2007, OUTRAM LINES, 1ST FLOOR, OPPOSITE MUKHERJEE NAGAR POLICE STATION, DELHI-110009
HARYANA SSC MOCK TEST - 45 (SOLUTION)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19. 20.
21.
22.
23.
24.
25.
(C)
(B)
(B)
(A)
(D)
(B)
(D)
(A)
(B)
(A)
(B)
(C)
(A)
(D)
(C)
(A)
(D)
(A)
(A)
(A)
(B)
(D)
(D)
(C)
(D)
26. 27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
(C)
(B)
(A)
(B)
(D)
(A)
(D)
(D)
(A)
(B)
(B)
(D)
(D)
(C)
(B)
(C)
(B)
(B)
(A)
(B)
(A)
(D)
(D)
(A)
(C)
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72.
73.
74.
75.
Explanation:
41. (C) Except option (C), all are used as home
appliance.
G
L
R
Y
G
42. (B) C
+4
43. (B)
+5
+6
+7
MIKE:OGMC::CIAD:EGCB
+2
P R O C E S S
opp.
opp.
opp.
opp.
opp.
opp.
opp.
S E R V I C E
49. (A)
C
opp.
opp.
opp.
opp.
opp.
opp.
opp.
5 km
K I L X V H H
Similarly,
H V I F R X V
I

Middle
-2
-2
(A)
(B)
(A)
(D)
(C)
(B)
(B)
(C)
(A)
(C)
(B)
(C)
(A)
(C)
(B)
(C)
(B)
(B)
(B)
(C)
(C)
(D)
(C)
(B)
(B)
46. (A) p o n m l k j i h g f e d c b a
47. (D) MONITER
48. (D) Sagar
Arjun Aman Biplab
+8
+2
44. (A)
76.
77.
78.
79. 80.
81.
82.
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
93.
94.
95.
96.
97.
98.
99.
100.
(B)
(C)
(D)
(D)
(A)
(A)
(D)
(B)
(A)
(D)
(A)
(C)
(A)
(B)
(D)
(A)
(B)
(C)
(D)
(B)
(D)
(D)
(B)
(D)
(A)
A
8 km
Ram + Mohan
B
2 kms
5 km
D
Required distance = EF
= (CB + BD) - (CF + DE)
= (5 + 5) - (4 + 4) = 10 - 8 = 2 km
50. (C)
(Here, x = 3)
45. (B) (6 × 5) + (3 × 3) = 39
(7 × 5) + (4 × 4) = 51
Similarly,
(5 × 5) + (3 × 4) = 37
Ph: 09555108888,
According to rule Middle Cube :12 (x – 2) or, 12 (3 – 2) or, 12 × 1 = 12
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2007, OUTRAM LINES, 1ST FLOOR, OPPOSITE MUKHERJEE NAGAR POLICE STATION, DELHI-110009
51. (B) Required no. = HCF of (38 – 2), (45 – 3)
and (52 – 4)
= HCF of 36, 42, 48 = 6
52.(C) Let the two digit number be 10x + y.
Then, we have;
x+y=8
.... (i) and
10y + x = 10x + y – 54
54
= 6 .... (ii)
9
From (i) & (ii),
or, x – y =
86
= 7 and y = 1
2
 The required number = 7 × 10 + 1 = 71
53.(D) 3a = 729
&
2b = 1024
3a = 36
and
2b = 210

a=6
and
b = 10

Now,
SP after 10% discount = 90% (100 + x)
SP at 20% profit
= 120% of 100
90% (100 + x) = 120% of 100

120  100
100(120  90)
x =
– 100 =
90
90
100  30
1
= 33 %
90
3
58. (B) Number of each type of coin
=
=
Total amount
Sum of values of each type of coins
=
35
35
=
= 20
1  0.5  0.25 1.75
x=
59. (A) Copper in 1st alloy =
5
7
3
7
Copper in (1st + 2nd) alloy
Copper in 2nd alloy =
4a  6b 4  6  6  10
24  60
84
=
=
=
=2
6b  3a 6  10  3  6
60  18
42
54.(D) Required percentage
1
1
=
1 1 2
By alligation rule,
400
=
× 100 = 26.67%
400  200  800  100
=
wife
son
3
3
= ; daughter =
wife
1
1
 son : wife : daughter = 9 : 3 : 1
son – daughter = ` 10,000
 9x – x = ` 10,000
76
1 3

q I alloy
2 7 = 14 = 1
= q alloy =
II
5 1 10  7 3

14
7 2
So, quantity of 1st alloy
qI
= q + q × 28
I
II
55.(A)
10,000
= ` 1,250
8
Total property of the man
= 9x + 3x + x =13x = 13 × 1250 = ` 16,250
56.(A) Let the cost price of the book = ` 100
 x=
`
120
× 100 = ` 120
100
If he had bought it at 20% less
 CP = 80% of 100 = ` 80
SP1 =
1
× 28 = 7kg
4
Quantity of 2nd alloy
= 28 – 7 = 21 kg
60. (D) Length of the train (when there were 12
boggies) = 12 × 15 = 180 m
125
= ` 100
 SP2 = ` 80 ×
100
SP2 – SP1 = ` 20,CP = ` 100
100
× 18
20
= ` 90
57.(D) Let the cost price of goods be ` 100 and it
is increased by x%.
Then,
MP = ` (100 + x)
SP2 – SP1
= ` 18,CP =
=
180
= 10 m/s
18
New length of the train (when two boggies
are detached)
= 10 × 15 = 150 m
Now, time taken by the train to cross
Its speed =
telegraph post =
150
= 15 seconds
10
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