SSC CHSL/CGL Mains - Aptitude Data Interpretation Questions (With Solutions) Set-34

SSC CHSL/CGL Mains - Aptitude Data Interpretation Questions
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)

 Check Below for Detailed Solutions of the above Aptitude Questions:


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