SBI PO 2019 Notification will be expected soon. It is one of the most expected recruitment among the banking aspirants. Every year the exam pattern for SBI PO has been changing. Depends upon the changing of exam pattern the questions are quite harder compare to the previous year. So the questions are in high level than the candidate’s assumption.
As per the latest trend, our IBPS Guide is providing the updated New Exam Pattern Quantitative Aptitude questions for SBI PO 2019 Day 17. Our Skilled experts were mounting the questions based on the aspirant’s needs. So candidates shall start your preparation and practice on daily basis with our SBI PO pattern quantitative aptitude questions 2019 day 17. Start your effective preparation from the right beginning to get success in upcoming SBI PO 2019.
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Directions (1 – 5): The questions below are based on the given Series-I. The series-I satisfy a certain pattern, follow the same pattern in Series-II and answer the questions given below.
1)
I) 221, 238, 204, 255, 187, 272
II) 116 ………. 82. If 82 is nth term, then what value should come in place of (n + 3)rd term?
a) 184
b) 192
c) 166
d) 178
e) 150
2)
I) 458, 230, 61.5, 17.25, 12.15625
II) 398 …… 12. If 12 is nth term, then find the value of n?
a) 6th term
b) 4th term
c) 5th term
d) 7th term
e) 8th term
3)
I) 382, 416, 457, 511, 586, 692
II) 543, 599…….. 963. If 963 is nth term, then what value should come in place of (n + 1)th term?
a) 1066
b) 1088
c) 1192
d) 1134
e) 1050
4)
I) 12, 7, 9, 21, 88, 709
II) 44…….. 1733. If 1733 is nth term, then find the value of n?
a) 5th term
b) 6th term
c) 7th term
d) 8th term
e) 4th term
5)
I) 1, 4, 36, 576, 14400
II) 3.5 …….. 2016. If 2016 is nth term, then what value should come in place of (n + 1)th term?
a) 50400
b) 42600
c) 38800
d) 45200
e) 48600
Directions (Q. 6 – 10) Study the following information carefully and answer the given questions:
The following table shows the total number of students enrolled for different computer courses and the percentage of total number of students enrolled for given courses from different colleges.
Courses Total students enrolled % of students enrolled for given courses from different colleges
6) Total number of students enrolled for the courses of .NET and Pascal from the college P and S together is approximately what percentage of total number of students enrolled for the courses of Java and Oracle from the college Q and R together?
a) 155 %
b) 170 %
c) 140 %
d) 125 %
e) 115 %
7) Find the difference between the total number of students enrolled for the course .PHP and Pascal from the college P and Q together to that of total number of students enrolled for the course Oracle and .NET from the college R and S together?
a) 2178
b) 3292
c) 1548
d) 1396
e) None of these
8) Find the ratio between the total number of students enrolled for all the given courses from college R to that of college S?
a) 2360 : 2155
b) 3952 : 3787
c) 4963 : 4366
d) 3145 : 2782
e) None of these
9) Find the total number of students enrolled for all the given courses from college P?
a) 4172
b) 3846
c) 2945
d) 3257
e) None of these
10) Total number of students enrolled for the course Java and Pascal is approximately what percentage less than the total number of students enrolled for the course PHP and Oracle?
a) 12 % more
b) 12 % less
c) 35 % more
d) 35 % less
e) 4 % less
Answers:
Directions (Q. 1 – 5):
1) Answer: a)
Series I pattern:
221 is the first term
221 + 17 = 238
238 – 34 = 204
204 + 51 = 255
255 – 68 = 187
187 + 85 = 272
272 is 6th term
Series II pattern:
116 is the first term
116 + 17 = 133
133 – 34 = 99
99 + 51 = 150
150 – 68 = 82 (nth term)
82 + 85 = 167 (n + 1)th term
167 – 102 = 65 (n + 2)nd term
65 + 119 = 184 (n + 3)rd term
2) Answer: c)
Series I pattern:
458 is the first term
(458/2) + 1 = 230
(230/4) + 4 = 61.5
(61.5/6) + 7 = 17.25
(17.25/8) + 10 = 12.15625
12.15625 is 5th term
Series II pattern:
398 is the first term
(398/2) + 1 = 200
(200/4) + 4 = 54
(54/6) + 7 = 16
(16/8) + 10 = 12
12 is 5th term
3) Answer: d)
Series I pattern:
382 is the first term. Then 692 is 6th term.
The difference of difference is, (22 + 3), (32 + 4), (42 + 5), (52 + 6),….
(Or)
The difference of difference of difference is, 6, 8, 10, 12,…
Series II pattern:
543 is first term.
nth term is 963. Then (n + 1)th term is 1134.
4) Answer: b)
Series I pattern:
12 is the first term
12*0.5 + 1 = 7
7*1 + 2 = 9
9*2 + 3 = 21
21*4 + 4 = 88
88*8 + 5 = 709
709 is 6th term
Series II pattern:
44 is the first term
44*0.5 + 1 = 23
23*1 + 2 = 25
25*2 + 3 = 53
53*4 + 4 = 216
216*8 + 5 = 1733
1733 is 6th term
5) Answer: a)
Series I pattern:
1 is the first term
1*22 = 4
4*32 = 36
36*42 = 576
576*52 = 14400
14400 is 5th term
Series II pattern:
3.5 is the first term
3.5*22 = 14
14*32 = 126
126*42 = 2016 (nth term)
2016*52 = 50400 (n + 1)th term
Directions (Q. 6 – 10):
6) Answer: a)
Total number of students enrolled for the courses of .NET and Pascal from the college P and S together
= > 5260*(45/100) + 4200*(41/100)
= > 2367 + 1722 = 4089
Total number of students enrolled for the courses of Java and Oracle from the college Q and R together
= > 1450*(48/100) + 3500*(56/100)
= > 696 + 1960 = 2656
Required % = (4089/2656)*100 = 153.95 % = 155 %
7) Answer: d)
Total number of students enrolled for the course PHP and Pascal from the college P and Q together
= > 2950*(46/100) + 4200*(45/100)
= > 1357 + 1890 = 3247
Total number of students enrolled for the course Oracle and .NET from the college R and S together
= > 3500*(50/100) + 5260*(55/100)
= > 1750 + 2893 = 4643
Required difference = 4643 – 3247 = 1396
8) Answer: c)
The total number of students enrolled for all the given courses from College R
= > 1450*(26/100) + 2950*(22/100) + 5260*(30/100) + 3500*(29/100) + 4200*(32/100)
= > 377 + 649 + 1578 + 1015 + 1344
= > 4963
The total number of students enrolled for all the given courses from College S
= > 1450*(28/100) + 2950*(32/100) + 5260*(25/100) + 3500*(21/100) + 4200*(23/100)
= > 406 + 944 + 1315 + 735 + 966
= > 4366
Required ratio = 4963 : 4366
9) Answer: b)
The total number of students enrolled for all the given courses from the
College P
= > 1450*(24/100) + 2950*(30/100) + 5260*(20/100) + 3500*(23/100) + 4200*(18/100)
= > 348 + 885 + 1052 + 805 + 756
= > 3846
10) Answer: b)
Total number of students enrolled for the course Java and Pascal
= > 1450 + 4200 = 5650
Total number of students enrolled for the course PHP and Oracle
= > 2950 + 3500 = 6450
Required % = [(6450 – 5650)/6450]*100 = 12.40 % = 12 % less
