Sol.Paper-2

WORKSHOP FOR NTSE TEST PAPER-2 Que s.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Ans.

D

C

D

C

A

C

A

C

B

B

A

B

C

B

B

C

D

D

D

A

Que s. 21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

A

A

A

A

A

B

B

D

A

B

D

A

B

C

A

C

C

C

D

B

Que s. 41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

B

B

B

A

C

B

A

D

D

B

A

B

A

A

A

B

B

A

C

C

Que s. 61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

A

A

C

B

A

C

C

C

B

D

B

D

A

C

B

B

B

C

C

D

Que s. 81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

C

B

C

B

A

C

C

A

D

A

C

C

B

A

C

B

A

C

D

C

Ans.

Ans.

Ans.

Ans.

1.

p+

1 1 = x + y and p – =x–y p p

(given) Adding both, we get 2p = 2x  p = x Now, subtracting both, we get

4.

5.

1 2 1 = 2y  =yp= y p p

2.

x=yz x+y+z=0 2x + z = 0 z = – 2x

2

z y 3.

3 or   5

xy = 1

z2  x 2 2

=

4x 2  x 2 4x 2  x 2

– 7 × 3x+1 + (3x+4)= 5x+3 – 5x+2 or 3x+1 [– 7 + 33] = 5x+2 [5–1] or 3x+1 [20] = 5x+2 [4] or 3x+1 [20] = 5x+1 [20]

3 or   5

1 Therefore, x = y 

1 2 3 n 1 1  × × .......... = 2 3 4 n n

=

3 5

x2 – 1is factor of f(x) = x4 + px3 + 2x2 – 3x + q  (x – 1) and (x + 1) are also factor f(x) By factor theorem f(1) = 0 1+p+2–3+q=0 p+q=0 ...(i) f(– 1) = 0 1–p+2+3+q=0 –p+q=–6 ...(ii) Add (i) and (ii) 2q = – 6 q = – 3, p = 3.

x 1

=1 x 1

0

3 =   x+1=0 x=–1 5

6.

(101)2 = have 3 non-zero digits (10101)2 = have 5 non-zero digits So, (101010101)2 = have 9 non-zero digits

7.

We have a + b + c = 6



...(i)

(a + b + c)2 = 36

[On squaring both sides of (i)]



a2 + b2 + c2 + 2(ab + bc + ac) = 36



a2 + b2 + c2 + 2 × 8 = 36

[ ab + bc + ac = 8]



a2 + b2 + c2 = (36 – 16) = 20.

Now, we have a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac) = (a + b + c) [(a2 + b2 + c2) – (ab + bc + ac)] = 6 × [(20 – 8)] = (6 × 12) = 72.

WORK SHOP_NTSE_ STAGE-I _PAGE # 1

8.

16.

a3  b3 a 2  ab  b 2

=a+b

× 32 × 35 × 40 × 42 × 45 × 50)

 8.73 + 4.27 = 13. 9.

10.

Zeroes in the product of 1 × 2 × 3 × 4 ............. 49 × 50 = 12 (2 × 5 × 10 × 12 × 15 × 20 × 22 × 25 × 30

a3 + b3 + a + b = (a + b) (a2 + b2 – ab) + (a + b) = (a +b) [a2 + b2 – ab + 1]

17.

x=

x2 = 2 +

1 x

a 2 x 2 (bx  ay ) b 2 y 2 (bx  ay ) 2abxy(bx  ay )   = (bx  ay ) (bx  ay ) (bx  ay )

4

4 x4

= a2 x2 + b2y2 + 2abxy = (ax + by)2

x4 + 11.

3

 

3

4 448 = 3 3 448 =

3

3

x

64 x

By compairision  64x = 448  x=7 19. 12.

2916 = 54 29.16 +

0.2916 +

14.

461 + 462 + 463 + 464 + 465 461 [1+ 4 + 42 + 43 + 44] 461 [1 + 4 + 16 + 64 + 256] 461  341  341 is divisible by 11 N = 323q + 125 N 323 q  125 = 19 19 Since 323q is divisible by 19. When we divide 125 by 19 we get 11 as remainder

15.

64 2

64 2 ×

64 2

=

64 2 4

=6–4 2 4 x4

= 6 + 4 2 + 6 – 4 2 = 12.

n2 = x (n + 1)2 = x + y n2 + 1 + 2n = x + y 2n = x + y – n2 – 1= x + y – x – 1 = y – 1 (x + 2)2 = n2 + 4 + 4n = x + 4 + 2 × 2n = x + 4 + 2(y – 1) = x + 4 + 2y – 2 = x + 2y + 2

[3(123 + n)]2, So, it is a number which is a multiple of 9. So, 145157 + n also as multiple of 9. So, as the sum of digit of 145157 is 23.  n = 4, So that it is divisible by 9. 5 x 9 y 4 13 z 8 1 12

Now, 1169 when divided by 585 gives remainder = 584.

0.002916

+ 0.00002916 = 5.4 + .54 + .054 + .0054 = 5.9994 13.

1 =

x

18. 3

2

x4 = 4 + 2 + 4 2 = 6 + 4 2

[bx(a2x2  2a2y2  b2y2 )  ay(a2x2  2b2x2  b2y2 )] bx  ay

4 16 Given, 3 16 =  3 27

2 2

20.

(36234) (33512) – (5429) (25123) = ......... 6× ........ 1 – (....... 4) (....... 5) = ......... 6 – .......... 0 so unit digit = 6 Ans.

71.

(B) 12 + 1, (2)2 +1, (3)2 +1,.......... = 2, 5, 10, 17.... (5)2 + 1 = 26

72.

(5 × 2) – 1, (9 × 2) – 2, (16 × 2) – 3, (29 × 2) – 4, (54 × 2) – 5 103 (D) no. is multiplied by 2 and then subtractedfrom 1, 2, 3, 4, 5....... respetively

74.

First letter of every term is moved three steps backward in each next term. Second number of every term is moved two steps backward and third letter is moved two steps forward. Hence, the next choice would be T11E.

WORK SHOP_NTSE_ STAGE-I _PAGE # 2

75.

76.

(B) In figure (A), (915 – 364) = 551. In figure (B), (789 – 543) = 246.  In figure (C), missing number = (863 – 241) = 622.

84.

14 Letters from the left A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 19 Letters from the right

By counting 19 letters from the right and 14 letters from the left, we get the following sequence in which K comes exactly middle. HIJKLMN

(B) Clearly

] 6 × 3 – 4 × 2 = 18 – 8 = 10 In figure B ] 9 × 5 – 5 × 3 = 45 – 15 = 30  In figure C ] 6 × 5 – 2 × 5 = 30 – 10 = 20 In figure A

86. 77.

Consider the English alphabet :

(B) The pattern of the series is ccc, bbb,

(C) Total up position number of letters and divide by the number of letters in each word.

aaa/ccc, bbb, aaa/c. 78.

87.

(C) mno / nopq / opqrs / pqrst

From statement (i) we know that the word FAMOUS has the letter M which is not common in any other words and from its coding

Directions :(79-82) Find the wrong term(s) —

we see that j is the letter which is not common 2

79.

(C)

5

11

+3

+6

20 +9

30 +12

47 +15

in any other coding. Hence, M stands for j.

65 +18

88. 80.

81.

From statements (ii) and (iv), ATE = hdn. From statement (vi), TE = dn. From statement (iii),

(D) 1 + 3 = 4, 3 + 4 = 7, 4 + 7 = 11,...... (instead of 28) 11 + 28 = 29, So, 28 is the wrong term

T = n. Hence, E = d. 89.

From statements (i) and (iii), F = c.

90.

From statements (ii) and (iv), ATE = hdn.

First letter of first, second, third,.........terms is moved three, four, five, ........steps forward

From statement (vi), TE = dn. Therefore, A =

respectively. Similarly, second letter is moved

h.

five, six, seven,......steps forward respectively and third letter is moved seven, eight,

91.

(C) According to the statement

nine,........steps forward respectively. The wrong term is NTA. So, the answer would be

North

MTA 82.

P3C,

R5F, +2

+3

+2

83.

Suresh

(B) Given series is : T8I, +2 +3

+3

17 V12L, X 18 O, Z23R +2 +4

+3

+2 +5

Letter in the given word alphabet

+3

+2

West

+3

Mukesh

Shyam

East

Ram

+6

South

Letter in the

(i) C R E

CDE

(ii) T I V

TUV

(iii) A T I V E

ABCDE

92.

(C) Y represents the finishing point of Rahim and it is to the South-East of point X.

WORK SHOP_NTSE_ STAGE-I _PAGE # 3

96.

On the basis of the information given in the question we have the sitting arrangements of the persons as per the fig. below. G or B

D

G or B

H

E F

A

C

97.

As per the question D A C B A C D A CB C A C B A C B A D C Thus, three such letters are there.

98.

As per the question 699969796979592989696999979 8 9 6 9 7 8. Thus, three such numbers are there.

99.

(D) The required numbers in ascending order are : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45. The 9th number from the top is 27.

100.

Number of students behind Anil in rank = (31 – 7) = 24. So, Anil is 25th from the bottom. Number of students behind Sunil in rank = (31 – 11) = 20 So, Sunil is 21st from the bottom.

WORK SHOP_NTSE_ STAGE-I _PAGE # 4