# AAO Exam-CT 6: Quant (Quadratic Equation)

## AAO Exam-CT 6- Quant (Quadratic Equation)

0  538

1. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. x2 – 23x + 132 = 0

II. y2 – 24y + 143 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or relationship between x and y cannot be established.

I. x2 – 23x + 132 = 0

⇒ x2 – 12x – 11x + 132 = 0

⇒ (x –11) (x – 12) = 0

⇒ x = 11, 12

II. y2– 24y + 143 = 0

⇒ y2 – 13y –11y + 143 = 0

⇒ (y – 11) (y – 13) = 0

⇒ y = 11, 13

 Value of x Value of y Relation 11 11 x = y 11 13 x < y 12 11 x > y 12 13 x < y

Hence, relationship between x and y cannot be established.

2. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. x2 – x – 2 = 0

II. y2 + 5y + 6 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or relationship between x and y cannot be established.

I. x2 – x – 2 = 0

⇒ x2 – 2x + x – 2 = 0

⇒ (x – 2)(x + 1) = 0

⇒ x = 2, –1

II. y2+ 5y + 6 = 0

⇒ y2 + 3y + 2y + 6 = 0

⇒ (y + 2)(y + 3) = 0

⇒ y = –2, –3

 Value of x Value of y Relation 2 –2 x > y 2 –3 x > y –1 –2 x > y –1 –3 x > y

Hence, x > y.

3. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. x2 + 3x + 2 = 0

II. y2 + 5y + 6 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or the relationship between x and y cannot be established.

I. x2 + 3x + 2 = 0

⇒ x2 + 2x + x + 2 = 0

⇒ (x + 2) (x + 1) = 0

⇒ x = –2, –1

II. y2 + 5y + 6 = 0

⇒ y2 + 3y + 2y + 6 = 0

⇒ (y + 3) (y + 2) = 0

⇒ y = –3, –2

 Value of x Value of y Relation –2 –3 x > y –2 –2 x = y –1 –3 x > y –1 –2 x > y

Hence x ≥ y

4. In the given question, two equations numbered I and II are given. You have to solve both the equations & mark the appropriate answer-

I) x2 + 6x - 40 = 0

II) y2 + 7y - 60 = 0

• x ≥ y
• x ≤ y
• x = y or relationship between x and y cannot be established
• x < y
• x > y

I) x2 + 6x - 40 = 0

⇒ x2 + 10x + 4x - 40 = 0

⇒ x(x + 10) - 4(x + 10) = 0

⇒ (x + 10)(x - 4) = 0

x = -10 or 4

II) y2 + 7y - 60 = 0

⇒ y2 - 5y + 12y - 60 = 0

⇒ y(y - 5) + 12(y - 5) = 0

⇒ (y + 12)(y - 5) = 0

y = -12 or 5

∴ No relation

5. In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer-

I. x2 – 4x + 4 = 0 II. y2 – 5y + 6 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or relation between x and y cannot be established

From (I)

x2 – 4x + 4 = 0

⇒ x2 - 2x - 2x + 4 = 0

⇒ x(x - 2) -2 (x - 2) = 0

⇒ (x - 2) (x - 2) = 0

Taking

(x - 2) = 0, then

x = 2

From (II)

y2 - 5y + 6 = 0

⇒ y2 - 3y - 2y + 6 = 0

⇒ y (y - 3) -2 (y - 3) = 0

⇒ (y - 3) (y - 2) = 0

 value of x value of y Relation 2 3 x < y 2 2 x = y

x ≤ y

6. In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer:

I. 3x2 – 11x + 10 = 0

II. 4y2 + 24y + 35 = 0

• x > y
• x ≥ y
• x < y
• x ≤ y
• x = y or the relation cannot be determined

Equation I.

3x2 – 11x + 10 = 0

⇒ 3x2 – 5x – 6x + 10 =0

⇒ x(3x – 5) – 2(3x – 5) = 0

⇒ (x – 2)(3x – 5) = 0

⇒ x – 2 = 0 or 3x – 5 = 0

⇒ x = 2 or x = 5/3

Equation II.

4y2 + 24y + 35 = 0

⇒ 4y+ 14y + 10y + 35 = 0

⇒ 2y(2y + 7) + 5(2y + 7) = 0

⇒ (2y + 5)(2y + 7) =0

⇒ 2y + 5 = 0 or 2y + 7 = 0

⇒ y = -5/2 or y = -7/2

 Value of x Value of y Relation 2 -5/2 x > y 2 -7/2 x > y 5/3 -5/2 x > y 5/3 -7/2 x > y

∴ x > y

7. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

i. x2 – 6x +  8 = 0

ii. y2 – 10y + 24 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or relationship between x and y cannot be established.

i. x2 – 6x + 8 = 0

⇒ x2 – 2x – 4x + 8 = 0

⇒ x(x – 2) – 4(x – 2) = 0

⇒ (x – 2) (x – 4) = 0

⇒ x = 2, 4

y2 – 10y + 24 = 0

⇒ y2 – 6y – 4y + 24 = 0

⇒ y(y – 6) – 4(y – 6) = 0

⇒ (y – 6) (y – 4) = 0

⇒ y = 6, 4

 Value of x Value of y Relation 2 6 x < y 2 4 x < y 4 6 x < y 4 4 x = y

∴ x ≤ y

8. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. 20x2 – 17x + 3 = 0

II. 8y2 + 10y – 3 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or relationship between x and y cannot be established.

I. 20x2 – 17x + 3 = 0

⇒ 20x2 – 5x – 12x + 3 = 0

⇒ 5x(4x – 1) – 3(4x – 1) = 0

⇒ (4x – 1) (5x – 3) = 0

⇒ x = 1/4, 3/5

II. 8y2 + 10y – 3 = 0

⇒ 8y2 + 12y – 2y – 3 = 0

⇒ 4y(2y + 3) – 1(2y + 3) = 0

⇒ (4y – 1) (2y + 3) = 0

⇒ y = 1/4, –3/2

 Value of x Value of y Relation 1/4 1/4 x = y 1/4 –3/2 x > y 3/5 1/4 x > y 3/5 –3/2 x > y

∴ x ≥ y

9. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. x2 – 58x + 840 = 0

II. y2– 53y + 700 = 0

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y or relationship between x and y cannot be established.

I. x2 – 58x + 840 = 0

⇒ x2 – 28x – 30x + 840 = 0

⇒ (x – 30) (x – 28) = 0

⇒ x =30, 28

II. y2– 53y + 700 = 0

⇒ y2 – 25y – 28y + 700 = 0

⇒ (y – 25) (y – 28) = 0

⇒ y = 25, 28

 Value of x Value of y Relation 30 25 x > y 30 28 x > y 28 25 x > y 28 28 x = y

Hence, x ≥ y.

10. In the following questions, two equations numbered I and II are given. You have to solve both the equations and find the relation between X and Y.

I. 2x2 – x – 10 = 0

II. 2y2 – 9y + 10 = 0

• x > y
• x ≥ y
• x < y
• x ≤ y
• x = y, Or the relationship cannot be established.

Calculations:

I. 2x2 – x – 10 = 0

⇒ 2x2 – 5x + 4x – 10 = 0

⇒ x(2x – 5) + 2(2x – 5) = 0

⇒ (2x – 5) (x + 2) = 0

⇒ x = 5/2 and x = -2

II. 2y2 – 9y + 10 = 0

⇒ 2y2 – 5y – 4y + 10 = 0

⇒ y(2y – 5) – 2(2y – 5) = 0

⇒ (y – 2) (2y – 5) = 0

⇒ y = 2 and y = 5/2

 Value of X Value of y Comparison 5/2 2 x > y 5/2 5/2 x = y -2 2 x < y -2 5/2 x < y

∴ Relationship cannot be established.