Quantitative Aptitude: Quadratic Equations Questions Set 58

Directions(1-10): Find the value of x and y and then compare their values.

1. I. x² + 3x – 4 = 0
II. y² + 10y + 24 = 0
y > x
x > y
No relation
x ≥ y
y ≥ x
Option D
𝐈. x² + 3x – 4 = 0
x² + 4x – x – 4 = 0
x(x + 4) – 1 (x + 4) = 0
(x − 1) (x + 4) = 0
x = 1, −4
II. y² + 10y + 24 = 0
y² + 4y + 6y + 24 = 0
y(y + 4) + 6(y + 4) = 0
(y + 6) (y + 4) = 0
y = −4, −6
x ≥ y

2. I. 48/x² − 14/x + 1 = 0
𝐈𝐈. 45/y² + 1/y = 2
y ≥ x
x ≥ y
y > x
x > y
No relation
Option D
𝐈. 48/x² − 14/x + 1 = 0
⇒ x^2 − 14x + 48 = 0
⇒ x^2 − 8x − 6x + 48 = 0
⇒ x(x − 8) − 6(x − 8) = 0
⇒ (x − 8)(x − 6) = 0
x = 8, 6
𝐈𝐈. 45/y² + 1/y = 2
⇒ 2y^2 − y − 45 = 0
⇒ 2y^2 − 10y + 9y − 45 = 0
⇒ 2y(y − 5) + 9(y − 5) = 0
⇒ (2y + 9)(y − 5) = 0
y = 5, − 9/2
x > y

3. I. x² + 2x – 35 = 0
II. y² + 15y + 56 = 0
No relation
x > y
y ≥ x
x ≥ y
y > x
Option D
I.x² + 2x – 35 = 0
=> x² + 7x – 5x – 35 = 0
=> x (x + 7) – 5 (x + 7) = 0
=> (x – 5) (x + 7) = 0
x = 5, –7
II. y² + 15y + 56 = 0
=> y² + 7y + 8y + 56 = 0
=>y (y + 7) + 6 (y + 7) = 0
=>(y + 8) (y + 7) = 0
y = – 8, – 7
x ≥ y

4. I.2x² + 7x + 5 = 0
II. 3y² + 12y + 9 = 0
x > y
y ≥ x
No relation
x ≥ y
y > x
Option C
I.2x² + 7x + 5 = 0
=> 2x² + 2x + 5x + 5 = 0
=> 2x (x + 1) + 5 (x + 1) = 0
=> (2x + 5) (x + 1) = 0
𝑥 = – 5/2 , – 1
II. 3y² + 12y + 9 = 0
=>3y² + 9y + 3y + 9 = 0
=> 3y (y + 3) +3 (y + 3) = 0
=>(3y + 3) (y + 3) = 0
y = –1, – 3
No relation can be established.

5. I.(x – 12)² = 0
II. y² – 21y + 108 = 0
y > x
x > y
y ≥ x
No relation
x ≥ y
Option E
I.(x – 12)² = 0
=> x – 12 = 0
x = 12
II. y² – 21y + 108 = 0
=> y² – 12y – 9y + 108 = 0
=>y (y – 12) – 9 (y – 12) = 0
=>(y – 9) (y – 12) = 0
y = 9, 12
x ≥ y

6. I.x² – 17x + 72 = 0
II. y² – 27y + 180 = 0
x ≥ y
y ≥ x
y > x
x > y
No relation
Option C
I.x² – 17x + 72 = 0
x² – 9x – 8x + 72 = 0
x(x – 9) – 8 (x – 9) = 0
(x – 8) (x – 9) = 0
x = 8, 9
II. y² – 27y + 180 = 0
y² – 12y – 15y + 180 = 0
y(y – 12) – 15 (y – 12) = 0
(y – 15) (y – 12) = 0
y = 15, 12
y > x

7. I.2x² – 5x + 3 = 0
II. 3y² – 4y + 1 = 0
y > x
No relation
x > y
x ≥ y
y ≥ x
Option D
I.2x² – 5x + 3 = 0
=> 2x² – 2x – 3x + 3 = 0
=>2x (x – 1) – 3(x – 1) = 0
=> (x – 1) (2x – 3) = 0
x = 1, 3/2
II. 3y² – 4y + 1 = 0
=>3y² – 3y – y + 1 = 0
=>3y(y – 1) –1 (y – 1) = 0
=> (3y – 1) (y – 1) = 0
=>𝑦 = 1/3 , 1
x≥y

8. I. (x − 2)² = 9
𝐈𝐈. (2y + 8)² = 16
y ≥ x
y > x
No relation
x > y
x ≥ y
Option C
𝐈. x² − 16x + 64 = 0
=> x² − 8x – 8x + 64 = 0
=>x(x − 8) − 8(x − 8) = 0
=> (x − 8) (x − 8) = 0
x = 8, 8
𝐈𝐈. y² − 16y + 63 = 0
=>y² − 7y – 9y + 63 = 0
=> y(y − 7) − 9(y − 7) = 0
=> (y – 9) (y – 7) = 0
y = 9, 7
No relation can be established between x & y

9. I. x² − 16x + 64 = 0
𝐈𝐈. y² − 16y + 63 = 0
y > x
x ≥ y
No relation
x > y
y ≥ x
Option D
𝐈. (x − 2)² = 9
⇒ (x − 2) = ± 3
⇒ x = 5, −1
𝐈𝐈. (2y + 8)² = 16
=>(2y + 8) = ± 4
⇒ y = −2, −6
x > y

10. I. 25/x² − 15/x + 2 = 0
𝐈𝐈. 40/y² + 1 = 13/y
x > y
No relation
y > x
y ≥ x
x ≥ y
Option D
𝐈. 25/x² − 15 x + 2 = 0
⇒ 2x^2 − 15x + 25 = 0
⇒ 2x^2 − 10x − 5x + 25 = 0
=>2x (x − 5) − 5(x − 5) = 0
=>(2x − 5)(x − 5) = 0
x = 5/2 , 5
𝐈𝐈. 40/y² + 1 = 13/y
⇒ y^2 − 13y + 40 = 0
⇒ y^2 − 8y − 5y + 40 = 0
⇒ y(y − 8) − 5(y − 8) = 0
=>(y − 5)(y − 8) = 0
y = 5, 8
y ≥ x