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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
Ph: 011-27607854, (M) 8860-333-333 1
1. (D) Since, divisor =a
Therefore, quotient =4
a a) b (a/4 a 2
and remainder =2
a
b =a4
a+
2
a
b =4
22 aa
( + 2)a ab
= 4
2. (C) Since, 738A6A is divisible by 11,
therefore
Sum of odd places Sum of even places
= either multiple of 11 or 0
(7 + 9 + 6) (3 + A + A) = either multiples
of 11 or 0.
21 2A + 3 = either multiple of 11 or 0
18 2A = 0 (it cannot be a multiple
of 11)
2A = 18
A = 93. (C) Let the numbers beaandb.
Then, ab = 1575
andb
a=
7
9
a =7
9b
Putting this value of Eq. (i), we get
7
9bb = 1575
b2 = 1575 9
7
b2 = 175 7
= 52 72
b = 35
a =7
9 35
= 45
Thus, sum of the numbers
=a+b
= 45 + 35
= 80
SSC Mains Test- 21 (SOLUTION)
4. (C)3)81()9()81(
2.4
7.6.3
= 18.16
4.54.14
)3()3(
)3()3(
= (3)14.4 + 5.4 16.8 1
= (3)2= 9
5. (D) ....666
Factorize 6 as the multiplication of twoconsecutive natural numbers. The greaterone will be the answer.
As, 6 = 2 3
Therfore, ....666 = 3
6. (D) Let the two consecutive odd numbers bexand (x+ 2).
Then, according to the question,(x)2+ (x+ 2)2 = 394
x2+x2+ 4 + 4x = 394
2x2+ 4x = 390
x2+ 2x = 195
x(x+ 2) = 195 = 13 15
x = 13
Thus, sum of the numbers=x+ (x+ 2)= 13 + 15= 28
7. (C) We know that, when (a 1)nis dividedbya, then remainder = (1)n
Now, 6767+ 67= (68 1)67+ 67
When (68 1)67is divided by 68, thenremainder = (1)67= 1
Thus, when 6767+ 67 is divided by 68,then remainder = 1 + 67 = 66
8. (A) Let the work is completed inndays.
Then,246 +
526n +
64n = 1
52
6n+
64
n= 1
4
1
16134
13)6(16
nn=
4
3
16n 96 + 13n= 3 13 16
29n= 624 + 96
29n= 720
n=
29
720
= 25 days
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
9. (B) ]1)75.0(75.0[)75.01(
)75.0( 23
=25.0
]1)75.0(75.0)[75.01()75.0( 23
=25.0
])75.0(1[)75.0( 333
=25.0
)75.0(1)75.0( 33
=25.0
1=
25
100= 4 = 2
10. (B) 4096 + 96.40 + 004096.0= 64 + 6.4 + 0.064 = 70.464
11. (C) Required number
= 3012 3011 (3011)2
= 3011(3012 3011)
= 3011
12. (D) (P + Q)s work =23
19... (i)
(Q + R)s work =23
8... (ii)
(P + Q + R)s work= 1 ... (iii)
Subtracting Eq. (i) from Eq. (iii), we get
Rs work = 1 23
19=
23
4
Subtracting Eq. (ii) and from Eq. (iii), we get
Ps work = 1 23
8=
23
15
Now Qs work = 1 (P + R)s work
= 1
23
15
23
4
= 1 23
19= 23
4
Qs share =23
4 5750
= 4 250
= 1000
13. (C) Let CP = x.
Then,
SP = 70% ofx= 0.7x
60% of MP = 0.7x
MP =60
7.0 x 100 =
6
7x
If article is sold at the marked price, then
SP =6
7x
Required profit per cent
=x
xx6
7
100%
=6
100% = 16
3
2%
14. (A) A ( )
B ( )
C ( )
3
4
1
4
3
12
5
12
Cistern fill till 5clock = 4 2 + 3 = 11
Cistern will empty in =5
11= 2 hr 12 min
or at 7 : 12 pm15. (D) Required number of days
= 164 = 8 days
16. (C) Here, M1= 250, D
1= 20, T
1= 5
M2= ?, D
2= 10, T
2= 8
Using M1D
1T
1= M
2D
2T
2, we have,
250 20 5 = M2 10 8
M2
=810
520250
= 312.5= 313
17. (A) 2m+ 5w =12
1... (i)
5m+ 2w =9
1... (ii)
Multiplying Eq. (i) by 5 and Eq. (ii) by 2and then subtracting it from Eq. (i), we get
10 + 25 =
10 + 4 =
21 =
m w
m w
w
512
512
2
9 29
=36
7
3w =36
1
3 womens 1 days work = 36
3 women can complete the work in 36days.18. (A) Let CP of the article = 100
Then SP = 125% of 100
= `
125
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
Also,4
3th of MP = SP
4
3 MP = 125
MP = 125 3
4= `
3
500
Required ratio =3
500: 100 = 5 : 3
19. (A) Let Earnings of A = 2xand Earnings of B =xExpenditure of A = 5yExpenditure of B = 3yFurther, suppose monthly savings of A = 4zAnd monthly savings of B = z
Also, total monthly savings of A and B= 5000
4z+z = 5000
5z = 5000
z = 1000Now using,
Income Expenditure = Savings2x 5y = 4000 ... (i)
and x 3y = 1000 ... (ii)Multiplying Eq. (ii) by 2,
2x 6y = 2000 ... (iii)Subracting EQ. (iii) from Eq. (i), we get
5y+ 6y= 2000
y= 2000Putting this value of Eq. (ii), we get
x= 1000 + 6000
= 7000Thus, monthly salary of B = `7000
20. (B) Let the greater and smaller beaandb,
respectively.
ba
ba
=
1
5
Applying componendo and dividendorule, we get
)()()()(
babababa
=
1515
b
a
2
2=
4
6
a:b= 3 : 221. (B) Required equivalent discount
100
50
100
80
100
901 100 = 64%
22. (D) I. Equivalent discount
=
100
10101010 %
= (20 1)%= 19%
II. Equivalent discount
=
100
812
812 %
= (20 0.96)%= 19.04%
III.Equivalent discount
=
100
515515 %
= (20 0.75)%= 19.25%
IV.Discount = 20%Selling price will be minimum wherediscount is maximumi.e., in IVcondition.
23. (D) In the mixture, acid = 80% =5
4
and water = 20% =5
1
Letxth part of the mixture be removed
and replaced by water.
Then,
xx
x
5
1
5
15
4
5
4
=3
4
xx
x51
44
=34
x
x
41
44
=
3
4
12 12x = 4 + 16x
8 = 28x
x=28
8=
7
2th
24. (B) Ratio of reduction is number of employees
= 9 : 8Ratio of increment in wages = 14 : 15
Now, ratio of reduction in the wage bill= 9 14 : 8 15= 126 : 120= 21 : 20
25. (D) Let maximum score of the cricketer =xThen, his minimum score =x 172Now, total score in 40 innings
= 40 50 = 2000
And, total scores in 38 innings
= 38 48 = 1824
Sum of remaining two innings= 2000 1824 = 176
Therefore, sum of maximum and minimum
scores = 176
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
x+x 172 = 176
2x = 348
x = 174
Thus, highest score of the cricketer is 174.
26. (C) Gold Copper
Alloy A9
7
9
2
AlloyB18
7
18
11
Since, alloys A and B are melted in the
ratio 1 : 1 to make the alloy, therefore
in the alloy C, the ratio of gold and copper.
2
1
18
7
2
1
9
7:
2
1
18
11
2
1
9
2
=
18
7
9
7:
18
11
9
2
=18
21:18
15= 21 : 15
= 7 : 5
27. (C) Acid Water
Bottle I7
2
7
5
Bottle II
10
7
10
3
Let the required ratio be x : 1. Then,
according to the question,
10
7
7
2x :
10
3
7
5x = 2 : 3
2 7
7 1 0
5 3
7 1 0
x
x
+
+
=3
2
7
6
x+ 10
21
= 7
10
x+ 10
6
7
4x=
10
15
x=10
15
4
7=
8
21
Hence, required ratio =x: 1 = 21 : 8
28. (A) Let third number =x
Then, second number = 2x
Then, the first number = 4x
Now, average of these three numbers = 154
x+ 2x+ 4x = 3 154
7x = 3 154
x = 3 22= 66
Thus, first number = 4 66 = 264
29. (D) Then, number of rest = (500 x)
500 5000 =x 14000 + (500 x) 4000
2500 = 14x+ 4(500 x)
2500 = 14x+ 2000 4x
500 = 10x
x =10
500= 50
Thus, number of officers is 50.
30. (D) Total marks of 40 students.
= 40 72 = 2880
Rectified total marks
= 2880 68 73 + 64 + 62 + 84
= 2884
Thus, rectified average =40
2884= 72.1
31. (B) Let first number =x
Second number = 3x
Third number =4
3x
x+ 3x+4
3x = 3 114
4
34 x = 342
x = 342 19
4
= 72
Thus, largest number = 3x
= 3 72
= 216
32. (D) Time taken to cover5
1of the distance
= 5
1
8
1
= 40
1
Time taken to cover10
1of the disteance
=10
1
25
1=
250
1
Time taken to cover rest of the distance
=
10
1
5
11
20
1
= 10
7
20
1
= 200
7
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
Therefore, total time taken
=40
1+
250
1+
200
1
=100
35425
=1000
64
Thus, average speed =
=
1000
64
1=
64
1000
= 15.625 km/h
33. (C) Letxmarbles must be added.
Then,x
x
40
10 100 = 60
x
x
40
10=
5
3
50 + 5x = 120 + 3x
2x = 70
x = 35
34. (A) Let the number bex.
Then according to the question,
x (25% ofx) =x+ (200% ofx)
x4
x= 3x
x2 = 12x
x = 12
35. (B) Let the value of the article 3 years ago
bex.
Then, 729 =x
3
100
101
729 =x
3
10
9
729 =x1000
729
x = 1000
36. (B) Required percentage decrease
=50100
50
100%
=150
50 100%
= 3331 %
37. (B) Let the required distance bedkm.
Then,3
d
4
d=
60
10
12
)34( d =61
d = 2 km
38. (A) CP per dozen =2
3040= 35
SP per dozen = 45
Profit per dozen = 45 35 = 10
But total profit = 480
Thus, required number of dozen
=10
480= 48
39. (D) Let CP of first chair = x
and SP of second chair = (900 x)
SP of first chair =5
4x
SP of second chair =4
5(900 x)
We know that,
SP CP = Profit
5
4x+
4
5(900 x) 900 = 90
20
)900(2516 xx = 990
16 +22500 25
20
x x= 990
9x+ 22500 = 19800
9x= 2700
x= 300
Thus, the cost of lowest price chiar is` 300.
40. (B) SP of 100 oranges CP of 100 oranges
= SP of 20 oranges
SP of 80 oranges = CP of 100 orangesLet CP of 1 orange = 1
CP of 100 oranges = 100SP of 80 oranges = 100
SP of 1 oranges = `80
100= `
4
5
Profit percent =
1
4
5 1 100%
=
4
1 100%
= 25%
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
41. (C) 60% of CP = 50% of SP
SP
CP=
6
5
Let CP = 5xand SP = 6xThen, there is a profit.
And profit per cent =x
xx
5
56 100%
= 20%42. (D) CP of two horses
= 40000 + 40000= 80000
SP of one horse = 115% of 40000= 46000
Let the second horse was sold at a lossofx%.
Then, SP of second horse
= (100 x)% of 40000= 40000 400x
Total SP = 46000 + 40000 400x= 86000 400x
Now, using CP SP = loss, we have80000 (86000 400x) = 3600
6000 + 400x = 3600
400x = 9600
x = 24Thus, SP of second horse
= (100 24)% of 40000= 76% of 40000 = 30400
43. (D) CP of 1 guava = ` x
y
SP of 1 guava = ` y
x
x>y, therefore SP > CP and hence thefruit seller will have a gain.
Gain per cent =CP
CPSP 100
=
x
y
x
y
y
x
100%
=xy
yx 22
y
x 100%
= 2
22
y
yx %
44. (B) According to the question,
100
max =
100
22 may
x =y am
x:y = am: 1
45. (D) Let A took xfrom B and `(1200 x)from C.
Then,
100
114x
+ 100
115)1200( x
= 172
14x+ 18000 15x= 17200
x= 800Thus, A borrowed 800 from B.
46. (B) Given,
n
n )42( 90 =
5
3 2 90
n
n )42( =
5
3 2
n
n 2=
5
3
5n 10 =5
3
2n= 10
n= 5Thus, the polygon has 5 sides.
47. (C) Side of given square = 200 m
= 210 m
Its diagonal = 210 m 2 = 20 m
which is the side of new square
Area of new square = (20)2
= 400 sq m48. (C) Let radii of cone, cylinder and hemisphereare 2r, 3r and r respectively.Volume of cone : Volume of cylinder :Volume of hemisphere
=3
1 (2r)2h : (3r)2h :
3
2 r3
=3
1 4h : 9h :
3
2r
(for hemisphere r = h)
=
3
4h : 9h :
3
2h
= 4 : 27 : 249. (A) Let rate of flow of river beukm/h.
Then,u10
91+
u10
91= 20
2100
1010
u
uu
=
91
20
2100
20
u=
91
20
100 u2 = 91
u2 = 9
u = 3 km/hr
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
50. (B) Let the original speed and time be S andT respectively.
Then, S T =4
3S
60
20T
3
4T =
3
1T
3
4T T =
3
1
3
T=
3
1
T = 1 hr = 60 minutes51. (C) Let speed of boat in still water
=ukm/hAnd speed of stream =vkm/hr
Then,vu
25+
vu
39= 8 ... (i)
andvu
35+
vu
52= 11 ... (ii)
Multiplying Eq. (i) by 4 and Eq. (ii) by 3and subtracting Eq. (ii) from Eq. (i), we get
vu
100
vu
105= 32 33
vu
5= 1
uv= 5 ... (iii)Now, from Eq. (i),
5
25+
vu
39= 8
vu
39= 8 5 = 3
u+v = 13 ... (iv)Now, subracting Eq. (iv) from Eq. (iii), we get
(uv) (u+v)= 5 13
2v= 8
v= 4 km/h
52. (C)Sum of money at compound interestbecomes 2 times in 5 years.
It will become 8 timesi.e.(2)3timesin 5 3 = 15 years.
53. (B) Let the share of elder and younger sonsbe `xand (120000 x) respectively.Amount got by elder son = Amount got
by youngerson
x+100
45 x
= (120000 x) +100
65)120000( x
x+5
x= (120000 x) +
10
3)120000( x
5
11x
10 = 120000 10 + 360000 3x
22x= 1200000 + 360000 3x
25x= 1560000
x= 62400
Thus, yonger sons share
= 120000 62400
= 57600
54. (C)a2+b2+c2= 2(abc) 3
a2+b2+c2 2a+ 2b+ 2c+ 1 + 1 + 1 = 0
(a2 2a+ 1) + (b2+ 2b+ 1) + (c2+ 2c+ 1) = 0
(a1)2+ (b+ 1)2+ (c+ 1)2= 0
a 1 = 0
a= 1
and b+ 1 = 0 b= 1
and c+ 1 = 0 c= 1
Thus, ab+c= 1 + 1 1 = 1
55. (A)x2+ 3x+ 1 = 0
x+ 3 +x
1= 0 [Dividing by x]
x+x
1= 3
On cubing both sides, we have
3
1
x
x = (3)3
x3+ 31
x+
xx
1= 27
x3+ 31
x+ 3(3) = 27
x3+ 31
x= 27 + 9 = 18
56. (D) xa.xb.xc = 1
xa+b+c = 1
xa+b+c =x0
a+b+c = 0
a3+b3+c3= 3abc(whena+b+c= 0, thena3+b3+c3= 3abc)
57. (C) Volume of pyramid = 1728
3
1 area of base height = 1728
3
1 24 24 height = 1728
height = 2424
31728
= 9 m
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
58. (B) Here,AOBand AOB are similar.Let OO=hand OB= r
AO'
AO
= B'A'
OB
h9
9=
r
3
3r = 9 h
h = 9 3rNow, volume of frustum = 44
3
1h(R2+r2+ Rr) = 44
3
1
7
22 (9 3r)(9 +r2+ 3r) = 44
(3 r)(9 +r2+ 3r) = 2 7
33r3= 14
r3= 27 14
r3= 13
r= 313 cm
59. (A) Let radii of first and second cylinders be2r and 3r, respectively and their heightsbe 5h and 4h respectively.
Then, ratio of their curved surface area2 (2r) 5h : 2 (3r) 4h
= 10 : 12= 5 : 6
60. (C) Curved surface area
=3
1 Total surface area
=3
1 462
= 154 sq cmNow, total surface area = 462 sq cm
2rh+ 2 r2= 462
154 + 2r2 = 462
2r2 = 308
r2 = 3087
222
r2 = 49
r = 7Again, 2rh= 154
h = r2
154
=7222
7154
=2
7cm
Thus, volume of cylinder
= r2h
=7
22 49
2
7
= 11 49
= 539 cm3
61. (D)Curved surface area of cylinder
Curved surface area of cone=
5
8
l
rh
2
2=
5
8
22 rh
h
=
5
4
22
2
rh
h
=
25
16
2
22
h
rh =
16
25
1 + 2
2
h
r=
16
25
2
2
h
r=
16
9
h
r=
4
3
r:h= 3 : 4
62. (D)
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
Surface area of conical part
= Surface area of hemisphere
22 rhr = 2r2
22 rh = 2r
h2+r2 = 4r2
h2 = 3r2
2
2
h
r=
3
1
h
r=
3
1
r:h = 1 : 3
63. (D) Volume of prism = Area of base height
3108 =4
3 (6)2 Height
Height =36
408
= 12 cm
64. (B) a+a
1+ 2 = 0
a2+ 1 + 2a= 0
(a+ 1)2 = 0
a= 1
a37 1001
a= 1
1
1
= 1 1 = 2
65. (A) Given lines
(k 1)x+y 2 = 0 ... (i)
and (2 k)x 3y + 1 = 0 ... (ii)
Since, the lines (i) and (ii) are parallel.
k
k
2
1=
3
1
3k 3 = 2 +k
2k = 1
k =2
1
66. (C)
Let height =h
Then , length of shadow =3
h
In the figure,
tan =
3
h
h
tan = 3
= 60
67. (D)
Given lines are x+ 2y = 3 ... (i)
and 3x 2y = 1 ... (ii)Put x = 0
Then, from Eq. (i)y=2
3
from Eq. (ii)y= 2
1
Distance between both the points
=2
3
2
1=
2
4= 2 units
68. (B) x+x16
1= 1
4x+x4
1= 4
Cubing both sides, we get
64x3+ 364
1
x+ 3.4x.
x4
1
xx
4
14 = 64
64x3+ 364
1
x+
xx
4
143 = 64
64x3+ 364
1
x= 64 3 4
4
4
14
xx
= 64 12
= 52
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Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
69. (B) a+b+c= 0
a+c= bOn squaring both sides,
(a+c)2 = (b)2
a2
+b2
+ 2ac=b2
a2+c2=b2 2ac
Now,cab
acba
2
22 2
=acb
acbb
2
22 2
=acb
acb
2
2 )(2= 2
70. (D) Given equations area4+a2b2+b4 = 8 ... (i)
a2
+ab+b2
= 4 ... (ii)Squaring Eq. (ii), we get(a2+ab+b2)2= (4)2
a4+a2b2+b4+ 2a3b+ 2a2b2+ 2ab3 = 16
8 + 2ab(a2+b2+ab) = 16[Using eq. (i)]
2ab(a2+b2+ab) = 16[Using eq. (ii)]
2ab 4 = 8 [using Eq. (ii)]
ab = 171. (D) We know the formula
a3+b3+c3 3abc= (a+ b+ c)(a2+ b2+ c2abbcca)
=2
1(a+b+c)(2a2+ 2b2+2c2 2ab 2bc2ca)
=2
1(a+b+c){(ab)2+ (bc)2+ (c+a)2}
Now, 222
333
)()()(
3
accbba
abccba
=2
1
})()(){(
})()()){((222
222
accbba
accbbacba
= 2
cba
=2
101525 =
2
30= 15
72. (D) InACT,
44
A
B
C
O
T40
ACB = 180 (CAT +ATC)= 180 (44 + 40)= 96
ACB = 180 ACT
= 180 96= 84Also,
ABC =CAT= 44
InABC,
BAC = 180 (ABC +ACB)= 180 (44 + 84)= 180 128= 52
BOC = 2BAC= 2 52= 104
73. (C)
O
P M5 5
1 2 c m
In OPM,OP2 = PM2+ OM2
= 25 + 144= 169
OP = 13
Diameter of circle= 2 OP= 2 13 = 26 cm
74. (C) Since, P, S, Q are the mid-points of ABand BC, therefore
A
P Q
B C
1 R 2
AB
AP=
BC
PQ=
2
1... (i)
Now,RQ
PR=
2
1
RQ
2=
2
1
RQ = 4
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==================================================================================
Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
PQ = 2 + 4 = 6 cmTherefore, from Eq. (i),
BC
6=
2
1
BC = 12 cm75. (A) Since. O is the circumference ofABC,
thereforeA
B C
O
35
OA = OB = OC
In BOC,OB = OC
OCB =OBC = 35
BOC = 180 (OCB +OBC)= 180 70= 110
BAC =2
1BOC =
2
1 100 = 55
76. (C) InBIC,A
B C
I
135
BIC = 135
2
1(A + C) = 180 135
2
1(B + C) = 45
B +C = 90
A = 90ABC is a right angled triangle.
77. (C) sin2 + sin2 = 2
sin = sin = 1 (sin < 1)
= = 90
cos
2
= cos 90 = 0
78. (D) cot20
cot
20
3cot
20
5cot
20
7cot
20
9
= cot 20
cot 20
3cot 4
cot
20
3
2
cot
202
= cot20
cot
20
3cot
4
tan
20
3tan
20
= 1
79. (D) sin + cos = 13
17
(sin + cos )2 = 2
2
)13(
)17(
sin2 + cos2 + 2sin cos =169
289...(i)
1 + 2sin cos =169
289
2sin cos =169
289 1 =
169
120... (ii)
Again, from Eq. (i),
sin2 + cos2 2sin cos + 4sin cos
=169
289
(sin cos )2+ 4sin cos =169
289
(sin cos )2+ 2 169
120=
169
289
[using eq, (ii)]
(sin cos )2= 169
289 169
240= 169
49
sin cos =13
7
80. (C) tan .tan2 = 1
tan .
2tan1
tan2
= 1
2tan2 = 1 tan2
3tan2 = 1
tan = 3
1
= 30
Now, sin2 2 + tan2
= (sin 60)2+ (tan 60)2
=
2
2
3
+ 2)3(
=4
3+ 3
= 4
123
= 4
15
= 3 4
3
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==================================================================================
Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
81. (C) Sales in 2008 = 5 croreSales in 2006 = 6 crore
Required percentage decrease
= 6
56
100%
=6
100% =16
3
2%
82. (B) Sales in 2002 = 2 croreSales in 2007 = 6 crore
Required ratio = 2 : 6 = 1 : 383. (A) Average sale of company
=Total sales of the company
Number of years
=5
661043
=5
29= 5.8 crore
84. (D) Sales in 2005 = 10 croreSales in 2004 = 4 crore
Required percentage income
=5
410 100%
=4
6 100% = 150%
85. (B) Total sales of company from 2005 to 2008= ` (10 + 6 + 6 + 5) crore
= 27 crore86. (C) Range of number of students in activity IV
= 438 105 = 333Average of number of students in activityIII per college
=7
1532205407542013065
=7
1603= 229
Required difference = 333 229 = 10487. (D) Number of students in activity II
= 100 + 200 + 200 + 100 + 100 + 100+ 100
= 900Number of students in activity IV
= 317 + 155 + 438 + 105 + 385 + 280+ 120= 1800
Required percentage =1800
900 100%
= 50%88. (B) The average number in the student in
activities III
=7
1532205407542013065
=7
1603 = 229
89. (A) Number of student inCollege D = 100 + 100 + 75 + 105 = 380
College G = 200 + 100 + 153 + 120 = 573College F = 300 + 100 + 220 + 280 = 900
College A = 200 + 100 + 65 + 317 = 682Thus, college D has minimum number ofstudents participate in extra-curricularactivities.
90. (B) Total number of students in activity II= 100 + 200 + 200 + 100 + 100 + 100+ 100= 900
Total number of students in activity I= 200 + 300 + 500 + 100 + 400 + 300 + 200
= 2000
Thus, Required ratio = 900 : 2000
= 9 : 20
91. (A) Total area under Bajra
18 300 acresTotal area under Rice and Barely
(72 + 36) 18
300 (72 + 36)
=18
300 108 = 1800 acres
92. (A) Angle covered by Wheat, Rice and Maize
= 72 + 72 + 45
= 189
Which is greater than 180
Hence, area covered by these three crops
is more than 50% of the total area.93. (C) Required ratio
=Land used for Rice
Land used for Barely
=Angle covered by Rice
Angle covered by Barely
=36
72=
1
2= 2 : 1
94. (B) 10% of the land reserved for Rice
= 10% of 72 = 7.2
It is distributed to Wheat and Barely in
the ratio 2 : 1. Therefore, angle increased
corresponding to Wheat =3
2 7.2 = 4.8
Now angle corresponding to Wheat= 72 + 4.8
= 76.8
95. (A) Production of Rice = 5 production of Jowar
Production of Bazra = 2 production of Jowar
Required ratio
= Production of Rice : Production ofBazra
= 5 : 2
96. (C) Total production of Rice and Wheat
in state B = 10 + 2 = 12 lakh tonnes
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Centres at:
==================================================================================
Param
ountCoaching
Centr
e
MUKHERJEE NAGAR
MUNIRKA
UTTAM NAGAR
DILSHAD GARDEN
ROHINI
BADARPUR BORDER
JAIPUR
GURGAON
NOIDA
1. (D)
2. (C)
3. (C)
4. (C)
5. (D)
6. (D)
7. (C)
8. (A)
9. (B)
10. (B)
11. (C)
12. (D)
13. (C)14. (A)
15. (D)
16. (C)
17. (A)
18. (A)
19. (A)
20. (B)
SSC Mains Test- 21 (ANSWER KEY)
21. (B)
22. (D)
23. (D)
24. (B)
25. (D)
26. (C)
27. (C)
28. (A)
29. (D)
30. (D)
31. (B)
32. (D)
33. (C)34. (A)
35. (B)
36. (B)
37. (B)
38. (A)
39. (D)
40. (B)
41. (C)
42. (D)
43. (D)
44. (B)
45. (D)
46. (B)
47. (C)
48. (C)
49. (A)
50. (B)
51. (C)
52. (C)
53. (B)54. (C)
55. (A)
56. (D)
57. (C)
58. (B)
59. (A)
60. (C)
Note: If your opinion differs regarding any answer please
message the mock test no. and question no. to 8860330003
61. (D)
62. (D)
63. (D)
64. (B)
65. (A)
66. (C)
67. (D)
68. (B)
69. (B)
70. (D)
71. (D)
72. (D)
73. (C)74. (C)
75. (A)
76. (C)
77. (C)
78. (D)
79. (D)
80. (C)
81. (C)
82. (B)
83. (A)
84. (D)
85. (B)
86. (C)
87. (D)
88. (B)
89. (A)
90. (B)
91. (A)
92. (A)
93. (C)94. (B)
95. (A)
96. (C)
97. (D)
98. (D)
99. (B)
100. (B)