AAO Exam-CT 6: Quant (Quadratic Equation)
AAO Exam-CT 6- Quant (Quadratic Equation)
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
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 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 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
-
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 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 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
⇒ 4y2 + 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 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 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 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
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.
What's Your Reaction?
