IBPS SO Prelims Exam Model Paper 1

I. 15x2 + 14x – 8 = 0
⇒ 15x2 + 20x – 6x – 8 = 0
⇒ 5x(3x + 4) – 2(3x + 4) = 0
⇒ (3x + 4) (5x – 2) = 0
Thus, x = –

4

3

or

2

5

II. 15y2 – 26y + 8 = 0
⇒ 15y2 – 20y – 6y + 8 = 0
⇒ 5y(3y – 4) – 2(3y – 4) = 0
⇒ (3y – 4) (5y – 2) = 0
Thus, y =

4

3

or

2

5

So, when x = – 4/3, for y =

4

3

, then x < y and when x = –

4

3

, for y =

2

5

, then x < y
And when x =

2

5

, for y =

4

3

, then x < y and when x =

2

5

, for y =

2

5

, then x = y
∴ We can observe that x ≤ y