LIC AAO/SBI PO Prelims Quantitative Aptitude Questions 2019 (Day-20)

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Quantitative Aptitude Questions For LIC AAO/SBI PO Prelims (Day-20)

maximum of 10 points
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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

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