Patna Centre Hint & Solution Part I MT 1. Hints:- 10 40 3 10, 60, 180 60 3 180, 90, 90 3 70. Hints:- 3, 6, 9,1, 15, 18 3. Hints:- a b c d 100 4 a b c d 4 100 400 a b d 400 70 330 330 Mean 110 3 4. Hints:- Perimeter of circle = r New perimeter of circle = 3r Percentage increase of perimeter = 3 r r 100% r 100% 50% r r 5. 4 6. Hints:- +1 +1 +1 +1 +1 G L I B H B M J C +1 +1 +1 +1 +1 T I G E R U J H F S 7. 4 8 Hints:- brother wife T P Q R father sister S son 9. Total horticulture 006-07 3.74% 007-08 4.1% 008 09.47% 009 10 0.96% 10. Hints:- Fruits = 63 100% 30.14% 09
Vegetables = 79 100% 37.79% 09 Flowers= 100% 95% 09 11. [3] Hints:- Increase % of Fruits = 63 53 100% 18.86% 53 Increase % of vegetable = 79 7 100% 8.86% 7 Increase % of flowers = 1 100% 100% 1 Increase % of horticulture = 09 187 100% 11.76% 187 1. Hints:- nswer Question number 11 13. Hints:- Percentage area = 138 100% 68.31% 0 14. Hints:- 006 07 56 53 100% 5.66% 53 007 08 58 56 100% 3.57% 56 008 09 61 58 100% 5.17% 58 009 10 63 61 100% 3.7% 61 15. 16. Hints:- - - - - - - - M T H U R - J X Q E R O X - - - - - - H O T E L S - E L Q B I P 17. 18. 3 19. 3 0. Hints:- Let present age of son s = x years Present age of father s = 4x years Present age of mother s (4x 3) years fter 3 years Son s age = x + 3 = 15 x = 15 3= 1 year. Father s age = 1 4 = 48 years Mother s age = 48 3 = 45 years fter 5 years mother s age = 50 year.
1. Hints:- (a) (d) Poets Philosophers Poets Philosophers. Hints:- 1 5,5 1 9, 9 1 19,19 1 37, 37 1 75 3. 1 4. Hints:- 5 km HOME 10 km 10 km 5 km 5. Hints:- verage (41 45) years 35 verage (46 50) years 50 verage (51-55) years 53.33 verage (36 40) years 37.5 6. Hints:- verage = 340 4.5 8 7. Hints:- Percentage salary 40,000 per months = 9 0 5 100% 45% 8. Hints:- verage = 775 38.75 0 11 % employees 100% 55% 0 9. Hints:- verage age = 808 40.4 30. Hints:- Percentage (30 35) years employees 7 100% 35% 0 31. Hints:- +9 +9 +9 +9 +9 +9 +9 4 5 8 13 14 17 3 6 31 3. Hints:-
BB, FE, IT, ML, PP, TS 65 99 13 1 16 16 0 19 +4 +3 +4 +3 +4 +3 +4 +4 +3 +3 33. Hints:- HOME 9 km 6 km 3 km 5 km 34. Hints:- Rs. 0 10 Rs. 3 8 7 Rs. 9 3 7 1 8 35. Hints:- QUESTION DOMESTIC (Code Similar letters) RESPONSE OMESICEM 36. 4 37. Hints:- 50 digits 18 digits = 3 digits 5,57,6,67,70,71,7,73,74,75,76,77,78,79,8,87,9,97 38. 1 39. 3 40. 41. Hints:- B, ED, IH, NM 4. Hints:- + + + + STREMERS - UVTGLIDQR -1-1 -1-1 KNOWLEDGE ++ + + -1-1 -1-1 MPQYLDCFD 43. Hints:- E C brother B brother Father husband D
44. Hints:- 3x 9 1 5x 9 3 69x 07 60x 108 69x 60x 07 108 9x 99 x 11 3x 33 5x 55 45. Hints:- Let present age of son s = x year Present age of father s = y year x y 7 x y 54.. i 46. Hints:- 47. 3 48. Hints:- y 18 x 18 y x 18... ii equation i & ii x 1 y 4 3 3 3 3 3 3 1 1, 1, 3 1, 4 1, 5 1, 6 1 6, 4, 1,,, 8, 7, 4,, 1, 5, 3, 8, 6,,, 7, 1, 4, 1, 3, 5, 8, 6 49. Hints:- Total marks of remaining 0 students 65 40 45 0 1700 1700 mean 85 0 50. Hints:-Let Present age of Sunita = x years Present age of nil = x years Before 3 years:- x 3 3 x 3 x 3 3x 9 x 6 x 1
Part II SCHOLSTIC PTITUDE TEST II (b) OTHER SUBJECTS () SCIENCE DISCIPLINE ST Physics 1. s wave carries energy and momentum.. By equation of motion v = u + gh v 0 gh v gh v 10 0 0m / s 3. It retraces its path. 4. v o 4000 5. Convex lens in used to correct long sightedness. 6. In sun proton-proton cycle is given as energy released = 6 MeV. 1 4 4 H He e v 6MeV 7. By equation of mass-energy equibalance E mc 1.66 10 3 10 7 8 = 934 MeV 8. Time period of pendulum T g T 9. s outward flux is increasing that s why according to Lenz s law current will be clockwise. V 0 10. For Ist bulb resistance R1 968 P 50 For IInd bulb resistance 1 0 V R 484 P 100 When bulbs are connected in series 50W bulb will glow brighter and when connected in parallel 100W bulb will glow more. 11. Hearing range for a man is 0Hz to 0000Hz. 1. From given options (3) is correct as voltmeter and ammeter are connected correctly. 13. 0 to is divided into 0 parts. So the least count of ammeter is 0.1.
Chemistry V 14. rms 3PV 3RT 3P M M d 15. DO 16. Neopentane 17. n K K RT P 18. Li, Mg C Where (I) n > 0 (II) n = 0 (III) n < 0 19. O OH CH C CH CH C CH 3 3 3 O OH CH C O C H CH C O C H 3 5 5 O O OH O CH C CH C O C H CH C CH C O C H 3 5 5 OH O or CH C CH C O C H 5 an P v nb nrt v 0. an P.v P a v n = atm L / mol. 1. Mg + 1s s p 6. lkene 3. PV = nrt PV n.rt N N PV nkt N PV nn.kt PV NKT CaO 4. (1) CH COONa NaOH CH Na CO 3 4 3 LilH4 () CH I CH 3 4 (3) l C 1H O 3CH 4lOH 4 3 4 3 5. NO 6. NaOH & SO3
Biology 7. Homologous organs are same in origin but are different in function. 8. 3 9. Chromoplast contains xanthophyll and carotene pigment which are responsible for different colour. 30. It was used for digestion of cellulose before discovery of fire but now is functional only in ruminants. 31. 4 3. 1 33. Medula oblongala controls all the involuntary activities of body. 34. 4 35. In asexual reproduction fertilization do not take place. 36. 1 37. 1 38. Bile does not contain any enzyme. 39. 40. 3 (B) Mathematics 41. Sample space will contain 36 elements in which only 10 are favorable namely {(1, ), (, 1), (, 3), (3, ), (3, 4), 10 5 (4, 3), (4, 5), (5, 4), (5, 6), (6, 5)}. So required probability 36 18 b... 1 a c.. () a q 4. /q, lso k, using (1) we have k p and (), we have c b r a a p 43. Since x 3 and xy = 1 y 3 b a q p.. (3) and k k k k. Now using (3) and on simplification, we have s x, y > 0, so x 4 3 & y 4 3 x 1 3 & y 3 1 x 1 3 &. y 3 1 s x y x y x y x y 3 3, on simplification 1 3 3 1 44. Given equation cause simplified as a b x a 3b 1 x a b 1 0 (1) r, expanding and using (1) p b 4ac a q 4pr p
If (1) is an identify, then a + b = 0. () a 3b + 1 = 0 (3) and a b + 1 = 0. (4) 1 1 solving () and (3), we have, a, b, but these two values do not satisfy (4) so ordered pair (a, b) doesn t 5 5 exist. 45. Let first term = a, common difference = d. Then a/q, a + d = x y.. (1) and a + 4d = x + y.. () eliminating d from (1) & () and simplify for a, we have, 3a = (3x 5y) 46. Using the fact that i 4n = 1 and i 4n + = -1, n N. 1 1 1 1 1 3 1 3 4 1 1 1 1 1 47. The possible triangles with side length are (, 6, 6), (3, 5, 6), (4, 4, 6), (4, 5, 5). Using the fact that sum of two sides is always greater than the third side in a triangle. 48. s median is a mid value (in general). Here median is 10. dditional two data are 7(< 10) and 0(> 10). So median will not change. x1 x... x49 x50 49. /q 169 50 x 1... x49 x50 50 169 Let us suppose that x50 = 134 Then x1 + x +.. x49 = 50 169 134 x x... x 143 50 169 134 143 Correct mean = 1 49 50 50 9 169 169.18 (in rupees) 50 x x... x 1 11 50. Let k say 11 x 1 x...x11 15 a/q k 1 11k 15 1k k 15, then 51. s sin 1 and sin1 sin sin3 3. This is possible only when 90 cos cos cos 0 o 1 3 1 3 11 11 5. Given cos ec cot, cos ec cot. On squaring and using 11 11 117 44 1 cot. cot cot 11cot tan. 4 4 117 53. s x sec tan, y cos ec cot xy 1 sec tancosec cos 1 sec cosec cosec sec 1 1 sec cos ec cos ec sec sin cos cosec sec tan cot cosec sec sincos = y x cosec 1 cot, we have
54. /q, c = a + b. (1) a cosb c sinb b ab ab a c ac b b Given b = ac a a cosb ac b lso sin B cos B 1 a a a a 1 1 0 b b b b a 1 1 4 5 1 cosb as 0 o < B < 90 o b b C 90 o a c B 55. /q 4.. (1) lso 5k 1 k 1. () 7k 1 k 1.. (3) Using () and (3) in (1) we have 5k 1 7k 1 4 5k 1 7k 1 4k 4 k 1 k 1 8k = 4 k = 1/ (-1, 1) K P( B (5, 7) 56. Only option (4), for which O = OB = B 57. If a, b, B 0,0, C a,b, D a,ab Here slope of D = slope of C = slope of B = b/a So they are collinear 58. Let radius of the conical heap = r a/q 1.r.4.18 18 3 r 36cm 3 59. Perpendiculars from B to C and from C to B meet at the point. So orthocenter is the point itself. 60. s, BD 4 CD 3 4 BD BC 7 3 CD BC 7 pply cosine formula in BC & CD D BD 16 cos (1) D.BD o D CD 9 cos cos180. () D.CD Dividing (1) by (), we have D BD 16 D.CD 1 D.BD D CD 9 BD D BD 16 4 D BD 16 CD D CD 9 3 D CD 9 B 30 o 30 o 4 3 180 o - 4 D 3 C
4D 4CD 36 3D 3BD 48 7D 3BD 4CD 84 16 9 49 49 84 84 BC 49 But BC = 16 + 9.4.3. cos 60 o = 5. 4. 3. 1/ = 13 3. BC 4. BC 84 7D 13 36.3.6 = 841 D 1 49 49 7 1 3 D 7