IBPS SO Prelims Exam Model Paper 2

I. x2 – 11x + 24 = 0
⇒ x2 – 8x – 3x + 24 = 0
⇒ x(x – 8) – 3(x – 8) = 0
⇒ (x – 8)(x – 3) = 0
Then, x = (8) or x = (3)
II. 2y2 – 9y + 9 = 0
y2 – 6y – 3y + 9 = 0
⇒ 2y(y – 3) – 3(y – 3) = 0
⇒ (2y – 3)(y – 3) = 0
Then, y = (

3

2

) or y = (3)
So, when x = (8), x > y for y = (

3

2

) and x > y for y = (3)
And when x = (3), x > y for y = (

3

2

) and x = y for y = (3)
∴ x ≥ y