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MOCK TEST PAPER - 12 (T-II) 2013 (SOLUTION)
1. (B)21
1
+
32
1
+
43
1
+ ...
121120
1
=)12(
)12(22
+
)23(
)23(22
+
)34(
)34(22
..... + )120121(
)120121(22
= ( 2 1 ) + 3 2 ) + 4 3 )
+.... ( 120 119 ) + ( 121 120 )
= 1 + 121 = 11 1 = 10 Ans.
2. (A)2
5
5
10 + 125 =
2
5
5
10 +
1
55
=52
510205 =
22362
35
=
472.4
35
= 7.826 Ans
3. (D)
3/2
216
1
3/2
36
1
=
2
6
1
= (6)2 = 36 Ans.
4. (C) Required measurement
= HCF of 403 kg, 434 kg and 465 kg
= 31 kg Ans.
5. (C) Let R = Age of Randheer,
A = Age of Anup
So, ATQ,18
6R = A
Given, Mahesh = 5 yrs
Anup = (5 2) yrs = 3 yrs
R = 18 3 + 6 = 60 yrs Ans
6. (B) Let x be the number of buffaloes and
y be the number of ducks.
Then, total number of legs = 4x + 2y
and total number of heads = x + y
So, 4x + 2y= 2(x + y) + 24
2x = 24
x = 12 Ans.
7. (A) Cost of 9 lemons = 4 48 paise
= 144 paise
= cost of 4 mangoes
So, cost of 3 mangoes = 4
144 3
= 108 paise
= Cost of 5 apples
Cost of 9 oranges = 108 paise
Cost of1 orange =9
108
= 12 paise Ans.
8. (A) Sum of digits from numbers 1 to 10 = 46,
Sum of digits from numbers 11 to 20 = 56
&Sum of digits from numbers 21 to 29 = 63
Sum of digits of the given numbers
= 46 + 56 + 63 = 165
Again,
Sum of digits of the number 165 = 1 + 6 + 5 = 12
Now, when 12 is divided by 9, we get
remainder = 3 Ans.
9. (A) Let x be the number.
Then calculation done by the student
= 6
12x=112
x = 660
So, the correct answer = 6
660 + 12
= 122 Ans
10. (D) Let x be the number of students and
w be their average weight.
ATQ,
1
50
x
xw= w + 1
x + w = 49 .... (i)
Again,2
5050
x
xw= w + 1.5
1.5x + 2w = 97 .... (ii)
From equations (i) and (ii),
we get, w = 47 Ans
11. (A) Let x be the first number
x x x x x + ( + 1) + ( + 2) + ( + 3) + ( + 4)
5 = 41
or, 5x + 10 = 205
5x = 195
x = 39
So, Product of A and E = x (x + 4)
= 39 43
= 1677 Ans.
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12. (A) Let Sunil, Sumant and Surat get x,
y and z respectively.
So, 100
15x=
100
25y =
100
35z
x = 2y = 3z
x : y : z = 6 : 3 : 2
Required difference= 236
26
7700
= 2800 Ans
13. (C) Let the speed of train on level terrain
= x km/h
So, the speed of train through mountainous
terrain = (x 10)km/h
Mountainous Level299 km
111 km 188 km
ATQ,
x
188 +
10
111
x = 7
188 1880 111
( 10)
x x
x x
= 7
7x2 369x + 1880 = 0
x = 47 km/h Ans.
14. (D)TV Newspaper
65% 40%
Number of people who either watch TV
or read newspaper = 65% + 40% 25%
= 80%
Number of people who neither watch
TV nor read newspaper.
= (100 80)%
= 20% Ans.
15. (D) Let required amount of second solution
to be added = x L
So,
x
x
20
52015= 10
300 + 5x = 20 + 10x
x = 5
100 = 20L Ans.
16. (A) Milk in mixture
= 6 100
25 + 4
100
30 =
100
270
Required percentage = 100
)46(270
= 27% Ans.
17. (B) Let x be the total score in the innings.
So, the highest score = 9
2x
And, the next highest score
= 9
2 of the remaining runs
= 9
2
xx
9
2
So, ATQ, 9
2x
9
2
xx
9
2= 8
x x + 9
2x =
2
98
x = 22
998
= 162 Ans.
18. (B) (13 + 22 + 33 + ... 153) (1 + 2 + 3 + ... + 15)
= 15 2
)15313(
2
1615
= 1245 120
= 1125 Ans.
19. (B) Let total marks be 100.
So, the minimum marks required to be
pass = 40% of 100 = 40 marks
ATQ, Marks obtained by A = 40 40 100
10
= 36 marks
&, Marks obtained by B = 36 9
100
100
36
= 36 4
= 32 marks
So, Marks obtained by C
= (36 + 32) (36 + 32) 10017
700
= 68 28 = 40 marks Ans.
20. (B) Let x be the monthly salary of Mohan.
ATQ,
10
4x +
10
6x
100
50+
10
3x
100
30 + saving = x
10
4x+
10
3x +
100
9x + 630 = x
630 = 100 70 9
100
x x x
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630 = 100
21x
x = 3000 Ans
21. (B) the total number of men employed =50
So, total number of wommen employed = 100
& total number of children employed = 150
Now, let the men, women and children
wages are 6x, 3x and 2x respectively.
50 6x + 100 3x + 150 2x = 4500
or, 900x = 4500
x = 5
Per day wages paid to a man, a woman
and a child are 30, 15 and 10.
Weekly wages paid to a man, a woman
anda child are 210, 105 and 70 Ans.
22. (B) Let the present and last year salaries of
Mahesh and Suresh are x, x' and y, y'
respectively.
'
'
y
x=
5
3,
x
x'=
3
2
andy
y' =
5
4
y
yx
x
'
'
=
5
43
2
'
'
y
x
x
y=
12
10
5
3
x
y=
12
10
x
y=
36
50
Also, x + y = 43000
x + 36
50x = 43000
x = 86
3643000= 18000 Ans.
23. (D) Actual amount to be paid
= 20000
3
100
101
= 26620
Amount that Rohit have to pay now
= 20000
2
100
101
100
151
= 27830
So, required extra amount
= 27830 26620
= 1210 Ans.
24. (A) Let the value of second rate of interest
be x% and equal amounts be P each.
So, ATQ,
P
6
100
51
= P
3
1001
x
2
100
51
=
1001
x
100100
105105
=
100
100 x
110.25 = 100 + x
x = 104
1% Ans
25. (B) Upstream speed = 2 km/h
Downstream speed = 10
1 60 = 6km/h
Speed in stationary water
= 2
62 = 4 km/h
Required time = 4
5 = 1 hr 15 min Ans.
26. (D) Let the man went up stream for x km.
He turned back downstream for (x 2) km.
So,)5.15.4(
x +
2
(4.5 1.5)
x
= 2 h 10 min
6
22 xx =
6
12 hour
3x 2 = 13
x = 5 km Ans.
27. (A) Let x% be the other discount.
So, 65 100
90
100
)100( x = 56.16
100 x = 56.16 100 100
65 90
100 x = 96
x = 4% Ans.
28. (A) Let cost price of the article = x.
So,
x 100
5.117 + 11.55 = x
100
92
100
130
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x = )5.11713092.0(
55.11
100
= 550 Ans.
29. (B) 12.5% of profit= 100
5.12 880 = 110
Remaining 770 is divided in the ratio= 5000 : 6000 = 5 : 6
Profit to Anu = 11
5 770 + 110
= 460
Profit to Bimla = 6
11 770
= 420
Required profits are 460 and 420 Ans.
30. (B) Ratio to capitals= 1
3 :
3
2 = 1 : 2
Ratio of profits = 5
3 :
5
2 = 3 : 2
Let Y's money was used for x months.
(1 9) : (2 x) = 3 : 2
x = 3 months Ans.
31. (B) A's 1 day work = 40
1
So, A's 5 day work = 40
1 5 =
8
1
So, remaining work= 1 8
1 =
8
7
B completed the remaining work in 21 days
So, 1 day's work of B = 218
7
=
24
1
Both (A + B)'s 1 day work = 40
1 +
24
1
= 120
53 =
120
8=
15
1
Hence, both complete the work in 15 days. Ans.
32. (D) Let (A + B) together complete the work inx hours.
A completes the work in (x + 8) days.
& B completes the work in
2
9x days.
Using formula,
2
252
2
9)8(
x
xx
= x
x2 + 8x + 2
9x + 36 = 2x2 +
2
25x
x2 = 36 x = 6 days Ans.
[* Read days as hours]
33. (C) Let the present population of male and
female be x and y respectively.
ATQ, x + y = 5500
and100
111x +
100
120y = 6330
111x + 120y = 633000
From equations (i) and (ii), we get
y = 2500 Ans.
34. (C) Let the ouput be x and percentage be a.
Then,
x
2
1001
a= 2x
1 + 100
a = 2
100
a = 2 1
a = ( 2 1) 100%
= 100( 2 1)% Ans.
35. (A) Let the cost of the product = x.
So, ATQ,
x 100
110
100
115
100
125 = 1265
x = 1265 100
110
100
115
125
100
= 800 Ans.
36. (C) Total work = 124 120 = 14880 men-days
Work completed in 64 days
= 3
2 14880 men-days
= 9920 men-days
Remaining work for remaining 60 days
= (14880 9920) men-days
= 4960 men-days
Now, 1
11
W
DM=
2
22
W
DM
9920
64120=
4960
602 M
M2
= 64
Number of workmen who can be
reduced =120 64 = 56 Ans.
37. (C) Till 3 pm, part of tank filled
= 8
2 +
12
1 =
4
1 +
12
1 =
12
4 =
3
1part
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Remaining part = 1 3
1 =
3
2
Now, let x be the time taken by all three
pipes work to fill the remaining part of tank
1 1 1
8 12 6
x = 3
2
48
)846( x=
3
2
48
2x=
3
2
x = 16 hour
Hence, the tank will be filled at 1 pm
+ 16 hour = 5 am. Ans.
38. (D) Let leakage alone can empty the full
cistern in x h.
Then,
2
92
9
x
x
= 5
2
9x = 5x
2
45
2
1x =
2
45
45 h Ans.
39. (D) ATQ, original (prime) cost of article = 6.
Original raw material cost = 2
So, New cost of raw material = 2 5
12
= 4.80
Also, original manufacturing expenses
= (6 2) = 4
So, new manufacturing expenses
= 4 4
5 = 5
New cost of article = 4.80 + 5
= 9.80 Ans
40. (A) Let the shares of Anita, Bindu and
Champa are 11x, 18x and 24x respectively.
So, 1105 = 11x + 10 + 18x + 20 + 24x + 15
1105 = 53x + 45
x = 20
Amount received by Champa
= 24x + 15
= 24 20 + 15
= 495
41. (A) Total marks got by the students in
8 subjects = 8 57 = 696
Total marks got by the student in
6 subjects = 6 85 = 510
Remaining marks in 2 subjects
= 696 510 = 186
Let the highest marks be x, then the
second highest marks will be x 2.
x + x 2 = 186
2x = 188
x = 94
Highest marks = 94 Ans.
42. (C) Total score of first three friends
= 15 3 = 45
and total score of last three friends
= 16 3 = 48
Total score of four friends
= 45 + 19 = 64
Score of first friend
= 64 48 = 16
So, required percentage
= 48
16 100% = 33
3
1% Ans.
43. (B) When a value is first increased and then
decreased by the same percentage, then
the initial value is always decreased by
100
2x%. (irrespective of initial value)
So, loss percent = 100
)15( 2= 2.25% Ans.
44. (B) Let rate of popualtion increase = R% per annum
So, 4800 = 3600
3
1001
R
3
4=
3
1001
R .... (i)
Now, the population after 3 years
= 4800
3
1001
R
= 4800 3
4[from (i)]
= 6400 Ans.
45. (D) Let the two numbers are x and y.
Then,
their AM =2
yx = 5
x + y = 10 ... (i)
and their GM= xy = 4
xy = 16
Now, (x y)2 = (x + y)2 4xy
= 100 64 = 36
or, x y = 6 ... (ii)
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From equations (i) and (ii),
x = 8 and y = 2
So, the required numbers are 2 & 8 Ans.
46. (B) Let the first term and common term of
the AP be a and d respectively.
ATQ, (a + 5d) + (a + 14d) = (a + 6d) +
(a + 9d) + (a + 11d)
2a + 19d = 3a + 26d
a + 7d = 0
8th term is 0. Ans.
47. (A) Let M stands for Mohan and R stands for
Ram.
Case : I (M + 30) = 2(R 30)
M 2R= 90 ... (i)
Case : II (R + 10) = 3(M 10)
R 3M = 40 ... (ii)
Solving Eqs. (i) and (ii), we get
M = 34 and 62 Ans.
48. (B) Given,
zyx
x
2 =
zyx
y
2 =
zyx
z
2 = a
x = a(2x + y + z), y = a(x + 2y + z)
z = a(x + y + 2z)
x + y + z = a(4x + 4y + 4z)
4a = 1 a = 4
1 Ans.
49. (C) Roots of a quadratic equation
ax2 + bx + c = 0 are real if b2 4ac > 0
Option (a): 3x2 4x + 5 = 0
b2 4ac = (4)2 4(3)(5)
= 44 < 0
Hence, roots are not real.
Option (b): x2 + x + 4 = 0
b2 4ac = (1)2 4(1)(4)
= 1 16 = 15 < 0
Hence, roots are not real.
Option (c): (x 1) (2x 5) = 0
x = 1 and x = 2
5> 0
Hence, roots are real.
Option (d): 2x2 3x + 4 = 0
b2 4ac = (3)2 4(2)(4)
= 9 32 = 23 < 0.
Hence, roots are not real. Ans.
50. (C) 2x2 7xy + 3y2 = 0
2
2
y
x7
y
x + 3 = 0
y
x=
a
acbb
2
42
= 7 49 24
2 2
= 4
57 = 3,
2
1
y
x=
1
3and
y
x=
2
1Ans.
51. (A) Let f(x) = 2x3 ax2 (2a 3)x + 2
(x + 1) is a factor of the above expression,
Remainder f (1) = 0
f (1) = 2(1)3 a(1)2 (2a 3) 1+ 2 = 0
2 a + 2a 3 + 2 = 0
a 3 = 0
a = 3 Ans.
52. (B) Let the common base be x m.
Then, area of triangle = area of parallelogram.
2
1 x Altitude = x 100
Altitude = 200 m Ans.
53. (D) For two similar triangles ABC and PQR.
A
B C
P
Q R
10 cm
PQ
AB=
24
36
AB =24
36 10 = 15 cm Ans
54. (C) R
S
N M
OQ SQ = 23 cm
17
cm
17cm
P
8
In PQO,
(17)2 = (8)2 + (OQ)2
(OQ)2 = 289 64 = 225
OQ = 15
OS = 23 15 = 8
Now in ORS,
(RS)2 = (17)2 (8)2
= 289 64 = 225
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RS = 15 cm
Hence, length of other chord
= 15 2 = 30 cm Ans.
55. (D) Let the three points be A
3
8,0 , B(1, 3)
and C(82, 30)
Then, AB =
22
3
83)01(
=
3
10
BC = 22 )330()182(
= 7296561
= 7290 = 1027
CA =
22
3
830)082(
=9
)82(10 2 = 10
3
82
Now, AB + BC = 3
10 + 1027 =
3
1082
Since, AB + BC = AC, it means points A,
B and C are collinear. Ans.
56. (A) Given, AD2 = AB2 + BC2 + CD2
A
BC
D
?90
In ABC,
AC2 = AB2 + BC2
AD2 = AC2 + CD2
C = 90 Ans.57. (A) Let the length and breadth of a cuboid be
8x cm and 7x cm respectively.
Volume of cuboid = l b h
1120 = 8x 7x 5
x2 = 578
1120
= 4
x = 2
Length and breadth of a cuboid are 16 cm
and 14 cm by 2 cm.
Hence, length exceeds its breadth by 2 cm. Ans.
58. (C) Let length and breadth of blackboard
be x m and (x 8) m respectively.
ATQ, x (x 8) = (x + 7) (x 12)
x2 8x = x2 5x 84
x = 3
84= 28 cm
Length = 28 cm
Breadth = x 8 = 20 m Ans.
59. (C) Let the outer radius of pipe be r.
Then, volume of the metal
= 28 (r2 82)
1496 = 28 (r2 82)
r2 64 = 2228
1496
7
r2 = 17 + 64 = 81
r = 9 cm Ans.
60. (D) Let dimensions of the rectangular stone
be 3x, 2x, x respectively.
Volume of stone= 10368
3x 2x x = 10368
6x3 = 10368
x3 = 1728
x = 12
Dimensions are 36 dm, 24 dm, 12 dm.
Entire surface area of a stone.
= 2(lb + bh + hl)
= 2(36 24 + 24 12 + 12 36)
= 2(864 + 288 + 432)
= 3168 dm2
Total cost of polishing
= 3168 0.02
= 63.36 Ans.
61. (A) Total area of two parks
= (82 + 62) = 100
Area of bigger park
= 100
r2 = 100
r2 = 100
r = 10 m Ans
62. (D) Total area of playground
= 750 2 rh
= 750 2 7
22
2
70 150
= 2475 104 cm2
= 2475 m2
Total cost of levelling = 2475 2
= 4950 Ans.
63. (C) Since AD bisects A.
DC
BD
AC
AB
x
x
89
6
9x = 48 6x
x = 15
48
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x = 3.2 cmA
B CD
9 cm40
406 cm
8 cm
x cm
DC = 8 3.2 = 4.8 cm Ans.
64. (A)
90 m
A
B
C
60
Let AB be the string.
AB
BC = sin 60
BC = 90 2
3= 45 3 m Ans.
65. (D) cos6 x + sin6 x + 3sin2 x cos2 x
=(cos2 x)3 + (sin2 x)3 + 3sin2 x cos2 x(sin2x
+ cos2 x)
= (cos2 x + sin2 x)3
= 1 Ans.
66. (A) Given, a sec + b tan =1 ... (i)
and, a2 sec2 b2 tan2 = 5 ... (ii)
From Eqs. (i) and (ii),
a sec b tan = 5 ... (iii)
From Eqs. (i) and (iii),
a sec = 3 ... (iv)
b tan = 2 ... (v)
From Eq. (iv),
a2 sec2 = 9
a2(1 + tan2 ) = 9
a2
2
41
b= 9
a2b2 + 4a2 = 9b2 Ans
67. (A)
A B
C
D E
60 m
30
x m
150
m
Let the height of the shorter tower be x m.
Then, from CDE,
tan 30 = 60
)150( x
3
1=
60
150 x
(150 x) = 3
60= 2 3
x = 150 20 3
= 116 m (approx) Ans.
68. (D) x = 2 + 22/3 + 21/3
The value of x is greater than 2, let us try
it for x = 2 and 3.
For x = 2, x3 6x2 + 6x = 4
For x = 3, x3 6x2 + 6x = 9
x3 6x2 + 6x < 0 for x = 2 and 3
Hence, option (d) is the answer. Ans.
69. (A) t 2 = 3 4 + 3 2
(t 2)3 = 4 + 2+ 3 3 4 3 2 ( 3 4 + 3 2 )
t3 23 3 2 t(t 2) = 6 + 6t 12
t3 6t2 + 6t 2 = 0 Ans.
70. (B) Sum of roots = a + b = 6
Product of roots = a b = 6
Now, a2 + b2 = (a + b)2 2ab
= 36 12 = 24 Ans.
71. (A) Using AM > GM
2
33 xx > xx 3.3
3x + 3x > 2 Ans.
72. (B) a + b + c = 0
a2 = (b + c)2 or a = b c
Given expression
bca
a
2
2
+ 2
2
b
b ca +
abc
c
2
2
= bccb
cb
2
2
)(
)( +
2
2 ( )
b
b c b c +
2
2 ( )
c
c b b c
=bccb
cb
22
2)(+
bccb
b
22
2
+ bcbc
c
22
2
= bccb
cbbccb
22
2222 2=
bccb
bccb
22
22 )(2
= 2 Ans.
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73. (C) Sum of all the exterior angles of the
polygon is 360.
x
75
105
65
180 x
90
90 115
x + 75 + 90 + 115 = 360
x = 80 Ans.
74. (B)
D
C
A
B
Given A = 2C
Also, A + C = 180
( ABCD is cyclic)
C = 60
and A = 120
Now, B D = 3
1A = 40
B = 110
D = 70
Minimum difference between any two
angles is 10. Ans.
75. (C) Given, OBC = OCB = 37
[ Angles opposite to equal
sides of a triangle are equal]
COB = 180 (37 + 37)
= 106
BAC = 2
1COB =
2
106
= 53 Ans.
76. (C) In OC1C
2,
(OC1)2 = (OC)2 + (CC
1)2
(r + 1)2 = (PC OP)2 + 1
(r + 1)2 = (2 r)2 + 1
r2 + 1+ 2r = 4 + r2 4r + 1
6r = 4
r =3
2Ans.
77. (B)AB
BD=
AC
DC
A
B CD
DC
BD =
5
3
BD : DC = 3 : 5
Now, BD = )53(
3
6
= 8
18 =
4
9 = 2.25 cm Ans.
78. (C) Area of four walls = 2h(l + b)
l = 2b and h = 4 m
120= 8 3b
b = 24
120 = 5 and l = 10
Area of floor = l b = 10 5 = 50 m2Ans.
79. (C) Initial length of rectangle = L
New length of rectangle = 2
3L
Initial breadth of rectangle = B
New breadth of rectangle=3
1B
Initial Area = L B
New Area = 2
3L
3
1B =
2
1LB
Hence, change in area is 2
1. Ans.
80. (D) Let water in the tank is h deep.
3 2
3 h = 8 2.5
h = 33
25.28
= 4.44 ft Ans.
81. (A) Let the water level rises by x m.
90 40 x = 150 8
x = 4090
8150
= 0.3333 m
= 33.33 cm Ans.
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82. (B) Curved Surface Area of 50 pillars
= 50 2 3.14 2
50 400
= 314 104 sq. cm. = 314 sq. cm.
Labour charges= 314 0.50
= 157 Ans.
83. (A) Total surface area of remaining solid
= curved surface area of the cylinder
+ area of base + curved surface area
of cone
= 2 rh + r2 + rl
= 2 8 15 + (8)2 + 8 17
= 240 + 64 + 136 = 440 cm2 Ans.
84 . (C)Let the breadth and height of b and h
respectively.
Then, 70 2.2 = 14 b
b = 11 m
Also, 14 b h = 70 11
h =1114
1170
= 5m Ans.
85. (B) Total cost of plastering inner surface
= 2 7
22
2
7 22.5 3
= 1485 Ans.
86. (C)
A B
C
D
2h
d
Let BC = h be the height of the pillar, then
CD = 2h.
Also, let BAC = CAD = and AB = d
In ABC,
tan = d
h
and in ABD,
tan2 =d
h3
2tan1
tan2=
d
h3
21
2
d
h
d
h
= d
h3
2
3= 1
2
d
h
d
h=
3
1Ans.
87. (D) Given,
A
A
cos
sin=
7
4
2
cos
sin7
3cos
sin7
A
AA
A
=
47 3
74
7 27
=
24
34
=
6
1
88. (C)
30 45
m
DC A
B
h
x
)13(50 -
Distance travelled in 2h, CD = )13(50 m
Let AB be the building and its height be h
and AD be x.
x
h = tan 45 h = x .... (i)
50( 3 1)
h
x = tan 30
h = 3
x+
50( 3 1)
3
.... (ii)
From eqs. (i) and (ii),
h 3
h =
50( 3 1)
3
h( 13 ) = 50( 3 1)
h = 50 m Ans.
89. (B)Let AB = h m and DC = EB = x m
A
B
CD
E30
6010 m
x m
(10 + )mh
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Here, pole DE = 10 mAC = tower = 10 + h
In AEB, we have
tan 30 = EB
AB=
3
1 =
x
h... (i)
x = h3
In ACD, we have,
tan 60 = CD
AC =
x
h10
3 = x
h10
x = 3
10 h ... (ii)
From eqs. (i) and (ii), we get
3 h = 3
10 h
3h = 10 + h
h = 5 cm
AC = 10 + h = 15 m Ans.[Remember the result in the above : heightof tower = 1.5 height of the pole]
90. (C) When x = 0sin x + cosec x = 0 + =
When x = 2
sin x + cosec x = 1 + 1 = 2
When2 < sin x + cosec x <
Hence, sin x + cosec x > 2 Ans.
91. (C)
19861987198819891990
44.5046.0057.0082.00
138.00
44.5090.50
147.50229.50367.50
YearTotal Sales ofall company
Cumulative Sales
Required ratio = 50.367
50.90 =
735
181 =
4
1Ans
92. (A)Average sales of Celia
= 5
505.145.161550.18
= 5
50.114 = 22.90 Ans.
93 (A) Growth rate for
Alpha = 20
2025 100 = 25%
Baron = 50.11
50.1100.15 100 =30.43%
Celia = 5.14
5.1450 100 = 244.8%
Dowby = 36
3648 100 = 33.33%
Growth is least for Alpha. Ans.94. (B) For the year 1986, total sales = 44.5
For the year 1987, total sales = 46For two years show the closest sales.
95. (A)Required ratio = 2
50 = 25 : 1 Ans.
96. (A)
A
Candidate who joinedMBA = 8550
Candidate who completedMBA = 5700
B
C
D
E
F
G
8550 = 1881 5700 = 1026
8550 = 1282 5700 = 969
8550 = 855 5700 = 741
8550 = 1454 5700 = 912
8550 = 684 5700 = 513
8550 = 1026 5700 = 855
8550 = 1368 5700 = 684
22
100
18
100
15
100
17
100
10
100
13
10017
100
16
1008
100
9
10012
100
15
10016
100
12
100
Required percentage of
A = 1881
1026 100 = 54.55%
B = 1282
969 100 = 75.59%
C = 855
741 100 = 86.67%
D = 912
1454 100 = 62.72%
E = 684
513 100 = 75%
F = 1026
855 100 = 83.33%
G= 1368
684 100 = 50%
97. (C) Required percentage = 75% Ans.(See solution of last question)
98. (B) Required ratio = 1368
684= 1 : 2 Ans.
99. (D) Required percentage
=8551282
741969
100
= 2137
1710 100 = 80% Ans.
100. (C) Required difference= (A + D) (C + E)
[Completed joined]= (1026 + 912) (855 + 684)= 1938 1539= 399 Ans.
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1. B
2. A
3. D
4. C
5. C
6. B
7. A
8. A
9. A
10. D
11. A
12. A
13. C
14. D
15. D
16. A
17. B
18. B
19. B
20. B
21. B
22. B
23. D
24. A
25. B
26. D
27. A
28. A
29. B
30. B
31. B
32. D
33. C
34. C
35. A
36. C
37. C
38. D
39. D
40. A
41. A
42. C
43. B
44. B
45. D
46. B
47. A
48. B
49. C
50. C
51. A
52. B
53. D
54. C
55. D
56. A
57. A
58. C
59. C
60. D
61. A
62. D
63. C
64. A
65. D
66. A
67. A
68. D
69. A
70. B
71. A
72. B
73. C
74. B
75. C
76. C
77. B
78. C
79. C
80. D
SSC MAINS (MATHS) MOCK TEST -12 (ANSWER SHEET)
81. A
82. B
83. A
84. C
85. B
86. C
87. D
88. C
89. B
90. C
91. C
92. A
93. A
94. B
95. A
96. A
97. C
98. B
99. D
100. C
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