Patna Centre. Hint & Solution , 90, , 60, 2. 4 a b c d a b d Mean

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1 Patna Centre Hint & Solution Part I MT 1. Hints: , 60, , 90, Hints:- 3, 6, 9,1, 15, Hints:- a b c d a b c d a b d Mean Hints:- Perimeter of circle = r New perimeter of circle = 3r Percentage increase of perimeter = 3 r r 100% r 100% 50% r r Hints: G L I B H B M J C T I G E R U J H F S Hints:- brother wife T P Q R father sister S son 9. Total horticulture % % % % 10. Hints:- Fruits = % 30.14% 09

2 Vegetables = % 37.79% 09 Flowers= 100% 95% [3] Hints:- Increase % of Fruits = % 18.86% 53 Increase % of vegetable = % 8.86% 7 Increase % of flowers = 1 100% 100% 1 Increase % of horticulture = % 11.76% Hints:- nswer Question number Hints:- Percentage area = % 68.31% Hints: % 5.66% % 3.57% % 5.17% % 3.7% 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 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.

3 1. Hints:- (a) (d) Poets Philosophers Poets Philosophers. Hints:- 1 5,5 1 9, , , 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 verage (36 40) years Hints:- verage = Hints:- Percentage salary 40,000 per months = % 45% 8. Hints:- verage = % employees 100% 55% 0 9. Hints:- verage age = Hints:- Percentage (30 35) years employees 7 100% 35% Hints: Hints:-

4 BB, FE, IT, ML, PP, TS Hints:- HOME 9 km 6 km 3 km 5 km 34. Hints:- Rs Rs Rs Hints:- QUESTION DOMESTIC (Code Similar letters) RESPONSE OMESICEM Hints:- 50 digits 18 digits = 3 digits 5,57,6,67,70,71,7,73,74,75,76,77,78,79,8,87,9, Hints:- B, ED, IH, NM 4. Hints: STREMERS - UVTGLIDQR KNOWLEDGE MPQYLDCFD 43. Hints:- E C brother B brother Father husband D

5 44. Hints:- 3x 9 1 5x x 07 60x x 60x x 99 x 11 3x 33 5x 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: Hints:- y 18 x 18 y x ii equation i & ii x 1 y , 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, Hints:- Total marks of remaining 0 students mean 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

6 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 m / s 3. It retraces its path. 4. v o Convex lens in used to correct long sightedness. 6. In sun proton-proton cycle is given as energy released = 6 MeV H He e v 6MeV 7. By equation of mass-energy equibalance E mc = 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 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 to is divided into 0 parts. So the least count of ammeter is 0.1.

7 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 < O OH CH C CH CH C CH O OH CH C O C H CH C O C H O O OH O CH C CH C O C H CH C CH C O C H 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 LilH4 () CH I CH 3 4 (3) l C 1H O 3CH 4lOH NO 6. NaOH & SO3

8 Biology 7. Homologous organs are same in origin but are different in function 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 Medula oblongala controls all the involuntary activities of body In asexual reproduction fertilization do not take place Bile does not contain any enzyme (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 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 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

9 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 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 x /q x 1... x49 x Let us suppose that x50 = 134 Then x1 + x +.. x49 = x x... x Correct mean = (in rupees) 50 x x... x 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 Given cos ec cot, cos ec cot. On squaring and using cot. cot cot 11cot tan 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

10 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 b b b b a 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 r 36cm 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 o - 4 D 3 C

11 4D 4CD 36 3D 3BD 48 7D 3BD 4CD BC 49 But BC = cos 60 o = / = BC 4. BC 84 7D = 841 D D 7

MT - GEOMETRY - SEMI PRELIM - I : PAPER - 4

MT - GEOMETRY - SEMI PRELIM - I : PAPER - 4 07 00 MT A.. Attempt ANY FIVE of the following : (i) Slope of the line (m) 4 y intercept of the line (c) 0 By slope intercept form, The equation of the line is y m + c y (4) + (0) y 4 MT - GEOMETRY - SEMI