CGL Tier I - Paper 4 Solution

10M

Here, CD = AE= 10M

5 Ii So, required distance'e-------'D AE= 10M

151!

24. (0) B+6 +is

16. (8)17. (A)18. (8HA) 7 x 11 - 3 = 74

(8) 9 x 11 - 3 = 96 m7I(C) 4 x 11 - 3 = 41(0) 6 x 11 - 6 = 63

19. (8) Given set = I 5 1211ill~

Similarly, in option (8)4824Iu.d~

nn-l\nn20. (A) t3 na:I M ~ ~ T+11

+1 +1I i'I +

21.(8) f;l~'X- ~,;-2ti'W-t3S+ri'sL±LJ +9 +11

22.(01 ftijrtMj:tM¥fff23 (B11 IIj fL1L__'f'

+1 . LJL1HUU

(C)

15.(0) (A)MI.!! • ll114

11. (A)12. (A)a 'L ~ b/abba/a _ abb/a b_ ba13. (A)AZ 8 '!_,AZ ~ Y,A~8Y,~Z8Y14. (C) r.uJ

102 101 98 93 86 74 66 53-1 -3 -5 ·7 ..g

1.7' - IJ

48•7I "80I

10. (A) 9

(9" - 1)

9.(BI~"~

626 - 7 = 619

Similarly,x- 28 = 38

x = 38 + 28 = 667.(0)216-7=209 209-7 = 202 and

522-7=515 515-7=508Similarly,633- 7 = 626

8. (8)

6. (0)54 - 30 = 24 and 112 - 42 = 70

1-<L- ~

Given expression A I8 x C A < 8 = Cbecause option (8) follows the givenexpression here because C - 8 + A means

C 8> A, one of which meaning is A<8= C2. (C)2 M C 3 K

(2 M = N) (2 N Jo-3 K)(2 N ... 3 K)

2 x 2 M ... 3 K (since N = 2 M)4M Jo-3K4 M is not more than 3 K, then 2 M is lessthan 3 K.Hence 2 M C 3 K is correct.

3. (8)(15 +12) .. 9 = 3 and (44 + 28) + 9 = 8Similarly,(64 + 53) + 9 = 13

4. (8) 95 115 145 155 175

(lO!D-5)(loxt'5) (lOJ4+5) (10*6-61 11JI6-5)

5. (C) 1 3 8 19 42 89!lx2 +1)+ J(lI.2 +2)+ IIx2 +3)+ IIx2 +41+ 1(X2+51+

x_=

1. (8) Given + _ >

- ...--<t

SSC MOCK TEST 4 (SOLUTION)

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A+-3km-+E

(Here = OA = C8 = 2 km)Required distance AE = DE - OA

=3-2= 1 km

52. (8)

45. (A) 46. (8) 47. (C) 48. (Al49. (D)50. (D) The numerical groups of PLAY will be

P-15,43L- 36,65A-42,46, 62Y-45

51. (8)Straight line 4 x + 3y = 12 passesthrough 1st, 2nd & 4th quadrant.

44. (A)

43. (8)

Conclusion I- ./II_x

39. (8) 40. (8) 41. (A)

42. (C)

According to given direction the faces ofRani and Sarita will be East and Southdirection with respect to x.

37. (D)

38. (8)

36. (C)

3km

Clf--IA---:a.B

3km

35. (A)

F RAN... ..; ..; {-Place V.a1ue to 18 1 14

~ ~!Oj:fi

Similarly,

+ rIS 14.!!2.!!.a:!O 2B

Similarly,42 x 17 4 x 2 x 1 x 7 = 56

32. (C) 12 P 6 M 15 T 16 8412x6+15-16+412x6+15-472+15-487 - 4 = 83

33. (8) Y M LOS Bel~.,.,.,.".,,,12345678Meaningful wordSYMBOLI C

t "·{t·~~·1264387~.(C) LON D..; + + +PIat:e Value12 15 14 4

!!.a~.!!..:l.!!.3.24E 28 8

2 x 9 x 4 x 8 = 5763 x 5 x 1 x 6 ;: 90,2 x 2 x 4 x 6 = 96 and

31. (A) 29 x 4835 x 1622 x 46

Similarly,

Son's age = 10 1 44 = 4 years

26. (8) Veni > Smith> Raju > Salim27. (8) G ENE RAT E28. (C) QUA I N T29. (8) According to question, the odd numerical

value of HOTEL will be -HOT E L.++ • ..;lIS ::1989 9 28HOTEL = 15 + 29 + 39 + 9 + 23

= 115

30. (A) !1~i£ ~ t 1iiI

Father: Son10 1

25. (8) Total age of father and son = 22 x 2= 44 years +E

S

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45° 35° 80°

oaA 45°

In Aoa,OA=OB

OABIn AOC,

OA=OCOAC OCA 35°BAC OAB OAC

59. (B)

Aliff:\BEe

BD =AD-AD= 10 - 4 = 6 cm

BD: DA = BE : EC6: 4= BE : EC BE :CE = 3 : 2

58. (D)

Again,CN2 + ON2 = OC282 + ON 2 = 102

,.....,..".--

ON = .J100 36

= .[36 = 6 cmThe distance between the chords

= MN = OM + ON= 8 + 6= 14 cm

OM = .J100 36 - J64 = 8 cm

1- x 16 = 8 cm2=OA2=102

Similarly CN =

AM2+0M262 + OM 2

Let AB & CD are two parallel chords of

= .!_X 12 = 6 cm2

(perp. from centre to any chord,bisects the chord)

57. (C)

94" must be added in 16 a2-12atomakeita

perfect square.

3 2 3 2"2 = 4a "23

2 _ 2.4ax2= a)1

33x- - =2x

1 2x- - = -x 3

1 2x2 7 = x 2

x

2 2

= 23

4 2 22 4= - - = --= 2-

9 1 9 956. (A)a16 '-12a

x=3x- 23x=32

55. (B) x 3 -x

=4 cm= 5 cm= PT' + TO'= PT' + 42

= 3 cm

OTPOPO'5'

PT

54. (A)

In ADC,AC is the hypotenuse, which is thelongest side of triangle.AC>AD

Similarly, AC> CFAB>ADAB>BEBC>CF

and BC> BEOn adding above inequalities, we have

2(AB + BC + CAl > 2(AD(AB + BC + CAl > (AD +

53. (B) (A B) (a C) =65~ ... ,40

(A a C) a = 205°

180° + a = 205°a = 205°-180°

=25°

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15Work done by Hari & Ram in 1 day

1= --

20Work done by 2[Ram + Shyam + Hari)

1 1 1=- -12 15 205 4 3 12= = --= -

60 60 5Work done by (Ram + Shyam + Hari)

1in 1 day = 10Work done by Ram alone in 1 day

1 1= 5" ;0-

3 2 1=30= 30

Ram can do the whole work in 30 days.

= -

1= 12

Work done by Shyam & Hari in 1 day1

15 2a =../3 cm

30 10J3../3Hence Be = ../3 - .j3 = 10../3 cm

.[3Area of ABC = - 824

J3 10../3 10./34

= 75$ ern".63. (A)Work done by Ram & Shyam in 1 day

COB (byRHS)CBO (byepeT)

Now, AOBABO

CBo

CAD CBo (Angles in the samesegment of a circle)

61. (e)

1000 J3= --=-,--2

= 500 J3= 500 J3 km = .!.J3 km

1000 2

h

= 1000

hBD = J3cm

In CBA

tan300 =CBBA

1 h h73 = = hBo oA 1000

$h 1000 = hJ313hJ3

h

J3 = 1000

CBtan600 =Bo

h./3= Bo

ALet h m be the height of the ballon.In CBo.

hm

./3= -82

15

8AB = a & BD = "2

height AD = JAB2 Bo2

15 = R

Let ABC is equilateral & AD is its height.Let 'a' unit is the side of the ABC .

62. (C)~

ABO CBoABC = ABO CBO 2

c60. (A)

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b: 9 =1002: 1

= -1

=210 20b

9

= 85Y = 100-85 = 15

2nd discount = 15%72. (8) Let 'g' stands for the no. of girls

& 'b' stands for the no. of boys1

10%ofg= 200fb

10 1100 x 9 = 20 x b

122.40 100 10090 160

(100-y) =

= Rs. 122.40160 ~ 100 Y100 100

'" Rs. 122.40

SP after two successive discounts= Rs. 122.40

1stdiscount = 10%ATQ,

160 100 10 100 Y100 100

= Rs. 16071. (D) MP

90xCP ill 45

Now, MP =-x-= 56

112=---90x

SP 100CP = 100 12%

9x 100_ 10- 112

% profit = 12%

9x10SP = 90%JO#:Rs.

189 18991701198

Greatest 4-digit perfect square number= 9999 -198= 9801

70. (A) MP '" (Say)

9 99999 81

.!.= (7 7 7 7 10 10 10 10)4= 7 x 10 = 70

69. (C) Greatest 4-digit number = 999999

1

= (24010000 )4

68. (D) 4 '" root of24010000

a'-h=V4 •

Volume = V

r2h = V

2r

a= -

=a2 r

67. (D)

X2r =--

1) = X2r(

Work done by 2(A + 8 + C) in 1 day1 1 1= - - --12 15 205 4 3 12 1

= :--=-60 60 5

Work done by (A + 8 + C) in 1 day1 1=--=-

5 2 10Work done by A alone in 1 day

1 3 2 1= - -= --=-10 15 30 30

Hence A can complete the work alonein 30 days.

66. (A) Perimeter - Diameter = X2 r 2r = X

Work done by (C + A) in 1 day =

Work done by (8 + C) in 1 day =

12

115

120

(C) Work done by (A + 8) in 1 day =65.

15= 5days

15 women complete a work in

6 men + 5 women = (10 + 5) women5 women complete a work in 15 days.1 woman completes a work in 15 x 5

15 5

5= "3 x 6 = 10 women6 men

5= "3women1 man

64. (A) 3 men = 5 women

.

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z, 4 1- = 20x 3

Y = 20 x 3 = 60x

ATQ,

:: y =20-x x4

x mlminy m/min

79. (0) let the ususal speed of the man =let the distance of his office =

3New Speed = -x m/min4

x = 24 km60 10

25

60x 35x= 60

10 25x= ---

1 60

2 9= x -x --x15 20

8x 27x10 = x

60

Remaining journey

x km (say)

9-x20

78. (8)TotaljxJrney oovered by theman=2

Joumey by train = --x & by bus =15

416 100 100x = ---:-:--.,-,---'--'- = Rs. 400

80 130

21x 16x20 = 100

5x = 100 x 20

100 20x = 5 - Rs. 400

ep = Rs. 40077. (e)let the original price of the article = Rs. x

Then,

x(100 20 )(100 30) = 416100 100

21x 4x20 5= 100

% gain = 5%105x

SP=105%ofx = 100ATQ,

4x 100 = 105x5 100

4x 1005New SP = Rs.

4x= Rs. 5

=% gain.3.!.. 100100~~- =2%

xthe ep of the article = Rs. x

SP at 20% loss = 80% of76. (e) let

102x 2x= -:;oo-x= 100Gain

102x=Rs. 100

SP after 15% discount = 85% of

the MP of the article = Rs.

h' = 1 ernthe ep of the article = Rs. x75. (e) let

3 62

4 27= -- =1

3 36

h' =274

=

=

r, : 3cm [r. ~ rad. of sJ>h,.ere ]r. - 6cm r. ~ rad. of cylinder

Volume of the water raised in the cylinder= Vol. of the sphere

4 3fc2h' = '3 fs

Usual time = distance = r. = 60 minapeed x

80. (0) a, b, C, d, e are consecutive odd numbers.b = a + 2, C = a + 2 + 2 = a + 4d=a+2+2+2= a+6e=a+8

x

.

120x100

120x100

74. (8)

73. (0) (25) 2.6 : (5)3(52)25 : 535 s : 5325 :1

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Hence 90°

90. (D) sin2 3sin 2=0sin 2 2sin sin 2 = 0sin (sin 2) 1(sin 2) 0(sin 1)(sin 2) 0sin 1,2

....(ilx sin y cos 4 .... (ii)

On squaring and adding (i) and (ii), we have

(x cos ysin'l (xsin yeos'l=22+42

x2 cos2 y2 sin 2 2xy cos sin xx2 sin 2 y2 cos2 2xy cos sin =4+ 16

X2(COS2 sin 2 ) y2(sin2 cos2) = 20x2 + y2 = 20

4tan 2 cot 2 2tan cottan 2 cot2 2 4tan 2 cot2 2

89. (A) x cos ysin 2

Cost of 2 mugs + 3 buckets=2)(4+3)(9= 8 + 27 = Rs. 35

88. (A) (tan cot r = 22

351= -=9

3992 5 ( 77) 8

Y = 8 8 5 5460 616

=39

156= -=4

39

=

8 8 5 5385 73664 25

x =

100= Rs. 2400

87. (A) let the cost of one bucket & one mug= Rs. x & Rs. y respectively.

Then,8x+5y=92

& x + !li!y = 77By cross multiplication.

5 77 8 92

=

100

10000 12 2

=SI

2544 25 25= (28 25 )(28 25)= Rs. 10000

P R T

24015

28 2

2544 = P 25

12 286. (A)CI = P 1 100

2P=4P=2

85. (Al 4 3J3 4(7 4/3) 3/3(7 4/3)7 4/3 49 48

= 28 16/3 21/3 36

= 8 5../3

4 6-- 2P= 123

=1284. (Cl4x 2P3

x2 L Z2

9 25 16

9 25 16= -- 16=39 25

Now,

1 1 11--; 1b :;-;;= 4

83. (C) *-3) 2+(y-5)2+(z-4)2=The above equality is possible only whenx - 3 = 0, y - 5 = 0 & z - 4 = 0

x = 3, Y = 5, Z = 4

a b c1 = 4

1 a 1 b 1 c

a 1 a b 1 b c 1 c1 1 b

=4a 1 c

abc1 a b :;-;;=1(C)82.

15= 16

=Average of the rest =300 60

5a 20= =a+4

581. (A) Sum of 20 numbers = 20 )( 15 = 300

Sum of 1 et five numbers = 12 )( 5 = 60

5a (a 2) (a 4) (a 6) (a 8)

=

bed e5

aAverage=

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100. (D) Cost of material =

170200 13= --% = 11--%17 17

=% passed 10020

3 180 3""5 rad = ""5 = 108°94. (D) No. of students passed in 1st division

in 2008 = 20Total students = 170

1 rad =

2 1= cos 2 A sin 2 A = 2 1" 2

92. (D) 3cos 80° cosec tO?+ 2cos 59°.cosec 31°3cos (90°-10°) (cosec 100+ 2cos(900-310)cosec 31°3sinl 0° cosec 10° + 2sin 31° x cosec 31°

= 3+2=593. (C) rad = 180°

180

91. (C) carl A(slnA cosA) sin2A(sinA cosA) x

cose<fA(sinA cosA) sec!A(sinA cosA)[sec? A - cosec? A)

co~ A(sinA cosA) cart A(sinA cosA)= cose<fA(sinA cose<fA(sinA

xcosA) cosA)

[sec! A - cosec? A)

cos2 A (sin A casAr (sinA cosAr= cosec2A sin2 A cos2 A

x

cos2 A sin2 A

cos2 A 2(sin2 A cos2 A) sin2A co~A= cossc/A sin2 A cos2 A sin2Acort A

144 96000360

= Rs. 38400

C t f dl 43.2 96000as a trect expense = 360

= RS.11520Difference = 38400 - 11520 = Rs. 26880

6= 82-%17

96. (A)% of pass students in 2009

= 140 100190

=73.68%% of pass students in 2008

=82.35%% of pass students in 2010

160= -- 100 = 80%200

In 2008 % of pass students is highest.97. (C) No. of students pass in 3 r. division in

2008 = 6098. (A)No.of students in 2 no division in 2010= 60

% of students passed in 2 no division60

= 200 100 = 30%

115.299. (A) Cost of labour = 360 96000 = Rs. 30720

95. (D) % pass students in 2008

= 140 100170

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SSC MOCK TEST 4 (ANSWER KEY)

1. B 26. B 51. B 76. C 101.B 126.0 151.0 176.02. C 27. B 52. B 77. C 102. C 127.B 152. A 177. C3. B 28. C 53. B 78. B 103. A 128.0 153. B 178.C4. C 29. B 54. A 79. 0 104. A 129.C 154. B 179.C5. C 30. A 55. B 80. 0 105. C 130.0 155.C 180.B6. 0 31. A 56. A 81. A 106. A 131. A 156. C 181. B7. 0 32. C 57. C 82. C 107. B 132.0 157. A 182.B8. B 33. B 58. 0 83. C 108. A 133.A 158. A 183. C9. B 34. C 59. B 84. C 109.B 134.A 159. C 184.B10. A 35. A 60. A 85. A 110. B 135.0 160.A 185. A11. A 36. C 61. C 86. A 111. B 136.B 161. A 186.C12. A 37. 0 62. C 87. A 112.0 137.B 162.0 187.013. A 38. B 63. A 88. A 113. B 138.0 163. 0 188.A14. C 39. B 64. A 89. A 114.0 139.C 164.C 189. A15. 0 40. B 65. C 90. 0 115.0 140.B 165.0 190.C16. B 41. A 66. A 91. C 116.0 141.B 166.C 191. C17. A 42. C 67. 0 92. 0 117. B 142.A 167.A 192. B18. 0 43. B 68. 0 93. C 118. A 143.0 168. B 193.A19. B 44. A 69. C 94. 0 119.0 144.B 169.C 194.020. A 45. A 70. A 95. 0 120.0 145.C 170.A 195.B21. B 46. B 71. 0 96. A 121. A 146.A 171. C 196.C22. 0 47. C 72. B 97. C 122. A 147.A 172.C 197.B23. B 48. A 73. 0 98. A 123. C 148.B 173. B 198.A24. 0 49. 0 74. B 99. A 124.0 149.C 174. B 199.025. B 50. 0 75. C 100.0 125.0 150. C 175. B 200.0

188 (A); Remove 'have'189. (A); Change 'is' into 'has been'190. (C); Remove 'up'191. (C); Change 'is tasting' into 'tasted'192. (B); Change 'casted' into 'cast'www.ss

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CGL Tier I - Paper 4 Solution