NTSE STAGE 2 - SAT SOLUTION-2009

NTSE STAGE-II (YEAR-2009) HINTS AND SOLUTIONS (SAT + MAT)

NTSE STAGE-II (2009) CLASS-VIII [SAT]

HINTS & SOLUTIONS ANSWER KEY Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans.

1 B 16 C 31 D 46 A 61 D 76 D 91 3

2 A 17 D 32 B 47 C 62 B 77 A 92 7

3 C 18 A 33 B 48 C 63 D 78 B 93 5

4 C 19 B 34 A 49 D 64 C 79 B 94 8

5 B 20 D 35 C 50 B 65 B 80 D 95 A

6 A 21 B 36 A 51 C 66 C 81 D 96 C

7 A 22 C 37 C 52 B 67 B 82 C 97 C

8 B 23 C 38 D 53 C 68 B 83 D 98 C

9 D 24 A 39 D 54 B 69 A 84 C 99 1

10 B 25 D 40 B 55 C 70 C 85 D 100 B

11 D 26 C 41 B 56 C 71 C 86 B

12 A 27 A 42 B 57 B 72 D 87 A

13 B 28 A 43 C 58 C 73 A 88 D

14 C 29 C 44 B 59 C 74 A 89 A

15 A 30 D 45 A 60 A 75 A 90 A

13.

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

35.

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

15.

Altaination of ignition temperaute is necessary for burning.

36.

18.

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

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 + b2 = 13  a = 2, b = 3 x3 + y3 = 65  x = 4, y = 1 {(ax + by) + (ay + bx)} {a(x + y) + b(x + y)} {(a + b)(x + y)} {(5)(5)} = 25

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.

23.

34.

Flame produce by the burning of gaseous matter. More reactive metal displaces less reactive metal from its salt solution. The correct increasing order of reactivity increasing is Copper < Iron < Zinc < Magnesium frequency N = 500 Time peroid, T = T = 0.0025

1

38.



a2  a 1 a

a

1 2

+

1

1

a +

1 1  N 500

1 2

1 a +

1

1 a a × 1 a 1 a

1

1 a 

1 a a



1 a

1 a a

1 a

a

1

1 a

1 a

2 a 

1 a

1 a 2 . 1 a

Page # 37

39.

original bill Pr esent wage bill

43.

 original Salary   original labour    ×  =   Pr esent Salary   Pr esent labour 

R   1   100  

original bill 22 15 = × 50,000 25 11

41.

1 –

z

15

R    = y 1   100 

2

3

3 3 3 x  = y  = z  2 2 2 x=y

y=

2 K 3

z=

4K 9

x:y:z=K:

3

    

2 4 K: K 3 9

44.

SP =

2 MP 3

...(i)

3 SP 4 From (i) × (ii) CP =

 n= 5 A

42. p3

CP =

...(ii)

2  3 ×  MP  3  4

MP 2 CP : MP = 1 : 2

p2

CP =

p1 B

C 2a

2

45.

3 (2a)2 4

1 × 2a (p1 + p2 + p3) = 2

s=

3 a2

3a

25   25    1   p' = p 1   100   100  2

1 3 (2a × p1 + 2a × p2 + 2a × p3) = × 4a2 2 4

a(s) =

4

9:6:4

x  z3/5 x  zn

arABC =

20

3 9 =z =K 2 4

From (i) × (ii)   1 x   –1  5 z

R    = z 1   100 

x=K

...(ii)

1 5

5

3 2

= 10

22 15 × × 50000 = 60000 25 11 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 x  y3 ...(i)

y

5

R    x 1   100 

Original bill = 40.

R   3  p = p 1  2  100 

2

3a

5 3 = p    4 4 =p×

2

2

25 9 × 16 16

225p 256

% chang =

p' p × 100 p

 225   1 × 100 =   256 

Page # 38

=

225  256  100 256

=

31  100  –12.10 256

n=

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

 12% decrease 46.

b a

x=

when julie made a mistake reading 'b' since a × b are coprime  a = 3. In same way b = 8 correct solution is =

2540  x 180

49.

H F

B

8 3 D

8 1 7  8  –   3 2 3 5 

G

3 3 (side) = × 4 = 2 3 cm 2 2

AF =

8 1  59  –   3 2  15 

So, EF = 2AF = 4 3 HI = 2 + FH + E

8 59 – 3 30

=2+4 3 +2

80  59 21 7 = = 30 30 10

= 4(1 + 50.

a+b+c=0 Squaring both sides (a + b + c)2 = 0 a2 + b2 + c2 + 2ab + 2bc + 2ca = 0 a2 + b2 + c2 = – (2ab + 2bc + 2ca) Again squaring both sides  a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2 = 4a2b2 + 4b2c2 + 4c2a2 + 4ab2c + 4abc2 + 4a2bc  a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2c2a2 + 4abc(b + c + a)  a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2c2a2

48.

2 2

2 2

2 2

a b b c c a

= 2.

Let remaing are angle is x. (n – 2)180 = 2180 + x n=

2180  x +2 180

D

b

C

a 2x A

x

B

E

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

a 2b 2  b 2 c 2  c 2a 2

2(a 2b 2  b 2c 2  c 2a 2 )

3)

2x

a4  b4  c 4

=

E

ABC, ADE is equilateral  of side 2r i.e. 2 × 2 = 4 cm

8 1  35  24  –   3 2  15 



C

A

 correct value – mean of wrong value

47.

2540  160 2700 = = 15 180 180

n=



51.

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

1 2

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

bh = 2(b – x) hm x=b–

bh 2(h  m)

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2bh  2bm  bh 2(h  m)

x=

55.

7  3  8  4  x  7  9  7  12 9

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,

bh  2bm = 2(h  m) =

7=

b(2m  h) 2(h  m)

th

52.

h' = h +

 9  1  term = 5th term = 8 median =   2 

h 11h = 10 10

difference between median and mode 8 – 6 = 2

r 9r r' = r – = 10 10

C.S.A' = 2r' h'  9r   11h   = 2     10   10 

 99   = 2rh   100 

100.

3

+

2

–

4

(3 + 4)2 + 2 + 4 – 42 + 4 = 49 + 6 – 16 + 4 = 35

CSA' = 0.99 CSA  Corved surbace area is decreased by 1%

2( 2  6 ) 53.

x=

3 2 3

squaring on both side x2 =

54.

4(2  6  2 12 ) 9(2  3 )

84 3    24 3   

=

4 9

=

2 3  4   × 4  9 2 3 

x2 =

16 9

x=

4 . 3

Mean age of two group is =

40  5  48  3 53

=

200  144 8

=

344 = 43 8

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