Mains Maths 21

8/11/2019 Mains Maths 21

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Param

ountCoaching

Centr

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MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

Ph: 011-27607854, (M) 8860-333-333 1

1. (D) Since, divisor =a

Therefore, quotient =4

a a) b (a/4 a 2

and remainder =2

a

b =a4

a+

2

a

b =4

22 aa

( + 2)a ab

= 4

2. (C) Since, 738A6A is divisible by 11,

therefore

Sum of odd places Sum of even places

= either multiple of 11 or 0

(7 + 9 + 6) (3 + A + A) = either multiples

of 11 or 0.

21 2A + 3 = either multiple of 11 or 0

18 2A = 0 (it cannot be a multiple

of 11)

2A = 18

A = 93. (C) Let the numbers beaandb.

Then, ab = 1575

andb

a=

7

9

a =7

9b

Putting this value of Eq. (i), we get

7

9bb = 1575

b2 = 1575 9

7

b2 = 175 7

= 52 72

b = 35

a =7

9 35

= 45

Thus, sum of the numbers

=a+b

= 45 + 35

= 80

SSC Mains Test- 21 (SOLUTION)

4. (C)3)81()9()81(

2.4

7.6.3

= 18.16

4.54.14

)3()3(

)3()3(

= (3)14.4 + 5.4 16.8 1

= (3)2= 9

5. (D) ....666

Factorize 6 as the multiplication of twoconsecutive natural numbers. The greaterone will be the answer.

As, 6 = 2 3

Therfore, ....666 = 3

6. (D) Let the two consecutive odd numbers bexand (x+ 2).

Then, according to the question,(x)2+ (x+ 2)2 = 394

x2+x2+ 4 + 4x = 394

2x2+ 4x = 390

x2+ 2x = 195

x(x+ 2) = 195 = 13 15

x = 13

Thus, sum of the numbers=x+ (x+ 2)= 13 + 15= 28

7. (C) We know that, when (a 1)nis dividedbya, then remainder = (1)n

Now, 6767+ 67= (68 1)67+ 67

When (68 1)67is divided by 68, thenremainder = (1)67= 1

Thus, when 6767+ 67 is divided by 68,then remainder = 1 + 67 = 66

8. (A) Let the work is completed inndays.

Then,246 +

526n +

64n = 1

52

6n+

64

n= 1

4

1

16134

13)6(16

nn=

4

3

16n 96 + 13n= 3 13 16

29n= 624 + 96

29n= 720

n=

29

720

= 25 days


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Param

ountCoaching

Centr

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MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

9. (B) ]1)75.0(75.0[)75.01(

)75.0( 23

=25.0

]1)75.0(75.0)[75.01()75.0( 23

=25.0

])75.0(1[)75.0( 333

=25.0

)75.0(1)75.0( 33

=25.0

1=

25

100= 4 = 2

10. (B) 4096 + 96.40 + 004096.0= 64 + 6.4 + 0.064 = 70.464

11. (C) Required number

= 3012 3011 (3011)2

= 3011(3012 3011)

= 3011

12. (D) (P + Q)s work =23

19... (i)

(Q + R)s work =23

8... (ii)

(P + Q + R)s work= 1 ... (iii)

Subtracting Eq. (i) from Eq. (iii), we get

Rs work = 1 23

19=

23

4

Subtracting Eq. (ii) and from Eq. (iii), we get

Ps work = 1 23

8=

23

15

Now Qs work = 1 (P + R)s work

= 1

23

15

23

4

= 1 23

19= 23

4

Qs share =23

4 5750

= 4 250

= 1000

13. (C) Let CP = x.

Then,

SP = 70% ofx= 0.7x

60% of MP = 0.7x

MP =60

7.0 x 100 =

6

7x

If article is sold at the marked price, then

SP =6

7x

Required profit per cent

=x

xx6

7

100%

=6

100% = 16

3

2%

14. (A) A ( )

B ( )

C ( )

3

4

1

4

3

12

5

12

Cistern fill till 5clock = 4 2 + 3 = 11

Cistern will empty in =5

11= 2 hr 12 min

or at 7 : 12 pm15. (D) Required number of days

= 164 = 8 days

16. (C) Here, M1= 250, D

1= 20, T

1= 5

M2= ?, D

2= 10, T

2= 8

Using M1D

1T

1= M

2D

2T

2, we have,

250 20 5 = M2 10 8

M2

=810

520250

= 312.5= 313

17. (A) 2m+ 5w =12

1... (i)

5m+ 2w =9

1... (ii)

Multiplying Eq. (i) by 5 and Eq. (ii) by 2and then subtracting it from Eq. (i), we get

10 + 25 =

10 + 4 =

21 =

m w

m w

w

512

512

2

9 29

=36

7

3w =36

1

3 womens 1 days work = 36

3 women can complete the work in 36days.18. (A) Let CP of the article = 100

Then SP = 125% of 100

= `

125


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Param

ountCoaching

Centr

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MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

Also,4

3th of MP = SP

4

3 MP = 125

MP = 125 3

4= `

3

500

Required ratio =3

500: 100 = 5 : 3

19. (A) Let Earnings of A = 2xand Earnings of B =xExpenditure of A = 5yExpenditure of B = 3yFurther, suppose monthly savings of A = 4zAnd monthly savings of B = z

Also, total monthly savings of A and B= 5000

4z+z = 5000

5z = 5000

z = 1000Now using,

Income Expenditure = Savings2x 5y = 4000 ... (i)

and x 3y = 1000 ... (ii)Multiplying Eq. (ii) by 2,

2x 6y = 2000 ... (iii)Subracting EQ. (iii) from Eq. (i), we get

5y+ 6y= 2000

y= 2000Putting this value of Eq. (ii), we get

x= 1000 + 6000

= 7000Thus, monthly salary of B = `7000

20. (B) Let the greater and smaller beaandb,

respectively.

ba

ba

=

1

5

Applying componendo and dividendorule, we get

)()()()(

babababa

=

1515

b

a

2

2=

4

6

a:b= 3 : 221. (B) Required equivalent discount

100

50

100

80

100

901 100 = 64%

22. (D) I. Equivalent discount

=

100

10101010 %

= (20 1)%= 19%

II. Equivalent discount

=

100

812

812 %

= (20 0.96)%= 19.04%

III.Equivalent discount

=

100

515515 %

= (20 0.75)%= 19.25%

IV.Discount = 20%Selling price will be minimum wherediscount is maximumi.e., in IVcondition.

23. (D) In the mixture, acid = 80% =5

4

and water = 20% =5

1

Letxth part of the mixture be removed

and replaced by water.

Then,

xx

x

5

1

5

15

4

5

4

=3

4

xx

x51

44

=34

x

x

41

44

=

3

4

12 12x = 4 + 16x

8 = 28x

x=28

8=

7

2th

24. (B) Ratio of reduction is number of employees

= 9 : 8Ratio of increment in wages = 14 : 15

Now, ratio of reduction in the wage bill= 9 14 : 8 15= 126 : 120= 21 : 20

25. (D) Let maximum score of the cricketer =xThen, his minimum score =x 172Now, total score in 40 innings

= 40 50 = 2000

And, total scores in 38 innings

= 38 48 = 1824

Sum of remaining two innings= 2000 1824 = 176

Therefore, sum of maximum and minimum

scores = 176


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MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

x+x 172 = 176

2x = 348

x = 174

Thus, highest score of the cricketer is 174.

26. (C) Gold Copper

Alloy A9

7

9

2

AlloyB18

7

18

11

Since, alloys A and B are melted in the

ratio 1 : 1 to make the alloy, therefore

in the alloy C, the ratio of gold and copper.

2

1

18

7

2

1

9

7:

2

1

18

11

2

1

9

2

=

18

7

9

7:

18

11

9

2

=18

21:18

15= 21 : 15

= 7 : 5

27. (C) Acid Water

Bottle I7

2

7

5

Bottle II

10

7

10

3

Let the required ratio be x : 1. Then,

according to the question,

10

7

7

2x :

10

3

7

5x = 2 : 3

2 7

7 1 0

5 3

7 1 0

x

x

+

+

=3

2

7

6

x+ 10

21

= 7

10

x+ 10

6

7

4x=

10

15

x=10

15

4

7=

8

21

Hence, required ratio =x: 1 = 21 : 8

28. (A) Let third number =x

Then, second number = 2x

Then, the first number = 4x

Now, average of these three numbers = 154

x+ 2x+ 4x = 3 154

7x = 3 154

x = 3 22= 66

Thus, first number = 4 66 = 264

29. (D) Then, number of rest = (500 x)

500 5000 =x 14000 + (500 x) 4000

2500 = 14x+ 4(500 x)

2500 = 14x+ 2000 4x

500 = 10x

x =10

500= 50

Thus, number of officers is 50.

30. (D) Total marks of 40 students.

= 40 72 = 2880

Rectified total marks

= 2880 68 73 + 64 + 62 + 84

= 2884

Thus, rectified average =40

2884= 72.1

31. (B) Let first number =x

Second number = 3x

Third number =4

3x

x+ 3x+4

3x = 3 114

4

34 x = 342

x = 342 19

4

= 72

Thus, largest number = 3x

= 3 72

= 216

32. (D) Time taken to cover5

1of the distance

= 5

1

8

1

= 40

1

Time taken to cover10

1of the disteance

=10

1

25

1=

250

1

Time taken to cover rest of the distance

=

10

1

5

11

20

1

= 10

7

20

1

= 200

7


5/14

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==================================================================================

Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

Therefore, total time taken

=40

1+

250

1+

200

1

=100

35425

=1000

64

Thus, average speed =

=

1000

64

1=

64

1000

= 15.625 km/h

33. (C) Letxmarbles must be added.

Then,x

x

40

10 100 = 60

x

x

40

10=

5

3

50 + 5x = 120 + 3x

2x = 70

x = 35

34. (A) Let the number bex.

Then according to the question,

x (25% ofx) =x+ (200% ofx)

x4

x= 3x

x2 = 12x

x = 12

35. (B) Let the value of the article 3 years ago

bex.

Then, 729 =x

3

100

101

729 =x

3

10

9

729 =x1000

729

x = 1000

36. (B) Required percentage decrease

=50100

50

100%

=150

50 100%

= 3331 %

37. (B) Let the required distance bedkm.

Then,3

d

4

d=

60

10

12

)34( d =61

d = 2 km

38. (A) CP per dozen =2

3040= 35

SP per dozen = 45

Profit per dozen = 45 35 = 10

But total profit = 480

Thus, required number of dozen

=10

480= 48

39. (D) Let CP of first chair = x

and SP of second chair = (900 x)

SP of first chair =5

4x

SP of second chair =4

5(900 x)

We know that,

SP CP = Profit

5

4x+

4

5(900 x) 900 = 90

20

)900(2516 xx = 990

16 +22500 25

20

x x= 990

9x+ 22500 = 19800

9x= 2700

x= 300

Thus, the cost of lowest price chiar is` 300.

40. (B) SP of 100 oranges CP of 100 oranges

= SP of 20 oranges

SP of 80 oranges = CP of 100 orangesLet CP of 1 orange = 1

CP of 100 oranges = 100SP of 80 oranges = 100

SP of 1 oranges = `80

100= `

4

5

Profit percent =

1

4

5 1 100%

=

4

1 100%

= 25%


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Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

41. (C) 60% of CP = 50% of SP

SP

CP=

6

5

Let CP = 5xand SP = 6xThen, there is a profit.

And profit per cent =x

xx

5

56 100%

= 20%42. (D) CP of two horses

= 40000 + 40000= 80000

SP of one horse = 115% of 40000= 46000

Let the second horse was sold at a lossofx%.

Then, SP of second horse

= (100 x)% of 40000= 40000 400x

Total SP = 46000 + 40000 400x= 86000 400x

Now, using CP SP = loss, we have80000 (86000 400x) = 3600

6000 + 400x = 3600

400x = 9600

x = 24Thus, SP of second horse

= (100 24)% of 40000= 76% of 40000 = 30400

43. (D) CP of 1 guava = ` x

y

SP of 1 guava = ` y

x

x>y, therefore SP > CP and hence thefruit seller will have a gain.

Gain per cent =CP

CPSP 100

=

x

y

x

y

y

x

100%

=xy

yx 22

y

x 100%

= 2

22

y

yx %

44. (B) According to the question,

100

max =

100

22 may

x =y am

x:y = am: 1

45. (D) Let A took xfrom B and `(1200 x)from C.

Then,

100

114x

+ 100

115)1200( x

= 172

14x+ 18000 15x= 17200

x= 800Thus, A borrowed 800 from B.

46. (B) Given,

n

n )42( 90 =

5

3 2 90

n

n )42( =

5

3 2

n

n 2=

5

3

5n 10 =5

3

2n= 10

n= 5Thus, the polygon has 5 sides.

47. (C) Side of given square = 200 m

= 210 m

Its diagonal = 210 m 2 = 20 m

which is the side of new square

Area of new square = (20)2

= 400 sq m48. (C) Let radii of cone, cylinder and hemisphereare 2r, 3r and r respectively.Volume of cone : Volume of cylinder :Volume of hemisphere

=3

1 (2r)2h : (3r)2h :

3

2 r3

=3

1 4h : 9h :

3

2r

(for hemisphere r = h)

=

3

4h : 9h :

3

2h

= 4 : 27 : 249. (A) Let rate of flow of river beukm/h.

Then,u10

91+

u10

91= 20

2100

1010

u

uu

=

91

20

2100

20

u=

91

20

100 u2 = 91

u2 = 9

u = 3 km/hr


7/14

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Param

ountCoaching

Centr

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MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

50. (B) Let the original speed and time be S andT respectively.

Then, S T =4

3S

60

20T

3

4T =

3

1T

3

4T T =

3

1

3

T=

3

1

T = 1 hr = 60 minutes51. (C) Let speed of boat in still water

=ukm/hAnd speed of stream =vkm/hr

Then,vu

25+

vu

39= 8 ... (i)

andvu

35+

vu

52= 11 ... (ii)

Multiplying Eq. (i) by 4 and Eq. (ii) by 3and subtracting Eq. (ii) from Eq. (i), we get

vu

100

vu

105= 32 33

vu

5= 1

uv= 5 ... (iii)Now, from Eq. (i),

5

25+

vu

39= 8

vu

39= 8 5 = 3

u+v = 13 ... (iv)Now, subracting Eq. (iv) from Eq. (iii), we get

(uv) (u+v)= 5 13

2v= 8

v= 4 km/h

52. (C)Sum of money at compound interestbecomes 2 times in 5 years.

It will become 8 timesi.e.(2)3timesin 5 3 = 15 years.

53. (B) Let the share of elder and younger sonsbe `xand (120000 x) respectively.Amount got by elder son = Amount got

by youngerson

x+100

45 x

= (120000 x) +100

65)120000( x

x+5

x= (120000 x) +

10

3)120000( x

5

11x

10 = 120000 10 + 360000 3x

22x= 1200000 + 360000 3x

25x= 1560000

x= 62400

Thus, yonger sons share

= 120000 62400

= 57600

54. (C)a2+b2+c2= 2(abc) 3

a2+b2+c2 2a+ 2b+ 2c+ 1 + 1 + 1 = 0

(a2 2a+ 1) + (b2+ 2b+ 1) + (c2+ 2c+ 1) = 0

(a1)2+ (b+ 1)2+ (c+ 1)2= 0

a 1 = 0

a= 1

and b+ 1 = 0 b= 1

and c+ 1 = 0 c= 1

Thus, ab+c= 1 + 1 1 = 1

55. (A)x2+ 3x+ 1 = 0

x+ 3 +x

1= 0 [Dividing by x]

x+x

1= 3

On cubing both sides, we have

3

1

x

x = (3)3

x3+ 31

x+

xx

1= 27

x3+ 31

x+ 3(3) = 27

x3+ 31

x= 27 + 9 = 18

56. (D) xa.xb.xc = 1

xa+b+c = 1

xa+b+c =x0

a+b+c = 0

a3+b3+c3= 3abc(whena+b+c= 0, thena3+b3+c3= 3abc)

57. (C) Volume of pyramid = 1728

3

1 area of base height = 1728

3

1 24 24 height = 1728

height = 2424

31728

= 9 m


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Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

58. (B) Here,AOBand AOB are similar.Let OO=hand OB= r

AO'

AO

= B'A'

OB

h9

9=

r

3

3r = 9 h

h = 9 3rNow, volume of frustum = 44

3

1h(R2+r2+ Rr) = 44

3

1

7

22 (9 3r)(9 +r2+ 3r) = 44

(3 r)(9 +r2+ 3r) = 2 7

33r3= 14

r3= 27 14

r3= 13

r= 313 cm

59. (A) Let radii of first and second cylinders be2r and 3r, respectively and their heightsbe 5h and 4h respectively.

Then, ratio of their curved surface area2 (2r) 5h : 2 (3r) 4h

= 10 : 12= 5 : 6

60. (C) Curved surface area

=3

1 Total surface area

=3

1 462

= 154 sq cmNow, total surface area = 462 sq cm

2rh+ 2 r2= 462

154 + 2r2 = 462

2r2 = 308

r2 = 3087

222

r2 = 49

r = 7Again, 2rh= 154

h = r2

154

=7222

7154

=2

7cm

Thus, volume of cylinder

= r2h

=7

22 49

2

7

= 11 49

= 539 cm3

61. (D)Curved surface area of cylinder

Curved surface area of cone=

5

8

l

rh

2

2=

5

8

22 rh

h

=

5

4

22

2

rh

h

=

25

16

2

22

h

rh =

16

25

1 + 2

2

h

r=

16

25

2

2

h

r=

16

9

h

r=

4

3

r:h= 3 : 4

62. (D)


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Param

ountCoaching

Centr

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MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

Surface area of conical part

= Surface area of hemisphere

22 rhr = 2r2

22 rh = 2r

h2+r2 = 4r2

h2 = 3r2

2

2

h

r=

3

1

h

r=

3

1

r:h = 1 : 3

63. (D) Volume of prism = Area of base height

3108 =4

3 (6)2 Height

Height =36

408

= 12 cm

64. (B) a+a

1+ 2 = 0

a2+ 1 + 2a= 0

(a+ 1)2 = 0

a= 1

a37 1001

a= 1

1

1

= 1 1 = 2

65. (A) Given lines

(k 1)x+y 2 = 0 ... (i)

and (2 k)x 3y + 1 = 0 ... (ii)

Since, the lines (i) and (ii) are parallel.

k

k

2

1=

3

1

3k 3 = 2 +k

2k = 1

k =2

1

66. (C)

Let height =h

Then , length of shadow =3

h

In the figure,

tan =

3

h

h

tan = 3

= 60

67. (D)

Given lines are x+ 2y = 3 ... (i)

and 3x 2y = 1 ... (ii)Put x = 0

Then, from Eq. (i)y=2

3

from Eq. (ii)y= 2

1

Distance between both the points

=2

3

2

1=

2

4= 2 units

68. (B) x+x16

1= 1

4x+x4

1= 4

Cubing both sides, we get

64x3+ 364

1

x+ 3.4x.

x4

1

xx

4

14 = 64

64x3+ 364

1

x+

xx

4

143 = 64

64x3+ 364

1

x= 64 3 4

4

4

14

xx

= 64 12

= 52


10/14

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Centres at:

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Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

69. (B) a+b+c= 0

a+c= bOn squaring both sides,

(a+c)2 = (b)2

a2

+b2

+ 2ac=b2

a2+c2=b2 2ac

Now,cab

acba

2

22 2

=acb

acbb

2

22 2

=acb

acb

2

2 )(2= 2

70. (D) Given equations area4+a2b2+b4 = 8 ... (i)

a2

+ab+b2

= 4 ... (ii)Squaring Eq. (ii), we get(a2+ab+b2)2= (4)2

a4+a2b2+b4+ 2a3b+ 2a2b2+ 2ab3 = 16

8 + 2ab(a2+b2+ab) = 16[Using eq. (i)]

2ab(a2+b2+ab) = 16[Using eq. (ii)]

2ab 4 = 8 [using Eq. (ii)]

ab = 171. (D) We know the formula

a3+b3+c3 3abc= (a+ b+ c)(a2+ b2+ c2abbcca)

=2

1(a+b+c)(2a2+ 2b2+2c2 2ab 2bc2ca)

=2

1(a+b+c){(ab)2+ (bc)2+ (c+a)2}

Now, 222

333

)()()(

3

accbba

abccba

=2

1

})()(){(

})()()){((222

222

accbba

accbbacba

= 2

cba

=2

101525 =

2

30= 15

72. (D) InACT,

44

A

B

C

O

T40

ACB = 180 (CAT +ATC)= 180 (44 + 40)= 96

ACB = 180 ACT

= 180 96= 84Also,

ABC =CAT= 44

InABC,

BAC = 180 (ABC +ACB)= 180 (44 + 84)= 180 128= 52

BOC = 2BAC= 2 52= 104

73. (C)

O

P M5 5

1 2 c m

In OPM,OP2 = PM2+ OM2

= 25 + 144= 169

OP = 13

Diameter of circle= 2 OP= 2 13 = 26 cm

74. (C) Since, P, S, Q are the mid-points of ABand BC, therefore

A

P Q

B C

1 R 2

AB

AP=

BC

PQ=

2

1... (i)

Now,RQ

PR=

2

1

RQ

2=

2

1

RQ = 4


11/14

Ph: 011-27607854 M 8860-333-333 11

Centres at:

==================================================================================

Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

PQ = 2 + 4 = 6 cmTherefore, from Eq. (i),

BC

6=

2

1

BC = 12 cm75. (A) Since. O is the circumference ofABC,

thereforeA

B C

O

35

OA = OB = OC

In BOC,OB = OC

OCB =OBC = 35

BOC = 180 (OCB +OBC)= 180 70= 110

BAC =2

1BOC =

2

1 100 = 55

76. (C) InBIC,A

B C

I

135

BIC = 135

2

1(A + C) = 180 135

2

1(B + C) = 45

B +C = 90

A = 90ABC is a right angled triangle.

77. (C) sin2 + sin2 = 2

sin = sin = 1 (sin < 1)

= = 90

cos

2

= cos 90 = 0

78. (D) cot20

cot

20

3cot

20

5cot

20

7cot

20

9

= cot 20

cot 20

3cot 4

cot

20

3

2

cot

202

= cot20

cot

20

3cot

4

tan

20

3tan

20

= 1

79. (D) sin + cos = 13

17

(sin + cos )2 = 2

2

)13(

)17(

sin2 + cos2 + 2sin cos =169

289...(i)

1 + 2sin cos =169

289

2sin cos =169

289 1 =

169

120... (ii)

Again, from Eq. (i),

sin2 + cos2 2sin cos + 4sin cos

=169

289

(sin cos )2+ 4sin cos =169

289

(sin cos )2+ 2 169

120=

169

289

[using eq, (ii)]

(sin cos )2= 169

289 169

240= 169

49

sin cos =13

7

80. (C) tan .tan2 = 1

tan .

2tan1

tan2

= 1

2tan2 = 1 tan2

3tan2 = 1

tan = 3

1

= 30

Now, sin2 2 + tan2

= (sin 60)2+ (tan 60)2

=

2

2

3

+ 2)3(

=4

3+ 3

= 4

123

= 4

15

= 3 4

3


12/14

Ph: 011-27607854 M 8860-333-333 12

Centres at:

==================================================================================

Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

81. (C) Sales in 2008 = 5 croreSales in 2006 = 6 crore

Required percentage decrease

= 6

56

100%

=6

100% =16

3

2%

82. (B) Sales in 2002 = 2 croreSales in 2007 = 6 crore

Required ratio = 2 : 6 = 1 : 383. (A) Average sale of company

=Total sales of the company

Number of years

=5

661043

=5

29= 5.8 crore

84. (D) Sales in 2005 = 10 croreSales in 2004 = 4 crore

Required percentage income

=5

410 100%

=4

6 100% = 150%

85. (B) Total sales of company from 2005 to 2008= ` (10 + 6 + 6 + 5) crore

= 27 crore86. (C) Range of number of students in activity IV

= 438 105 = 333Average of number of students in activityIII per college

=7

1532205407542013065

=7

1603= 229

Required difference = 333 229 = 10487. (D) Number of students in activity II

= 100 + 200 + 200 + 100 + 100 + 100+ 100

= 900Number of students in activity IV

= 317 + 155 + 438 + 105 + 385 + 280+ 120= 1800

Required percentage =1800

900 100%

= 50%88. (B) The average number in the student in

activities III

=7

1532205407542013065

=7

1603 = 229

89. (A) Number of student inCollege D = 100 + 100 + 75 + 105 = 380

College G = 200 + 100 + 153 + 120 = 573College F = 300 + 100 + 220 + 280 = 900

College A = 200 + 100 + 65 + 317 = 682Thus, college D has minimum number ofstudents participate in extra-curricularactivities.

90. (B) Total number of students in activity II= 100 + 200 + 200 + 100 + 100 + 100+ 100= 900

Total number of students in activity I= 200 + 300 + 500 + 100 + 400 + 300 + 200

= 2000

Thus, Required ratio = 900 : 2000

= 9 : 20

91. (A) Total area under Bajra

18 300 acresTotal area under Rice and Barely

(72 + 36) 18

300 (72 + 36)

=18

300 108 = 1800 acres

92. (A) Angle covered by Wheat, Rice and Maize

= 72 + 72 + 45

= 189

Which is greater than 180

Hence, area covered by these three crops

is more than 50% of the total area.93. (C) Required ratio

=Land used for Rice

Land used for Barely

=Angle covered by Rice

Angle covered by Barely

=36

72=

1

2= 2 : 1

94. (B) 10% of the land reserved for Rice

= 10% of 72 = 7.2

It is distributed to Wheat and Barely in

the ratio 2 : 1. Therefore, angle increased

corresponding to Wheat =3

2 7.2 = 4.8

Now angle corresponding to Wheat= 72 + 4.8

= 76.8

95. (A) Production of Rice = 5 production of Jowar

Production of Bazra = 2 production of Jowar

Required ratio

= Production of Rice : Production ofBazra

= 5 : 2

96. (C) Total production of Rice and Wheat

in state B = 10 + 2 = 12 lakh tonnes


13/14


14/14

Centres at:

==================================================================================

Param

ountCoaching

Centr

e

MUKHERJEE NAGAR

MUNIRKA

UTTAM NAGAR

DILSHAD GARDEN

ROHINI

BADARPUR BORDER

JAIPUR

GURGAON

NOIDA

1. (D)

2. (C)

3. (C)

4. (C)

5. (D)

6. (D)

7. (C)

8. (A)

9. (B)

10. (B)

11. (C)

12. (D)

13. (C)14. (A)

15. (D)

16. (C)

17. (A)

18. (A)

19. (A)

20. (B)

SSC Mains Test- 21 (ANSWER KEY)

21. (B)

22. (D)

23. (D)

24. (B)

25. (D)

26. (C)

27. (C)

28. (A)

29. (D)

30. (D)

31. (B)

32. (D)

33. (C)34. (A)

35. (B)

36. (B)

37. (B)

38. (A)

39. (D)

40. (B)

41. (C)

42. (D)

43. (D)

44. (B)

45. (D)

46. (B)

47. (C)

48. (C)

49. (A)

50. (B)

51. (C)

52. (C)

53. (B)54. (C)

55. (A)

56. (D)

57. (C)

58. (B)

59. (A)

60. (C)

Note: If your opinion differs regarding any answer please

message the mock test no. and question no. to 8860330003

61. (D)

62. (D)

63. (D)

64. (B)

65. (A)

66. (C)

67. (D)

68. (B)

69. (B)

70. (D)

71. (D)

72. (D)

73. (C)74. (C)

75. (A)

76. (C)

77. (C)

78. (D)

79. (D)

80. (C)

81. (C)

82. (B)

83. (A)

84. (D)

85. (B)

86. (C)

87. (D)

88. (B)

89. (A)

90. (B)

91. (A)

92. (A)

93. (C)94. (B)

95. (A)

96. (C)

97. (D)

98. (D)

99. (B)

100. (B)

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Mains Maths 21