Download - Ssc Tier II Maths Solution Paper 2
SSC (T-II) 2013 (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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=
����
�
�
���
���
��
�
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 + 1www.ssc-cgl2014.in
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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
24��x +
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
51��x +
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 = 8x
and 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 cba
111
�� .
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%www.ssc-cgl2014.in
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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 =
a�1
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 � = 2125
�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 = 2
1211 )()( 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 = 90º
� OK = 25
Area common to both circle = 2 × area ofsegment OXK (or OYK)Using the formula, for central angle = 90ºArea 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 45º = AE
DE=
3
DE
� DE = 3tan 45º = 3 m
In right �CAE,
tan 60º = 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 � � = 60º
Now, sin2 � + cosec2 � – 2
1cot2 �
= sin2 60º + cosec2 60º – 2
1cot2 60º
=
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 = 60º = � , r = 15
� area of sector = 2
º360r��
�
= 360
º60×
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 = 30º
sin 30º = l
200
� l = 200 × 2 = 400 m
��
�
��
2
1º30sin
cos 30º= l
r
� r = l cos 30 º = 400 × 2
3 = 3200 m
���
�
��
2
3º30cos
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
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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 x = 2, x = –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(a2 + 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
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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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�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 � = a2
� (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))º75º90º360sin(2)º75º90cot(2
)º75º360tan()º75º180cos(
����
���
= º75cos2º75tan2
º75tanº75cos
��
�� =
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 = Aº (angle of depression)
�ZTQ = Bº (angle of drepression)
Then, � ZTA = � TPO = Aº
� ZTQ = � TQO = Bº
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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86. (A)
In 11DAP� ,
tan 60º= 1
11
AD
DP=
1
32000
AD
3 = 1
32000
AD
� AD1
= 2000 m
In 22DAP� ,
tan 30º= 2
22
AD
DP=
2
32000
AD
3
1=
2
32000
AD
� AD2
= 6000 m�D
1D
2= 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 30º = ED
KE=
ED
dh �... (i)
In �K1ED,
tan 60º = 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 = 120ºd = 5
� Sn
= 2
n[2 × 120º + (n – 1)5º] ... (i)
But, also, using the formula,Sum of interior angles for n – sidedploygon = (2n – 4) × 90ºFrom (1) and (2)
(2n – 4) × 90º =2
n[2 × 120º + (n – 1)5º]
� n2 – 25n + 144 = 0� (n – 16)(n – 9) = 0
n = 16 or 916th interior angle
= a + (n – 1)d, where a = 120º, n = 16= 120 + (16 – 1) × 5= 195º, which is greater than 180º.
Since no interior angle of a regularpolygon can exceed 180º.
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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In �OXZ, �X = 90º
� 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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= 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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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 (ANSWER SHEET)
81. B82. *83. C84. D85. C86. A87. C88. B89. A90. C91. A92. B93. D94. B95. D96. C97. C98. B99. D100. C
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