Kvpy Sa Stream 2007

HINTS & SOLUTIONS (YEAR-2007) ANSWER KEY  Ques.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

B

C

D

C

B

D

D

B

A

B

D

C

D

D

B

Ques.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

B

C

B

C

C

B

C

C

A

C

 A

C

B

A

C

Ques.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

B

A

C

C

D

B

A

B

B

D

A

C

C

A

D

Ques.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans.

 A

A

A

D

B

A

D

A

B

D

D

D

D

B

B

Ques.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

Ans.

C

A

B

A

C

A

A

D

A

B

D

D

A

C

C

PART-I (1 Mark) MATHEMATICS

1.

n [2.8 + (n – 1) 4] 2 s1(n) = s2(n) s1(n) =

s2(n) =

n [2.17 + (n – 1)2] 2

n n [2.8 + (n – 1) 4] = [2.17 + (n – 1)2] 2 2 2(n – 1) = 18 n–1=9 n = 10 s1(10) = 5[16 + 36] = 2 60 = s 2(10)

2.

(sin(2x)) 4 = 1/8 Range   x    [0, 2] Let 2x = y given range 0   x    2  0  2x   4   0   y   4  4 (sin(2x))  = 1/8 From above graph we can say, say, Total Eight solution 3.

3.r 2  – 8.r  5

0 4.r 2  – 3.r  7  3r 2 – 3r – 5r + 5 > 0 3r(r – 1) –5(r – 1) > 0   (3r – 5)(r – 1) > 0 (3r – 5)(r – 1) > 0   (– , 1)  (5/3, ) r  

and

4r2  – 3r + 7 > 0

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as

D 0 can be

O

So option (D) is right 8.

Case - I Here a = 1

Case – II

3

 Altitute

a=

9.

2 3

2

a=1

.

Let AB = a then BE = a tan CE

= tan tan CF CF = a cot  – a Now, In GHF tan =

HF GH

=

1  tan  2  cot 

Solving we get tan =

 10.

2 3

sin =

2 13

R = 5, c = 6, b = 6 S=

= R=

12  x  ,    = 2

s(s – a )(s – b)( s – c )

12  x x x 12 – x . . . 2 2 2 2 abc 4

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

5

⇒

36 x 4 (12  x )x 2 (12 – x ) 4

(12)2  – x 2  =

⇒

36 5

2

11.

⇒

 36    = x2 (12)  –    5  

⇒

x = 48/5

2

Let AP = x 1, PQ = x 2 Now,

ar (  ADE) ar (  ABC)

 =

1 2

2

 =

2

    

12.

x1

( x1  x 2 )

1 2

2x12 = x 12 + x 22 + 2x 1x2 x12 – x22 – 2x1x2 = 0

x1 x2

 =

2 44 2

x1  = x2

2  1 1 1 4

cos2  + cos2   = 3/2

and

sin   sin    =

1  cos 2 1  cos 2  +  = 3/2 2 2

and

2sin   sin    =

cos 2  + cos 2  = 1

and

cos (  –  ) – cos ( +  ) =

1 2 1 2

...(ii)

2 cos (  +  ) cos (  –  ) = 1 ...(i) From (i) and (ii) cos ( +  ) cos (  –  ) = cos ( –  ) – cos (  +  ) ⇒

 +    = 13.

cos ( +   ) =

1 2

cos (  –  ) = 1

 3

1  Area = [x + 2x + 3x] = 2

3x2 4

3x2

⇒

3x =

⇒

x= 4 3

⇒

and

4

 Area = 12 3

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14.

Let wealth A, B, C are x, y, z Given : A = B + C x=y+z ....(i)  A distribute half of his wealth wealth to B & C

 x  2 1    2 : 1 i.e.  & 2 3 3     Now, A =

x 2

B=y+

 x    3  2 

C=z+

 x    3  2 

2

1

Now, B = A + C y+

x 3

x x  + z + 6 2

 = x

y–z=

3

3y – 3z = x

. . . ( i i)

Let t be fraction that A should distribute and the ratio of distribution is 1 : 2 i.e.

1 3

 <

2 3

.

Now, A = (1 – t)x tx

B=y+

3 2tx

C=z+ z+ z+

2tx 3 tx 3

3

 = y +

tx 3

 + (1 – t) x

 = y + x – tx

3z + tx = 3y + 3x – 3tx 4tx = 3 (y – z) + 3x 4tx = 3x + x 4tx = 4x t = 1. So, A would distribute his whole wealth t o B and C.  Fraction is 1. 15.

Put 3 times water of 110 ml to container and and take 13 times 25 ml water from container then container  has 5 ml water.

16.

Distance = speed × time Distance covered in car = 50 × 4 = 200 km He move for 1 hr = 20 km Time take while retraining from town to village time =

Dis tan ce 200  =  = 5 hr  Speed 40

Total journey = 200 + 200 + 20 Total time taken = 4 + 1 + 5 = 10 average speed =

420 10

 = 42 km/hr 

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17.

Given 4 × 4 × 4 cubes is mode faces 64 1 × 1 × 1 cubes total cubes = 64, white = 20, Red = 44 To find minimum number of visible white box Counting total visible faces of unit cube Total number of f aces of small cube on bigger cube except boundry cubes = 4 × 6 = 24 Counting boundry cube = 16 + 8 + 8 = 32 Total visible faces = 56  But we have 44 Red cube minimum of number of white faces  cubes which are visible = 56 – 44 = 12

18.

Multiple of 3   3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 36 then n = 5, 8, 11, 14, 17, 20, 23...... Multiple of 5   5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 then n = 8, 13, 18, 23, 28 ...... Now, common two digit numbers are 23, 38, 52, 67, 82 , 97.

19.

if start with H1

 2 ways

if start with H 2  2 ways if start with W 1

 2 ways

if start with W 2

 2 ways

m=8 in case of circle H1

W2

H1

H2

H2

W2

W1 20.

W1

Total students = 300 One student read = 5 newspapers Number of newspapers read by 300 students = 5 × 300 = 1500 newspaper  Number of different newspapers =

1500 60

 = 25.

PHYSICS

T

21.

Hence n = 2

a

mg–T = ma T = m(g–a)

mg

= 6000 (10–2) = 4.8 × 104 N

22.

 A

u

C v

5 5 Between A and B v2 = u2 = +2a(2s) Between A and C v2 = u2 + 2as

from (i) and (ii) v = 24.

v

B .....(i) .....(ii)

u2

 v2 2

238 92 U

206  82 Pb  824 He  601e

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CHEMISTRY

31.

Using the formula, N = N 0

 1     2 

n

where, No = initial amount of radioactive substance N = Amount of substance left after ‘n half lives

Total time (t) No. of half lives (n) =

n=

Half  life period

21.2 years 5.3 years

n=4

 1  so, N = 20    2  32.

4

 =

20  = 1.25g 16

Zn(s) + H2SO4(aq) Zinc

  ZnSO4(aq) + H2(g)

Sulphuric acid

Zinc sulphate

Hydrogen gas

(salt)

Zn(s) + 2NaOH (aq)  Na2ZnO2 (aq) + H2(g) Zinc

Sodium hydroxide

33.

Sodium

Hydrogen gas

zincate

 Acetylene (CH   CH) has a triple bond.

34.

Ne has 10 electrons and oxygen has 8 electrons.

35.

KMnO4 being strong oxidising agent will oxidise Cl – ion present in HCl to form chlorine gas. 2KMnO4 + 16 HCl   2KCl + 2MnCl2 + 8H2O + 5Cl2

36.

Order of strength for halogen acids is H I > HBr > HCl > HF

38.

Concentrated sulphuric acid is a dehydrating agent and it burns the organic compounds like sucrose so the colour of the solution turns black. Conc . H2SO 4 C12H22O11(s)   11H2O(g)            12C + 11H

Sucrose

Carbon

W ater vapour  

  (Black)

Oxidation 39.

ZnO+C Zn + CO Reduction

40.

Monomer of Teflon is Tetrafluoroethene(CF 2=CF2)

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PART-II (2 Mark) MATHEMATICS 51.

x = x3 + y4 y = 2xy Two cases

;

y = 2x y

1.

case -1

y = 0 & x    R

2.

case - 2

x=

1 & y   R 2

Now, equation x = x 3 + y4 Taking case - I x = x3 Solution x = 0, 1, –1 ∴ pairs = (0, 0), (1, 0), (– 1, 0) Taking case ca se 2

1  = 2 3 8

 3  y=±   8   52.

3

 1    + y4  2 

 = y4

1/ 4

 1  3 1/ 4    Solution set =  2 ,  8   ,        

f(x)  degree (n)

 1  3 1/ 4   ,      2  8   

f(1) =

2 f(3) =  f(x) = axn + bxn – 1  + cx n – 2 ....... If degree 0 f(x) = a = constant f(1) = f(3) but it is not t rue If degree 1 f(x) = ax + b f(1) = a + b =

2

f(3) = 3a + b =   Two variables & two unknown a & b can be found uniquly one polynomial used only  For n>1 Let n = 2 ax 2 + bx + c We have 3 variable & only 2 equations can be formed from given condition Hence infinite such polynomial can be formed 53.

p(x) = Q(x) . (x2 – 3x + 2) + (2x – 3) p(1) = – 1 p(2) = 1 at least one root between 1 & 2

1  –1

1

2

 –1

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N

x

D

x

C y

2y

M

54.

y B

2x

 A

In MCN, tan  =

y 2y and In  ADN, tan (90 – ) = x x

cot  =

2y x

tan  =

x 2y

2y2 = x 2

x  = y

2 1

B ( 2 , 3) S

R

x

x

x P

O(0, 0)

55.

x

(6, 0)  A

Q

a 2

tan  =

x=

3 2

 =

3a

x a

 =

3 – x

....(i)

2  – a

and tan   =

x 6 – a – x

=

3 6 – 2

....(ii)

[from (i)]

2

2x 3



a=

and

6x –

2 x  = 18 – 3a – 3x



9x –

2 x  = 18 – 3 .



x= 2

Q’

[from (ii)]

2x 3

P

102º

56.

R’

r   A

r 

O

B

Since R can be any point between A & R & hence its corresponding point Q will lie on the arc AQ. Hence PRA can not be uniquely determined.

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y

3x 57.

 A

2x

B

C

D

5y

Given

 AB CD

 =

3 2

1 BC  = 5  AD Let AB = 3x CD = 2x , BC = y, y, AD = 5y 3x + y + 2x = 5y 5x = 4y To find AC : BD = (3x + y) : (y + 2x)

3x  y 3 x  5x / 4  = 2x  y 2 x  5x / 4 3  5/4

58.

12  5

17

2  5/4

=

Let and and

r = radius of sphere R = radius of cone H = height of cone

4 3

1

r 3  =

3

85

 =

13

R2H

4r 3 = R2H Now,

....(i)

k4r 2 =  R2 +  R R 2 k(4r 2) = R2 + R R 2

 H2

 H2

  16r 6   2  4kr 2 = R  R  R  R4      6     1  R 2  16r   r 2 Let 2  = x  R6  R    



  r 2     4k  2   = 1  R  



4kx = 1 +

 

16k 2x2 – 8kx + 1 = 1 + 16x 3 x=0 16k2x – 8k = 16x 2

k = 0,

1  16 x 3

2x2 – 2k 2x + k = 0 4k4 – 8k = 0 (2)1/3

sinc since e cone cone is uniq unique ue ther theref efo ore it has equa equall root roots s

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59.

Let original four digit number is = m2 and new 4-digit number is = n 2 then m 2 – n2 = 1111 (m + n)(m – n) = 1 × 1111 or 11 × 101

(

w

h

e

r

e

,

m

,

n

,

)

Case - I

m + n = 1111 m–n=1 m = 556 n = 555 But m 2 = is a 4-digit number  Case - II m + n = 101 m – n = 11 m = 56 and n = 44 So, there is only one such 4-digit number. 60.

m , n are integer  1/3 < m/n < 1 ...(i) m n

m n

 =

n  0

m  = +ve integer  m 1 m=2  Using equation (i) n = 3, 4, 5

 =

m



n

2

=

3

,

2 4

,

2 5

PHYSICS 61.

 = 0t –

1 2 t 2

 = 60 × 5 –

1  × 8 × 25 2

= 300 – 100  = 200 radian or n =

 100  2 

~  32 rev./s 62.

Energy released = mL V = volume of water × density ×L V = 100 × (10 3)2 ×1×103 × 22.5 × 10 5 J = 22.5 × 1016 J Number of bombs =

63.

Req  = 

22.5  1016 1014

 2250 ~ 2000

5r  5  1 5    6 6 6

V  10 volt

So, I =

V R



10  6 5

 12 A

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64.

time taken in rotataing angle   is t, then

45  103

t=

8

 15  10 5 s

3  10 t = 15 ×10 –5 second 360

T=

 1 T

N= 65.



N

27º

 15  10 5 1 4  10



3

1000  = 250 rev/s 4

Focal length, f = 1m, size of object, h 1 = 1000m, size of image, h 2 = 0.2 m image is real, so  –

h2 h1





v u

 v  2  10  4 u

1 1 1   f  v u

1=–

1 2  10

4

u



1 u

 u = – 5000 m or u = –5 km

CHEMISTRY 66.

N1V1 = N2V2 36 × V1 = 3 × V 2 V2 V1

 = 12

67.

4Fe + 3O2 + nH2O  2Fe2O3.nH2O Iron Iron Oxy Oxygen gen Water Water Hydr Hydrat ated ed Ferr Ferric ic oxid oxide e   (Rust)

68.

Solubility of KNO3 at 90º C = 200 g/100 m l Solubility of KNO3 at 30º C = 40 g/100 ml Decrease in solubility when KNO 3 solution is cooled from 90ºC to 30ºC = 200 – 40 = 160 g/100ml

69.

Yeast

4     2C2H5OH          CH3COOH C6H12O6     – 2CO

KMnO

2

70.

Zn + 2H 2SO4(conc.)     ZnSO4 + SO2(g) + 2H2O (CH3COO) 2 Pb + SO 2

    PbS (Black Precipitate)

Cu + SO 2

    CuS

+

O2

(Black mirror)

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