IBPS SO / NIACL AO Prelims – Quantitative Aptitude Questions Day- 55

Dear Readers, Bank Exam Race for the Year 2018 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for IBPS SO / NIACL AO Prelims 2018 Exams can practice these questions daily and make your preparation effective.

Click here to View Quantitative Aptitude Questions in Hindi

Directions (Q. 1 – 5): Study the following information carefully and answer the given questions:

The following pie chart shows the total number of scans taken by 6 different labs in 1^st week of December 2018 in a certain district.

1) If the ratio of total number of males to that of females taken a scan in lab R is 3: 2, then find the total number of males taken a scan in lab R?

a) 603

b) 578

c) 627

d) 555

e) None of these

2) If the total number of scans taken in 2^nd week of December 2018 is 20 % more than the previous week and the percentage distribution is same as 1^st week of December 2018, then find the difference between the total scans taken in lab P to that of lab S in 2^nd week of December 2018?

a) 132

b) 145

c) 97

d) 116

e) None of these

3) Find the average number of scans taken in lab P, Q and U together?

a) 856

b) 834

c) 813

d) 825

e) None of these

4) Find the central angle of lab T?

a) 82.4⁰

b) 79.2⁰

c) 85.6⁰

d) 77.5⁰

e) None of these

5) If 40 % of the scans taken in lab S is serious cases, then find the total number of normal scans taken in lab S?

a) 534

b) 462

c) 428

d) 556

e) None of these

Directions (Q. 6 – 10): In each of these 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 no relation can be established between x and y

6)

I) 7x² + 3x – 54 = 0

II) 9y² + 18y – 27 = 0

7)

I) 9x² + 22x + 13 = 0

II) y² – 19y – 92 = 0

8)

I) 3x² – 2x – 8 = 0

II) y² + 2y – 15 = 0

9)

I) 5x – 3y = -5

II) 3x + 2y = -22

10)

I) 3x² – 441 = 2x²

II) y = ∛9261

Answers:

Directions (1-5):

1) Answer: c)

Total number of peoples taken a scan in lab R

= > 5500*(19/100) = 1045

The ratio of total number of males to that of females taken a scan in lab R

= > 3: 2

The total number of males taken a scan in lab R

= > 1045*(3/5) = 627

2) Answer: a)

The total number of scans taken in 2^nd week of December 2018

= > 5500*(120/100) = 6600

The difference between the total scans taken in lab P to that of lab S in 2^nd week of December 2018

= > 6600*[(14 – 12)/100] = 132

3) Answer: d)

The average number of scans taken in lab P, Q and U together

= > (5500/3)*[(12 + 15 + 18)/100]

= > 825

4) Answer: b)

The central angle of lab T

= > (22/100)*360 = 79.2⁰

5) Answer: b)

The total number of normal scans taken in lab S

= > 5500*(14/100)*(60/100)

= > 462

Directions (6-10):

6) Answer: e)

I) 7x² + 3x – 54 = 0

7x² + 21x – 18x – 54 = 0

7x (x + 3) – 18 (x + 3) = 0

(7x – 18) (x + 3) = 0

X = 18/7, -3 = 2.57, -3

II) 9y² + 18y – 27 = 0

9y² – 9y + 27y – 27 = 0

9y (y – 1) + 27 (y – 1) = 0

(9y + 27) (y – 1) = 0

Y = -3, 1

Can’t be determined

7) Answer: e)

I) 9x² + 22x + 13 = 0

9x² + 9x + 13x + 13 = 0

9X (x + 1) + 13 (x + 1) = 0

(x + 1) (9x + 13) = 0

X = -1, -13/9 = -1, -1.44

II) y² – 19y – 92 = 0

y² – 23y + 4y – 92 = 0

y (y – 23) + 4 (y – 23) = 0

(y + 4) (y – 23) = 0

Y = -4, 23

Can’t be determined

8) Answer: e)

I) 3x² – 2x – 8 = 0

3x² – 6x + 4x – 8 = 0

3x (x – 2) + 4(x – 2) = 0

(3x + 4) (x – 2) = 0

X = -4/3, 2 = -1.33, 2

II) y² + 2y – 15 = 0

(y + 5) (y – 3) = 0

Y = -5, 3

Can’t be determined

9) Answer: a)

5x – 3y = -5 —> (1)

3x + 2y = -22 —> (2)

By solving the equation (1) and (2), we get,

X = -4, y = -5

X > y

10) Answer: d)

I) 3x² – 441 = 2x²

3x² – 2x² = 441

X² = 441

X = 21, -21

II) y = ∛9261

Y = 21

X ≤ y

Daily Practice Test Schedule | Good Luck