Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ QUESTION PAPER CODE 30/1/3 EXPECTED ANSWERS/VALUE POINTS SECTION - A Q.No. 1. 25 o 2. –9 4 Marks 3. 1:3 4. 21 26 1×4 = 4 m SECTION - B In Δs TPC and TQC TP TQ 5. TC TC 1 2 (TP and TQ are equally inclined to OT) 1m Δ TPC Δ TQC PC QC and 3 4 3 4 180 o 3 4 90 o OT is the right bisector of PQ ½m But 6. 7. ½m The given A.P. is 6, 13, 20, ---, 216 ABQ 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 1 AOQ 29o 2 ATQ 180o – ABQ BAT 180 o – 119o 61o 21 http://jsuniltutorial.weebly.com/ 1m 1m http://jsuniltutorial.weebly.com/ 8. 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 9. 1 2 3 8 K 2 3K 3 K 1 4 K1 5 Required ratio = 1 : 5 10. 1m Let P divide AB in the ratio of k : 1 2K 1m ½m 1m ½m The given quadratic equation can be written as (9x2 – 6b2x + b4) – a4 = 0 2 ½m 2 or (3x – b2) – (a2) = 0 or (3x – b2 + a2) (3x – b2 – a2) = 0 x b 2 – a 2 b2 a 2 , 3 3 1m ½m SECTION - C 11. Let OA = OB = r 40 22 r 22 r r 280 40r 7 2 7 r7 22 http://jsuniltutorial.weebly.com/ 1m http://jsuniltutorial.weebly.com/ 1 22 7 7 1 22 shaded area 7 7 cm 2 2 7 2 2 2 7 1m 5 385 1 77 or cm 2 96 cm 2 4 4 4 12. 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 ½+½ m ½m Cost of Paenting = Rs. (77 × 10) = Rs. 770 13. ½m ARQ ~ ADC ½m x 4 x 2 6 12 ½m 82 4 2 4 5 ½m QC Total surface area of frustum PQCB 2 2 π 6 2 4 5 6 2 1m 22 32 2.236 40 22 111.552 22 15.936 7 7 350.592 23 http://jsuniltutorial.weebly.com/ 1m 1m http://jsuniltutorial.weebly.com/ 14. Total surfacearea of solid cuboidal block = 2 (15 × 10 + 10 × 5 + 15 × 5) cm2 = 550 cm2 Area of two circular bases = 2 1m 22 7 7 77 cm 2 7 2 2 Area of curved surface of cylinder = 2rh 2 ½m 22 7 5 110 cm 2 7 2 1m Reqd - area = (550 + 110 – 77) cm2 = 583 cm2 15. ½m 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 ½m 16. Height of building is 10 3 m 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 17. 1m 1m The given quadratic eqn. can be written as (k + 1)x2 – 2(k – 1)x + 1 = 0 24 http://jsuniltutorial.weebly.com/ 1m http://jsuniltutorial.weebly.com/ 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 18. 1m 1 1 , 2 2 ½+½ m Number of redface cards removed = 6 1m Remaining cards = 46 19. 20 10 or 46 23 (i) P (a redcard) = (ii) P (a facecard) = (iii) P (a card of clubs) = 1m 6 3 or 46 23 1m 13 46 1m Getting x = 1, y = – 4 P(1, – 4) z = – 1, t = 2 1m R(– 1, 2) ½m Area PQR 1m 1 1 2 – 2 – 1 2 4 3 – 4 – 2 1 24 2 2 12 sq.u. 20. 1½ m Let a be the first term and d the common difference of the A.P. S30 = 15 [2a + 29d] = 30a + 435 d 1m S20 = 10 [2a + 19d] = 20a + 190 d ½m S10 = 5 [2a + 9d] = 10a + 45 d 3 (S20 – S10) = 3 (10a + 145d) = 30a + 435d = S30 25 http://jsuniltutorial.weebly.com/ ½m 1m http://jsuniltutorial.weebly.com/ SECTION - D 21. 22. Correctly stated Given, to Prove, Construction and correct figure 2m correct Proof 2m PR PQ o 180 – 30 PRQ PQR 2 75o SR | | QP and QR is a transversal SRQ 75o ORQ RQO 90o – 75o 15o 1m 1m QOR 180 – 2 15 150o QSR 75o o 1m RQS 180 o – SRQ SQR 30o 23. 1m 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 24. 3x x 5 or x 2.5 Height of Tower 2.5 m 1½ m ½m Money required for Ramkate for admission of daughter = Rs. 2500 A.P. formed by saving 26 http://jsuniltutorial.weebly.com/ 1m 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) 25. 1½ m She can get her doughter admitied ½m Value : Small saving can fulfill your big desires or any else 1m 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 26. 1m x = 10 7 10 1m Let x m be the internal radius of the pipe Radius of base of tank = 40 cm = 1m 2 m 5 Level of water raised in the tank = 3.15 or 315 100 27 http://jsuniltutorial.weebly.com/ http://jsuniltutorial.weebly.com/ 2.52 km/hour 1.26 km in half hour = 1260 m 1m Getting the equation π x 2 . 1260 π . 2 2 315 . 5 5 100 1m 4 315 1 1 . 25 100 1260 2500 1 x m 2 cm 50 x2 1½ Internal diameter of pipe = 4 cm 27. ½m Area ABC 1 – 4 – 4 5 – 3 – 5 – 8 0 8 4 2 1 2 – 4 39 35 2 1½ m Area of ACD 1 – 4 – 5 – 6 0 6 – 8 5 8 5 2 109 2 Area of Qurd. ABCD = 28. 1½ m 35 109 72 sq.u. 2 2 1 m. Volume of earth taken out after digging the well 22 3 3 21 cu.m 59 4 cu.m 7 Let h be the height of the platform 28 http://jsuniltutorial.weebly.com/ 1+1 m http://jsuniltutorial.weebly.com/ 27 × 11 × h = 594 h 1m 594 27 h 1m Height of platform = 2 m 29. i) Number of numbers dividible by 3 or 5 in numbers 1 to 25 (3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) : their number is 12 P (divisible by 3 or 5) = 12 25 1m 1m No. of favourable outcomes = 5 ii) P (a Perfect square number) = 30. Correct Construction 31. 3 4 29 x 1 x –1 4x – 1 5 1 25 5 (1, 4, 9, 16, 25) 1m 4m [3 (x – 1) + 4 (x + 1)] [4x – 1] = 29 (x2 – 1) 1m (7x + 1) (4x – 1) = 29 x2 – 29 1m 28x2 – 3x – 1 = 29 x2 – 29 or x2 + 3x – 28 = 0 1m (x + 7) (x – 4) = 0 x = – 7, 4 29 http://jsuniltutorial.weebly.com/ 1m