IBPS PO Quantitative Aptitude Quadratic Equations Quiz For Prelims: 7th March

Dear Aspirants,

Quantitative Aptitude Quiz For IBPS PO/Clerk Prelims

Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

Directions (1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer 

Q1. I. x²+5x+6=0 
II. y²+3y+2=0

if x>y

if x≥y

if x<y

if x≤y

if x = y or no relation can be established between x and y.

Solution:

Q2. I. 4x²-8x+3=0 
II. 2y²-7y+6=0

if x>y

if x≥y

if x<Y

if x≤y

if x = y or no relation can be established between x and y.

Solution:

Q3. I. 4x+2y=51 
II. 15y+13x=221

if x>y

if x≥y

if x<y

if x≤y

if x = y or no relation can be established between x and y.

Solution:

Q4. I. x²-14x+48=0 
II. y²+6=5y

if x>y

if x≥y

if x<y

if x≤y

if x = y or no relation can be established between x and y.

Solution:

Q5. I. x²-1=0 
II. y²+4y+3=0

if x>y

if x≥y

if x<y

if x≤y

if x = y or no relation can be established between x and y.

Solution:

Directions (6-10): Solve the two equations in each question and give answer 

Q6. I. x² + 10x + 24 = 0 
II. 4y² – 17y + 18 = 0

if x<y

if x>y

if x≤y

if x≥y

if x=y or no relation can be established

Solution:

I. x² + 6x +4x + 24 = 0
x (x + 6) + 4 (x + 6) = 0
(x + 4) (x + 6) = 0
x = –4, –6
II. 4y² – 8y – 9y + 18 = 0
4y (y – 2) – 9 (y – 2) = 0
(4y – 9) (y – 2) = 0
y = 9/4, 2
x < y

Q7. I. 16x² + 20x + 6 = 0 
II. 10y² + 38y + 24 = 0

if x<y

if x>y

if x≤y

if x≥y

if x=y or no relation can be established

Solution:

I. 16x² + 8x + 12x +6 = 0
8x (2x + 1) + 6 (2x + 1) = 0
(8x + 6) (2x + 1) = 0
x = -3/4, -1/2
II. 10y² + 30y + 8y + 24 = 0
10y (y + 3) + 8(y + 3) = 0
(10y + 8) (y + 3) = 0
y = -4/5,-3
x>y

Q8. I. 17x² + 48x = 9 
II. 13y² = 32y – 19

if x<y

if x>y

if x≤y

if x≥y

if x=y or no relation can be established

Solution:

I. 17x² + 51x – 3x – 9 = 0
17x (x + 3) –3 (x + 3) = 0
(17x – 3) (x + 3) = 0
x = 3/17,–3
II. 13y² – 13y – 19y + 19 = 0
13y (y – 1) –19 (y – 1) = 0
y = 1, 19/13
x<y

Q9. I. 4x + 7y = 209 
II. 12x – 14y + 9 = –38

if x<y

if x>y

if x≤y

if x≥y

if x=y or no relation can be established

Solution:

4x + 7y =209 ……(i) x (–2) = –8x – 14y = – 418……………..(i)
12x – 14y = – 47 ……………..(ii)
Subtracting (i) from (ii) and solutions
x = 371/20 = 18.55, y = 19.25
x<y

Q10. I. x² = 729 
II. y = √729

if x<y

if x>y

if x≤y

if x≥y

if x=y or no relation can be established

Solution:

x² – 729 = 0
(x – 27) (x + 27) = 0
x = 27, –27
y = √729 = 27
x ≤ y

Directions (11-15): In the following questions, two equations in x and y are given. Solve these equations and give answer 

Q11. I. 3x² – 11x + 10 = 0 
II. 2y² – 3y – 2 = 0

if x>y

if x<y

if x≥y

if x≤y

x=y or relation cannot be established between x and y

Solution:

Q12. 

if x>y

if x<y

if x≥y

if x≤y

x=y or relation cannot be established between x and y

Solution:

Q13. 

if x>y

if x<y

if x≥y

if x≤y

x=y or relation cannot be established between x and y

Solution:

Q14. 

if x>y

if x<y

if x≥y

if x≤y

x=y or relation cannot be established between x and y

Solution:

Q15. I. x² - 9x + 20=0 
II. 2y² - 13y + 18 = 0

if x>y

if x<y

if x≥y

if x≤y

x=y or relation cannot be established between x and y

Solution:

               

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