solution mat + sat.pdf

8/10/2019 Solution MAT + SAT.pdf

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SOL._NTSE OPEN TEST_PAGE # 1

1.

2.

3.

4.

5.

6. The all letter of each term are moved four steps backward to obtain the corresponding letters of the next

term. So the missing term would be DGJM.

ALL INDIA NTSE (STAGE-I) TEST SERIES

OPEN TEST

CLASS-X [MAT]

NSWER KEY

HINTS & SOLUTIONS

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

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

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

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

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

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

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

Ans. A C A A C C D D D B B B B D CQues. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

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

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

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


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7. All the letter of each term are moved three steps forward to obtain the corresponding letters of the next

term. So the missing term would be ORQ.

9. The all letter of each term are moved three steps forward to obtain the corresponding letters of the next term. So

the missing term would be OSVX.

10. First and third letters are moved four steps forward in each term. Second and fourth letters are moved four

steps backward in each term. So, the missing term would be QIRH.

11. 8 5 = 3 ; 13 7 = 6

12. 32= 9 , 42= 16

52= 25 , 62= 36

13. (C) Suppose X denotes the numbers in the first row and Y denotes the numbers in the second row.

Then, the pattern is X2X = Y.

Clearly, 323 = 9 3 = 6 ; 828 = 64 8 = 56 ; 10210 = 100 10 = 90 ; 222 = 4 2 = 2 ; 121

= 1 1 = 0. Similarly, 525 = 20. So, the missing number is 5.

15. 9 4 = 36 !"2

9 16 =144 !#$2

16 25 = 400 !$%2

16. c c ba/ c c b a/ ccb a /c c ba

17.

18.

19. B D C b d c F H G f h g

20. A C C E a c c e G I I K

Sol. (31 to 35) :

The best method to solve this question is to make a table and fill the place according to the information

given logically. On analyzing the information given in the question, we arrive at the following table

Student Compulsory Subject Optional Subject

M Geography (given) English (given)

N Geography (given) Biology (given)

O Geography (given) Physics (given)

Q Chemistry (given) Physics (given)

R Physics (given) Chemistry (given)

Geography (given)

Female student (given)English (given)P


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41.

B O Y&' X T D

B 'X, O 'T, Y' D,

42.

N E V E R&'O F W F S

43.

44. C A L C U T T A&' 5 3 7 5 9 8 8 3

A C T U A L &' 3 5 8 9 3 7

45.

48. Bhavan15 Ranks =27th

Prekshas Ranks = 20th

Prekshas Ranks from last = 36th

So, total students = (20 + 36) 1

= 55

49. Manan < vedant

Manan < pavan

ohm < pavan

ohm > Manan

So manan is the tallest

50.

Mother Son

51. MENTAIN


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52. 4 b's are there

53. Middle letter = I

Fourth letter form I is N

54. Total number of student

7 + 29 1 = 35

55. 64 ) 2 13 + 10

64 2 + 13 10

128 + 3 = 131

Sol. 61-65.

Letters L E I T P

Code m f s/q/j u/b s/q/j

! ! ! ! !


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OPEN TEST PAPERNTSE STAGE-I

CLASS-X [SAT]_DATE : 30-09-12

1. The given surds are

4 5 , 5 4 , 4 and 3

By taking L.C.M. 20 and make the order of

the surds equal

20 55 , 20 44 ,20 104 and

20 103

which is greater than 20 44

so 5 4 is smallest

2.!"

!#$

!%

!&'

((()

)(()22(

)22)(22(

)22()22(= 2

3. x = 0.uwuvuvuvuv.........

x = 0.uwuv ... (i)(i) 100

100x = uw. uv ... (ii)

(ii) 100

10000x = uwuv . uv ... (iii)

(iii)(ii)

uw

uwuvx9900

________________

uv.uwx100

uv.uwuvx10000

*

*

*

NSWER KEY

HINTS & SOLUTIONS

x =9900

uwuv100uw (+

=9900

uv)1100(uw (=

9900

uvuw99 (

4. Time taken by three wheels to complete 1 revo-

lution are 1 sec,3

5sec and

2

5sec. respec-

tively

L.C.M. of ,-

./0

1

2

5,

3

5,1 =

)2,3(.F.C.H

)5,5,1(of.M.C.L

So all the sprot will simultaneously touch theground again after 5 sec.

5. 64636261 4444 ((( = 614 (1 + 4 + 42+ 43) =

614 85 = 604 340, which is clearly divisibleby 10

6. Let the side of the square be x , then the totalnumber of bangles he had = x2+ 38If he increases the size of the square by one uniteach side , then the total number of bangles= (x + 1)225Thus x2+ 38 = (x +1)225

x2+ 38 = x2+ 1 + 2x 252 2x = 62

2 x = 31

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

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

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

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

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

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

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

Ans. A A A B A C A B B C D D B A BQues. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

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

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

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


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Thus total number of bangles = x2 + 38= 961 + 38 = 999

Alternatively : Go through options, consider asuitable value from the options and check thatin each case it must produce a perfect squarenumber.As 999- 38 = 961 is a perfect squareand 999 + 25 = 1024 is also a perfect square.

7. (10a + b) (10c + d) (10b + a) (10d + c)= (100 a.c + 10 b. c + 10a d + b.d)(100b.d + 10b.c + 10a.d + a.c)= 99 (a.c b.d)Now, inorder to the difference be maximum soa c will be maximum and b.d will be minimum,thus a c = 9 8 = 72 and b d = 1 2 = 2Hence 99(acbd) = 99(722) = 99 70 = 6930

8. Given 2a = b

and ax + by = 2a23b2

ax + 2ay = 2a2

12a2ax + 2ay = 10a2

x + 2y = 10a ....(i)

x + 2y = 2a 6b

x + 2y = 2a 12a

x + 2y = 10a ...(ii)

From equation (i) & (ii) we have2

1

a

a=

2

1

b

b=

2

1

c

c

or infinite number of solutions of the given pair

of equation

9. p(x) = x4+ 6x3+ 13x2+ mx + 4

q(x) = x2+ 3x + 2

q(x) = (x + 2) (x + 1)

x = 1, 2p(1) = 1 6 + 13 m + 4 = 0m = + 12

10. f (x) = x312x2+ 39x + k

Let the zeros are a

d, a and a + dSo, sum of zeros = a d + a + a + d = 123a = 12 2a = 4.Sum of product of zeros taken two at a time= (a d) a + a(a + d) + (a + d) (a d) = 3924(4 d) + 4 (4 + d) + (16 d2) = + 39216 4d + 16 + 4d + 16 d2= + 392d2 = 48 39 = 92d = 3Now, product of zeros = (ad) a (a + d) =k2a (a2d2) =k24 (16 9) = k2k = 28.

11. 01x1x *)((Taking square both side,

(x + 1) + (x 1) + 2 )1x( ( )1x( ) = 0

2x + 2 1x2 ) = 0

2 x + 1x2 ) = 0

2 1x2 ) = x

x21= x

2

0 = 1,which is not possible.

12. In 3CAB and 3CEDC

B

E

2

10D

8

A

X

9

7

y

y

4A = 4CED (Given)

4C = 4C (Common to both 3s)

5 3CAB ~ 3CED (AA critertion of similar

3s)

5CE

CA=

DE

AB=

CD

CB

ConsiderDE

AB=

CD

CB5

x

9=

8

210 (

Orx

9=

8

125 x =

12

98+= 2 3 = 6

Hence x = 6

13.

A

BD C

Given in 3ABC, AB = BC = CA

and AD 7BC

Now, in 3ABD, AB2= AD2+ BD2

!AB = BC = 2BD (!D is mid point of BC)

5(2BD)2= AD2+ BD2

24BD2= AD2+ BD2

or AD2= 3BD2


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14.

E D

CA

B

65

O

4EAC = 4EBC = 65 (angles in the same

segment )

In right-angled triangle AEC.

4ACE = 180 (90 + 65)

4ACE = 25

4DEC = 4ACE = 25 {alternate angle}

15. In quadrilateral ONAM

OR

aMa

N

A

a

4N = 90, 4M = 90

4A = 90, so 4O = 90

OM = ON = 22 aR )

So, ONAM is a square

Diagonal of square ONAM = OA

= 2 22 aR )

= )aR(2 22 )

16. Let vertex D be (x, y).Diagonals of a porallelogram bisect each other.So, E (mid point of AC)

= ,-

./0

1 ())

2

54,

2

63= ,

-

./0

1)

2

1,

2

3

Also E (mid point of BD) = ,-

./0

1 ()

2

1y,

2

2x

52

2x)=

2

3)2 x = 1

And

2

1y(=

2

1 2 y = 0

So, the fourth vertex be (1, 0).

17.8(

8(

8)

8

cos1

sin

cos1

sin= 4

2)cos1)(cos1(

)cos1(sin)cos1(sin

8(8)

8)8(8(8= 4

28)

82cos1

sin2= 4 2

8sin

2= 4

2 sin 8=2

12 8= 30

18. In 3ADC

60 30A B Cx

30 m

D

h

tan 30 =30

h

h =3

30m

In 3ADB

tan 60 =x

h

x

h3* 2 x =

3

h

=33

30

+

x=3

30= 10 m

5 length of shadow will be 10 m.

19. Let radius = r and height = h

New radius = r +2

r= r

2

3

New height = h

100

h20 =5h4


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Volume of original cylinder (V1) = 9r2h

Volume of new cylinder (V2) = 9

2

r2

3,

-

./0

1 ,

-

./0

1

5

h4

= 94

9r2

5

h4

=59 9r2h

Percentage change in Volume

=1

12

V

VV 100

= 100hr

hrhr5

9

2

22

+9

99

= 100hr

15

9hr

2

2

+9

:;

?9

= 1005

59+

=5

1004 += 80%.

20. Mode = 3 Median 2 Mean16 = 3 Median 2 (28)

3

5616 (= Median

Median =3

72= 24

27. From the conservation of linerar momentum,m

1u

1+ m

2u

2= m

1v

1+ m

2v

2

1000 50 + 0 = 1000 V + 250 V

1250

501000 += V

V = 40 km/h

28. mg B = ma

a =vd

vrgvdg

m

Bmg )*

)

a = g ,-

./0

1 )

d

rd

29. Time taken in hearing an echo

t =500

1

330

33.02*

+s

So, frequency = 500 hz

30. Power of the boat is = FV = 500 3 = 1500P = 1.5 KW

33. Electrolysis of fused NaCl occurs as -

NaCl(") @@@@ A@yElectricit

Na++ Cl-

Reaction at cathode -

Na+ + e- @A@ Na (s)

Reaction at anode -

2Cl @A@ Cl2 + 2e

Thus, chloride ions are oxidised at anode.

34. Cu(OH)2is a weak base and H

2SO

4is strong

acid. So CuSO4has acidic nature.

35. Artificial gold (rolled gold) is an alloy of 90%Cu, 9.5 % Al and 0.5% Sn.

36. CH3COONa + NaOH @@ A@

CaOCH

4+ Na

2CO

3

37. The given compound can be written as follows.

38. H3O+, CH

3and NH

3all have same no. of

electrons, i.e. 10.

39. Noble gases have stable electronicconfiguration. Hence, it is difficult to remove anelectron from them. So, their I.E is maximum.

40. Suppose % of isotopes having atomic weight35 & 37 are X and (100X) respectively -

average atomic weight =100

)x100(MxM 21 (

35.5 =100

)X100(37X35 (

3550 = 35X + 370037X35503700 = 35X37X150 =2XX = 75percentage of heavier isotopes = (10075)= 25

41. No. of neutrons in X = 70 34 = 36 = No. ofneutrons in Y (being isotonic).Mass no. of Y = 725No. of protons in Y = 72 36 = 36 = Z

42. Kwof pure distilled water is = [H

3O+] [OH]

and in pure distilled water [H3O+] = [OH]

At 80 C [H3O+] =1 106mol L1

So value of Kwat this temperature will be -

Kw= [1 106] [1 106] = 1 1012M2

43. 44 g CO2= N

Amolecules

54.4 g CO2= 10

Nmolecules

22.4 litre H2at STP = N molecules

52.24 litre H2at STP =

10

Nmolecules

Thus total molecules =10

N+

10

N=

5

N

solution mat + sat.pdf

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