SSC CHSL/CGL Mains - Aptitude Data Interpretation Questions (With Solutions) Set-34:
List of Practice Aptitude Questions for Upcoming SSC CGL Mains / SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.
Directions (Q.1-5): Study the data given
below and answer the given questions.
Analyzing capacity and samples analyzed
for the given years
Year
|
No. of
Fertilizer
Control
labs
|
Analyzing
capacity
|
No. of
Samples
analyzed
|
% Non-
Standard
Samples
|
2006-07
|
45
|
104375
|
76665
|
5.4
|
2007-08
|
64
|
107625
|
93834
|
5.2
|
2008-09
|
62
|
108430
|
95450
|
5.5
|
2009-10
|
62
|
108520
|
78451
|
5.8
|
2010-11
|
65
|
121400
|
92456
|
6.5
|
1).What is the approximate number of
non-standard samples analyzed together during 2007-08 and 2010-11?
a)
10689
b)
10282
c)
10889
d)
10589
e)
10780
2).What is the difference between per lab
analyzing capacity during 2008-09 and 2007-08? (You are not expected to
calculate the exact value.)
a)
67
b)
58
c)
79
d)
62
e)
63
3).What is the average of the standard
analyzed samples during the given five-years? (Take approx values.)
a)
82414
b)
82492
c)
82390
d)
82406
e)
82682
4).What is the approximate average of the
annual average analyzing capacity of a fertilizer control lab during the given
five years?
a)
1678
b)
1874
c)
1768
d)
1797
e)
1886
5).What is the average number of samples
analyzed every year for the given years? (Answer in approx.)
a)
83371
b)
84252
c)
82271
d)
87371
e)
87571
Directions (Q-6-10): Study the given
information carefully and answer the questions.
Total number of
candidates applied (in lakh), appeared (in lakh), % passed (% of appeared
candidates), % selected (% of passed candidates) in IBPS (SO) Exam 2014 in five
different states
State
|
Applied
|
Appeared
|
Passed
|
Selected
|
Jharkhand
|
5.4
|
4.7
|
23
|
15
|
MP
|
7.2
|
5.8
|
25
|
35
|
Delhi
|
3.4
|
2.9
|
29
|
25
|
Haryana
|
5.8
|
5.1
|
40
|
6.25
|
Kerala
|
2.3
|
1.7
|
32
|
18.75
|
6).What is the total number of candidates
selected in IBPS (SO) Exam from all states together?
a)
162940
b)
110940
c)
141940
d)
145440
e)
None of these
7).What is the ratio of the number of
candidates applied from Jharkhand and MP together to the number of candidates
passed from Haryana, Kerala and MP in 2014?
a)
630:217
b)
6300:2067
c)
6300:2017
d)
2017:630
e)
1260:219
8).The number of candidates selected in IBPS
(SO) Exam from Jharkhand and Delhi together is approximately what per cent of
the total number of candidates passed from the same states?
a)
17.5%
b)
21.5%
c)
22%
d)
19.5%
e)
None of these
9).What is the difference between the number
of candidates applied from Delhi, Haryana and Kerala together and the number of
candidates passed from Jharkhand, MP and Haryana together?
a)
592900
b)
690900
c)
692900
d)
792900
e)
None of these
10).The number of candidates selected from MP
is what per cent of the number of candidates passed from Delhi and Haryana
together?
a)
17.61%
b)
15.61%
c)
30.21%
d)
19.41%
e)
None of these
Answers:
1).
c) 2). a) 3).
d) 4). b) 5).
d) 6). b) 7).
c) 8). d) 9).
c) 10). a)
1.
Non-standard samples analyzed =
5.2% of 93834+6.5% of 92456
=
10889.006 ≈ 10889
Answer: c
2.
Per lab analyzing capacity for the
year
2007-08
= 107625/64 = 1681.64
Per
lab analyzing capacity for the year
2008-09
= 108430/62 = 1748.87
Difference
= 1748.87-1681.64 = 67.23 ≈ 67
Answer: a
3.
Number of standard analyzed
samples in 2006-07 = 76665-5.4% of 76665
=
76665-4139.91 = 72525 ≈ 72525
Similarly
in 2007-08 = 93834-5.2% of 93834 = 93834 – 4879.36 ≈ 88954.632 ≈ 88955
In
2008-09 = 95450 – 5.5% of 95450
=
95450 – 5249.75 ≈ 90200
In
2009-10 = 78451 – 5.8% of 78451
=
78451 – 4550.15 ≈ 73901
In
2010-11 = 92456 – 6.5% of 92456
=
92456 – 6009.64 ≈ 86446
Average
=
(72525+88955+90200+73901+86446) / 5
=
412027 / 5
=
82405.4 ≈ 82406
Answer: d
4.
Average analyzing capacity of a
fertilizer lab in 2006-07 = 104375 / 45 = 2319.44 ≈ 2319
In
2007-08 = 107625 / 64 = 1681.64 ≈ 1682
In
2008-09 = 108430 / 62 = 1748.87 ≈ 1749
In
2009-10 = 108520 / 62 = 1750.32 ≈ 1750
In
2010-11 = 121400 / 65 = 1867.69 ≈ 1868
Average
= (2319+1682+1749+1750+1868) / 5
=
1873.6 ≈ 1874
Answer: b
5.
Average number of samples analyzed
=
(76665+93834+95450+78451+92456) / 5
=
87371.2 ≈ 87371
Answer: d
Questions(6-10):
State
|
Passed
|
Selected
|
Jharkhand
|
(470000×23)/100
= 108100
|
(108100×15)/100
= 16215
|
MP
|
(580000×25)/100
= 145000
|
(145000×35)/100
= 50750
|
Delhi
|
(290000×29)/100
= 84100
|
(84100×25)/100
= 21025
|
Haryana
|
(510000×40)/100
= 204000
|
(204000×6.25)/100
= 12750
|
Kerala
|
(170000×32)/100
= 54400
|
(54400×18.75)/100
= 10200
|
6.
Required number of candidates
selected from all states together
=
16215+50750+21025+12750+10200 = 110940
Answer: b
7.
Required ratio
=
[(5.4+7.2)×100000] / [204000+54400+145000]
=
1260000/403400
=
12600/4034 = 6300/2017
=
6300:2017
Answer: c
8.
Required % = {(16215+21025) / (108100+84100)}×100
=
(37240×100)/192200 = 19.37% ≈ 19.5%
Answer: d
9.
Required difference =
(3.4+5.8+2.3)×100000-(108100+145000+204000)
=
1150000-457100 = 692900
Answer: c
10. Required
% = [ 50750 / (84100+204000) ] ×100
=
5075000/288100 = 17.61%
Answer: a
