ssc cgle 2012 english answer key

8/13/2019 ssc cgle 2012 english answer key

1/8

S f f f K l u? m PIT LEDUCATIONwww.capitaleducation.in

T H I N Kli UI SUC:CESS THINKIH J SUCCESS

DearStudentsWe areprovidingPaper Answerkey andDetailedSolution

forpaper code223 MN 4.Questionsaresamein allsetsso youcaneasily checkyourperformence andquestionno. 58 is notexactlysame inHindiand english. If 4 decimeter is, consideredas areaof antilethenthereis noanswer.If 4decimeter iscon-sideredaslengthofsideof antilethenanswerwill300.

Team CapitalMOTE:DETAILEDSOLUTIONWILLBEUPLOADEDON OURFACEBOOK PAGESHORTLY

PAPER 1 QUANTITATIVE ABILITIES)

LI. C) 12. D) 13. D) 14. B) 15. C) 16. A) 17. B) 18. B) 19. C) 20. C)

1. C) 32. C) 33. C) 34. B) 35. B) 36. D) 37. B) 38. C) 39. D) 40. A)

>1. B) 52. C) 53. B) 54. A) 55. D) 56. B) 57. A) 58. *) 59. C) 60. C)

1. C) 72. B) 73. D) 74. D) 75. C) 76. A) 77. B) 78. A) 79. C) 80. A)

91. (A) 92. (D) 93. (C) 94. (A) 95. (B) 96. (D) 97. (A) 98. (A) 99. (C) 100.(C)

5:.cioottt: tfe .aiWefi nlsz inT j jj

ffQ[] W/; :::ira

U OS- 17,MANGALMARG BAPUNAGAR JAIPUR M)+919782519669, 0141-4009669


2/8

luialian

EDUCATIONwww.capitaleducation.m

THINKHUJSUCCESS

SSCCGL2013TIERII PAPERCODE223 MN 4

- Let 30+ 30+ 30 V 3 0 +X = x3 0 + x = x 2x2-x-30= 0After solving or satisfyingvaluesfromoptionwegetresult6.

2. In the givenexpression eachtermis one lessthancubeof position.So,217 is oddterm inexpansion.

23. Xy1 1 1 1 14. +++ = 8 12 16 x 3

5

1111i:3:w-6 4:S S-ShareofFourth person=x 200 =225.16

= . . . . . . i)

o oSolving(i) and(ii)we get

4x = 25

25 1Daysrequired= = 6 days.

_L 2 l i _20 30 x - >x " avs>7. 335 + 5A7 = 8B2

.'.4 + A=B10 + B isdivisibleby 3.So ,B cantake8max.So, A = 4.

8. LCMof10,15,20is60.So, answeris of 60ktype.9960 isexactlydivisibleby 60.

9. Numberwhichisdivisibleby 25.Containslasttwodigitsas 25 or 50 or 75 or 00So,only303375 satisfies it.

10. 1 + 1n+ jn n-1 n-2-H 1 h

n+1

l n+nn(n+ l)n1n n1 + 2 + 3 + 4........n__

11.Partofgirls=1--12. LCM of6,8,12,il6is96.13. 3 x 8 x 7 x 6 = 8 isunitdight.14.2(1+ 2 + 3 +12)=156.15. Letthirdproportions x,then12:18::18 : x

18)2So, x = 27

16.E :M:S=2:3:1.Science=-x 80=306

17. Let the numberbe 2x, 3x and 4x,then accordingquestion. 2 x )2+ 3 x )2 + 4x )2 = 1856

185629 = 64

x= 8So,Numbersare 16, 24, 32.18. x : y : z = 9:6 : 4A'srun=x342 =162.19So,onlyone choice satisfiesthis.A B C19.= =2 3 4 =2:3

B: C = 3:4 >A : B:C = 2:3:4Soansweris200, 300, 400.

20.A : B C = 2:5 : 4Total=llx=126.50and we havetofindou

126.50valueof3x=x3=34.50. 17,MANGAL MARG, BAPU NAGAR, JAIPURM)+919782519669,0141 4009669


3/8

maaMna CAPITALEDUCATIONwww.capitaledgcation.in

21. Let the numberare a, b, c, da+b + cthen,-=2D

a + b + c = 6 Da+b+c+d uPuttingvalue of a, b, c in equation (ii).

48d=T22. Sunil= 4days Dinesh= 6day s .

4 8Ram es h== 6 + 4 + 9 _ 194 6 8 24 ~24

24 5Daysrequired= =1.23.6x18= 12 xx => . . x = 924. Discount = 5 SP =100 xx .=121-100 100 8

Profit =21-885 8026. C P =600xx=108100 100

New CP = 408 + 28 = 436545-436Profit =x l OO =25

43680 90 9027. ISOOOxxx=9720100 100 10028.Applying alligation

-6x-10 : 16

x-10 2 2B ut- => X =20-Ib o

30.17CP-17SP=5CP17CP-720=5CP12CP= 720

31. MeanPrice=30x1035x11 13765 13S.P.= 137 130

3 2 . x x -13 100

122.5100 =98=> . . x=80

10 15 3000 =300 2000 =3000100 ' 100Considerorder whichis3000, 2000.

34. D=320 C =320x=40010090B=400x_=360,.r 25A= 360x 360x =450.

100 x =20andx+ 10 = 30Difference= 10

2xy 2x20x3036.Avg.Speed==-24kmph.x+ y 20 + 30100x ( 2 4 + 6 x 2 . 537. NewMean= = 75

10038.120=x+12x5

x= 60(OldAverage.)Newaverage= 60 + 5 = 65.39.49x6+52x6-50x11= 56

40. 25 P = 5SP

SP 25 CPP 1Profit =xlOO =-xlOO= 25CP 4

xxllO 13041. ... XTTT = 2 8 6 0 = > x=?2000

42. P

100 1005Y PxllS

100P=32000

100 = 24 4

S- 17,MANGALMARG,BAPU NAGAR JAIPUR(M) 919782519669, 0141-4009669


4/8

id n PIT LEDUCATIONwww.capitaleducation.in

43. 3362= 3200 1 +3362 f4lY 1681 4lY3200 UOj 1600n = 2. So,hereare two cyclesofquarterwhich

means6monthor - year.

1100

So, R = 2045. Letlengthandbreathbe 3k and 4k

10000

100 100=>K= ^>Breadth=3 x = 2 5

46. -xx(1.6fx3.6=xx(1.2) 2xhh = 6.4 cm.47. Let the sideof squarebe a.20x-a = 3 x a2 = > a =10cm

48. R equired = 40x100 + 50x90 + 60x8040 + 50 + 60

49. Let distance co vere d by cycl ist be d but joggerdcover indoublethen so insametimejogger

would have covered distance.4. . Ratioofspeedisd =499950.-= 18hours

So,train would have reach at 12 but there isrestof 1hour20min.So,time is1:20 am.

D D 10 + 10^ 3 4 ~ 60 12

2060

= 4 k m .52

53.

54.

18 18 9Q =5000 l+I-5000I loo j .

II 11 11= S O O O x xx 500010 10 10= 6655-5000=?1655.

xx r2 x 2 4 =1232

55.

56.

Curvedsurface Area = rcr l =x7x25 = 550. 22 7

x x 2 xx- = 22000x= 1000

Median= 6 - -

57. Precent increase= x + y +

cmxy100

58.

r 5 0 x 5 0= 50 + 50+ =125100Whe n we take 4 dec imeter as length o f squaretile then only answer ispossible.

4 4nxx =8x610 10

A,59. 49

27S-17 MANGALMARG,BAPU IMAGAR,J IPUR M)+919782519669, 0141-4009669


5/8

PIT LEDUCATIONwww.capitaleducation.in

60. x3+ y3+z3- 3 x yz x+ y + z )= x - y ) 2 + y - z ) 2 + x - z )2 )^333+ 333 + 334

= 261. Byusing optionsand letvaluea = 1, b = 2, c = 3,

an d substitutingwe g et a + b + c)2 as aresult.62. 37wh3- c2h 2+9v2

7t2 r2h 4- r h2- r2h4 r 4 h 2 )7I 2 0) = 0

6 -*r2h=i232 i )T i r 2 = 154Dividing (ii)by (i) we getPuting h = 24 in (i) we get r = 7

Curvedsurface Area=nrl= x7x25 = 550Area 550Length== = 275 mfa width 2

4 22 f l V 1 22 f d V64 . 3 2 0 0 0 x xx =- xx x d3 7 d = 4 =height.

3 7

4 22 565 11x10x5=xx-xx 66. In rhombus

p2+ q2- 4 a2122+162=4a2.'. a = 10 cm

4 22 567. 21x77 x24=~xxr3 r = 21

68.x=

x=8400

Cubeon bothsidex 3-8-6x(x-2)=5x3-6x 2+12x-13= 0

69. According question

C 100 D 60 B

In AABCtan ot= 160In A A B Dtan2a= 60W eknowthattan2a= 2tana1-tarraAfter substituding andsolvingwe get h = 80 m.So ,heightof the toweris 80 m.

70. This question can be solved quickly using op-tions.Weuseoption(C),806535.71. According question

A

we know that, In any equilateraltriangleper-pendicular drawn fromvertex to the oppositeside, bisectsthe opposite side.S o,B D : B C=1:2AB= BC, S o , A B : B D=2:l.

72. According questionA

2a 2a

Weknowthat,Ingiven isocelestriangle, drawnperpendicularADfromvertexA,willbisecttheBC.So,BD= - .Now, In A A B D AD )2=(2a)2 -[- AD = -a unit

73. According questionWeknow that,lengthoftangents drawnfromout sidepointof acircle,areequalinlengthasshown indiagram.

D k_r_z ck

S-17,MANGAL MARG, B P UNAGAR,J IPUR M)+919782519669,0141-4009669


6/8

i K I M iax; tnaa s m PIT LEDUCATIONwww.capitaleducation.in

THINKMfaflMSUCCESS

So,x = 7.5

Add eqat ion ( i ) and iii)x + y + z + k = 97.5 + x + k = 9AD =x + k =9-7.5= 1.574 . x-a) x-b) =l,a-b + 5 =0,a-b= -5

1x-a = x-bSo,we cansayx - a) isreciprocal x - b).x - a ( x -a ) = 5(xa) ~ 1

75 . Vx =V3-V5Square on both s ide.x 3 5-2^15

Again square onboths idex 2+64-16x= 6 0x2-16x+ 6=2

76.Let 2 2 /4 = x:Square onboth side

/55x =

Cube on both sidex 64x = ^>x5= 32=>x =2

Hence the reasultwill be 2 .77.Multiplying eachwith conjugate anddividingalso we get the result 0 asfollowing.

-3 -1

-3 -1

=078.

a 2- 2a 1 b 2 2b 1 c2 2 c 1= 0a1=0,b+=0,c+=

79.

4xl-3x-l+5x-l4+3-5=2

2-=x 2 x +=31 3 3 1 T s , 3 27 9x+ -= f x3+ = -3x- = ---x 2 x3 UJ 2 8 2

_ 27-36 _-98 ~ 83 1 - 9 2 7So,x ++2= + - = -x 3 8 1 880. A ccord ing quest ion, th e g iven expre ss ion(a 2 -b z ) 3+ ( b 2 - c 2 ) 3 + - ( c 2 - a 2 ) 3 -W eknowthatifa+ b + c=0

So ,g iven express ionwillbe3(a2- b2 ) (b 2- c2(c 2-a2)So ,a 2- b2= a - b) (a+ b) will be afactor.81. AP = a ,A Q = bA \

/.InAAPDZAD P = 75In AD AQA Qcos30= A D

2b

360 36082 . InteriorAngle=180 =180x 8n 135= 483. t anZABC =3.6Let B O = x. So, AO =3.6x

A

AAOC ~ ADEC 17,MANGALMARG,BAPU N AGAR JAIPUR M)+919782519669, 0141-4009669


7/8

HIhaaoiI]JITRttKIJK

EDUCATIONwww.capitaleducation.inTHINKHasan S U C C E S S

AC _PC_ DECD~CE~AO

AC_y+5_3.6xy= x-5. . 0 is the midpointof BC

A C O C B C / 2 C D C E C E

84. tanZACB=6tanZDBEA

=B C :2C E

B D CAD . ED =6xCD BDButAE:ED= 5:1.'.AD = 6ED. . CD - BD. So, D ismidpointOr we can saythatAD isperpendicular bisec-tor. So,A A D BandA A D Carecongruent.But ZB = ZC. So, inA A C Bwe have ZACB= 60

85. sin9+cos9=^ cos6Squaringwe get=> sin29 +cos 29+2s in9cos9=2cos29=>2sin9 cos9=2cos291cos9-sin9= P Squar ing we getl-(2cos29-l)=P2. . P2=2cos29=>PV 2co s a2+b2-2ab = 0 >(a - b)2= 0.'.a = b. So,AABCisisosceles rightangled tri-angle andothertwo anglesare 45each.

.'.BC = 4A B=2 x B C=8cm

88 . ABEC isequilateralA

90ZOBC= =45InAOBC

ZC=18 + 60 = 180= 75

89. A D2=A C2+CD2A

C DA D 2=AB2- BC2+CD2A D 2+BC2=A B2+C D2

90.ZAQC=ZAPB= 90

91

Angle in semi circle isright angle).'. AAQC ~A A P BSo, Q C | |P B

sinA sinA1+cosA 1-cosA

sinA(l-cosA)+sinA(l+cosA)l +cosA)(l-cosA)

sinA[l-cosA++c o sA ] l - c o s2 A )

2s inAsin2A = 2c o se c A

92.rsin9=lrcos9=. 3Squaring Equation (i) and ii)

Sosin9=-,9 = 30the resultwillbe 2.

- 17,MANGALMARG,BAPUNAGAR,JAIPUR M)4-919782519669,0141 4009669


8/8

iiiat;t.-m- PIT LEDU TIONwww.capitaleducation.in

93. We Know that In a f requency distributionogives are graphica l representa t ion ofCumu-lative frequency.x cosecBO 0 x)4.- = =>- sin45 y

So, = 64 = 4395 . TowerAB ,subten ds angle30 atpointC asfol-lowing d iagrams .

50mh 1 50

96. Accord ing ques t ionL e t A B i s S x a n d BC is 2x.CP)2 = P B )2 + B C)2

A 3x P 3x

2x

(CP)Z_ C P )2=97. A)98. A)99. C)100. C)

CP =

W e doitteachC T W e d i

f W eachSmart-33Batches

S-17,MANGALMARG, BAPU NAGAR, JAIPUR M)+919782519669, 0141-4009669

ssc cgle 2012 english answer key

Documents

Transcript of ssc cgle 2012 english answer key