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ABQ 1 AOQ 29o 2 1m ATQ 180o – ABQ BAT 180 o – 119o 61o 6. The given quadratic equation can be written as 4x 2 – 4a 2 x a 2 – b 4 0 ½m – b 1m or 2x – a 2 2 2 2 2x – a 2 b 2 7. 1m x 0 2x – a 2 – b2 2 0 a 2 – b2 a 2 b2 , 2 2 In Δs TPC and TQC TP TQ TC TC 1 2 (TP and TQ are equally inclined to OT) ½m 1m Δ TPC Δ TQC PC QC and 3 4 2 http://jsuniltutorial.weebly.com/ ½m http://jsuniltutorial.weebly.com/ 3 4 180 o 3 4 90 o OT is the right bisector of PQ But 8. 9. The given A.P. is 6, 13, 20, ---, 216 Let n be the number of terms, d = 7, a = 6 ½m 216 = 6 + (n – 1) .7 ½m n = 31 Middle term is 16th ½m a16 = 6 + 15 × 7 = 111 ½m ABC is right triangle 2 2 2 AC = BC + AB 2 2 AB 2 5 – 2 2 2 25 AB 5 2 2 2 BC 2 2 2 t 2 16 t 2 2 2 2 2 AC 5 2 2 – t 49 2 – t 49 2 – t 41 t 2 2 t 22 – 2 – t 2 2 8 4 2t 8 t 1 10. ½m 1m Let P divide AB in the ratio of k : 1 1 2 3 8 K 2 3K 3 K 1 4 K1 5 2K Required ratio = 1 : 5 3 http://jsuniltutorial.weebly.com/ 1m ½m 1m ½m http://jsuniltutorial.weebly.com/ SECTION - C 11. P is the mid-point of AB x+1=4 x=3 similarly y = 2 B (3, 2) 1m similarly finding C (–1, 2) 12. Area Δ ABC ½m 1 1 2 – 2 3 2 4 – 1 – 4 – 2 1 24 12 sq.u. 2 2 1½ m The given quadratic eqn. can be written as (k + 1)x2 – 2(k – 1)x + 1 = 0 1m For qual roots 4(k – 1) 2 – 4(k 1) 0 or k 2 – 3k 0 k 0, 3 Non-zero value of k = 3 : Roots are 1 1 , 2 2 13. 14. 1m ½+½ m Fiqure ½m (i) 30 tan 45o 1 y 30 y 1m (ii) x 1 y 30 tan 30 o x 10 3 y 3 3 3 1m Height of building is 10 3 m ½m The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) : Number : 4 1m Total possible out comes = 36 (i) Required Probabilit y 4 1 36 9 4 http://jsuniltutorial.weebly.com/ ½m http://jsuniltutorial.weebly.com/ (ii) The possible outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) 1m their number is 9 15. 16. Required Probabilit y 9 1 36 4 Let a be the first term and d the common difference S12 = 6 [2a + 11d] = 12a + 66d 1m S8 = 4 [2a + 7d] = 8a + 28d ½m S4 = 2 [2a + 3d] = 4a + 6d ½m 3 (S8 – 34) = 3 (4a + 22d) = 12a + 66d = S12 1m Let OA = OB = r 40 22 r 22 r r 280 40r 7 2 7 r7 1 22 7 7 1 22 shaded area 7 7 cm 2 2 7 2 2 2 7 5 385 1 77 or cm 2 96 cm 2 4 4 4 17. ½m ARQ ~ ADC 1m 1m 1m ½m x 4 x 2 6 12 ½m 82 4 2 4 5 ½m QC 5 http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ Total surface area of frustum PQCB π 18. 6 2 4 5 6 2 2 2 22 32 2.236 40 22 111.552 22 15.936 7 7 350.592 1m 1m Volume of solid wooden toy 5 2 22 7 7 7 1 22 7 7 h 6 3 7 2 2 2 3 7 2 2 1001 22 7 7 or 7 h 6 7 2 2 166 7 h 1001 7 13 22 7 h 6 cm Area of hemispherical part of toy 2 22 7 7 cm 2 7 2 2 77 cm 2 19. 1m Cost of Paenting = Rs. (77 × 10) = Rs. 770 ½+½ m ½m ½m Total surfacearea of solid cuboidal block = 2 (15 × 10 + 10 × 5 + 15 × 5) cm2 = 550 cm2 Area of two circular bases = 2 22 7 7 77 cm 2 7 2 2 Area of curved surface of cylinder = 2rh 2 22 7 5 110 cm 2 7 2 Reqd - area = (550 + 110 – 77) cm2 = 583 cm2 6 http://jsuniltutorial.weebly.com/ 1m ½m 1m ½m http://jsuniltutorial.weebly.com/ 20. Area of Sq. ABCD = 142 or 196 cm2 ½m Area of Small Sq. = 42 or 16 cm2 ½m Area of 4 semi circles 4. 1 3.14 (2) 2 cm 2 2 25.12 cm 2 Reqd. area (196 – 16 – 25.12) cm 2 154.88 cm 2 1m 1m SECTION - D 21. Let the fraction be x –3 x ½m By the given condition, new fraction x – 3 2 x –1 x2 x2 x – 3 x – 1 29 x x 2 20 20 x – 3 x 2 x x – 1 29 x 2 2x 2 2 2 20 x – x – 6 x – x 29x 58x ½m 1m or 11x2 – 98x – 120 = 0 or 11x2 – 110x – 12x – 120 = 0 (11x + 12) (x – 10) = 0 The Fraction is 22. 1m x = 10 7 10 1m 1m Money required for Ramkate for admission of daughter = Rs. 2500 A.P. formed by saving 1m 7 http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ (i) = 100, 120, 140, --- upto 12 terms 12 2 100 11 20 6 420 2 Rs. 2520 Sum of AP (i) 23. She can get her doughter admitied ½m Value : Small saving can fulfill your big desires or any else 1m 2 3 23 x 1 2x – 2 5x or 5x 4 x – 2 3x 3 46 x 1x – 2 1½ m 5x 7x – 5 46 x 2 – x – 2 11x 2 – 21x – 92 0 x 25. 1m 21 441 4048 21 67 22 22 1m – 23 11 ½m 4, 24. 1½ m Correctly stated Given, to Prove, Construction and correct figure 2m correct Proof 2m PR PQ PRQ PQR 180 – 30o 2 75o SR | | QP and QR is a transversal SRQ 75o ORQ RQO 90o – 75o 15o 1m 8 http://jsuniltutorial.weebly.com/ 1m http://jsuniltutorial.weebly.com/ QOR 180 – 2 15 150o QSR 75o o 1m RQS 180 o – SRQ SQR 30o 26. 1m Correctly drawn ABC 1½ m Correctly drawn a triangle similar to ABC of given scale factor 2½ m 27. figure 1m (i) x 1 tan 30o y 3x y 3 1m (ii) x 5 x 5 tan 60 o 3 or 3 y 3x Writing the trigonometric equations 28. (i) 3x x 5 or x 2.5 Height of Tower 2.5 m Required Probabilit y 13 20 Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 : 8 in number 29. ½m Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15 Their number is 13 (ii) 1½ m Required Probabilit y 8 2 or 20 5 Area ABC 1 – 4 – 4 5 – 3 – 5 – 8 0 8 4 2 9 http://jsuniltutorial.weebly.com/ 1m 1m 1m 1m http://jsuniltutorial.weebly.com/ 1 2 – 4 39 35 2 1½ m Area of ACD 1 – 4 – 5 – 6 0 6 – 8 5 8 5 2 109 2 1½ m 35 109 72 sq.u. 2 2 1 m. 22 2 2 14 cu.m 176 cu.m ................. (i) 7 1m Area of Qurd. ABCD = 30. Volume of earth taken out after digging the well Let x be the width of embankment formed by using (i) Volume of embankment 22 2 x 2 – 22 40 176 7 100 x2 + 4x – 140 = 0 31. (x + 14) (x – 10) = 0 x = 10 1½ m 1½ m Width of embankment = 10 m Let x m be the internal radius of the pipe Radius of base of tank = 40 cm = 2 m 5 Level of water raised in the tank = 3.15 or 315 100 2.52 km/hour 1.26 km in half hour = 1260 m 10 http://jsuniltutorial.weebly.com/ 1m http://jsuniltutorial.weebly.com/ Getting the equation 2 2 315 π x 2 . 1260 π . . 5 5 100 1m 4 315 1 1 . 25 100 1260 2500 1 x m 2 cm 50 x2 Internal diameter of pipe = 4 cm 11 http://jsuniltutorial.weebly.com/ 1½ ½m