ssc mains maths - 53 (solution)...meerut varanasi rohtak panipat sonepat bahadurgarh patna agra...
TRANSCRIPT
1
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
1.(C) Let Price = ` 100After two successive discounts, price will be
= 100 × 10070
×10083
= ` 58.1
Single Discount (D) = 1001.58100
× 100
= 41.9%Shortcut:–Equivalent single discount (D)
=
100xyyx
= 30 + 17 –30 17
100
= 47 – 5.1= 41.9%
2.(C) Let the original price = ` x per kg
New price = `34x
2520 43x
–
2520x
= 15
3360 2520
x = 15
x = 15840
x = ` 56
New price = 34x
= `34
× 56
= ` 42 per kg.3.(B) Cost of 126 toffees = 64
after giving discount of 25%
Actual cost = 64 × 75
100 = ` 48
126 toffees cost = ` 48
1 toffee costs = `48
126
1170414 toffee cost = 48
126 ×1170414
= ` 445872
SSC MAINS MATHS - 53 (SOLUTION)
4.(B)S.P. C.P. 100
36
= C.P.
252 10036
= C.P.
C.P. = 700
S.P. = 136 700100
= ` 952
5.(C) Total present age = 26 × 7= 182 years
and5 years before total age = 152 – 5 × 7
= 147 yearsNumber of persons = 6
So, average age = 147
6= 24.5 years
6.(A) Total quantity of milk= 4 × 0.8 + 7 × 0.7 + 10 × 0.06= 3.2 + 4.9 + 6.0 = 14.1
Total quantity of water = 21 – 14.1 = 6.9
Milk : water = 9.61.14
= 69141
= 2347
7.(C) Let height of the tower = h cm AD = x cm
In ACD
tan 60º =hx 3 x = h
x = 3
h
In ACD
tan 30º = 30h
x
13 = 30
hx
Put the value of x
3h
+ 30 = 3 h
h + 30 3 = 3h
15 3 cm = h
Height of the tower = 15 3 cm
2
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
8.(D)
C is mid point of AB
Co–ordinates of C = 2
2x
, 3
2y
22
x= 1 and
32
y= + 2
x = 0 and y =1 Coordinates of other end = (0,1)
9. (A) If x2 + 1 = 0 x2 = –1Put value in given polynomial.x4 + x3 + 8x2 + ax + b = 0(–1)2 + (–1)x + 8(–1) + ax + b = 01 – x – 8 + ax + b = 0x(a – 1) + b – 7 = 0
So, a = 1 and b = 7.10. (A) 2(cos2 – sin2 )= 1
cos2 – sin2 = 12
1 – sin2 – sin2 = 12
2sin2 =12
sin2 =14
sin2 = +12
For acute angle
sin = 12
= 30º11. (B) cos2 A – sin2 A = tan2 B
cos2 A – sin2 A+1 = tan2 B + 1 2cos2 A = sec2 B
Asec
22 =
Bcos1
2
2 cos2 B = sec2 A 2 cos2 B –1 = sec2 A –1 cos2 B – sin2 B = tan2 A
12. (A) Let the side of longer square = a unitSide of smaller square = m unit
ATQ,4a – 4m = 100 a – m = 25 .... (i)
Again from question:-a2 – 3m2 = 325 .... (ii)
From (i) and (ii):-m = 30, –5(side cannot be negative)
m = 3013. (B) Let OABCD be the right pyramid on a rect-
angular base ABCD
AB = CD = 32 cmBC = AD = 10cmSince it is a right pyramind,OXZ = 90º andOX = h = 12cmX is the midpoint of the base.Now, in OXZ , OXZ = 90ºOZ = 2 2OX XZ
= 2 212 16
(since XZ = 12 DC = 16)
= 20cmIn OXY , OXY = 90º
OY = 2 2OX XY
= 2 212 5 (XY = 12 BC = 4)
= 13 cmNow , since the base is a rectangle ,OBC = OAD and OAB = ODC
OBC =12 × BC × OZ =
12 × 10× 20 = 100cm2
OAB = 12 AB × OY =
12 × 32 ×13 = 208cm2
Slant surface of pyramid= 2 ( OBC+ OAB)= 2 (100 + 208) = 616 cm2
whole surface of pyramid= Slant surface + Base area= 616 + (32 × 10) = 936 cm2
3
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
14. (C) Let A = (x1, y1) = (t, t – 2)B = (x2, y2) = (t + 2, t + 2)C = (x3, y3) = (t + 3, t)
The vertices of the vertices of the giventriangle.
Area of ABC = 21
|x1(y2 – y3) + x2(y3 – y1)
+ x3(y1 – y2)|
Area of ABC = 21
|{t(t + 2 – t) + (t + 2)
(t – t + 2) + (t + 3)(t – 2 – t – 2)}|
Area of ABC = 21
[{2t + 2t + 4 – 4t – 12)}]
= |–4| = 4 sq. unitsClearly, area of ABC is independent of t.
15. (C) 22
11
nn
nn
ba
ba
is the AM between a & b.
2ba
= 22
11
nn
nn
ba
ba
for n = 0 LHS = RHS
Hence, n = 016. (D) Cannot be determined.17. (D) Cannot be determined.18. (B)
Only agriculture and commercial increasesby more than 50% during the same period.
19. (B) Electricity consumption in traction in
2001-2002 = 10013
× 100 = 13
Electricity consumption in traction in
2014-2015 = 1002
× 1300 = 26
% increase = 131326
× 100 = 100%
20. (B) Industrial + Agriculture will be more willbe more than 50%.
21. (A) 20%.
22. (B)
squaring both side squaring both side
2 2
2 2
a b c a b c
a c b a b c
a c ac b a b ab c
a b c ac a b c ab
2a b ca b c
=
222
abac
= bc
23. (A) (875321 × 875319 + a) is a perfect squareSo, (875320 + 1) (875320 – 1) + a
= (875320)2 – 1 + a
So, a – 1 = 0 a = 1
24. (B) sin3 (1 + cot ) + cos3 (1 + tan )
= sin3cos1sin
+ cos3
sin1cos
= sin2 (sin+ cos ) + cos2 (sin+ cos )= (sin + cos ) (sin2 + cos2 )= sin + cos
25. (B) Let speed of boat = x km/hrspeed of current = 4 km/hrA.T.Q.
104x +
104x =
8060
10(x + 4 + x – 4) = 43 (x2 – 16)
15x = x2 – 16 x2 – 15x – 16 = 0
So x = 16, –1So speed of boat in still water = 16 km/hr
26. (C) sin + sin = 2
So, = 90° and = 90°
So, cos 2
= cos 90°
= 027. (A)Rise in water level
= 50
100 ×
2 1000 1000100
100 10
= 10m
4
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
28. (A) A.T.Q. ABC DEF
So, ALDM
= ABDE
= 12
29. (D) p = cos x – sin x
q = 31 sin
1 sinx
x
= 1 + sin2x + sinx
r = 31 cos
1 cosxx
= 1 + cos2x – cosx
p + q + r = cosx – sinx + 1+ sin2x + 1 + cos2x – cosx
p + q + r = 3squaring both side(p + q + r)2 = 9
30. (D) Cost price = `2400
Selling price = 2400 × 120100 = ` 2880
So, marked price = 2880 × 10090 = ` 3200
If discount on selling price = ` 288Required difference= `(320 – 288)
= ` 3231. (C) A.T.Q.
Area of BDG = 16 × Area of ABC
= 24 cm2
32. (C) Ratio of their profit
= 81
: 31
:
31
811
= 81
: 31
: 2413
= 3 : 8 : 13Now, for A & C
A × 4 : C × 8 = 3 : 13A × 4 : 1560 × 8 = 3 : 13
A = 134381560
= ` 720
For B & C,B × 6 : C × 8 = 8 : 13B × 6 : 1560 × 8 = 8 : 13
B = 136881560
= ` 1280
Contribution by A & B together= 720 + 1280= ` 2000
33. (B) Let the present age of my son = x yrs.Then, the present age of mine = 3x yrs.
(3x + 5) = 2 21
(x + 5)
(3x + 5) × 2= 5(x + 5)6x + 10 = 5x + 25
x = 15 yrs.Father's age = 45 years
Son's age = 15 years
34. (B) Average area = n
n2222 ...321
= nnnn
6)12)(1(
= 6)12)(1( nn
=
312
21 nn
35. (A)
104 – 38 = 66 Students speak Marathi104 – 30 = 74 Students speak English104 – 40 = 64 Students speak HindiBy Venn diagram(n) + (44 – n + 50 – n + 40 – n)
+ (n – 18 + n – 20 + n – 26) = 104n = 34
No. of students who speak exactly twolanguage= 40 – n + 44 – n + 50 – n
= 134 – 3n= 134 – 3 × 34= 134 – 102= 32
5
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
36. (C) Let the Ist part be ` x.Then, 2nd part = ` (4350 – x)
SI1 = SI2
10019 x
= 100210)4350( x
9x + 20x = 4350 × 20
x = 29204350
= ` 3000Ist part = ` 30002nd part = ̀ 1350
37. (A) Amount = `400 5 2
400100
= ` 440
Amount returned by Ramu to Arun= 2% of 440
= 2 440
100
= ` 8.838. (C) Radius of the semicircle AQC
= 2 21 14 142
= 7 2 cmArea of the shaded region= Area of semi-circle AQC
+ Area of triangle ABC– Area of Quarter circle APCB
=12 ×
722
× 27 × 27 + 21
× 14 × 14
– 41
× 722
× 14 × 14=154 + 98 – 154= 98 cm2
39. (C) xax + yb
y + zc
z
=axa
xa2.
+ byb
yb2.
+ czc
zc2.
= axczbyax
+ byaxczby
+ czbyaxcz
= czbyaxczbyax
= 140. (A) h = 45 cm, r1 = 28 cm and r2 = 7 cm
Volume of the bunket
= 13 (r1
2 + r1r2+ r22)×h
= 13 ×
227
(282 + 28 × 7 + 72) × 45
= 13 ×
227
(784 + 196 + 49) × 45
= 13 ×
227
×1029 × 45= 48510 cm3
41. (D) Total change = 25 + 10 + 1001025
= 37.5%
So, revenue increased by 37.5%.
42. (C)54
11 13 94
= 18 [On solving]
Time taken in completing 18 th = 10 mins
Time taken in completing 35 =10 × 8 ×
35
= 48 mins
43. (B) Time 1
cross-sectional area of plpe
Time 2
1
4d
Time 21
d
2
1
tt =
21
22
( )( )dd
t2 + t1
21
2
dd
t1 = 40 mins, d1 = d, d2 = 2d
t2 = 40 2
2dd
t2 = 40 21
2
t2 = 10 mins
44. (B) 1hr 40 min 48 sec = 1hr 484060
min
= 1hr 4405
min = 1hr 2045 min
= 2041300
hr = 504300
speed = 42
504300
= 25 kmph
57
× usual speed = 25
Usual speed = 25 7
5
= 35 kmph
6
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
45. (B) Let speed of the stream = x km/hrATQ,
Related Speed = (12 + x) + (15 – x)= 27 km/hr
Required time = 27108
= 4 km/hr
46.(C) Surface area of one cube = 6a2 when 6cubes are fixed on the 6 faces of a cubethen only 5 faces of a cube are visible ofeach cube. Since, central cube iscompletely covered. So, the only 6 cubesare visible each with 5 faces. Hence, thetotal surface area of this solid = 5a2 × 6
Required Ratio = 2
25 6
6a
a
=51
= 5 : 1
47.(B) 2p + 1p = 4
p + 1
2p = 2
3
12
pp
= p3 + 3
18p + 3p
12p
12
pp
8 = p3 + 31
8p + 32 × 2
p3 + 31
8p = 8 – 3
= 548.(B) Let the each side of cube be a
CD = 2 a
CQ = 2a
Let the radius of cone be x and height beh, then r = h 2In APO and CQO ( triangle)
APAO =
CQOQ =
rh
= / 2
( )ah a
/ 2
( )ah a = 2
a = 2 (h – 2)
h = 32a
r = 23a
× 2 and h = 32a
Volume of cone = 13
23 2
2a
× 32a
= 394
a
and volume of cube = a3
Required Ratio = 394
a : a3
= 94 :1
= 2.25 :149.(C) x2 + y2 – 4x – 4y + 8 = 0
x2 – 4x + 4 + y2 – 4y + 4 = 0 (x – 2)2 + (y – 2)2 = 0 x = 2 and y = 2 x – y = 2 – 2
= 0
50. (C) Volume of new ball = 3 44 3 (r1
3 + r23 + r3
3)
= (13 + 23 + 33)= 36 cm3
ATQ,43 r3 = 36
r3 = 27
r = 3 27 = 3 cm.
51. (A) Average of the batsman after 12th innings= 63 – 11 × 2= 41
52. (C) BDE = 115°,ADF = 65° and AED = 75°AED ~ ABC
DE AE ADBC AB AC
2 103 AB AB = 15 cm.
7
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
53. (D) Go through options considering option (D)No. of sides = 5 : 10Ext. angle = 72° : 36°Int. angle = 180° – 72° : 180° – 36°
= 108° : 144°= 3 : 4
54. (A)
Now, in ABD,
tan 60° =hy 3 =
hy ...(i)
In ABC,
tan 30° =h
x y13
...(ii)
From eqn. (i) and (ii), we get
yx y
3 13
x = 2y y =
x2
Required time = 210
= 5 minutes
55. (C)
Required ratio = 8 : 7
56. (B) Required percentage = 20
100 20 × 100
=20
120 × 100
= 50 216 %3 3
57. (A) Let the CP of 1 orange = ` 1 SP of 10 oranges = `13
gain % = 13 10
10
× 100 = 30%
58. (B) Let the Principal is ` x
A = PTR
1100
8x = x3R
1100
23 =3R1
100
24 = 4R1
100
Required time = 4 years.
59. (C) Ratio = 1 : 1 1:3 6
= 6 : 2 : 1
Sum of the ratios = 6 + 2 + 1 = 9
Middle part = 29
× 78 = 523
= 1713
60. (B) Women = 8343
× 311250 = 161250
Men = 311250 – 161250 = 150000Total number of literature person
= 161250 × 1008
+ 150000 × 10024
= 4890061. (C) Amount Left = ` 1400
Amount before gift = `(1400 + 120)= ̀ 1520
Amount spent on transport= 951520
× 5
= ` 8062. (A) If distance between stations be x km.
Then, speed of train =
6045x
= 34
x km/hr
ATQ,
534
xx
= 6048
1543
xx
= 54
15x = 16x – 60 x = 60 km
63. (C) C.P. of article = ` x
100108x
– 10092x
= 28
10016x
= 28 x = 1610028
= ` 175
64. (A)
OE AB and OF CDAE = EB = 5 cmCF = FD = 12 cmAO = OC = 13 cm
From AOE
OE = 22 513 = 12 cmFrom CDF,
OF = 22 1213 = 5 cm
EF = OE + OF = 17 cm
8
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
65. (D)
CBE = 50ºBAC + BCA = 90ºABE = 90º – 50º = 40º
ABE = ACE = 40º [ angle on same chord]ACE = DEC = 40º [ opposite angle]
66. (C)
AB = AC = tangents from the same point.OB = OC = 3 cmOA = 12 cmABO = 90º
AB = 22 312 = 153
OAB = 21
× 3 × 153 = 2159
cm2
So, area of ABOC = 159 cm2
67. (C) 1º360
n – 2º360
n = 6º
360
)2)(1(12
nnnn
= 6
(n – 1)(n + 2) = 180n2 + n – 182 = 0
(n + 14)(n – 13) = 0n = 13
68. (D) Area of canvas cloth required
= rl = 722
× 2.5 × 314
= 3110
m2
Length of canvas cloth
= 3110
× 25.11
= 388
m
Cost of canvas for tent = 388
× 33 = ` 968
69.(B) l = 15 +15 = 30 cmb = 15, h = 15
Total Surface Area = 2(lb + bh + hl )= 2[30 × 15 + 15 × 15 +
30 × 15]= 2(450 + 225 + 450)= 2250 cm2
70. (D)
Area of equilateral triangle = 234
a
16 3 = 34
a2 a = 8 cm
Height of the equilateral triangle
= 3 82
= 4 3 cm
Radius of the circumcircle
= 23 × 4 3 =
83
cm
Area of the circle = × 83 ×
83
= 643
= 1213 cm2
71. (C) Area of square = 121 m2
Perimeter of square = 1214 m
= 44 mSo, perimeter of circle = 44 m
Area of circle = 4
)2( 2r
= 22474444
= 154 m2
9
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
72. (A)
Area of 4 isosceles triangle = 400 m2
Area of an isoscles triangle = 100 m2
ATQ, 21
× AP × AS = 100
AP = 210 [AP = AS]
Length of greater square = 220 m
Length of new square = 2
220
= 20 mArea of new circle= 400 m2
73. (D) Price of tea in 2014 = 280680
lakh/tonne
Price of tea in 2015 = 200500
lakh/tonne
= 2.5 lakh/tonne
Required percentage =
280600
2806005.2
× 100
= 600100
× 100
= 16 32
%
74. (B) Price of tea in 2013 = 240600
lakh/tonne
= 2.5 lakh/tonne Value of quantity will be 2010 = (250) × 2.5
= 625 lakhs
75. (C) Price of tea in 2010 = 200520
× 100100000
= ` 260076. (C) In 2010 price was maximum
i.e. ` 2600/kg77. (D) L.C.M. of 15, 20, 36 and 48 = 720
Required number = 720 + 3 = 723
78. (A)12a b b a
At a = 64 and b = 289
=1264 289 289 64
=128 17 17 8
=125 3 =
122
79. (A) L.C.M. = (x2 + 6x + 8)(x + 1)= (x + 1)(x + 2)(x + 4)
H.C.F. = (x + 1)First expression= x2 + 3x + 2
= (x + 1)(x + 2)ATQ,(second expression) × (x + 1)(x + 2)
= (x + 1)(x + 1)(x + 2)(x + 4)Second expression = (x + 1)(x + 1)
= x2 + 5x + 480. (B) Rate = 10%
Time = 2 years
C.I. = P 1100
tr
– P
525 = P2101
100
– P
P =525 100
21
= ` 2500
S.I.=
12500 (2 2) 102
100
= ` 50081. (C) Side of equilateral triangle = a
Height of prism = hRequired ratio= Volume of prism : Volume of new prizm
= 234
a
× h : 232
4a
× 2h
= 1 : 2
82. (B) Work done by A and B =35
Work done by C = 1 – 35 =
25
Amount paid to C = 30,000 × 25
= `12,000
10
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
83. (C) (x + 2)2 = 9 (y + 3)2 = 25(x + 2)2 = (±3)2 (y + 3)2 = (±5)2
x = 1, – 5 y = 2, – 8
Possible value of xy
are 1 1 5 5, , ,2 8 2 8
Maximum Value of xy
= 58
84. (B)
ATQ,
PR = 2 2PQ QR = 2 25 12 = 13 cm
Length of median in right angle triangle
= PR2 =
132 cm
OQ = 23 × OM =
132 ×
23 = 4
13
85. (B)
In PQSQPS = 180° – 90° – 60°
= 30°In PQR, if C is circumcentre than
QPR =12 × QCR = 65°
RPS = QPR – QPS= 35°
86. (C) Given, 4096 = 64
40.96 + 0.4096 + 0.004096
+ 0.00004096= 6.4 + 0.64 + 0.064 + 0.0064= 7.1104
87. (C) Expression
=
.....100 2 100 200100 1
100 99 98 ..... 3 2 1
=
(100 1).....(100 100)....(100 200)
100 99 98 .....3 2 1= 0
88. (C)
Required ratio = 4 : 389. (B) Let expenditure per student =` x
Total expenditure = ` yATQ,
40 × x = y ...(i)(40 + 8)(x – 2) = y + 4840x + 8x – 96 = y + 48
y + 8x – 96 = y + 488x = 144 x = 18
So, original expenditure of mess =40 × 18 = ` 720
90. (B) Cost price of land = ` 96,000
Loss on 25
th land = `2 6960005 100
= ` 2304
Overall profit = `1096000100
= ` 9600Required profit percentage
=9600 2304
3960005
× 100
= 2023 %
Short Cut :
11
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
91. (C) Income = 9520 × 100 100 10080 85 70
= ̀ 20000
92. (A)
So, units of work done by A and C in a day= 4 – 1 = 3
Time required by A and C = 243 = 8 days
93. (B)1
1x +
22y +
10091009x
= 1 given
1x
x + 2
yy + 1009
zz
= 11
1x
+ 21
2y
+ 10091
1009z
= 1 + 1 + 1 – 1 2 1009
1 2 1009x y z
= 3 – 1 = 2
94. (A) sin = 2 2
2 2
m nm n
BC = 2 2 2 2 2 2( ) ( )m n m n = 2mn
cos = 2 2
2mnm n
95. (C)2sin68 2cot15cos22 5 tan75
–
3 tan20 .tan 40 tan 45 .tan50 .tan705
= 2cos22 2tan75cos22 5 tan75
–
3.tan20 .cot20tan40 .cot 40
5
= 2 – 25 –
35 =
10 2 35
= 1
96. (B)
AB = 108 m, CD = x m
From ABC, tan 60° = ABBC
3 = 108BC BC =
1083 = 36 3 m
From AED, tan 30° = AEED
AE = 13 × 36 3 = 36 m
Height of pole (x) = 108 – 36 = 72 m
97. (B)
AB = temple = 54 mCD = temple = h mBC = width of river = x m
From ABC,
tan 60° =ABBC 3 =
54x
x =543 = 18 3 m
From ADE,
tan 30° =AEDE
13 =
5418 3
h
54 – h = 18h = 54 – 18 = 36 m
98. (A) Greatest number that will divide 148, 246and 623 leaving remainders 4, 6 and 11respectively= H.C.F. of (148 – 4),(246 – 6) and (623 – 11)= H.C.F. of 144, 240 and 612= 12
Required number = 12
12
==================================================================================
Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD
1. (C)2. (C)3. (B)4. (B)5. (C)6. (A)7. (C)8. (D)9. (A)10. (A)
11. (B)12. (A)13. (B)14. (C)15. (C)16. (D)17. (D)18. (B)19. (B)20. (B)
31. (C)32. (C)33. (B)34. (B)35. (A)36. (C)37. (A)38. (C)39. (C)40. (A)
SSC MAINS MATHS - 53 (ANSWER KEY)
41. (D)42. (C)43. (B)44. (B)45. (B)46. (C)47. (B)48. (B)49. (C)50. (C)
21. (A)22. (B)23. (A)24. (B)25. (B)26. (C)27. (A)28. (A)29. (D)30. (D)
51. (A)52. (C)53. (D)54. (A)55. (C)56. (B)57. (A)58. (B)59. (C)60. (B)
Note:- 1. For any issue related to RESULT Processing, kindly contact uson 8287200200.
2. If your opinion differs regarding any answer, please message themock test and question number to 8287500500.
99. (B) ATQ,15 th of the full tank + 22 litres =
34
th of full tank
3 14 5
th of the full tank = 22 litres
1120
th of the full tank = 22 litres
Capacity of the full tank =202211
litres
= 40 litres
100.(D) Let cost price = ` 100Profit earned = 26%
Selling price =126100100
= ` 126
Discount = 10%
Marked price = `10012690
= ` 140
Marked price is more then cost price by
= 140 100
100
=40%
61. (C)62. (A)63. (C)64. (A)65. (D)66. (C)67. (C)68. (D)69. (B)70. (D)
71. (C)72. (A)73. (D)74. (B)75. (C)76. (C)77. (D)78. (A)79. (A)80. (B)
81. (C)82. (B)83. (C)84. (B)85. (B)86. (C)87. (C)88. (C)89. (B)90. (B)
91. (C)92. (A)93. (B)94. (A)95. (C)96. (B)97. (B)98. (A)99. (B)100. (D)