NTSE STAGE 2 - SAT SOLUTION-2009

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NTSE STAGE-II

(YEAR-2009)

HINTS AND SOLUTIONS

(SAT + MAT)

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13. Quick time (CaO) is basic in nature. It neutralizes the acidic nature of soil.

15. Altaination of ignition temperaute is necessary for burning.

18. Rusting of iron occurs in the presence of moisture and oxygen. So rusting is fastest in rainy season.

20. Rayon Absorbs less water and take lesser time to dry.

21. C + O₂ CO2

Charcoal Carbon dioxide C + H₂O H₂CO₃

Carbonic acid acids trun blue litmus into red.

22. Flame produce by the burning of gaseous matter.

23. More reactive metal displaces less reactive metal from its salt solution. The correct increasing order of reactivity increasing is Copper < Iron < Zinc < Magnesium

34. frequency N = 500 Time peroid, T = 500 1 N 1  T = 0.0025

ANSW ER KEY

HINTS & SOLUTIONS

NTSE

STAGE-II (2009)

CLASS-VIII [SAT]

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. B A C C B A A B D B D A B C A Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. C D A B D B C C A D C A A C D Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 Ans. D B B A C A C D D B B B C B A Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. A C C D B C B C B C C B C C A Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 Ans. D B D C B C B B A C C D A A A Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 Ans. D A B B D D C D C D B A D A A Ques. 91 92 93 94 95 96 97 98 99 100 Ans. 3 7 5 8 A C C C 1 B

35. Inclination of earth to the plane of rotation = 90º – 235º = 66.5º

36. When we put any positive value of n.

We get answer in such a way that we get two one at the start and one at the end and in the middle we get two.

So sum of digit is always be 4. 37. a2_{+ b}2_{= 13}

 a = 2, b = 3

x3_{+ y}3_{= 65}

 x = 4, y = 1

{(ax + by) + (ay + bx)} {a(x + y) + b(x + y)} {(a + b)(x + y)} {(5)(5)} = 25 38. a 1 a a 2 1 2 1    + a 1 a 1 2 1    a 1 a 1 a   + a 1 a 1 1   × a 1 a 1   a 1 a 1 a   + a 1 1 a 1 a 1     a 1 a 1 a 2 a 1 a      2

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Page # 38 39. bill wage esent Pr bill original =        labour esent Pr labour original ×        Salary esent Pr Salary original 000 , 50 bill original = 11 15 × 25 22 Original bill = 11 15 × 25 22 × 50000 = 60000 40. If 6xy5 is divisible by 55 (5 × 11) i.e. divisible by 5, 11 since it is divisible by 11 6 + y = 5 + x y – x = 5 – 6 = – 1 41. x  y3 ...(i) y  5 1 – z 1 ...(ii)

From (i) × (ii)

x  3 5 1 – z 1           x  z3/5 x  zn  n = 5  42. A B C 2a p2 p1 p3 arABC = 4 3 (2a)2 2 1 (2a × p1 + 2a × p2 + 2a × p3) = 4 3 × 4a2 2 1 × 2a (p1 + p2 + p3) = 3a 2 a(s) = 3a2 s = ₃ a 43. 2 3 p = p 5 100 R 1        5 100 R 1        = 2 3 x 10 100 R 1        = y 15 100 R 1        = z 20 100 R 1        x 2 2 3       = y 3 2 3       = z 4 2 3       x = y 2 3 = z 4 9 = K x = K y = 3 2 K z = 9 K 4 x : y : z = K : 3 2 K : 9 4 K 9 : 6 : 4 44. SP = 3 2 MP ...(i) CP = 4 3 SP ...(ii) From (i) × (ii)

CP = 4 3 ×       MP 3 2 CP = 2 MP CP : MP = 1 : 2 45. p' = p 2 100 25 1        2 100 25 1        = p 2 4 5       2 4 3       = p × 16 25 × 16 9 256 p 225 % chang = p p ' p × 100 =       1 256 225 × 100

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= 100 256 256 225   = 100 256 31   –12.10  12% decrease 46. x = a b

when julie made a mistake reading 'b' since a ×

b are coprime  a = 3. In same way b = 8 correct solution is = 3 8

 correct value – mean of wrong value

3 8 – 2 1        5 8 3 7 3 8 – 2 1        15 24 35 3 8 – 2 1       15 59 3 8 – 30 59 30 59 80 = 30 21 = 10 7 47. a + b + c = 0

Squaring both sides (a + b + c)2_{= 0}

a2_{+ b}2_{+ c}2_{+ 2ab + 2bc + 2ca = 0} a2_{+ b}2_{+ c}2₌

– (2ab + 2bc + 2ca)

Again squaring both sides

 a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2 = 4a2b2 + 4b2_c2_{+ 4c}2_a2_{+ 4ab}2_{c + 4abc}2_{+ 4a}2_bc  a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2c2a2 + 4abc(b + c + a)  a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2c2a2  2 2 2 2 2 2 4 4 4 a c c b b a c b a     = ₂ ₂ ₂ ₂ ₂ ₂ 2 2 2 2 2 2 a c c b b a ) a c c b b a ( 2     = 2. 48. Let remaing are angle is x.

(n – 2)180 = 2180 + x n = 180 x 2180 + 2 n = 180 x 2540

x should be like that

The value of x in such a way that 2540 + x is divisible by 180 and it should not be greatly than 180 because polygon is convex

 x = 160  n = 180 160 2540 = 180 2700 = 15 49. D C B A E H F G

ABC, ADE is equilateral  of side 2r

i.e. 2 × 2 = 4 cm AF = 2 3 (side) = 2 3 × 4 = 2 3 cm So, EF = 2AF = 4 3 HI = 2 + FH + E = 2 + 4 3 + 2 = 4(1 + 3) 50. A D _C B E 2x _x 2x a b Draw CE || AD  AECD is ||gm EC = AD = a, AE = DC = b AEC = ADC = 2x AEC = EBC + BCE  BCE = 2x – x

= x.

 BE = EC = a

 AB = AE + EB = a + b

51. Let x be taken from base so the area is half of the original triangle

2 1       h b 2 1 = 2 1 (b – x)(h + m) m h bh  = 2(b – x) x = b – ₂₍_h _m₎ bh 

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Page # 40 x = ) m h ( 2 bh bm 2 bh 2    = ) m h ( 2 bm 2 bh   = ) m h ( 2 ) h m 2 ( b   52. h' = h + 10 h = 10 h 11 r' = r – 10 r = 10 r 9 C.S.A' = 2r' h' = 2       10 r 9       10 h 11 = 2rh       100 99 CSA' = 0.99 CSA  Corved surbace area is decreased by 1% 53. x = 3 2 3 ) 6 2 ( 2  

squaring on both side x2₌ ) 3 2 ( 9 ) 12 2 6 2 ( 4    = 9 4           3 4 2 3 4 8 = 9 4 × 4 _         3 2 3 2 x2₌ 9 16 x = 3 4 .

54. Mean age of two group is = 3 5 3 48 5 40     = 8 144 200 = 8 344 = 43 55. 7 = 9 12 7 9 7 x 4 8 3 7        63 = 57 + x x = 6 12, 10, 8, 10, 6, 7, 6, 8, 6 mode = 6

for median we arranged data in ascending order 6, 6, 6, 7, 8, 8, 10, 10, 12, median = th 2 1 9        term = 5th_{term = 8} difference between median and mode 8 – 6 = 2

100. 3 + 2 – ₄

(3 + 4)2_{+ 2 + 4}

– 42 + 4

= 49 + 6 – 16 + 4