ntse stage i rajasthan last 4 years solutions

8/16/2019 NTSE Stage I Rajasthan last 4 Years Solutions

1/29Page # 1

ANSWER KEY

HINTS SOLUTIONS

RAJASTHAN -NTSE_ STAGE-I (2013)

CLASS-X [MAT]

1. The sequence is a combination of two series

I 2, 6, 12, ?

II 30, 20, 12

The pattern followed in I is 12 + 1, 22 + 2,32 + 3,42 + 4,...

missing number = 42 + 4 = 20.

2. The terms of the series are

12 x 6, 22 x 5, 32 x 4, 42 x 3 ,52 x 2 ,62 x 1 ,72 x 0.

Hence, the missing term would be62 x 1 = 36.

3. The terms of the series are

previous term x2 + 1, x2 –2, x2 + 3, x2 – 4, x2 + 5.... and so on.

Hence, the next term would be

114 x2 + 5 = 233.

4. The term of the series are

previous term + 24,+ 40,+56,72,88, and so on.

Hence, the next term would be

217+ 88 = 305.

5. The sequence is a combination of three series

I 0, 3, 8, ?

II 2, 3, 4,5

III 2, 5, 10, 17

The pattern followed in I is + 3,+ 5,+ 7

missing number = 8+ 7 = 15.

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans C A B A D C B D A C B D C D B

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans A C A D A C B C B A C B A D C

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans B A D D B B D A A C C A D C B

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans C A C B B D A D A D C D B C B

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans B A D B C A A D D A B C C B A

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans D C C D B B D C B D A C B C B


2/29Page # 2

9. 8 × 2 – 1 = 15

15 × 2 + 1 = 31

31 × 2 – 1 = 61

61 × 2 + 1 = 123

123 × 2 – 1 = 245

245 × 2 + 1 = 491

10. 22 –1= 3

22 + 2= 6

52 –1= 24

5

2

+ 5= 3082 –1= 63

82 + 8= 72

112 –1= 120

112 + 11= 132

15. ( + 5, + 5, 1, + 5, + 5) Ans. (2) SUNQS

16. ( + 2, + 1, – 1, + 2, – 2) Ans (1) ZJSUM

17. (+3, –2, +3, –2, + 3) Ans. (3) MQNRO

19. stq logic + 5

20. (7 + 5 + 9) × 3 = 63

(11 + 3 + 6) × 2 = 40

(7 + 18 + 11) × 4 = 144

21.9

6812 = 64

8

10125 = 75

1412921

= 162

22. (7 × 8 ) + (6 + 4) = 66

(3 × 9 ) + (7 + 4) = 38

(11 × 9 ) + (7 + 2) = 108

23.

16

2534 = 7.5

16

3265 = 11.25

16

9225 = 11.25

27. The pattern of numbers are as followed12 – 1 = 0

22 – 2 = 2

32 – 3 = 6


3/29Page # 3

42 – 4 = 12

52 – 5 = 20

62 – 6 = 30

72 – 7 = 42

82 – 8 = 56

92 – 8 = 72

28. In column first 91 + 84 + 25 = 200

In column second 64 + 76 + 60 = 200

In column thrid 73 + 61 + x = 200

So, x = 66

31. 20 ÷ 4 = 5 52 = 25

81 ÷ 27 = 3 32 = 9

44 ÷ 11 = 4 42 = 16

36. Ans. – 2

37.

38.

(41to 45)

Letter Code

G u

Z n

C q

U z

S oX v

B l

U e

W y

T x

O b

A s

R w

E t

H f


4/29Page # 4

M d

J a

I p

N k

L g

P r

41. Ans. (3) bpxg

42. Ans. (1) boqx

43. Ans. (4) gteqw

44. Ans. (3) ptok

45. Ans. (2) wfsq

46. Ans. 3

So B in Uncle of C

47.

So R in mother of P. Ans. 1

48. 32 – 1 : 33 + 1 : : 42 – 1 : 43 + 1

OR

22 × 2 : 33 + 1 : : 32 × 2 : 43 + 1

49. 23 + 33 : 33 + 43 : : 43 + 53 : 53 + 63

50.23

19::

17

13:

11

7

Next Prime no after 23 is 29, 31 :31

29

OR

7 + 17 = 11 + 13 and 19 + 33 = 23 + 29

So, Ans.33

29

51. Ans.

52.

53.


5/29Page # 5

PHYSICS

9. R =220

)60(

P

V 22

= 806.66

I =

V

P

=220

60

= 0.27

CHEMISTRY

16. Isotopes have same atomic number, but diffferent

mass number. They have different number of

neutrons.

17. 2 Cu + O2

2CuO

(x) (y)

Brown & Br ight Black

CuO + H2 Cu + H

2O

ANSWER KEY

HINTS SOLUTIONS

NTSE STAGE-I (2013)

CLASS-X [SAT]

21. Atomic number of argon is 18. Hence electronic

configuration

Ar = 2, 8, 8

22. Element aluminium has atomic number 13.

Electronic configuration is 2, 8, 3

Valency = 3

2Al + 3Cl2 2 AlCl3

23. Structure of methane H H

H

H

C

It has 4 covalent bonds.

MATHEMATICS

36. y = x2 – 3x – 4

x2 – 3x = y + 42

2

3x

= y + 4 +

4

9

2

2

3x

= y +

4

25

Graph of equation will be parabola.

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans 3 4 1 1 4 1, 3 4 3 1 3 1 3 3 1 2

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans 3 1 4 1 2 2 3 1 2 4 4 1 2 1 2

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans 4 3 2 4 4 2 1 4 3 3 2 1 3, 4 4 1

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans 4 1 2 4 4 3 1 3 2 1 1 2 3 4 2

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans 3 4 2 1 1 1 2 1 4 1 2 1 2 3 4Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans 2 4 1 4 1 1 4 4 2 3 3 2 1 3 1


6/29Page # 6

37. 2x + 3y = 7

(a – b)x + (a + b)y = 3a + b – 2

For infinite solution

2

1

a

a =

2

1

b

b =

2

1

c

c

ba

2

=

ba

3

=

2ba3

7

ba

2

=

ba

3

2a + 2b = 3a – 3b a = 5b

ba

3

=

2ba3

7

9a + 3b – 6 = 7a + 7b 2a – 4b = 6 2(5b) – 4b = 6 [a = 5b] 10b – 4b = 6 6b = 6 b =1 a = 5

38. Roots will bea2b

a2ac4b2 .

39. a3 + a

7 = 6

a + 2d + a + 6d = 6 2a + 8d = 6 a + 4d = 3 a = 3 – 4d(a

3) (a

7) = 8

(a + 2d) (a + 6d)= 8 [3 – 4d + 2d] [3 – 4d + 6d] = 8 (3 – 2d) (3 + 2d) = 8

9 – 4d

2

= 8 4d2 = 1

d2 =4

1

d = 21

.

40.OC

OA =

OD

OB

5

2 =

5x2

2x

4x + 10 = 5x – 10 x = 20

41. (x, 0), (0, y) and (1, 1) are coll inear

Then, area of ABC = 0

2

1 [x (y – 1) + 0 + 1(0 – y)] = 0

xy – x – y = 0

x + y = xy

42. sin (A + B) =2

3

sin (A + B) = sin 60°

A + B = 60° .... (1)

cos(A – B) =2

3

cos (A – B) = cos30°

A – B = 30° ... (2)

solving (1) & (2)

A = 45°, B = 15°

43. Length of shadow decreases and will be zero.

44. Perimeter of square = Perimeter of circle

4a = 2r

r =2

a4

Given : a2 = 121

a = 11

r =222

114

7

r = 7 m

Area of circle = r 2 = 49 m2

45. Volume of frustum =3

1 h( 21r +

22r + r 1r 2)

=3

1

7

22 14 (22 + 12 + 2)

=3

1 22 2 7

=3

308 = 102

3

2 cm3

46. Median = 525, Mode = 500

Mode = 3 median – 2mean500 = 3(525) – 2mean

2 mean = 1575 – 500

mean =2

1075 = 537.5

47. The author of the book "The book on games of

chance is given by J. Cardon


7/29Page # 7

48.53

5

=53

5

53

53

=53

)53(5

= –

2

5( 3 + 5 )

49.2

2100 = 2100–1 = 299

50. 1 line

51.

P Q

S T

130°

50°110°

R

QRS = 110° – (50°) = 60°

52. Rectangle.

53. Diameter

54. S =2

324024 =

2

96 = 48 m

Area = )cS)(bS)(aS(S

= )3248)(4048)(2448(48

= 1682448

= 16824224

= 24 16 = 384 m2

55. 2rh = 4.4

2 7

22 0.7 h = 4.4

h = 1 m.


8/29Page # 8

1. (1) In the given series, every letter of every

term is moved two steps forward to obtain the

corresponding letters of the next term. So, the

missing term is JMO.

2. (4)The first letter of each term is moved two

steps backward, the second and third letter is

moved one steps backward to obtain the

corresponding letters of the next term.So, the

missing term is EBJ.

3. (1) Series is cab/ cab/ cab/ cab. So, pattern

cab is repeated.

4. (4)The first,third and forth letter of each term is

moved two steps forkward, the second letter is

moved one steps forward to obtain the

corresponding letters of the next term.So, the

missing term is PFSK.

5. ( 3 ) 2 2 + 1 , 3 2 + 1 , 4 2 + 1 , 5 2 + 1 ,

62+1,72+1.82+1..................

6. ( 2 ) 2 2 – 2 , 3 2 – 2 , 4 2 – 2 , 5 2 – 2 ,

62 –2,72 –2.................

7. (1) 22+1,22 –1,32+1,32 –1,42+1,42 –1,52+1,52 –1,...

8. Series : + 4, + 6, + 8, + 10, + 12,......

16. (3) Rest of all has the same difference between

ANSWER KEY

HINTS SOLUTIONS

RAJASTHAN_NTSE STAGE-I (2014)

CLASS-X [MAT]

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans 1 4 1 4* 3 2 1 2 4 2 3 1 * 1 1

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans 3 2 2 4 3 3 2 4 2 3 2* 1* 2 4 4

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans 3 3 3 3 3 2 3 3 2 2 2 4 1 2 4

Ques. 46 47 48 49 50

Ans 1 * 2,3 3 3

them.

17. (2) Only option 2 have the same difference

between them.

24-27.

Letters E F H I N O R S T U V W X

Code y z l m c d g r/h s t f g h/r

33. 23 * 52 = 23 + 25 = 48

43 * 35 = 43 + 53 = 96


9/29Page # 9

ANSWER KEY

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans 1 4 4 2 4 2 3 1 3 1 2 3 1 4 3

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans 3 2 1 4 4 3 2 2 3 4 1 2 4 1 1

Ques. 31 32 33 34 35 36 37 38 39 40

Ans 3 2 1 1 3 4 2 4 1 2

HINTS SOLUTIONS


CLASS-X

LANGUAGE COMPREHENSIVE TEST (LCT )


10/29Page # 10

PHYSICS

2. F = 221

r

mmG = 6.675 × 10 –111 × 2)1(

)1)(1( = 6.675 ×

10 –11N

4. Req.

=21

21

RR

RR

= )222(6)222(6

=12

36 = 3

Itotal

=eqR

V =

3

12 = 4 Amp.

I =2

totalI =

2

4 = 2 Amp.

Voltage drop across 2 resister isV = IR = 2×2 = 4 Volt

7. P1 =

R3

V2

5 =R3

V2

R

V2 =15 .... (i)

ANSWER KEY

HINTS SOLUTIONS


CLASS-X [SAT]

Now P2 =

)3/R(

V2

=

R

V3

2

= 3(15) = 45W

9. u = 160 m/s

t = 0.02 sec

v= 0

v = u + at

0 = 160 + a(0.02)

a = – 8000 m/s2

v2 = u2 + 2as

s =a2

u –v 22=

)8000(–2

)160( –0 2

=16000 –

25600 – = 1.60 m

CHEMISTRY

13. Ammonium chloride (NH4Cl) salt consists of non-

metallic elements only. It contains N, H and Cl whichare non metals.

14. Gases and liquids are different from solids in Fluid-

ity property. In solids particles are bind together by

means of strong forces of attraction hence they are

not able to move i.e. show no fluidity.

15. The action of cleaning of oily dirt by soap is based

on presence of hydrophilic end (i.e. ionic part) and

hydrophobic end (i.e. hydrocarbon past)

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans 3 2 3 1 4 1 4 1 2 * 2 3 2 4 4

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans 4 4 3 1 2 1 2 2 1 2 4 2 4 3 4

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans 4 3 2 1 2 3 1 4 3 4 2 4 1 3 2

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans 1 1 3 3 1 1 2 2 4 2 3 2 4 2 1

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans 4 3 2 1 2 2 3 4 2 3 2 3 1 4 4

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans 2 1 3 4 1 2 3 4 1 3 3 1 3 2 1


11/29Page # 11

16. Na2311 + He42

Al2713 +

Addition of alpha particle forms aluminium atom.

17. Benzene has 15 covalent bonds.

18. In second period element no. of neutrons can never

be less than the no. of protons in atoms.

19. Copper is present below hydrogen in metal reactiv-

ity series hence cannot displace hydrogen from di-

lute acids.

20. At room temperature, liquid non metal is bromine.

21. Hydrogen atom does not contain any neutrons.

22. Pb(s) + CuCl2(aq) PbCl

2(aq) + Cu(s)

Pb is more reactive than Cu. It displace Cu from

CuCl2 solution.

23. Soda-water contains dissolved CO2 gas. Hence its

pH is less than 7.

BIOLOGY

24. Wheat, Rice and Maize are the major food grains

produced.

25. Mitochondria Name was given by Benda

26. Abscisic acid is a kind of growth inhibitor. It

promote abscission of leaves, flowers and fruits.

So it controls fall of leaf.

28. The use of disposable paper cups is more

benificial because its recycling process is

possible. So it will degrade early.

29. The endosperm produced after the fusion of male

gamete (n) with polar nuclei (2n) so it will be 3n.

endospesm provide nourishment to the growing

embryo.

30. “Systema Naturae” is written by Carolus

Linnaeus. In this book we study about Taxonomy

and Nomanclature.

31. Leech is in Annelida phylum the do not have

jointed appendages.

Jointed appandags are the unique feature of

arthropoda.

32. Allergy is a non communicable disease.

33. Pepsin is secreted by gastric gland used to

digest protein into peptones.

34. Tenden are made up of white enelastic fibre. Thisconnects muscle with the bone

35. Plasma membrane is selectively permeable that

allows certain substance to enter and exit from

the cell.

MATHEMATICS

36. cbcabcbaacab xx1

1

xx1

1

xx1

1

=

a

c

a

b

x

x

x

x1

1

+

b

c

b

a

x

x

x

x1

1

+

c

b

c

a

x

x

x

x1

1

=cba

a

xxx

x

+

cba

b

xxx

x

+

cba

c

xxx

x

=cba

cba

xxx

xxx

= 1.

Ans [3]

37. Given :

Least prime factor of a = 3Least prime factor of b = 7

Least prime factor of (a + b) = 2 a + b will be always an even number.Ans [1]

38. 9, a, b, – 6 are in A.P.

a – 9 = b – a = – 6 – b

by a – 9 = – 6 – b

a + b = 3

Ans [4]

39. Given equation x2 + bx + 12 = 0

one root is 2So put x = 2

4 + 2b + 12 = 0

2b = – 16

b = – 8

Now roots of equation x2 + bx + q = 0 are equal

So D = 0

b2 – 4q = 0

(–8)2 – 4q = 0

4q = 64 q = 16Ans [3]


12/29Page # 12

40. Given sin – cos = 2 sin (90 – )

sin – cos = 2 cos

sin = cos (1 + 2 )

tan = ( 2 + 1)Ans [4]

41. Given acos – bsin = c ...(1)now let asin + bcos = k ...(2)

by squaring eqaution (1) and equation (2) and thenaddinga2 (sin2 + cos2) + b2(sin2 + cos2) = c2 + k2a2 + b2 = c2 + k2

k2 = a2 + b2 – c2

k = cba 22 Ans [2]

42. In ABE

45°

7 7

h

60° A

D C

B

E

tan60° = AB

h

h = AB 3 ...(1)

In ABC :

tan 45° = AB

7

AB = 7 ..(2)By equation (1) and equation (2)

h = 7 3

So height of tower is= h + 7

= 7( 3 +1)

Ans [4]43. For Infinitely many solution

2

1

a

a =

2

1

b

b =

2

1

c

c

12K =

K3 =

K)3K(

By (1st) and (2nd ) ratio K2 = 36 K = 6..(i)By (1st ) and (3rd) ratio3K = K2 – 3KK2 – 6K = 0K (K – 6) = 0K = 0 or K = 6 ..(ii)So by equation (i) and (ii)K = 6 Ans [1]

44. First 12 prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37Number of observations are even = 12

So median =

2

obs12

nobs

2

nthth

median =

2

obs7obs6thth

median =2

1713 =2

30 = 15

Ans [3]

45. Total cases = 36Favourable cases = 25So probability that 5 will not come up either of the

time =36

25

Ans [2]

46. Curved surface are of sphere = 4r 2

diameter after decreased by 25% A

D = d2

d1 =

100

75d =

4

d3

A1 =

2

4

d3

=

16

9d2

% decrease =2

22

d

dd16

9

100

= – 43.75%Ans [1]

47.30°125°

E

A

B C

D

CDE + DCE = 125°CDE + 30° = 125°CDE = 125° – 30° = 95°

48. B C

A

E

Dx

2x

ABC Area

BDE Area

=

2

x2

x

=

4

1

ABC : BDE = 4 : 1


13/29Page # 13

49.

B

A C12cm

6cm6 3 90°

AB2 + BC2

= (6 3 )2 + 62 = 144

AC2 = 144

So B = 90°

50.

14 214

14 C

B

A

Area of quadrant =4

1r 2

=

4

1 (14)2

= 49 = 154 cm2

Area of semi circle =2

r 2 =

2

)27( 2 = 154cm2

Area of triangle =2

1 14 14 = 98

Area of shaded portion = 154 – (154 – 98) = 98cm2

51. az = 2r

3

3

r 3

4

a

=

6

52. f(x) = x2 – px – p – c

+ = + p = – p – c( + 1) ( + 1) = + + + 1 = 1 – c

53.

A B

C y=x+3

(2,1)(3,-2) Area of triangle = 5

by options only B option satisfy the given condition

54.

3 0 °

90°

60°120°

40° y

M

Q S R

P

T

InPTM

MTP + TPM = TMS

30° + 90° = TMS

TMS = 120°

In QMS

Q + M + S = 180°

y = 180° – (100°)

y =80°

55. (1 – tan + sec) (1 – cot + cosec)

=

cos

1

cos

sin1

sin

1

sin

cos1

=

cos

1sincos

sin

1cossin

=

cos

1)cos(sin

sin

1)cos(sin

=

sincos

1)cos(sin 22

=

sincos

1)cossin2cos(sin 22

=

cossin

cossin2 = 2.


14/29Page # 14

RAJASTHAN NTSE STAGE-I (2014-15)

CLASS-X [SAT]

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 2 4 2 3 2 1 4 3 2 4 1 4 3 4 2

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 2 3 1 4 2 3 2 4 3 3 4 3 1 2 3

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 2 1 2 3 1 1 4 2 3 All are

correct3 4 1 2 1

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. 1 3 4 2 1 3 1 4 2 3 1 2 3 3 1

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans. 2 3 4 2 1 3 4 2 1 4 1 3 3 1 3

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans. 3 3 1 3 4 2 1 4 3 1,4 4 4 1 4 3

ANSWER KEY

CLASS-X [LANGUAGE TEST]

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 2 3 1 4 1 3 4 2 1 3 1 2 2 4 1

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 3 4 4 2 1 4 2 3 1 2 4 3 2 3 2

Ques. 31 32 33 34 35 36 37 38 39 40

Ans. 2 1 4 2 2 1 2 4 1 2

CLASS-X [MAT]

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 1 2 2 3 4 1 2 4 3 4 1 1 3 3 2

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 4 1 3 4 2 3 4 2 2 2 3 3 3 4 2

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 3 1 2 4 1 1 2 1 4 2 4 2 1 4 4

Ques. 46 47 48 49 50

Ans. 2 4 3 3 4



15/29Page # 15

PHYSICS

1. Distance covered , d = 2Rtime taken is t

so speed will be =time

distance=

t

R2

Ans. (2)

2. Object is moving with uniform velocity

so Resultant force is zero.

Ans (4)

3. Pressure P = Area

)thrust(forceNormal

so. it’s S.I. unit N/m2

Ans (2)

4. Frequency n = 50 Hz

It means it will complete 50 vibrations per second.

Number of cycles in one minutes

will be = 50 × 60 = 3000

Ans. (3)

5. Height h = 40 × 15 = 600 cm = 6m

Power P =t

mgh=

6

61050 = 500 w

Ans. (2)

6. Focal length of mirror f =2

R

it doesnot depend on medium

so correct option is

Ans (1)

7. When object is between focus point and optical centre then image will be virtual, erect & magnified.

Ans. (4)

8.

R R A

B

R

R

R

R

A

B

R

R

3R

R

HINTS SOLUTIONS


16/29Page # 16

A

B

R

R

4

R3

Req = R +4

R3+ R

= 4

R11

Ans.(3)

9. For Myopiau = v = –75 cm

f

1=

v

1 –

u

1

f

1= –

75

1 –

1 f = –75 cm

Power P = – f 100

= –75

100

= –1.33D

Ans.(2)

10. By fleming’s left hand rule

Force on electron will be perpendicular into the page.

Ans. (4)

11. Potential Difference v =q

w=

1

1= 1 volt.

Ans (1)

12. 1 unit = 1kwh = 3.6 × 106 J

so 200 unit = 200 × 3.6 × 106 J

= 7.2 × 108 Joule.

Ans (4)

CHEMISTRY

13. CuSO4(aqs)

+ Ag no reactionCuSO

4(aqs)

+ Hg no reactionCuSO

4(aqs) + Au no reaction

blue

aqs4CuSO + Zn

Colourless

ZnSO )aqs(4 + Cu(s)

14. IUPAC name of the first member of homologous series of Ketone is (Propanone)CH – C – CH3 3

O

15. When Na2CO

3 dissolve in H

2O the nature of solution will be basic due to the formation of strong basic

solution


17/29Page # 17

Na2CO

3 + 2H

2O

baseStrong

NaOH2 +

acidWeak

COH 32

16. Valence electron = two General electronic configuration for third period and second group is i.e.

Mg –1s22s22p63s2

17. Since HNO3 is acid and KOH is base it is a neutralisation and double displacement reaction

18. Since pH = – log[H+]

and pH – 0 – 6 acidic7 – neutral8 – 14 – basic

hence nature of solution will be acidic

19. gypsam = CaSO4 . 2H

2O

Plaster of paris = CaSO4 .

2

1H

2O

difference =

2

1 –2 H

2O

= 2

3

H2O

20. Alkye H–C – C – C–H

H

Hif no. of carban is three then no. of Hydrogen = 4

21. mwf of CO = 28 gm

28 gm CO contain no, of CO molecule is = 6.022 × 1023

14 gm CO contain no, of CO molecule is =1

14

28

10028.6 23

= 3.011 × 1023 molecule

22. B.P. = – 80°C

Temperature in K = degree centigrade + 273

= – 80°C + 273

= 193 K

23. Suger solution does not show Tyndall effect since it is a true solution.

but milk, starch and ink form a colloidal solution therefore these all are show Tyndall effect.

MATHEMATICS

36. bc1

cbx ca1

acx ab1

bax

= abc)ba(c)ac(b)cb(a

x

= abc0

x = x° = 1

37. option (4) is correct


18/29Page # 18

38. 10,15,20,......,95

so total no are 18

39. ,

1

×

1 =

2

BB = 2

40. sin(A + B) = cos (A – B)

A + B + A – B = 90°

2A = 90°

A = 45°if B = 0 then A + B = /4if B = 45º then A + B = /2 . So option (1) & (2) both are possible.

41. cos2 + cos4= cos2 + sin2= 1

42.

D

CB

A

60° 30°x

Tower 30M

Building

tan60° =x

30

3 =x

30

x =3

30

tan30 =x

AB

3

1 =

30

3 AB

AB =3

30 = 10

43. For No solution

2

1

a

a =

2

1

b

b

2

1

c

c

1k2

3

=

1k

1

1k2

1

3k – 3 = 2k – 1 and 2k + 1

k – 1

k = 2 k – 2

44. X = 10

102.....2212

=10

)10......321(2

=102

11102

= 111


19/29Page # 19

45. Req. Probability =36

18 =

2

1

46.

Vc : V

H

31 r 2h :

32 r 3

3

1r 3 :

3

2r 3

1 : 2

47.

3 3MC D

B A

O

N4

5

4

OC = 22 43 = 5

ON = 22 45 = 3

48. 120°

120°60°

60°

60°

So side of hexagon =3

1 side of

=3

1 6 = 2

area of hexagon = 6 43 (2)

2

= 6 3

49.

BAC = BDC= 180 – (22 + 78)

= 180 – (100)

= 80


20/29Page # 20

50.

A

D

B E C

F

By MPT

DF =2

1BC

=2

12 3 = 3

DEF is also eq. rs

4

3(2 3 )

2 = r3 3

4

3

33

34 = r

r = 1

shaded area = (1)2 –4

3( 3 )

2

= –4

33 =

4

1(4 – 3 3 )

51. (3)2 10 = N3

4

3

2

1

4

8390 = N

540 = N52. option (3) is correct

53.O

B C

A (7,–5)

(3,3)

(x,y)

(3,–7)

OA = OB = OCOA2 = OB2 OB2 = OC2

(x – 7)2 + (y + 5)2 = (x – 3)2 + (y + 7)2 (x – 3)2 + (y + 7)2

–8x – 4y = – 16 = (x – 3)2 + (y – 3)2

2x + y = 4 20y = – 402x – 2 = 4x = 3 y = – 2 x = 3, y = – 2

54. OBA = 80° (A I A)115 = 80 + OABOAB = 35

55. tan43° tan45° tan47°

(tan43°tan47°)tan45°

(cot47°tan47°) 1

= (1)(1)

= 1


21/29

Page # 21

MENTAL ABILITY TEST (Year - 2016) MAT _ SOLUTION

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 3 2 1 4 1 2 3 2 3 1 4 3 3 1 2

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 4 3 4 4 3 4 2 4 1 2 1 3 1 1 2

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 4 1 4 3 4 3 2 4 4 1 4 3 4 1 2

Ques. 46 47 48 49 50

Ans. 3 3 2 3 2

Sol.126 23 19

-3 -4 -5

Sol.21 14 3 16 6 19 10 23

+2 +3 +4

+2 +3 +4

Sol.3

13 25 26 12 23 24 11 29 22 10 19 2

-2

-2 -2 -2

-1 -1 -1-2 -2

Sol.4

2 4 6 8 10 12 14 16 18 20 22 24

+6 +6 +6

+6 +6 +6 +6 +6 +6

Sol.5

2 , 3 , 4 5 , 6 73 3 3 3 3 3

Sol.6

+6 +8 +10 +12

Sol.7

5 -5 , 4 -4, 3 -3, 2 -2, 1 -13 33 3 3

Sol.8 19

181

1729

7299

819

99

19

19×9

, , , , ,

Sol.9

Pens

Markers

All pencils are markers

some pens one pencils both are true

Sol.20

251

3

6

so opposite to 3 4

Sol.21 One adjacent surface of 2 ( ) and 3 ( )

is 1 ( ) and other adjacent surface is 5( ) so opposite to 2 may be 4 or 6,according to option answer is 6.

Sol.22 No. of dvisions =2

6 = 3

Total no. fo cubes (no. of divisions)3

(3)3 = 27


22/29Page # 22


CLASS-X [SAT]

CLASS-X [LANGUAGE TEST]Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 2 1 4 1 3 1 4 3 2 3 1 4 1 3 1

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 3 3 1 3 4 3 3 4 3 2 3 3 Bonus 1 2

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 2 Bonus 1 2 2 1 3 2 1 2 1 2 4 3 4

Ques. 46 47 48 49 50

Ans. 3 1 2 3 2

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 4 2 3 4 3 2 2 1 1 3 1 4 3 1 2

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 3 3 3 1 1 4 4 3 3 1 4 2 1 2 4

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 1 3 3 1,2 1 2 Bonus 1 3 2 3 2 3 4 1

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. 3 2 1 2 4 3 3 2 4 2 1 2 2 3 1

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans. 2 3 2 1 2 1 2 1 4 2 3 4 2 2 2

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans. 3 3 3 3 4 4 3 Bonus 1 4 2 3 1 4 2

Ques. 91 92 93 94 95 96 97 98 99 100

Ans. 1 2 3 4 2 2 4 4 3 2



23/29Page # 23

PHYSICS

1. Vavg

=

2

2

1

1

21

v

S

v

S

SS

=40

40

80

40

4040

= 12

1

80

= 23

80

/= h/km

3

160 = 53.33 km/h Ans. (4)

2. Ans. (2) inertia

3. W = k =2

1 mv2 -

2

1mu2

= 2

1

× 5 × (10)2 - 2

1

× 5 × (6)2

= 250 - 90

= 160 J

Ans. (3)

4. Ans. (4) acceleration

5. P A

1

As A decreases, P increases

Ans. (3)

6. Distance b/w compression & adjacent rarefaction =2

=

2

3 = 1.5 m

Ans. (2)

7. For a conductor

R L

R A

1

& R increases with increase in temperature so Ans. (2)

8. I = . AR

V

eq 4

1

80

20

20202020

20

P.D. across a 20 any resistor is V1 = IR,

= 204

1

= 5 volt Ans. (1)

9. Ans. (1) Magnetic field lines form closed curves

10. Ans. (3) As light ray is bending towards normal, so it is going from a rarer medium to denser medium.

Speed & wavelength of light will decrease & its frequency will remain unchanged.

HINTS SOLUTIONS


CLASS-X [SAT]


24/29Page # 24

11.

Ans. (1) 0.2d, 1.8d

12. Ans. (4) 2 hectares

91. Ans. (1) Centre of curvature of a concave mirror lies in front of it whereas that of convex mirror lies

behind the mirror.

CHEMISTRY

Sol.13 Milk of magnesia is an example of sol, with dispersed phase solid and dispersion medium liquid.

Sol.14 Molar mass of Al2O

3= 102 g

so , 102 g Al2O

3contains = 2 mole Al3+ion

0.051g will contain =102

051.02

= 0.001 mole Al3+

Sol.15 Electronic configuration of

Cl = 2,8,7

No. of valence electrons = 7

Sol.16 Isotopes differ in number of neutrons.

e.g. 11H, 2

1H, 3

1H

No. of

neutrons = 0, 1, 2 respectively

Sol17. Addition reactions are characteristics of unsaturated compounds, like propene C3H

6.

Sol.18 While moving from left to right in a period, tendency of an atom to lose electrons decreases.

Sol.19 H—C—C—O—H

H

H

O

Thus, acetic acid has 8 covalent bonds.

Sol. 20. Calcium oxide (CaO) is an ionic compound , Thus , it has high melting point and is soluble in

water.

Sol.21 Metals are extracted by the reduction of their oxides.

Sol.22 Pb(s) + CuCl2(aq.) PbCl

2 (aq.) + Cu(s) is an example of single displacement reaction, where

more reactive Pb displaces less reactive Cu from its salt solution.

Sol.23 Na2311 + He42 Al2713

Sol.92 Elements of group 2 form compounds of formulae XCl2 with chlorine. Thus, the element will belong to

the same group as Mg.

Sol.93 32 g O2has 6.02 × 1023molecules

8g O2has

32

1002.6 23molecules

1.51 × 1023molecules


25/29


26/29Page # 26

41.

b

ax

yD

B A

C

ADC = 180° – bCBA = 180° – aIn quad. ABCD

A + B + C + D = 360º180 – a + 180 – b + x + y = 360

x + y = a + b

42.

x = 360 – 80 = 280

y =2

1x =

2

1 280 = 140

ROS = z = 180 – y = 180 – 140 = 40

43. Let a, b, c be the sides of the given triangle and s be its semi-perimeter.

s =2

1 (a + b + c) ...(i)

The sides of the new triangle are 2a, 2b and 2c.Let s’ be its semi-perimeter.

s’=2

1(2a + 2b + 2c) = a + b + c = 2s [Using (i)]

Let = Area of given triangle

= )cs)(bs)(as(s ... (ii)

And, ’ = Area of new triangle

’ = )c2's)(b2's)(a2's('s

= )c2s2)(b2s2)(a2s2(s2 [Using (i)]

= )cs)(bs)(as(s16

’ = 4 [Using (ii)]

'

=

4

1

44. x – y = 5

x2 – y2 = 300

(x – y) (x + y) = 300

5(x + y) = 300

x + y =5

300= 60


27/29Page # 27

45. ax2 + 2x – 2 = 0

for real & distinct root

D > 0

(2)2 – 4 (a) (– 2) > 0

4 + 8a > 0

2a + 1 > 0

2a > – 1

a >2

1

46. a + b + c = 0

ab

)ba( 2 +

bc

)cb( 2 +

ac

)ac( 2

a + b = – c, b + c = – 1 & c + a = – b

a + b = – c , b + c = – a & c + a = – b

ab

)c( 2 +

bc

)a( 2 +

ac

)b( 2

ab

c2 +

bc

a2 +

ac

b2

abc

bac333

=abcabc3 = 3 {if a + b + c = 0 then a3 + b3 + c3 = 3abc}

47.

X

P

O

R Y

S

T BQ

1

2

43

1 = 2 3 = 4

As XY || PQ

RAS + SBT = 180

(sum of conterior angle is 180°)

1 + 2 + 3 4 = 180°

2 2 + 3 = 180°

2 + 3 = 90°

In AOB

2 + 3 + AOB = 180°

90° + AOB = 180°

AOB = 90

48.

tan1

cos –

1cot

sin

tan1

cos –

tan1

tan.sin


28/29Page # 28

tan1

1[cos – sin.tan] =

tan1

1[cos –

cos

sin2]

tan1

1

cos

sincos 22

cos

sincos

1

cos

sincos 22

= cos + sin

49. Total no. of possible out come = 52

no. of red ace card = 2

probability =52

2 =

26

1

50. tan20° tan40°tan50°tan70°

tan20° tan40° tan (90° – 40°)tan(90° – 20°) (tan (90° – )= cot)= 1

51. Tn + T

n–1 = 60

a = 11

d = 2

a + (n–1) d + a + (n – 2) d = 60

2a + nd – d + nd – 2d = 60

2a + 2nd – 3d = 60

2 × 11 + 2 × n × 2 – 3 × 2 = 60

22 + 4n – 6 = 60

4n = 66 – 22

4n = 44

N = 1152. ATP

2r – 2r = 60

2r [ – 1] = 60

2r

1 –

7

22= 60

2r ×7

15= 60

R = 14

Area = r 2 = 722

× 14 × 14 = 196


29/29

53. a

2

c2

b2

L

B

H

LH = a2 ...(1)

BH = b

2

...(2)LB = c2 ...(3)

Eq. (1) (2) (3)(LBH)2 = (abc)2

V2 = (abc)2

V = abc

54.

O

CD

A B3x

2x

In AOB and COD AOB = COD (V.O.A)

OAB = OCD (Alternate interior angle)

by AA, AOB ~ COD

CODar

AOBar

2

2

CD

AB 2

2

)x2(

)x3(

4

9

55. (B)

6

y47x95 = 7

25 + x + y = 42

x + y = 17

ntse stage i rajasthan last 4 years solutions

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