Dear Aspirants,

Quantitative Aptitude Questions for IBPS Clerk Mains 2017

Quantitative Aptitude is a very important section you must prepare if you are aiming for a job in Bank or Insurance sector. So, these 15 questions can help you practice three very important topics of Quant Section.

Directions (1-15): In each of the following questions two equations are given. You have to solve the equations and
(a) if x < y
(b) if x ≤ y
(c) relationship between x andy cannot be determined
(d) if x ≥ y
(e) if x > y

Q1. I. 3x² + 10x + 8 = 0
II. 3y² + 7y + 4 = 0

S1. Ans.(b)
Sol.
I. 3x² + 10x + 8 = 0
⇒ 3x² + 6x + 4x + 8 = 0
⇒ (x + 2) (3x + 4) = 0
⇒ x = –2, –4/3
II. 3y² + 7y + 4 = 0
⇒ 3y² + 3y + 4y + 4 = 0
⇒ (y + 1) (3y + 4) = 0
⇒ y = –1, –4/3

y ≥ x

Q2. I. 2x² + 21x + 10 = 0
II. 3y² + 13y + 14 = 0

S2. Ans.(c)
Sol.
I. 2x² + 21x + 10 = 0
⇒ 2x² + 20 + x + 10 = 0
⇒ (x + 10) (2x + 1) = 0
⇒ x = –10, –1/2

II. 3y² + 13y + 14 = 0
⇒ 3y² + 6y + 7y + 14 = 0
⇒ (y + 2) (3y + 7) = 0
⇒ y = –2, –7/3
No relation

Q3. I. x² + x – 12 = 0
II. y² + 2y – 8 = 0

S3. Ans.(c)
Sol.
I. x² + x – 12 = 0
⇒ x² + 4x – 3x – 12 = 0
⇒ (x + 4) (x – 3) = 0
⇒ x = 3, –4

II. y² + 2y – 8 = 0
⇒ y² + 4y – 2y – 8 = 0
⇒ (y + 4) (y – 2) = 0
⇒ y = – 4, 2
No relation

Q4. I. 4x² – 13x + 9 = 0
II. 3y² – 14y + 16 = 0

S4. Ans.(c)
Sol.
I. 4x² – 13x + 9 = 0
⇒ 4x² – 4x – 9x + 9 = 0
⇒ (x – 1) (4x – 9) = 0
⇒ x = 1, 9/4

II. 3y² – 14y + 16 = 0
⇒ 3y² – 6y – 8y + 16 = 0
⇒ (y – 2) (3y – 8) = 0
⇒ y = 2, 8/3
No relation

Q5. I. 8x² + 18x + 9 = 0
II. 4y² + 19y + 21 = 0

S5. Ans.(e)
Sol.
I. 8x² + 18x + 9 = 0
⇒ 8x² + 12x + 6x + 9 = 0
⇒ (2x + 3) (4x + 3) = 0
⇒ x = –3/2, –3/4
II. 4y² + 19y + 21 = 0
⇒ 4y² + 12y + 7y + 21 = 0
⇒ (y + 3) (4y + 7) = 0
⇒ x = –3, –7/4
x >y

Q6. I. 3x² + 16x + 21 = 0
II. 6y² + 17y + 12 = 0

S6. Ans.(a)
Sol.
I. 3x² + 16x + 21 = 0
⇒ 3x² + 9x + 7x + 21 = 0
⇒ (x + 3) (3x + 7) = 0
⇒ x = –3, –7/3
II. 6y² + 17y + 12 = 0
⇒ 6y² + 9y + 8y + 12 = 0
⇒ 3y (2y + 3) + 4 (2y + 3) = 0
⇒ y = – 3/2, –4/3
y > x

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

S7. Ans.(e)
Sol.
I. 16x² + 20x + 6 = 0
⇒ 8x² + 10x + 3 = 0
⇒ 8x² + 4x + 6x + 3 = 0
⇒ (2x + 1) (4x + 3) = 0
⇒ x = –1/2, –3/4

II. 10y² + 38y + 24 = 0
⇒ 5y² + 19y + 12 = 0
⇒ 5y² + 15y + 4y + 12 = 0
⇒ (y + 3) (5y + 4) = 0
y = –3, –4/5
x > y

Q8. I. 8x² + 6x = 5
II. 12y² – 22y + 8 = 0

S8. Ans.(b)
Sol.
I. 8x² + 6x – 5 = 0
⇒ 8x² + 10x – 4x – 5 = 0
⇒ (4x + 5) (2x – 1) = 0
⇒ x = ½, –5/4

II. 12y² – 22y + 8 = 0
⇒ 6y² – 11y + 4 = 0
⇒ 6y² – 3y – 8y + 4 = 0
⇒ (2y – 1) (3y – 4) = 0
⇒ y = 1/2, 4/3
y ≥ x

Q9. I. 17x² + 48x = 9
II. 13y² = 32y – 12

S9. Ans.(a)
Sol.
I. 17x² + 48x – 9 = 0
⇒ 17x² + 51x – 3x – 9 = 0
⇒ (x + 3) (17x – 3) = 0
⇒ x = 3/17, – 3

II. 13y² – 32y + 12 = 0
⇒ 13y² – 26y – 6y + 12 = 0
⇒ (y – 2) (13y – 6) = 0
⇒ y = 2, 6/13
y > x

Q10. I. 8x² + 26x + 15 = 0
II. 4y² + 24y + 35 = 0

S10. Ans.(d)
Sol.
I. 8x² + 26x + 15 = 0
⇒ 8x² + 20x + 6x + 15 = 0
⇒ 4x (2x + 5) + 3(2x + 5) = 0
⇒ (2x + 5) (4x + 3) = 0
⇒ x = – 5/2, –3/4

II. 4y² + 24y + 35 = 0
⇒ 4y² + 10y + 14y + 35 = 0
⇒ 2y (2y + 5) + 7 (2y + 5) = 0
⇒ (2y + 5) (2y + 7) = 0
⇒ y = –5/2, –7/2
x ≥ y

Q11. I. 6x² + 19x + 15 = 0
II. 24y² + 11y + 1 = 0

S11. Ans.(a)
Sol.
I. 6x² + 19x + 15 = 0
⇒ 6x² + 9x + 10x + 15 = 0
⇒ (2x + 3) (3x + 5) = 0
⇒ x = –3/2, –5/3

II. 24y² + 11y + 1 = 0
⇒ 24y² + 8y + 3y + 1= 0
⇒ (3y + 1) (8y + 1) = 0
⇒ y = –1/3, –1/8
y > x

Q12. I. 2x² + 11x + 15 = 0
II. 4y² + 22y + 24 = 0

S12. Ans.(c)
Sol.
I. 2x² + 11x + 15 = 0
⇒ 2x² + 6x + 5x + 15 = 0
⇒ (x + 3) (2x + 5) = 0
⇒ x = – 3, –5/2

II. 4y² + 22y + 24 = 0
⇒ 2y² + 11y + 12 = 0
⇒ 2y² + 8y + 3y + 12 = 0
⇒ (y + 4) (2y + 3) = 0
⇒ y = –4, –3/2
No relation

Q13. I. 2x² + 9x + 9 = 0
II. 2y² + 17y + 36 = 0

S13. Ans.(e)
Sol.
I. 2x² + 9x + 9 = 0
⇒ 2x² + 6x + 3x + 9 = 0
⇒ (x + 3) (2x + 3) = 0
⇒ x = –3, –3/2

II. 2y² + 17y + 36 = 0
⇒ 2y² + 8y + 9y + 36 = 0
⇒ (y + 4) (2y + 9) = 0
y = – 4, –9/2
x > y

Q14. I. 5x² + 29x + 20 = 0
II. 25y² + 25y + 6 = 0

S14. Ans.(a)
Sol.
I. 5x² + 29 + 20 = 0
⇒ 5x² + 25x + 4x + 20 = 0
⇒ (x + 5) (5x + 4) = 0
⇒ x = –5, –4/5
II. 25y² + 25y + 6 = 0
⇒ 25y² + 15y + 10y + 6 = 0
⇒ (5y + 3) (5y + 2) = 0
⇒ y = – 3/5, –2/5
y > x

Q15. I. 3x² – 16x + 21 = 0
II. 3y² – 28y + 65 = 0

S15. Ans.(a)
Sol.
I. 3x² – 16x + 21 = 0
⇒ 3x² – 9x – 7x + 21 = 0
⇒ (x – 3) (3x – 7) = 0
⇒ x = 3, 7/3

II. 3y² – 28y + 65 = 0
⇒ 3y² – 15y – 13y + 65 = 0
⇒ (y – 5)(3y – 13) = 0
⇒ y = 5, 13/3

y > x