Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for LIC AAO/SBI PO 2019 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
[WpProQuiz 5843]
Click Here to Take LIC AAO Prelims Mock Test
Click Here for SBI PO Pre 2019 High-Quality Mocks Exactly on SBI Standard
Click Here to View Quantitative Aptitude Questions in Hindi
Directions (1 – 5): Study the following information carefully and answer the given questions:
Following table shows the percentage of girls and the difference between the girls and boys in different schools who appeared for final examination in different years.
1) Find the difference between the total number of boys who appeared for final examination in the year 2016 in School P to that of 2017 in School R?
a) 215
b) 278
c) 163
d) 174
e) None of these
2) Find the ratio between the total number of girls who appeared for final examination in School Q and S together in the year 2016 and that of School P and T together in the year 2017?
a) 386 : 629
b) 174 : 357
c) 52 : 115
d) 27 : 98
e) None of these
3) Find the average number of boys in School Q, R and S together who appeared for final examination in the year 2017?
a) 662
b) 606
c) 715
d) 740
e) None of these
4) Total number of students who appeared for final examination in School Q in the year 2016 is approximately what percentage of total number of students who appeared for final examination in School T in the year 2017?
a) 65 %
b) 52 %
c) 40 %
d) 78 %
e) 85 %
5) Find the sum of the total number of students who appeared for final examination in School P, Q and R together in the year 2016?
a) 5000
b) 4000
c) 5500
d) 6000
e) None of these
Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
 6)
I) 9x + 5y = -26
II) 3x – 4y = -20
7)
I) x2 – 19x + 48 = 0
II) y2 – 9y – 52 = 0
8)
I)  2x2 – 20x + 32 = 0
II) 4y2 – 25y + 36 = 0
9)
I) x + √961 = √2916
II) y = ∛9261
10)
I) x2 + 30x + 81 = 0
II) y2 + 16y + 48 = 0
Answers :
Direction (1-5) :
1) Answer: d)
The total number of boys who appeared for final examination in the year 2016 in School P
= > [(52 – 48)/100]*total students in School P = 64
= > Total students in School P = 64*(100/4)
= > Total boys = 64*(100/4)*(48/100)
= > 768
The total number of boys who appeared for final examination in the year 2017 in School R
= > [(55 – 45)/100]*total students in School R = 108
= > Total students in School R = 108*(100/10)
= > Total boys = 108*(100/10)*(55/100)
= > 594
Required difference = 768 – 594 = 174
2) Answer: a)
The total number of girls who appeared for final examination in School Q in the year 2016
= > [(52 – 48)/100]*total students in School Q = 56
= > Total students in School Q = 56*(100/4)
= > Total girls = 56*(100/4)*(48/100)
= > 672
The total number of girls who appeared for final examination in School S in the year 2016
= > [(54 – 46)/100]*total students in School S = 72
= > Total students in School S = 72*(100/8)
= > Total girls = 72*(100/8)*(54/100)
= > 486
The total number of girls who appeared for final examination in School P in the year 2017
= > [(53 – 47)/100]*total students in School P = 72
= > Total students in School P = 72*(100/6)
= > Total girls = 72*(100/6)*(47/100)
= > 564
The total number of girls who appeared for final examination in School T in the year 2017
= > [(51 – 49)/100]*total students in School T = 54
= > Total students in School T = 54*(100/2)
= > Total girls = 54*(100/2)*(49/100)
= > 1323
Required ratio = (672 + 486): (564 + 1323)
= > 1158: 1887
= > 386: 629
3) Answer: b)
The total number of boys in School Q who appeared for final examination in the year 2017
= > [(51 – 49)/100]*total students in School Q = 32
= > Total students in School Q = 32*(100/2)
= > Total boys = 32*(100/2)*(49/100)
= > 784
The total number of boys in School R who appeared for final examination in the year 2017
= > [(55 – 45)/100]*total students in School R = 108
= > Total students in School R = 108*(100/10)
= > Total boys = 108*(100/10)*(55/100)
= > 594
The total number of boys in School S who appeared for final examination in the year 2017
= > [(56 – 44)/100]*total students in School S = 120
= > Total students in School S = 120*(100/12)
= > Total boys = 120*(100/12)*(44/100)
= > 440
Required average = (784 + 594 + 440)/3 = 1818/3 = 606
4) Answer: b)
Total number of students who appeared for final examination in School Q in the year 2016
= > [(52 – 48)/100]*total students in School Q = 56
= > Total students in School Q = 56*(100/4) = 1400
Total number of students who appeared for final examination in School T in the year 2017
= > [(51 – 49)/100]*total students in School T = 54
= > Total students in School T = 54*(100/2) = 2700
Required % = (1400/2700)*100 = 52 %
5) Answer: a)
The total number of students who appeared for final examination in School P in the year 2016
= > [(52 – 48)/100]*total students in School P = 64
= > Total students in School P = 64*(100/4) = 1600
The total number of students who appeared for final examination in School Q in the year 2016
= > [(52 – 48)/100]*total students in School Q = 56
= > Total students in School Q = 56*(100/4) = 1400
The total number of students who appeared for final examination in School R in the year 2016
= > [(51 – 49)/100]*total students in School R = 40
= > Total students in School R = 40*(100/2) = 2000
Required sum = 1600 + 1400 + 2000 = 5000
Direction (6-10) :
6) Answer: c)
9x + 5y = -26 —> (1)
3x – 4y = -20 —> (2)
By substituting (1) and (2), we get,
x = -4, y = 2
x < y
7) Answer: e)
I) x2 – 19x + 48 = 0
(x – 16) (x – 3) = 0
x = 16, 3
II) y2 – 9y – 52 = 0
(y – 13) (y + 4) = 0
y = 13, -4
Can’t be determined
8) Answer: e)
I) 2x2 – 20x + 32 = 0
2x2 – 4x -16x + 32 = 0
2x (x – 2) – 16 (x – 2) = 0
(2x – 16) (x – 2) = 0
X = 16/2, 2 = 8, 2
II) 4y2 – 25y + 36 = 0
4y2 – 16y – 9y + 36 = 0
4y(y – 4) – 9 (y – 4) = 0
(4y – 9) (y – 4) = 0
Y = 9/4, 4 = 2.25, 4
Can’t be determined
9) Answer: a)
I) x + √961 = √2916
X = 54 – 31 = 23
II) y = ∛9261
y = 21
x > y
10) Answer: e)
I) x2 + 30x + 81 = 0
(x + 27) (x + 3) = 0
X = -27, -3
II) y2 + 16y + 48 = 0
(y + 12) (y + 4) = 0
Y = -12, -4
Can’t be determined
