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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.
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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)
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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
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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
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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
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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
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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
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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.
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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
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