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Dear Students
We are providing Paper, Answer key and Detailed Solution
for paper code 223 MN 4. Questions are same in all sets so you
can easily check your performence and question no. 58 is not
exactly same in Hindi and english. If 4 decimeter is, considered
as area of an tile then there is no answer. If 4 decimeter is considered as length of side of an tile then answer will 300.
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MOTE :— DETAILED SOLUTION WILL BE UPLOADED ON OUR FACEBOOK PAGE SHORTLY
PAPER - 1 (QUANTITATIVE ABILITIES)
LI.
(C) 12. (D) 13.
(D) 14.
(B) 15.
(C)
16.
(A) 17. (B)
18. (B)
19.
(C) 20. (C)
(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)
1.
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SSCCGL 2013 TIER II
PAPER CODE 223-MN-4
9.
-
Let ^30 + ^30 + ^30
«
V30 + X = x
30 + x = x2
x 2 -x-30 = 0
10.
After solving or satisfying values from option
we get result 6.
2.
In the given expression each term is one less
than cube of position.
So, 217 is odd term in expansion.
Number which is divisible by 25.
Contains last two digits as 25 or 50 or 75 or 00
So, only 303375 satisfies it.
+ 1-
1—
n + i; V
n
-H
n-1 n-2
1
n+ 1 n + l
n+ l
n + lj
l—
n+ l
1
h
n + l n+ l
1 + 2 + __
3 + 4.........n
n(n + l)
X2
3.
4.
11. Part of girls = 1--
y
12. LCM of 6, 8,12,il6is96.
13. 3 x 8 x 7 x 6 = 8 is unit dight.
14.2(1 + 2 + 3
+ 12) = 156.
15. Let third proportion is x, then 12 :18 :: 18 : x
1 1 1 1 1
— + — + — + — =—
8 12 16 x 3
1 1 1 1
So, x =
i:3!»:w-6!4:S!S-
(18)2
„
= 27
16. E : M : S = 2:3:1.
Share of Fourth person = —x!200 = 225.
16
= - ...... (i)
5.
= ...... oo
1856
Solving (i) and (ii) we get
x=
29
25
_L 2- l°-i _
7.
30
= 64
x =8
So, Numbers are 16, 24, 32.
18. x : y : z = 9 : 6 : 4
4
25
1
Days required = — = 6— days.
20
Science = -x!80 =30
6
17. Let the number be 2x, 3x and 4x, then accord
ing question.
(2x) 2 + (3x)2 + (4x)2 = 1856
x
" •->x"
avs>
335 + 5A7 = 8B2
.'. 4 + A=B
10 + B is divisible by 3.
So, B can take 8 max. So, A = 4.
8. LCM of 10,15,20 is 60.
So, answer is of 60k type.
9960 is exactly divisible by 60.
A'srun = —x 342 = 162.
19
So, only one choice satisfies this.
A B C
19. —=-=-
2 3 4
= 2:3
B : C = 3 : 4 => A : B : C = 2 : 3 : 4
So answer is 200, 300, 400.
20. A : B : C = 2 : 5 : 4
Total = llx = 126.50 and we have to find ou
126.50
value of 3x = - x3 =34.50.
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21. Let the number are a, b, c, d
a+b + c
then, —-— = 2D
30. 17CP-17SP = 5CP
17CP-720 = 5CP
12CP = 720
a+b+c=6D
30x10 + 35x11
137
65
13
31. Mean Price =
a+b+c+d
—— =u
Putting value of a, b, c in equation (ii).
S.P.=
48
d=T
137 130
13 100
122.5
= 98 => .-. x = 80
100
32. xx-
22. Sunil = 4 days
Dinesh = 6 days.
4
8
Ramesh = — = —
33. 3000 x
I I I 6 + 4 + 9 _ 19
4 6 8 "
24 ~ 24
10
15
=300, 2000 x
= 3000
100
'
100
Consider order which is 3000, 2000.
C= 320x - =400
100
34. D = 320
24
5
Days required = — = 1— .
90
B=400x_=360,
23. 6x18 = 12 xx => .'. x = 9
-.r,
125
24. Discount %=
A = 360x 360x
25. SP = 100 x — x^^. = 121-%
" "
100 100
8
.-. x = 20andx + 10 = 30
Difference = 10
100
=450.
2xy
2x20x30
36. Avg. Speed = —' = ———— - 24 kmph.
x +y
20 + 30
Profit % = 21-%
8
85
80
26. CP= 600x - x - =108
100 100
New CP = 408 + 28 = 436
545-436
Profit % = - xlOO = 25%
100 x (24+ 6 x2.5
= 75
• 100 '
38. 120 = x + 1 2 x 5
37. New Mean =
x = 60 (Old Average.)
New average = 60 + 5 = 65.
436
80 90 90
100 100 100
27. ISOOOx - x - x - =9720
39. 49x6 + 52x6-50x11 = 56
40. 25 P = 5SP
28. Applying alligation
SP
-6
\0
x -10 : 16
x-10 2
2
But -— » - => X = 20-%
Ib
o
6
CP
25
P
1
Profit % = —xlOO = -xlOO = 25%
CP
4
41.
xxllO 130
... XTTT =2860 =>x = ? 2000
100 100
5 Y
42. P
100
P=32000
PxllS
100
= 244
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43. 3362=3200 1 +
3362
3200
f4lY
UOj
D
^i
1681
1600
(4lY
n = 2. So, there are two cycles of quarter which
1
means 6 month or - year.
D
' 3 4 ~ 60
= 4km.
52.
53.
12
20
60
18
9
10 + 10
18
Q = 5000(l + — I -5000
I looj .
D
I
II 11 11
1 +
= SOOOx—x — x
100
10 10 10
= 6655-5000 = ? 1655.
54.
5000
-x — xr 2 x24 =1232
So, R = 20%
45. Let length and breath be 3k and 4k
10000
100
100
= > K = — ^> Breadth = 3 x — = 2 5
Curved surface Area = rcrl = — x 7 x 2 5 = 550
55.
x = 1000
46. -x — x(1.6fx3.6 = ix — x(1.2) 2 xh
h = 6.4 cm.
47. Let the side of square be a.
56.
20x-a =3xa 2 =>a = 10cm
48. Required % =
40x100 + 50x90 + 60x80
40 + 50 + 60
. 22 7
x x 2 x — x- =22000
Median = 6 -
57. Precent increase = x + y +
rn
n
= 50 + 50 +
49. Let distance covered by cyclist be d but jogger
d
cover — in double then so in same time jogger
-
cm
xy
100
50x50
=125%
100
58. When we take 4 decimeter as length of square
tile then only answer is possible.
4 4
nx — x — = 8 x 6
10 10
would have covered — distance.
4
.'. Ratio of speed is — '• d = l: 4
A,
999
50. —- = 18 hours
So, train would have reach at 12 but there is
rest of 1 hour 20 min. So, time is 1: 20 am.
59.
49
27
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In AABC
60. x3 + y3 + z3 - 3xyz
=
(x + y + z)
2
tan ot =
( ( x - y ) 2 + ( y - z ) 2 + (x-z) 2 )
^333 + 333 + 334
=
l
160
In AABD
tan2a= —
2
61. By using options and let value a = 1, b = 2, c = 3,
and substituting we get (a + b + c)2 as a result.
62. 37wh3 - c2h2 + 9v2
7t2(r2h4 - r"h2 - r2h4 + r4h2)
7I2 (0) = 0
63. -*r 2 h=i232
(i)
Tir2 = 154
Dividing (ii) by (i) we get
Puting h = 24 in (i) we get r = 7
Area
550
Length = —— =
= 275 m
fa
width
2
,
4 22 f l V 1 22 f d V ,
64. 32000x—x — x — = - x — x — xd
3
4 22 ( 5
65. 11x10x5= x x - x — x —
We know that tan2a =
2tana
1-tarra
After substituding and solving we get h = 80 m.
So, height of the tower is 80 m.
70. This question can be solved quickly using options.
We use option (C), 80°, 65°, 35°.
71. According question
A
Curved surface Area = nrl = — x 7 x 2 5 = 550
3 7
.-. d = 4 = height.
60
we know that, In any equilateral triangle perpendicular drawn from vertex to the opposite
side, bisects the opposite side.
So, B D : B C = 1:2
AB = BC, S o , A B : B D = 2:l.
72. According question
7
A
2a
2a
x = 8400
66. In rhombus
p2 + q2 - 4a2
122 + 162 = 4a2
.'. a = 10 cm
We know that, In given isoceles triangle, drawn
perpendicular AD from vertex A, will bisect the
4 22 5
67. 2 1 x 7 7 x 2 4 = ~ x — xr 3
BC. So, BD = - .
Now, In AABD
.-. r = 21
68. x =
(AD)2 = ( 2 a ) 2 - [ -
Cube on both side
x3-8-6x(x-2) =5
x 3 -6x 2 +12x-13 = 0
69. According question
AD =
-a unit
73. According question
We know that, length of tangents drawn from
out side point of a circle, are equal in length as
shown in diagram.
D k _ r _ z c.
k
C 100 D 60 B
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So,x
4xl-3x-l+5x-l
= 7.5
4+3-5=2
Add eqation (i) and (iii)
x+y+z+k=9
7.5 + x + k = 9
AD = x + k = 9-7. 5 = 1.5
74. (x-a)(x-b) = l,a-b + 5 = 0,a-b = -5
1
x-a =
x-b
So, we can say (x - a) is reciprocal (x - b).
x-a—
(x " a)
(x-a)
~
=5
79.
2
-=
2 x + - =3
x
1
3
1
3
Ts
, 3
27 9
x + - = - f> x 3 +— = - -3x- =
--x
2
x3
UJ
2 8 2
_ 27-36 _ -9
8
~ 8
3
1
- 9 2 7
So, x + — + 2 = — + - = x3
8 1 8
80. According question, the given expression
(a 2 -b z ) 3 + (b 2 -c 2 ) 3 +-(c 2 -a 2 ) 3 We know that ifa + b + c = 0
1
So, given expression will be 3(a 2 - b 2 )(b 2 - c2
(c2 -a2)
So, a 2 - b2 = (a - b)(a + b) will be a factor.
81. AP = a, A Q = b
75. Vx" = V3-V5
Square on both side.
A
x= 3 + 5-2^15
\
Again square on both side
x2 + 64-16x = 60
x 2 -16x + 6 = 2
76. Let ^2^2^/4
= x:
/55x =*
Square on both side
/. In AAPD
ZADP = 75°
In ADAQ
cos30°=
Cube on both side
AQ
AD
2b
x6
4x = — =^> x5 = 32 => x = 2
Hence the reasult will be 2.
77. Multiplying each with conjugate and dividing
also we get the result 0 as following.
-1
-3
360°
x
= 180°
360°
In
= 135°= —
4
83. tanZABC = 3.6
Let BO = x. So, AO = 3.6x
1
-3
82. Interior Angle = 180°
A
-1
=0
78.
a 2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0
a - 1 = 0, b + l = 0,c + l =
AAOC ~ ADEC
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88. ABEC is equilateral
AC _ PC _ DE
A
" CD ~ CE ~ AO
A C _ y + 5_3.6x
y = x-5
.-. 0 is the midpoint of BC
AC
OC
BC/2
= BC: 2CE
" C D CE
CE
84. tanZACB = 6tanZDBE
90°
ZOBC = — =45°
In AOBC
A
ZC=180°
+ 60° = 180°
= 75°
B
D
C
89. AD2 = AC2 + CD2
A
AD
. ED
— = 6x —
CD
BD
But AE : ED = 5 :1
.'. AD = 6ED
.-. CD - BD. So, D is midpoint
Or we can say that AD is perpendicular bisector. So, AADB and AADC are congruent.
C
D
AD2 = AB2 - BC2 + CD2
AD2 + BC2 = AB2 + CD2
90. ZAQC = ZAPB = 90°
But ZB = ZC. So, in AACB we have ZACB = 60°
85. sin9 + cos9 = ^2 cos6
Squaring we get
=> sin29 + cos29 + 2sin9cos9 = 2cos29
=> 2sin9 cos9 = 2cos29 -1
cos9-sin9 = P
Squaring we get
l-(2cos 2 9-l) = P2
.-. P2 = 2cos29 => P - V2cos<9
(Angle in semi circle is right angle)
.'. AAQC ~ AAPB
So, QC||PB
91.
sinA
sinA
1 + cosA
1-cosA
sinA(l-cosA) + sinA(l + cosA)
86. ab= —
•ab =
(l + cosA)(l-cosA)
sinA[l-cosA + l + cosA]
(l-cos 2 A)
2ab = a2 + b2 =5> a 2 + b2 - 2ab = 0 => (a - b)2 = 0
.'. a = b. So, AABC is isosceles right angled triangle and other two angles are 45° each.
2sinA
sin2 A
92. rsin9 = l
= 2 cosecA
r cos9 = .^3
Squaring Equation (i) and (ii)
So sin9 = -, 9 = 30°
.'. BC = 4
AB = 2 x B C = 8cm
the result will be 2.
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93. We Know that In a frequency distribution
ogives are graphical representation of Cumulative frequency.
96. According question
L e t A B i s S x a n d BC is 2x.
(CP)2 = (PB)2 + (BC)2
x
cosecBO0
x
)4. - =
=> V
sin45°
y
2x
So, — = 64 = 43
A
95. Tower AB, subtends angle 30° at point C as following diagrams.
1
W
3x
( C P) 2 =
CP =
50
We; doit teach CAT
f
. _
"P
97. (A)
98. (A)
99. (C)
100. (C)
50m
h
(CP)Z
3x
We di
each
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