ssc mains (maths) mock test-16 (solution)
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MOCK TEST for ssc exams you can practice these papers and crack ssc exams which are difficult to passTRANSCRIPT
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SSC (T-II) 2013, MTP - 16 (SOLUTION)
1. (A) Unit digits in (7)4 = 1, therefore unit digitin (74)8 i.e. 732 will be 1.Hence, unit digits in (7)35
= 1 7 7 7 = 3Again, unit digit in (3)4 = 1Therefore, unit digit in the expansion of (34)17
= (3)68 = 1
Unit digit in the expansion of (3)71
= 1 3 3 3 = 7and unit digit in the expansion of (11)35
= 1
Unit digit in the product of735 371 1155 = 1
2. (C) a = 100
1, b =
5
1, c =
10
1
or a = 0.01, b = 0.2, c = 0.1
b > c > a3. (A) Product of the numbers
= HCF LCM= 21 4641= 21 3 7 13 17= 3 7 3 7 13 17.
i.e. 273 and 357.4. (A) Let the number be x.
x = 765k + 42 17 45K + 17 2 + 8 17(45K + 2) + 8 Remainder = 8
5. (B) Distance traversed by the extremity of the
minute-hand in one hour = 2 7
22 10
Distance traversed by the extremity of theminute-hand in 3 days and 5 hour, i.e.in 77 hours
= 2 7
22 10 77
= 22 220 = 4840 cmDistance traversed by the hour-hand in12 hour
= 2 7
22 7 = 44 cm
Distance traversed by the hour-hand in77 hour
= 12
44 7 =
3
7711=
3
847= 282.33 cm
Reqd. difference = 4840 282.33= 4557.67 cm
6. (D) The series obtained= 4 2 + 3, 4 5 + 3, 4 8 + 3, 4 11 + 3 ...= 11, 23, 35, ...
7. (A)
2
1
2
1
2
12
1
2
1
2
1
=
4
32
12
2
1
= 2
1
3
4 =
3
2
8. (C) Let a1, d
2 represent the Ist term and
common difference of AP1.
and a2, d
2 represent the Ist term and
common difference of AP2.
ATQ,
'n
n
S
S=
])1(2[2
])1(2[2
22
11
dnan
dnan
274
17
n
n=
22
11
)1(2
)1(2
dna
dna
...... (i)
Now, 11
11
b
a=
22
11
10
100
da
da
=
22
11
302
302
da
da
= 22
11
)121(2
)121(2
da
da
= '21
21
S
S
= 27214
1217
= 111
148 Ans.
9.(B) Sn
= 0.4 + 0.44 + 0.444 + ... + to n terms= 4[0.1 + 0.11 + 0.111 + ... + to n terms
= 9
4[0.9 + 0.99 + 0.999 + ... + to n terms
= 9
4
termnto....
1000
999
100
99
10
9
=
=
=
10
11
10
11
10
1
n9
4
n
)r1(
)r1(aS
n
n
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2
=
10
910
11
10
1
n9
4 n
=
n10
11
9
1n
9
4
=
n10
11n9
9
4 Ans.
10. (C) Number of women who have either nosestuds or ear right = (15 3) = 12
Number of those who have both nose studsand ear rings = (8 + 7) 12 = 3
11. (C) Ashok's present age = 26 6 = 20 yrs
Pradeep's present age = 20 4
3= 15 yrs]
12. (A) Let the number of girls = x
boys = x 2According to question:-
x + x 2 = 52
x = 27
No. of bags = 25Again:-
Let average no. of girls = y kg42 25 + 27xy = 52 52
y = 27
25425252
= 61.25~61 kg (Approx.)13. (D) Wrong calculated marks
= 35 75
= 35
86362520 =
35
2570= 73.42
14. (C) If a container contains y units ofliquid and x units of liquids is taken out.If this operation is repeated n times. Thefinal quantity of the liquid in thecontainer is
n
y
xy
1
24 = 54
2
541
x
2
541
x=
54
24 =
9
4
541
x =
3
2
54
x =
3
1
x = 18 l
15. (C) Let the quantity of pure milk be x l.If 5 l of water is added to it,then,
Cost of (5 + x) l= Rs. (3x + 5)
Profit = Rs. 15Given, 20% of 3x = 15
5
3x= 15
x = 25 l16. (D) Let the amount be Rs. x and rate is r%.
Then, Simple interest
100
rx = 25
x r = 2500For true discount
100
)20( rx = 20
100
20rrx = 20
xr 20r = 2000From Eqs. (i) and (ii), we get
2500 20r = 2000
r = 25%From Eq. (i)
x 25 = 2500
x = 10017. (*) Let the CP price of pen = Rs. 1
CP of 40 pen = 36 1 = Rs. 36
Selling price of 40 pen = 40 3% of 40
= 40 100
3 40
= 10
12400
= Rs. 38.80
% profit =
36
3680.38 100
= 36
280=
9
70= 7
9
7%
18.(C) Length of bridge = 1000 mLength of train = 500 mTotal length = 1000 + 500 = 1500 m
Speed of train = 1000
1500
2
60= 45 km/h
19. (B) Distance (D) = Speed (S) Time (T)
D = 4
60
15T
D = 4T + 1
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and D = 6
60
10T
D = 6T 1Solving eqns. (i) and (ii),
T = 1 hD = 4 1 + 1 = 5 km
20. (D) It is clear from the question that when Acovers 500m, B covers 400 m i.e., A takesa lead of 100 m in every 500 m of distance.Therefore, a lead of 400 m will be takenin travelling a distance of 2000 m or inother words A passes B after every 2000m.Hence, total number of such pass
= 2000
5000=
2
5 = 2
2
1 times
21. (C) Let three numbers A, B and C are Rs. 12x,15x and 25x respectively.
12x + 15x + 25x = 312
x = 52
312 = 6
Required ratio = 615625
612615
= 610
63
= 10
3= 3 : 10
SHORTCUT METHOD:-
There is no need to calculate the value of x.
Required ratio = xx
xx
1525
1215
= x
x
10
3=
10
3= 3 : 10
22. (B) Ratio of investment of Sita, Gita and Rita is(5000 3 + 7000 9) : (4000 1 + 3000 11) : (7000 11)
= 78000 : 37000 : 77000= 78 : 37 : 77
Share of Rita in profit
= 773778
77
1218 = Rs. 488.47
23. (A) Let the speed of man and current be x km/hand y km/h respectively.Then,
yx
30 +
yx
44 = 10 ... (i)
andyx
40 +
yx
55 = 13 ... (ii)
Solving eqns.(i) and (ii),y = 3 km/h
24. (C) Let the tap can fill the cistern in x h.
8
8
x
x= 12
8x = 12x 96 x = 24 h
Capacity of cistern= 24 60 6 = 8640l
25. (A) Part of the cistern filled in 3 min
= 12
3 +
16
3 =
48
21 =
16
7
Let remaining 16
9 part was filled in x min
Then,12
x
8
7 +
16
x
6
5 =
16
9
x
96
57 =
16
9
x = 16
9
12
96 = 4.5 min
26. (A) Ratio of efficiencies of the three persons
= 6
24 :
8
24 :
3
24
8
24
6
24
= 4 : 3 : 1
Boy's share = )134(
1
600 = Rs. 75
27. (A) Let the cost price of geyser be Rs.x,then, x 1.1 1.15 1.25 = 1265
x = 58125.1
1265
= Rs. 80028. (B) Let his increased income be x.
Then,
(x 1200) 100
80
100
12= x
100
80
100
10
12x 14400 = 10%x = Rs. 7200
29. (D) Let the present value of what A owes to Bbe Rs. x.Then,
x + 1002
314
x= 1573
x + 100
21x = 1573
100
121x= 1573
x = Rs. 1300Let y be the present value of what B owes A.
Then, y + y 2
1
100
14 = Rs. 1444.50
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y + 100
7y = Rs. 1444.50
y = 107
10050.1444 = Rs. 1350
Hence, B pay Rs. 50 to A.30. (A) Let the amount given at 4% per annum
be Rs. x.Amount given at 5% per annum
= (1200 x)
100
24x +
100
25)1200( x= 110
100
120002 x= 110
x = Rs. 500Also, (1200 x) = 1200 500
= Rs. 70031.(A) Let money invested at 5% be Rs. x.
100
51x +
100
81)10000( x = 688.25
5x 8x + 80000 = 68825 3x = 11175 x = Rs. 3725
32. (C) Let principal amount be Rs. 100.
Then, SI = 100
320100 = Rs. 60
and CI = 100
3
100
201
100
= 100
3
5
6
100 =
5
364
CI SI = 5
364 60 =
5
64
If difference is Rs. 5
64, Principal = Rs. 100
If difference is Rs. 48,
Principal = 64
5100 48 = Rs. 375
33. (C)
100
Pr
1001
2
P
rP
= 5
6
1001
r=
5
6
r = 20%34. (D) Train with a speed of 54km/h passes the
man in 20s.
Length of the train = 54 18
5 20
= 300 mLet the length of platform be x m.
Then, (300 + x) = 54 18
5 36
x = 540 300 = 240 m35. (A) Length of train = 12 15 = 180 m
Time = 18 s
Speed = 18
180 = 10 m/s
New distance = 15 10 = 150 m
Required time = 10
150 = 15s
36. (B) Let the CP of the pen and book be Rs. xand Rs. y respectively.
0.95x + 1.15y = (x + y) + 7 0.15y 0.05x= 7 ... (i)and 1.05x + 1.1y = (x + y) + 13 0.05x + 0.1y = 13 ... (ii)Solving Eqs. (i) and (ii), we get
y = Rs. 8037. (D) Let the cost price of articles be Rs.x.
Then, selling price of article = 0.88x.Marked price of article
= 80
88.0 100 x = 1.1 x
New selling price of article = 1.045x
Profit per cent = x
xx 045.1 100 = 4.5%
38. (A) Let the time of meet = t h
Now,
60
2015 t + 20t = 450
t = 13h
Distance from A = 15
3
113 = 190 km
39. (A) Speed of Ramesh = 3x km/hrSpeed of Suresh = 4x km/hrLet the distance = DATQ,
x
D
3
x
D
4=
2
1
12
1
x
D=
2
1
D = 6x
Time of Ramesh = x
D
3=
x
x
3
6 = 2 h
Time of Suresh = x
D
4=
x
x
4
6= 1.5 h
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40.(*) Work done by A & B in 16 days = 30
16 =
15
8
Remaining work = 1 15
8 =
15
7
15
7 work is done by B alone in 44 days.
1 work is done by B alone in
= 157
44
= 7
1544 = 94.29 days
41. (C) SP = 700
70 MP
SP For Tarun = Rs. 8750
Labelled price = 100
125
100
70MP = 8750
MP = Rs. 1000042. (A) Let the CP of the article be Rs. x.
Then,SP = x 1.12 1.1
Given, x 1.12 1.1 = 616
x = 232.1
616= Rs. 500
43. (B) Let CP = Rs. x and SP = Rs. y. y 7% = x 8%and y 9% = x 10% + 1
and 100
9 y =
100
10 x + 1
7y = 8xand 9y = 10x + 100
9 7
8x= 10x + 100
x = Rs. 35044. (B) Let the numbers be 3x and 4x.
Then,16x2 = 8 (9x)2 224
16x2 = 72x2 224 56x2 = 224 x2 = 4 x = 2Hence, numbers are 6, 8.
45. (A) P = 49
362
2
x
x =
)7)(7(
)6)(6(
xx
xx
Q = 7
6
x
x
Q
P=
)7(
)6()7)(7(
)6)(6(
x
x
xx
xx
=7
6
x
x
46. (B) Sn
= pn + qn2
Sn 1
= p(n 1) + q(n 1)2
= pn p + qn2 2qn + q
an
= Sn S
n 1
= pn + qn2 (pn p + qn2 2qn + q)= p + 2qn q
Common difference= a
2 a
1
= (p + 4q q) (p + 2q q)= 2q Ans.
47.(C) a, b and c are in GP and zyx cba111
.
Let, kcba zyx 111
a = kx
b = ky
c = kz
Now:-b2 = ac [a, b and c are in G.P.]
or, (ky )2 = kx + z
2y = x + z
48. (C)
= 3
4
12
2
1
34
32
2
1
=
7
13
49. (C)zx
y
=
z
xy
yz = xy + x2 yz xz ... (i)
Also,y
x=
zx
y
x2 xz = y2 ... (ii)Using Eqs. (i) and (ii), we get
yz = xy yz + y2
2yz = xy + y2
2z = x + y ... (iii)Only option (C) satisfies the Eq. (iii).
50. (A) CP = x (say)34% of CP = 26% of SP
x26
34= x
13
17= SP
% profit = 10013
1317
x
xx
= 30.77%
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51. (B) Given roots are real and equal.
B2 4AC = 0
[2(a2 bc)]2 4(c2 ab)(b2 ac) = 0
4(a4 + b2c2 2a2bc c2b2 + ac3 + ab3 a2bc = 0
4a(a3 + b3 + c3 3abc) = 0
a3 + b3 + c3 = 3abc
52. (B)a
a
a
aa
1
)1(
1
2/12/12/1
)1(
1
)1)(1(
2 2/1
a
a
aa
a
aa
1
)1)(1(2 2/1
a
aa
1
112 2/12/1
a
aa
1
)(22 2/12/1
a
1
222 =
a1
2
53. (A) Both the expressions are divided by (x 2).Hence f
1(2) = f
2 (2).
( Both remainders are same)
p (2)4 3(2)3 + 20 = 4(2)2 + 7(2) p
16p 24 + 20 = 16 + 14 p
17p = 34
p = 2
Hence, the value of 'p' is 2.
54. (A) )(333 baabbax = 25
)2( 5
3)( bax = 2
125
35x =
352
x = 2
55.(A) Joining point O to three vertices A, B and C.
Now,
Area of ( OBC + OCA + OAB) = area of
ABC.
2
1(a 8 + a 7 + a 6) =
4
3a2 =
2
21a
a = 3
3
3
42
a = 314
Hence, area of triangle ABC.
= 4
3a2
= 4
3( 314 )2 m2
= 254.6 m2
56. (D) Let A1B
1 be the ladder placed against the
wall.A
1B
1= length of ladder = 25 m
B1C = 7 m
Now, ladder foot is drawn out from B1 to B
2
such that B1B
2 = x m, then the top of the
ladder comes down from A1 to A
2 and A
1A
2
= 2
xm (as per question),
So, A1B
1 = A
2B
2 = 25 m and A
2C = Y (say)
In A1B
1C
A1C = 21
211 )()( CBBA
A1A
2 + A
2C = 22 725
2
x + Y = 24 ... (i)
In A2B
2C
(A2C)2 + (B
2C)2 = A
2B
22
Y2 + (x + 7)2 = 252
2
224
x+ (x + 7)2 = 252 [from (i)]
576 + 2
2x 24x + x2 + 49 + 14x = 625
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4
5 2x = 10x
x = 8 m57. (C) Let the circumference of outer circle
C1 = 62.832 m
circumference of inner circleC
2 = 37.6992 m
The area between two circles = A1 A
2 = ?
Here, using the correlation formula forcircumference and area.
A = 4
2C
A1 A
2=
4
21C
4
22C
A1 A
2=
4
22
21 CC
A1 A
2=
4
))(( 2121 CCCC
A1 A
2=
1416.34
)6992.37832.62)(6992.37832.62(
A1 A
2= 201 m2
58. (A)
Let AOB be one quadrant of a circle radius.OA = OB = 10 m
Two circles OKB and OKA are made on theabove radiii as diameter. So, if E and D arecentres.
OD = BD = OE = EA = 5 mSince two circles are equal, k will bemidpoint of arc AKO and OKB.
KDO = 90
OK = 25
Area common to both circle = 2 area ofsegment OXK (or OYK)Using the formula, for central angle = 90Area of segment OXK = 0.285 r2
= 0.285 52
= 7.13 m2
Area common to both circle= 2 7.13= 14.26 m2
59. (B) Let A = 2 pens are red.B = 2 pens are blue.
than P(A) = 9
5
8
4 =
72
20 =
18
5
P(B) = 4
9
3
8 =
12
72 =
6
1
Now, A and B are mutually exclusive events.Hence,P(selecting either 2 red pens or blue pens)= P(A + B) = P(A) + P(B)
= 18
5 +
6
1=
4
9
Hence, the required probability is 4
9.
60. (B) sec2 cosec2 [tan2 + cot2 ]
= (1 + tan2 ) (1 + cot2 ) [tan2 + cot2 ]
[since sec2 = 1 + tan2 ]
= 1 + tan2 + cot2 + tan2 cot2 tan2
cot2 = 1 + 1= 2Hence, the required value is 2.
61.(B)Let AB be the man standing out side a housewhich has a window CD. C (top point) and D(bottom point) of window are viewed frompoint A.
In right DAE,
tan 45 = AE
DE=
3
DE
DE = 3tan 45 = 3 m
In right CAE,
tan 60 = AE
DE=
AE
DECD =
AE
CD 3
3 =
1
3
CD
Therefore, the length of the window is 2.2 m.
62. (C) (1 sin2 A) = Asec
8.0
cos2 A . sec A = 0.8
cos A = 0.8 = 5
4
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sin A = 5
3
tan A = 4
3
So, tan A + Acos
1=
4
3 +
4
5 =
4
8 = 2
63. (A) tan = 3 cot
tan = tan
3
tan2 = 3
tan = + 3
Since is an acute angle = 60
Now, sin2 + cosec2 2
1cot2
= sin2 60 + cosec2 60 2
1cot2 60
=
2
2
3
+
2
3
2
2
1
2
3
1
= 4
3 +
3
4
6
1 =
12
23
64. (C)
Let ABCD be a rectangular grass plot whichhas a gravel path of width = W = 2.5 m(shaded portion)length of plot = l = 112 mbreadth of plot = b = 78 mHere, the path is inside the rectangularplot.Using the formula:-
= 2W(l + b 2W)= 2 2.5 (112 + 78 2 2.5)= 925 m2
Now, cost of construction the path= 925 3.40 = Rs. 3145
65. (A)
Since the triangle is an equilateral, so
each angle = 60 = , r = 15
area of sector = 2
360r
= 360
60
7
22 225
= 117.85 m2.
66. (D)
Let ABCO be the right cone whose slantheight = l = AB = ACheight = h = 200 m
In ABO, ABO = 30
sin 30 = l
200
l = 200 2 = 400 m
2
130sin
cos 30= l
r
r = l cos 30 = 400 2
3 = 3200 m
2
330cos
Now,Area of curved surface = rl
= 7
22 200 3 400
= 435312 m2.67. (A) Let, the side of equilateral triangle is 'a' m.
Area (A) = 4
3a2 m2
Cost of paving = 10 4
3a2 rupees
Similarly, perimeter (P) = 3a m Cost of fencing = 25 a rupeesAccording to the question,
10 4
3a2 = 25 3a
a = 310
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a = 17.32 mHence, the side of the equilateral triangleis 17.32 m.
68. (C)
Let ABCD is a rectangle whosediagonal = d = 25 m
area = A = 168 m2 length = l = ? breadth = b = ?
Using the correlation formula
(l + b)2 = d2 + 2A [Refer 12.4]
and (l + b)2 = d2 2A
(l + b)2 = (25)2 2 + 168
l + b = 31 (i)
Similarly,
(l + b)2 = (25)2 2 168
l b = 17 (ii)
From (i) and (ii)
l = 24 m = length of the rectangle
b = 7 m = breadth of the rectangle
69. (A) The given points A, B and C are collinear.
So, these area of the triangle formedby these points will be zero.
(m + 1)(3 2m) + (2m + 1)(2m 1) +
(2m + 2)(1 3) = 0
2m2 3m 2 = 0
2m2 4m + m 2 = 0
2m(m 2) + 1(m 2) = 0
(2m + 1) (m 2) = 0
m = 2
1 or m = 2
70. (B) Let P(x, y) be the required point dividingthe line AB externally in the ratio 5 : 3.
BP
AB =
3
5 =
m
1 (say)
Using the formula (section)
x = ml
mxlx
12
where x1 = 2, x
2 = 3
35
2335
x =
2
21 = 10.5
and y = ml
myly
12
where, y1 = 3, y
2 = 7
y = 35
)3(375
= 22
Hence, the required point P is ( 10.5, 22).71. (D) Using :-
22
21 dd = 2(a
2 + b2)
a2 + b2 = 2
1[(10)2 + (17.78)2] = 208... (i)
and 2(a + b) = 40 (given) a + b = 20 ..... (ii)From (i) and (ii), we have
a = 12, b = 872. (B) Let ABCD be a rhombus whose
perimeter = p = 36 marea = A = 72 m2
perimeter p = 4 a 36 = 4a a = 9 m
Area, A = a h 72 = 9 h
h = 9
72m
h = 8 m
73. (C)
Given that:-
DA
BD=
DC
DA.... (i)
= DA2 = BD DC .... (ii)
In right triangle AOB and AOC:-
AB2 = AD2 + BD2 .... (i)
and, AC2 = AD2 + DC2 .... (ii)
Adding (A) and (B):-
AB2 + AC2 = 2AD2 + DC2
= 2 BD DC + BD2 + DC2
[from (ii)]
Thus in ABC;
AB2 + AC2 = BC2
Hence, ABC is a right angled at A.
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74. (D) Use length of the rope
= )12(2
ndn
Where d = 2 cm
n = 140
length = 2
]281[140214.3
= 1236.40 m
75. (C) Let ABCDEF be a regular hexagonalpyramid whose base
ABCDEF is a regular hexagon.
If a = each side of regular hexagon,
then, 6a = 30
a = 5 m
height of pyramid = OX = h = 20 m
Volume (V) = 3
1Ah
= 204
36
3
1 2
a
= 2310 a
= 25310
= 250 1.732 m3
= 433 m3
76. (B) It is given that
S = 2.5 k, where k stands for curvedsurface area.
2 r (h + r) = 2.5(2 rh)
h
rh = 2.5
7 + r = 2.5 7 [since h = 7 cm]
r = 10.5 cm
Now,
Volume V = r2h
= 7
22 (10.5)2 7
= 2425.5 cm3
77.(A)
Let ABCD be a rectangular grass plotwhose length = l = 80 m
breadth = b = 60 mTwo roads of width W = 10 m (shaded part)are crossing each other at the middle ofplot.Area of roads = W(l + b W)
= 10(80 + 60 10)m2
= 1300 m2
Cost of gravelling the roads= rate of gravelling / m2 area of roads= Rs. 2 1300= Rs. 2600
78. (C) Let r = radius of hemisphere bowl 2 r = 176 r = 28 cmVolume of the quantity in hemispherical
punch bowl = 3
3
2
2
1
= 328
3
1 cm3
Volume of the bowl in which food is to be
served = 32
3
2 cm3.
No. of persons served
= 3
3
23
2
283
= 1372
79. (D)22
3 yxyx = 81 = 34
x2 xy + y2 = 4 ... (i)
and x y3 3
2 = 256 = 28
x3 + y3 = 8 ... (ii)Dividing (ii) by (i) x + y = 2
80. (B) The given equation is
c + y
yd = e 1 +
y
f
c + y
d
y
y= e 1 +
y
f
c + y
d 1 = e 1 +
y
f
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11
y
d
y
f= e c
y
fd = e c
y = ce
fd
81. (B) The given equation is
bx2 ax + log2 my = 0
Now,
Sum of the roots = x1 + x
2
= b
a
= b
a... (i)
The given relation is
22
21 xx = a
2
(x1 + x
2)(x
1 x
2) = a2
b
a(x
1 x
2) = a2
[From equation (i)]
x1 x
2 = ab ... (ii)
From equation (i) and equation (ii)
x1
=
ab
b
a
2
1 =
b
ba
2
)1( 2
x2
=
ab
b
a
2
1 =
b
ba
2
)1( 2
Hence, the roots are
= b
a
2(b2 + 1),
b
a
2(1 b2)
82.(*) The given expression is
= 2)( lk + 2)( m + 2)( n + m2 lk
n2 m n2 lk
= 2nmlk
So, the square root of the given expressions
= + nmlk
83. (C))7590360sin(2)7590cot(2
)75360tan()75180cos(
= 75cos275tan2
75tan75cos
=
2
1
84. (D) Let OT = height of tower = h metres
PQ = width of the river
Where P = point of the near shore to tower.
Q = point of the far shore to the tower.
ZTA = A (angle of depression)
ZTQ = B (angle of drepression)
Then, ZTA = TPO = A
ZTQ = TQO = B
Now,
In TOP , tan A = OP
h
OP = h cot A .... (i)
In TQO, tan B = OQ
h
= OQOP
h
OP + PQ = h cot B .... (ii)
From (i) and (ii),
PQ = h(cot B cot A)
85. (C)
h2 = a2 + b2
since, a and h are consecutive integers,
h = a + 1
(a + 1)2 = a2 + b2
b2 = 2a + 1
a = 2
12 b
h =2
12 b
So, sin = h
a=
b
b
2
2
1
1
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12
86. (A)
In 11DAP ,
tan 60= 1
11
AD
DP=
1
32000
AD
3 = 1
32000
AD
AD1
= 2000 m
In 22DAP ,
tan 30= 2
22
AD
DP=
2
32000
AD
3
1=
2
32000
AD
AD2
= 6000 mD1D2 = 6000 2000 = 4000 m
Speed of the Jet = 18
4000
5
18
= 800 km/h87. (C)
Let K1 be the reflection of kite K.
Then AK = h (say) = AK1
D is a point above the surface AC of thelake such that CD = d = AE
Then, In KED,
tan 30 = ED
KE=
ED
dh ... (i)
In K1ED,
tan 60 = ED
EK1 =ED
dh ... (ii)
Dividing (i) by (ii), (ED get cancelled)
3
1=
dh
dh
h = 2d
88. (B) Let the no. of sides of the polygon = nSo, there are n interior angles which arein A.P.Now,Sum of n interior angles
= Sn =
2
n[2a + (n 1)d]
Where,a = 120d = 5
Sn = 2
n[2 120 + (n 1)5] ... (i)
But, also, using the formula,Sum of interior angles for n sidedploygon = (2n 4) 90From (1) and (2)
(2n 4) 90 =2
n[2 120 + (n 1)5]
n2 25n + 144 = 0 (n 16)(n 9) = 0
n = 16 or 916th interior angle
= a + (n 1)d, where a = 120, n = 16= 120 + (16 1) 5= 195, which is greater than 180.
Since no interior angle of a regularpolygon can exceed 180.
So, n = 16 is not valid.
Hence, no.of sides of the polygon is 9.
89. (A) Volume of earth dug out
= r2h = 7
22
2
2
5.3
12 = 115.5m3
The earth dug out is exactly spread is forma platform of height h m.
115.5 = 10.5 8 h
h = 1.375 m
90. (C)
Let OABCD be the right pyramid whoseAB = DC = 24 cm
BC = AD = 18 cm
If Z is the mid point of side BC, then OZ =slant height = l = 17 cm
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13
In OXZ, X = 90
OX = height = 22 XZOZ
h =
22
217
DC
h = 22 1217
h = 12.04 cmNow,
Volume = 3
1Ah
= 3
1 24 18 12.04
= 1733.76 cm3
= 1733.5 cm3 (approx.)
91. (A) Required difference
10012000
CD
10015000
EA
100
222512000
100
10515000
= 5640 2250 = 3390
92. (B) There is only one absolute decrease i.e. E.
93. (D) A + B = 360 + 1560 = 1920 i.e.approximately equal to F type of tyres in
October i.e. 1950.
94. (B) Required difference= (750 + 2250 + 3750 + 1950 + 1800) (360 + 1560 + 2640 + 960 + 1440)= 10500 6960 = 3540
[Note : D and E decreasing in sales]95. (D) Number of tyres D and E sold in September
= 3000 + 2040 = 5040Number of tyres D and E sold in October
= 3000 + 1500 = 4500Required percentage
= 4500
5040 100 = 112%
96. (C) Leather goods turnover in Tanzania
= 40 100
25 = $ 10 million
Leather goods turnover in Africa= 30 million
Rest of Africa turnover for leather goods= $ 30 million 10 million= $ 20 million
Required percentage = 30
20 100
= 66.67% = 67%97. (C) Jewellery items turnover in Tanzania
= 40 100
20 = $ 8 million
Jewellery items turnover in Africa= $ 10 million
Jewellery items turnover in rest ofAfrica
= $ 10 million $ 8 million= $ 2 million
Garments items turnover in Tanzania
= 40 100
30= $ 12 million
Garments items turnover in Africa= $ 40 million
Garments items turnover in rest of Africa= 40 12 = $ 28 million
Total of Jewellery items and Garmentsitems turnover in rest of Africa
= 2 + 28 = $ 30 millionTurnover from Tanzania from Electricaland leather goods
= 40 100
10 + 40
100
25
= 4 + 10 = $ 14 million
Ratio = 14
30 = 2.14 : 1
98. (B) Turnover from Tanzania for Electricalgoods and Handicraft together
= 40 100
10 40
100
15
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14
= 4 + 6 = $ 10 millionTurnover of Garment in Tanzania
= 40 100
30 = $ 12 million
Turnover of Garment in Africa= $ 40 million
Turnover from rest of Africa for Garments= $ 40 million $ 12 million= $ 28 million
According to question,28 x = 10
x = 28
10= 0.36 times
99. (D) Turnover from Jewellery and Garmentstogether from Tanzania
= 40 100
20 + 40
100
30
= $ 20 millionTurnover from the rest of Africa forElectrical and Leather goods together
= (15 + 30)
100
251040
= 45 14 = $ 31 millionRequired percentage
= 31
20 100 = 65%
100. (C) Turnover from rest of Africa for Electrical
goods = 15 40 100
10
= $ 11 million
Now, 11 100
120 = $ 13.2 million
Turnover from Tanzania for handicrafts
items = 40 100
15 = $ 6 million
Required ratio = 2.13
6
= 6 : 13
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15
1. A2. C3. A4. A5. B6. D7. A8. C9. B10. C11. C12. A13. D14. C15. C16. D17. *18. C19. B20. D
21. C22. B23. A24. C25. A26. A27. A28. B29. D30. A31. A32. C33. C34. D35. A36. B37. D38. A39. A40. *
41. C42. A43. B44. B45. A46. B47. C48. C49. C50. A51. B52. B53. A54. A55. A56. D57. C58. A59. B60. B
61. B62. C63. A64. C65. A66. D67. A68. C69. A70. B71. D72. B73. C74. D75. C76. B77. A78. C79. D80. B
SSC MAINS (MATHS) MOCK TEST -16 (ANSWER SHEET)
81. B82. *83. C84. D85. C86. A87. C88. B89. A90. C91. A92. B93. D94. B95. D96. C97. C98. B99. D100. C