Quantitative Aptitude

LIC AAO/SBI PO Prelims Quantitative Aptitude Questions 2019 (Day-12)

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Directions (Q. 1 – 5): 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

 

1)

I) x2 + ∛59319 = (529 ÷ 23) × 8 – 24

II) y2 – 26y + 169 = 0

 

2)

I) x2 – 19x + 78 = 0

II) y2 + 14y – 51 = 0

 

3)

I) 35x2 – 3x – 2 = 0

II) 49y2 + 7y – 6 = 0

 

4)

I) x2 – 12x + 35 = 0

II) y2 – 8y +15 = 0

 

5)

I) 5x2 + 19x + 12 = 0

II) 32y2 – 28y + 3 = 0

Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:

The following table shows the total number of candidates (In lakhs) enrolled for the JEE examination from 6 different cities and also shows the ratio of passed and failed candidates among them.

6) Find the total pass percentage in Chennai?

a) 40 %

b) 25 %

c) 50 %

d) 35 %

e) None of these

7) Find the total number of failed candidates in Bangalore, Delhi and Pune together?

a) 284000

b) 256000

c) 320000

d) 340000

e) None of these

8) Find the difference between the total number of passed candidates from Mumbai and Kolkata together to that of total number of candidates enrolled from Chennai?

a) 48000

b) 35000

c) 52000

d) 26000

e) None of these

9) If the ratio of male to that of female candidates enrolled for the examination from Chennai is 3 : 2, then find the ratio between the total number of male candidates enrolled for the examination from Chennai to that of total number of candidates enrolled for the examination from Bangalore?

a) 29: 60

b) 35: 72

c) 44: 91

d) 57: 113

e) None of these

10) Total number of candidates enrolled for the examination from Delhi and Mumbai together is approximately what percentage of total number of candidates enrolled for the examination from Kolkata and Pune together?

a) 70 %

b) 50 %

c) 84 %

d) 100 %

e) 115 %

Answers :

Direction (1-5) :

1) Answer: c)

I) x2 + ∛59319 = (529 ÷ 23) × 8 – 24

x2 + 39 = (529/23)*8 – 24

x2 = 184 – 24 – 39

x2 = 121

x = 11, -11

II) y2 – 26y + 169 = 0

(y – 13) (y – 13) = 0

y = 13, 13

x < y

2) Answer: a)

I) x2 – 19x + 78 = 0

(x – 13) (x – 6) = 0

X = 13, 6

II) y2 + 14y – 51 = 0

(y + 17) (y – 3) = 0

Y = -17, 3

X > y

3) Answer: e)

I) 35x2 – 3x – 2 = 0

35x2 + 7x – 10x – 2 = 0

7x (5x + 1) – 2 (5x + 1) = 0

(7x – 2) (5x + 1) = 0

X = 2/7, -1/5 = 0.285, -0.2

II) 49y2 + 7y – 6 = 0

49y2 – 14y + 21y – 6 = 0

7y (7y – 2) + 3 (7y – 2) = 0

(7y + 3) (7y – 2) = 0

Y = – 3/7, 2/7 = – 0.428, 0.285

Can’t be determined

4) Answer: b)

I) x2 – 12x + 35 = 0

(x – 7) (x – 5) = 0

X = 7, 5

II) y2 – 8y + 15 = 0

(y – 5) (y – 3) = 0

Y = 5, 3

X ≥ y

5) Answer: c)

I) 5x2 + 19x + 12 = 0

5x2 + 15x+ 4x + 12 = 0

5x (x + 3) + 4 (x + 3) = 0

(5x + 4) (x + 3) = 0

x = -4/5, -3 = -0.8, -3

II) 32y2 – 28y + 3 = 0

32y2 – 28y + 3 = 0

32y2 – 24y – 4y + 3 = 0

(8y – 1) (4y – 3) = 0

y = 1/8, ¾ = 0.125, 0.75

x < y

Direction (6-10) :

6) Answer: a)

Total candidates enrolled = 145000

Total number of passing candidates = 145000*(2/5) = 58000

Required % = (58000/145000)*100 = 40 %

(Or)

The ratio of passed to that of failed candidates = 2: 3 (2x, 3x)

Required % = (2x/5x)*100 = 40 %

7) Answer: d)

The total number of failed candidates in Bangalore, Delhi and Pune together

= > 180000*(5/9) + 165000*(7/11) + 225000*(3/5)

= > 100000 + 105000 + 135000 = 340000

8) Answer: b)

The total number of passed candidates from Mumbai and Kolkata together

= > 240000 *(5/12) + 260000*(4/13)

= > 100000 + 80000 = 180000

The total number of candidates enrolled from Chennai = 145000

Required difference = 180000 – 145000 = 35000

9) Answer: a)

The ratio of male to that of female candidates enrolled for the examination from Chennai = 3: 2 (3x, 2x)

The total number of male candidates enrolled for the examination from Chennai

= > 145000*(3/5) = 87000

The total number of candidates enrolled for the examination from Bangalore

= > 180000

Required ratio = 87000: 180000 = 29: 60

10) Answer: c)

Total number of candidates enrolled for the examination from Delhi and Mumbai together

= > 1.65 + 2.4 = 4.05 lakhs = 405000

Total number of candidates enrolled for the examination from Kolkata and Pune together

= > 2.6 + 2.25 = 4.85 lakhs = 485000

Required % = (405000/485000)*100 = 83.5 % = 84 %

 

This post was last modified on May 4, 2019 12:14 pm