seen through a turbulent stream of hot air rising above a fire or a radiator.

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 35.
Draw the ray diagrams for the image formation in a convex lens when an object is placed.
Answer:
i) at focus F1
Karnataka SSLC Science Previous Year Question Paper March 2019 - 10
ii) beyond 2F1
Karnataka SSLC Science Previous Year Question Paper March 2019 - 11
Question 36.
i. Write the differences between saturated and unsaturated hydrocarbons.
Answer:
Saturated hydrocarbons :

  • In carbon compounds, carbon atoms are satisfied by a single bond between them.
  • These compounds are normally not very reactive.

Unsaturated hydrocarbons :

  • In carbon compounds, carbon atoms have double or triple bonds between them.
  • They are more reactive than the saturated carbon compounds.

ii. Write the molecular formula and structural formula of an alkene having five carbon atoms.
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 12

OR

i. Carbon atom does not form C4- anion and C4+ cation. Why?
Answer:

  • Carbon can gain four electrons. But it would be difficult for the nucleus with six protons to hold on to ten electrons, that is four extra electrons.
  • It can lose four electrons but it would require a large amount of energy to remove four electrons leaving behind a carbon cation with six protons in its nucleus holding on to just two electrons.

ii. How can ethanol be converted into ethanoic acid?
Answer:
Alkaline potassium permangnate or acidified potassium dichromate is added to ethyl alcohol. When it is heated it oxidises to form ethanoic acid.
Karnataka SSLC Science Previous Year Question Paper March 2019 - 13

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 37.
Draw the diagram showing the sectional view of the human heart. Label the following parts.
(i) Aorta
(ii) Chamber of the heart that receives deoxygenated blood.
Answer:
Karnataka SSLC Science Previous Year Question Paper March 2019 - 14

Question 38.
i. Name the major constituent of biogas. Write the properties of biogas which make it a good fuel.
Answer:

  • Methane / CH4
  • Leaves no residue like ash
  • It bums without smoke.
  • Its heating capacity is high
  • Bio-gas is also used for lighting.

ii. Name the two devices that work using heat energy of the sun.
Answer:

  • Solar water heater
  • Solar cooker

OR

i. Write the advantages of solar cells.
Answer:

  • They have no moving parts.
  • Require little maintenance and work quite satisfactorily without the use of any focusing device.
  • They can be set up in remote and inaccessible hamlets also.
  • Very sparsely inhabited areas in which laying of a power transmission line may be expensive and not commercially viable.

ii. Write any two hazards of nuclear power generation.
Answer:

  • Improper nuclear waste storage and disposal result in environmental contamination.
  • There is a risk of accidental leakage of nuclear radiation.

Question 39.
Observe the given table and answer the following question :
Karnataka SSLC Science Previous Year Question Paper March 2019 - 15
Identify the two elements that belong to the same period and the two elements that belong to the same group. Give reason for your conclusion.
Answer:

  • Element B and element D are in same period because their atoms have two shells.
  • Element A and element E are in the same , group because their outermost shell has one electron.

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 40.
i) How does overload and short- circuit occur in an electric circuit? Explain. What is the function of fuse during this situation?
Answer:

  • Overloading can occur when the live wire and the neutral wire come into direct contact.
  • This occurs when the insulation of wires is damaged or there is a fault in the appliance.
  • In such a situation, the current in the circuit abruptly increases.
  • The heating that takes place in the fuse melts it to break the electric circuit and prevents the electric appliances from possible damage.

ii) Mention two properties of magnetic field lines.
Answer:

  1. No two field lines are found to cross each other.
  2. The density of the magnetic field lines are more in their poles.
  3. The magnetic field lines emerge from north pole and merge at south pole.
  4. Inside the magnet, the direction of field lines is from its south pole to its north pole.
  5. Thus the magnetic field lines are closed curves.

Question 41.
Give reasons :
i) Ionic compounds in solid state do not conduct electricity, whereas in molten state are good conductors of electricity.
Answer:

  • In the solid state ionic compounds do not conduct electricity because movement of ions in the solid is not possible due to their rigid structure.
  • In solid state, they are hard because of the strong force of attraction between the positive and negative ions.
  • In molten state, electrostatic forces of attraction’ between the oppositely charged ions are overcome due to the heat.
  • thus the ions move freely and conduct electricity.

ii) Silver articles when exposed to air gradually turn blackish.
Answer:
Silver reacts with sulphur in the air to form a coating of silver sulphide.

iii) Chemical reaction does not take place when copper is added to iron sulphate solution.
Answer:
Reactivity of copper is less than that of iron.

OR

Give reasons :
i) “Alloys of iron are more useful when compared to pure iron.”
Answer:

  • Pure iron is very soft.
  • Streches easily when hot.
  • Alloys are hard.
  • The properties of iron can be changed if it is mixed with other substances.

ii) Copper loses its brown layer gradually when exposed to air.
Answer:
Copper reacts with moist carbon dioxide in the air and slowly loses its shiny brown surface and gains a green coat.

iii) Aluminium oxide is called amphoteric oxide.
Answer:
Aluminium oxide (Al2O3) reacts with both acid as well as bases to produce salt ans water.

Karnataka SSLC Science Previous Year Question Paper 2019 1

Question 42.
(i) Write the differences between homologous organs and analogous organs.
Answer:
(i) Differences between homologous organs and analogous organs:
Homologous organs:

  • Organs of different organisms have common origin
  • They have similar structure and perform different function
  • Ex: Forelimbs of frog and forelimbs of bird

Analogous organs:

  • Organs of different organisms have different origin
  • They have different structure and perform similar function
  • Ex: Wings of bird and wings of bat.

(ii) Write the differences between the sex chromosomes of man and sex chromosomes of woman.
Answer:
Woman has a perfect pair of sex chromosomes, both called X. Man has a normal sized chromosome X and another short sized chromosome Y.

(iii) Sex of a child is determined by the father. How ?
Answer:
A child who inherits X chromosome from her father will be a girl and a child who inherits Y chromosome from his father will be a boy. Both the girl and the boy inherit only X choromosome from the mother. Therefore sex of a child is determined by the father.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Students can Download Karnataka SSLC Maths Model Question Paper 5 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Maths Model Question Paper 5 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.
The pair of linear equations 3a + 4b = k, 9a + 12b = 6 have infinitely many solutions when,.
a) K = -2
b) K = 3
c) K = 2
d) K = -3
Answer:
c) K = 2
Solution:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 1

Question 2.
n2 – 1 is divisible by 8, if n is
a) Prime numbers
b) Odd integer
c) Even integer
d) Natural number
Answer:
b) Odd integer
Solution:
n2 – 1
If n is an odd. integer, 1,3,5,
Ex: 12-1 = 1- 1= 0 divisible by 8.
32 – 1=9-1=8 divisible by 8.
52 – 1 = 25 – 1 = 24 divisible by 8.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 3.
\(\sqrt{1+\tan ^{2} \theta}\) = , where 0<θ< 90°
a) secθ
b) cosec θ
c) cos θ
d) sin θ
Answer:
a) secθ
Solution:
\(\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=\sec \theta\)

Question 4.
If Q divides the line A(3, 5) and B (7,9) internally in the ratio 2:3, then the co ordinates of Q are.
Karnataka SSLC Maths Model Question Paper 5 with Answers - 2
Answer:
c) \(\left(\frac{23}{5}, \frac{33}{5}\right)\)
Solution:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 3

Question 5.
Karnataka SSLC Maths Model Question Paper 5 with Answers - 4
If Area of the sector  OPRQ = \(\frac{5}{18}[ /latex] of Area of circle. Then the value of x
a) 25°
b) 50°
c) 75°
d) 100°
Answer:
d) 100°
Solution:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 5
Karnataka SSLC Maths Model Question Paper 5 with Answers - 6

Question 6.
If 1 + 2 + 3 + n terms = 28, then n is equal to
a) 28
b) 7
c) 8
d) 56
Answer:
b) 7
Solution:
[latex]\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)
n(n + 1) = 28 x 2 = 56
n(n + 1) = 7 x 8
∴ n = 7

Question 7.
If E1E2 E3…….E10 are the possible elementary events of a random experiment, then P(E1) + P(E2) + P(E3) + ………P(E10) is equal to
a) 0
b) 1
c) 2
d) 3
Answer:
b) 1

Question 8.
If we express sec A in terms of sin A, then sec A is equal to
Karnataka SSLC Maths Model Question Paper 5 with Answers - 7
Answer:
\(\frac{1}{\sqrt{1-\sin ^{2} A}}\)
Solution:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 8

Karnataka SSLC Maths Model Question Paper 5 with Answers

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

Question 9.
What is the \(\frac{p}{q}\) form of 43.123456789 ?
Answer:
\(\frac{43123456789}{999999999}\)

Question 10.
Write the quadratic equation formed by the roots √3 + √5 and √3 – √5
Answer:
x2 -(α + β)x + αβ = 0
X2 – [3 + √5 + 3 – √5]x + (3 + √5)(3 – √5) = 0
x2 – [6] x + (9 – 5) = 0
x2 – 6 + 4 = 0

Question 11.
Find the value of \(\frac{\sin 26}{\sec 64}+\frac{\cos 26}{\csc 64}\)
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 9

Question 12.
What is the distance between the points p(cos0, sin0) and Q(sin0, -COS0)?
Answer:
Karnataka SSLC Maths Model Question Paper 5 With Answer - 10

Question 13.
If a number “x” choosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that x2 < 3 ?
Answer:
Clearly “x” can take any of the five given values.
∴ n(x) = 5.
If x2 < 3, then x can take the values -1, 0, 1.
∴ This even A = ,{-1, 0, 1}
n(A) = 3.
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{5}\)

Question 14.
What is the Area of a circle whose perimeter is 44 cms.
Answer:
2πr = 44
2 x \(\frac{22}{7}\) x r = 44
r = \(\frac{44 \times 7}{22 \times 2}\)
r = 7
A = πr² = \(\frac{22}{7}\) x 7 x 7 = 154cm2

Question 15.
A toy was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is h cm and base radius is r cms, find the total surface area of the toy.
Answer:
Total surface Area of the toy = C.S.A of cylinder + 2 [Surface Area of Hemisphere]
= 2πrh + 2πr²
= 2 πr (h + r)
Karnataka SSLC Maths Model Question Paper 5 With Answer - 11

Question 16.
The circumference of a circle exceeds the diameter by 15cm. Find the raidus of the circle.
Answer:
Circumference = 2r + 15
2πr = 2r + 15
2πr – 2r = 15
2 x \(\frac{22}{7}\)r – 2r = 15
Multiply by 7
2 x 22r – 14r = 105 .
30r = 105
r = 3.5

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

Question 17.
Prove that √5 + √3 is an irrational number.
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 12
∴ √3 is rational, is a contradiction
∴ √5 + √3 is irrational.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 18.
AX and BY are perpendiculars to segment XY. If AO = 5cm, BO = 7cm and Area of ∆ AOX = 150 cm2, find the area of ∆ BOY.
Answer:
Karnataka SSLC Maths Model Question Paper 5 With Answer - 13
In ∆ OYB and ∆ OXA,
We have ∠X = ∠Y
So, by AA – criterion of similarity, we have
∆ YOB ≅ ∆ XOA
Karnataka SSLC Maths Model Question Paper 5 With Answer - 14

OR

In the given figure, BD ⊥ AC. Prove that AB2 + CD2 = AD2 + BC2
Answer:
Soln: In ∆ BDC,
∠ BDC = 90°
Karnataka SSLC Maths Model Question Paper 5 With Answer- 15
BC2 = BD2 + DC2 (By Pythagoras)
In ∆ BDA, ∠ APB = 90°
AB2 = AD2 + BD2 (By Pythagoras)
AB2 – BC2 = AD2 + BD2 – BD2 – DC2
AB2 + CD2 = AD2 + BC2
Hence proved.

Question 19.
If zeroes of the polynomial p(y) = y3 – 3y2+ y + 1 are a – b, a, a+ b. Find a and b.
Answer:
Sum of the zeroes of p(y) = y3 – 3y2+ y + 1
Karnataka SSLC Maths Model Question Paper 5 With Answer - 16
(a-b)(a)(a + b)= \(\frac{-1}{1}\) = -1
(a2 – b2) a = -1
(1 – b2) (1) = -1
– b2 = -1
-b2 = – 1 – 1
-b2 = -2
b2 = 2
b = ±√2

Question 20.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.
Answer:
Karnataka SSLC Maths Model Question Paper 5 With Answer - 17

Question 21.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 16y + 80. If zeroes are α, -α and β.
Answer:
Sum of the zeroes
Karnataka SSLC Maths Model Question Paper 5 With Answer - 18
α + (-α) + β = 5
α – α + β = 5
β = 5
Karnataka SSLC Maths Model Question Paper 5 With Answer - 19
(α )(-α )(β) =-80
2β = -80
α2β = 80
α2(5) = 80
α = 80/5 = 16
α = √16
= ±4

Question 22.
Two pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60 and 30 at a point on the road between the pillars. Find the position of the point between the pillars and height of each pillars.
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 20
In ∆ PAB,
tab 60 = \(\frac{\mathrm{AP}}{\mathrm{AB}}\)
√3 = h/x
h = √3x
In ∆ BCQ,
tan 30 = \(\frac{\mathrm{CQ}}{\mathrm{BC}}\)
\(\frac{1}{\sqrt{3}}=\frac{h}{100-x}\)
h√3 -= 100 – x
(√3x)(√3x) = 100-x
4x = 100
x = 25
∴ h= √3 X 25 = 25√3
= 25 x 1.73 = 43.3

OR

The three metallic spheres of radii are in the ratio of 3 : 4 : 5 are melted to form a single solid sphere of radius 12cm. Find the radius of the three small metallic spheres.
Answer:
Volume of the solid sphere = Sum of the volumes of each sphere
\(\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right)\)
4 x 123 = 4[(3x)3 + (4x)3 + (5x)3]
1728 = 27x3 + 64x3 + 125x3 1728 = 216 x3
x = \(\frac{1728}{216}\)
x3 = 8 x = 3√8 = 2
∴ R1 = 6, R2 8, R3 = 10

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 23.
In the given fig. OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 21
Karnataka SSLC Maths Model Question Paper 5 With Answer - 22

Question 24.
Find the volume of the largest right circular cone that can be cut of a cube of edge 7cm.
Answer:
Given radius of the base of cone = \(\frac{7}{2}\) cm
h = height of cone = 7cm
Volume of the cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7\)
= \(\frac{22 \times 7 \times 7}{12}\)
= 89.83cm3

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

Question 25.
The sum of a two digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Answer:
Let the digits at units and tens place of the given number be y and x respectively, then
Given number = 10x + y
Number obtained by reversing the order of the digits = 10y + x
(10x + y) + (10y + x) = 165
10x + y+ 10y + x = 165
(11x+11y=165) = 11 ’
x + y = 15 ……(1)
Digits differ by 3, if x > y
x – y = 3 ……..(2)
Solve (1) and (2)
Karnataka SSLC Maths Model Question Paper 5 With Answer - 23
Consider x + y =15
9 + y = 15
y = 15 – 9
y = 6
∴ The number is 96, if y >x, number is 69.

OR

Ten years ago Sudhir was twelve times as old as his son Raghav and ten years, hence, he will be twice as old as his son will be. Find their present ages.
Answer:
Let the present ages of Sudhir and Raghav be x years and y years respectively. Ten years ago, Father ’s age = (x -10) years Son’s age = (y – 10) years,
x- 10= 12(y – 10)
=> x – 10 = 12y – 120
x – 12y = 120+ 10
x – i2y = – 110
x – 12y + 110 = 0. …….(1)
Ten years later, Father’s age = (x + 10) years
Son’s age = (y + 10)
x + 10 = 2(y + 10)
⇒ x+ 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10
x – 2y – 10 = 0 ……(2)
Solve (1) and (2)
Karnataka SSLC Maths Model Question Paper 5 With Answer- 24
y = 12
x – 12y = 110
x – 12(12) = -110
x – 144 = -110
x = -110 + 144 = 34
∴ Present age of Sudhir = 34 years and present age of Raghav = 12 years.

Question 26.
The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Answer:
Let the base BC be = x cm
Then its height AB = (x – 7) cm
Hypotenuse AC = 13 cm
In ∆ABC, AC2 = AB2 + BC2
132 = (x – If + (x)2
169 = x2 + 49 – 14x + x2
169 = 2x2 – 14x + 49
2x2 – 14x + 49 – 169 = 0
2x2 – 14x-120 = 0
x2 – 7x – 60 = 0
x2– 12x + 5x-60 = 0
x(x- 12) + 5(x – 12) = 0
(x-12) (x + 5) = 0
(x – 12) = 0 or x + 5 = 0
x = 12 or x = -5
The base of the triangle = 12 cm Length of the altitude = (12 – 7)cm = 5 cm.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 25

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 27.
X(2, 5), Y (-1, 2) and Z(5, 8) are the co – ordinates of the vertices of A XYZ. The point A and B lie on XY and XZ respectively such that XA : AY = XB: BZ =1:2. Calculate the co – ordinates of X and Y.
Answer:
Karnataka SSLC Maths Model Question Paper 5 With Answer - 26
Karnataka SSLC Maths Model Question Paper 5 With Answer - 27
B(X2 y2)= [3, 6]
∴ Co – ordinates of A = (1,4)
Co – ordinates of B = (3, 6)

OR

Find the Area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0, -1) (2, 1) and
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 28
Co – ordinates of A
\(\frac{0+2}{2}, \frac{-1+1}{2}\)
= [1,0]
Co – ordinates of B
\(\frac{0+0}{2}, \frac{-1+3}{2}\)
= [0,1]
Co – ordinates of B
\(\frac{2+0}{2}, \frac{1+3}{2}\)
= [1,2]
∴ Area of A ABC = \(\frac{1}{2}\) Σ x1 (y2 – y3)
Karnataka SSLC Maths Model Question Paper 5 With Answer - 29
= \(\frac{1}{2}\) [1(1-2) + 0(2 – 0) + 1(0-1)]
= \(\frac{1}{2}\) [-1 + 0 + 1]
∴ Area of ∆ ABC = 1 sq. unit

Question 28.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contacts.
Answer:
Data : AB is the tangent drawn to a circle centered at O
‘C’ is the point of contact.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 30
To prove that :
OC ⊥AB
Construction : Mark D .on AB. Join OD.
Let it cut the circle at E
Proof: OE < OD
OE is the radius of the circle.
OE = OC (radii of the same circle)
⇒ OC < OD
⇒ OC is the shortest distance between centre of the circle and tangent AB
∴ OC ⊥ AB
Hence proved.

Question 29.
A hollow sphere of internal and external radii are 6cm and Scm respectively is melted and recast into small cones of base radius 2cm and height 8cm. Find the number of cones.
Answer:
Inner radius of a hollow sphere (r) = 6cm
and outer radius (R) = 8cms.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 31
Radius of one small cone (r,) 2cm and
heigh (h) = 8 cm.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 32

OR

A medicine capsule is in the shape of a cylinder with two hemispheres stuch to each of its ends. The length of the entire capsule is 14mm and the diameter of the capsule is 5 mm. Find its surface area.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 33
Answer:
Let r mm be the radius and h mm be the height of the cylinder.
r = \(\frac{5}{2}\) mm = 2.5mm
h = (14 – 2 x \(\frac{5}{2}\) mm
= (14-5) mm
= 9 mm
The radius of hemisphere = r = \(\frac{5}{2}\) mm
Now, the surface area of the capsule. = Curved surface Area of the cylinder surface Area of two hemispheres.
= 2πrh + 2 x 2 πr²
= 2πr² (h + 2r)
Karnataka SSLC Maths Model Question Paper 5 With Answer- 34

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 30.
Solve graphically
y = \(\frac{1}{2}\) x and 3x + 4y – 20 = 0
y = \(\frac{1}{2}\) x
Karnataka SSLC Maths Model Question Paper 5 With Answer - 35
3x + 4y – 20 = 0
3x + 4y = 20
4y = 20 – 3x
Karnataka SSLC Maths Model Question Paper 5 With Answer - 36
Karnataka SSLC Maths Model Question Paper 5 With Answer - 37

Question 31.
The following distribution gives the daily income of 65 workers of a factory. Convert this into less than type cumulative frequency distribution and draw its ogive.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 38
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 54
Karnataka SSLC Maths Model Question Paper 5 With Answer - 39

Question 32.
The fourth term of an AP is 11, and 8th term exceeds twice the fourth term by 5. Find AP and find the sum of first 100 terms.
Answer:
Given : T4 = -11 and T8 = 2T4 + 5
T8 = 2(11) + 5
a + 7d = 22 + 5 = 27
Karnataka SSLC Maths Model Question Paper 5 With Answer - 40
d = 4
a + 7d = 27
a + 7(4) = 27
a+ 28 = 27
a = 27-28 = -1
a = -1
A.P = -1,3, 7, …..
Sn = \(\frac{n}{2}\) [2a + (n-1)d] = \(\frac{100}{2}\) [2(-1) + (99)4]
= 50 [-2 + 396]
= 50 [394]
= 19700

Question 33.
A person on tour has ₹ 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by ₹ 3. Find the original duration of tour.
Answer:
Let the original duration of the tour be x days.
Total expenditure on four = 360 ₹
Expenditure per day = \(\frac{360}{x}\)
Duration of the extended tour = (x +4) days
∴ Expenditure per day according to new schedule = \(\frac{360}{x+4}\)
It is given that daily expenses are cut down by 3₹
\(\frac{360}{x}-\frac{360}{x+4}\) = 3
360 (x + 4) – 360x = 3(x) (x + 4)
360x + 1440 – 360x = 3x (x + 4)
1440 = 3x2+ 12x
3x2 + 12x – 1440 = 0
x2 + 4x – 480 = 0
x2 + 24x – 20x – 480 = 0
x(x + 24) – 20(x + 24) = 0
(x + 24) (x – 20) = 0
x + 24 = 0 and x – 20 = 0
x = -24 and x = 20
∴ Duration of tour = 20 days.

OR

Two pipes running together can fill a cistern in 3\(\frac{1}{13}\) minutes. If one pipe
takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Answer:
Let the faster pipe takes x minutes.
Let the smaller pipe take (x + 3) minutes to fill.
Portion of the cistern filled by the faster 1
pipe in one minute = \(\frac{1}{x}\)
⇒ Portion of the cisterin filled by the faster pipe in \(\frac{40}{13}\) minutes
Karnataka SSLC Maths Model Question Paper 5 With Answer - 41
|||ly portion of the cistern filled by the slower pipe in \(\frac{40}{13}\) minutes.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 42
It is given that the cistern is filled in \(\frac{40}{13}\) min.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 43
[40(x + 3) + 40x] = [(x) (x +3)] 13
40x + 120 + 40x = (x2 + 3x) 13
80x + 120= 13x2 + 39x
13x2 + 39x – 80x – 120 = 0
13x2 – 41x-120 = 0
13x2 – 65x + 24x – 120 = 0
13x (x – 5) + 24(x – 5) = 0
x – 5 = 0, 13x + 24 = 0
x = 5, 13x = -24, x = \(\frac{-24}{13}\)
Faster pipe fills the cistern in 5 minutes and the slower pipe takes 8 min to fill the cistern.

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

Question 34.
In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that T20 = -112. Find S20.
Answer:
a = 2, T1 + T2 + T3 + T4+ T5
= \(\frac{1}{4}\) (T6 + T7 + T8 + T9 + T10)
a + a + d + a + 2d + a + 3d + a + 4d
= \(\frac{1}{4}\)(a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)
5a+ 10d = \(\frac{1}{4}\) (5a + 35d)
5(a + 2d) = \(\frac{1}{4}\) x 5( a + 7d)
a + 2d = \(\frac{1}{4}\) (a + 7d)
4(a+2d) = a+7d
4a + 8d – a – 7d = 0
3a + d = 0
3(2) + d = 0
d = -6
T20 = a + 19d
= 2 +19(-6)
= 2 – 114
T20 = -112
Sn = \(\frac{n}{2}\) [2a + (n-1)d]
Sn = \(\frac{20}{2}\) [2(2) + (20-1)(-6)]
= 10 [4- 114]
= (10) (-110)
S20 = -1100.

OR

A man repays a loan of ₹ 3250 by paying 7 20 in the first month and then increases the payment by ₹ 15 every month. How long will it take to clear his loan?
Answer:
Here, a = 20, c.d = 15, n = ?
Sn = \(\frac{n}{2}\)[2a + (n-1)d]
3250 = \(\frac{n}{2}\) [2 x 20 + (n-1)15]
3250 x 2 = n(40 + 15n- 15)
6500 = n (25 +15n)
6500 = 25n + 15n2
(15n2 + 25n – 6500 = 0)÷ 5
3n2 + 5n – 1300 = 0
3n2-60n + 65n- 1300 = 0
3n(n – 20) + 65 (n – 20) = 0
(n – 20) (3n + 65) = 0
(n – 20) = 0, 3n + 65 = 0
n = 20, 3n = -65
n = \(\frac{-65}{3}\)
∴ Total amount will be paid in 20 months.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 35.
Find the mean, median and mode for * the following frequency distribution.
Karnataka SSLC Maths Model Question Paper 5 With Answer - 44
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 45
Karnataka SSLC Maths Model Question Paper 5 With Answer - 46

Question 36.
P.T \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1+ secθ cosecθ
Answer:
Karnataka SSLC Maths Model Question Paper 5 With Answer - 47
Karnataka SSLC Maths Model Question Paper 5 With Answer - 48

Question 37.
Draw a right triangle in which the sides (other than hypotenuse) arc of lengths 8cm and 6cm, then construct another triangle whose sides are 5/3 times the corresponding sides of given triangle.
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 49
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.
State and prove pythagoras theorem.
In a right angled triangle, square on the hypotenuse is equal to sum of A the squares on the other sides.
Answer:
Karnataka SSLC Maths Model Question Paper 5 with Answers - 50
Data : ∆ ABC, ∠B = 90°
T.P.T : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof: In ∆ ABC and ∆ ABD
∠ ABC = ∠ADB = 90 [∵ Data & construction]
∠A = ∠A (∵ common Angle)
∠ACB =∠ ABD [ ∵ Re maining angle]
∴ ∆ ABC and ∆ ABD are equiangular
=> ∆ ABC ~ ∆ABD
Karnataka SSLC Maths Model Question Paper 5 with Answers - 51
AB2= AC x AD ……… (1)
In ∆ABC and ∆ BDC
∠ ABC = ∠ BDC = 90 [∵ Data & construction]
∠C = ∠C (∵ common Angle)
∠BAC =∠ DBC [ ∵ Re maining angle]
∴ ∆ ABC and ∆ DBC are equiangular
∆ABC ~ ∆BDC
Karnataka SSLC Maths Model Question Paper 5 with Answers - 52
AB2 + BC2 = AC (AD + DC)
AB2 + BC2 = AC(AC)
AB2 + BC2=AC2
AC2 = AB2+ BC2