BANK & INSURANCE (QUADRATIC EQUATION PART 1)

Total Questions: 50

1. Directions: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.

(a) x < y
(b) x ≤ y
(c) x = y or No relation
(d) x > y
(e) x ≥ y

I. x² - x - 42 = 0

II. y² - 16y + 63 = 0

Correct Answer: (b) x ≤ y
Solution:

From I we get:
x² - x - 42 = 0
x² + 6x - 7x - 42 = 0
x(x + 6) - 7(x + 6) = 0
(x + 6)(x - 7) = 0
x = - 6, 7

From II we get:
y² - 16y + 63 = 0
y² - 9y - 7y + 63 = 0
y(y - 9) - 7(y - 9) = 0
(y - 9)(y - 7) = 0
y = 9, 7

2. I. x² + 3x - 18 = 0

II. y² - 9y + 20 = 0

Correct Answer: (a) x < y
Solution:

From I:
x² + 3x - 18 = 0
x² + 6x - 3x - 18 = 0
x(x + 6) - 3(x + 6) = 0
(x - 3)(x + 6) = 0
x = - 6, 3

From II:
y² - 9y + 20 = 0
y² - 4y - 5y + 20 = 0
y(y - 4) - 5(y - 4) = 0
(y - 4)(y - 5) = 0
y = 4, 5

Compare

x values: −6, 3
y values: 4, 5

  • −6 < 4

  • −6 < 5

  • 3 < 4

  • 3 < 5

Final Answer: x < y

3. I. x² + 17x + 72 = 0

II. y² + 13y + 42 = 0

Correct Answer: (a) x < y
Solution:

From I:
x² + 17x + 72 = 0
x² + 8x + 9x + 72 = 0
x(x + 8) + 9(x + 8) = 0
(x + 8)(x + 9) = 0
x = - 8, - 9

From II:
y² + 13y + 42 = 0
y² + 7y + 6y + 42 = 0
y(y + 7) + 6(y + 7) = 0
(y + 7)(y + 6) = 0
y = - 7, - 6

x values: −8, −9
y values: −6, −7

  • −8 < −6

  • −8 < −7

  • −9 < −6

  • −9 < −7

Final Answer: x < y

4. I. x² - 4x - 77 = 0

II. y² + 2y - 143 = 0

Correct Answer: (c) x = y or No relation
Solution:

From I:
x² - 4x - 77 = 0
x² + 7x - 11x - 77 = 0
x(x + 7) - 11(x + 7) = 0
(x + 7)(x - 11) = 0
x = 11, - 7

From II:
y² + 2y - 143 = 0
y² + 13y - 11y - 143 = 0
y(y + 13) - 11(y + 13) = 0
(y + 13)(y - 11) = 0
y = - 13, 11

x values: 11, −7
y values: 11, −13

Check combinations:

  • 11 = 11

  • 11 > −13

  • −7 < 11

  • −7 > −13

Both x > y and x < y occur.

Final Answer: x = y or No relation

5. I. x² - 28x + 192 = 0

II. y² - 24y + 140 = 0

Correct Answer: (c) x = y or No relation
Solution:

From I:
x² - 28x + 192 = 0
x² - 12x - 16x + 192 = 0
x(x - 12) - 16(x - 12) = 0
(x - 12)(x - 16) = 0
x = 12, 16

From II:
y² - 24y + 140 = 0
y² - 10y - 14y + 140 = 0
y(y - 10) - 14(y - 10) = 0
(y - 10)(y - 14) = 0
y = 10, 14

x values: 12, 16
y values: 10, 14

Check:

  • 12 > 10

  • 12 < 14

  • 16 > 10

  • 16 > 14

Since both x > y and x < y occur.

Final Answer: x = y or No relation

6. I. (x + 3)² = 15x - 11

II. (y - 6)² = 4y - 27

Correct Answer: (a) x < y
Solution:

From I:
(x + 3)² = 15x - 11
x² + 6x + 9 = 15x - 11
x² - 9x + 20 = 0
x² - 4x - 5x + 20 = 0
x(x - 4) - 5(x - 4) = 0
(x - 4)(x - 5) = 0
x = 4, 5

From II:
(y - 6)² = 4y - 27
y² - 12y + 36 = 4y - 27
y² - 16y + 63 = 0
y² - 7y - 9y + 63 = 0
y(y - 7) - 9(y - 7) = 0
(y - 7)(y - 9) = 0
y = 7, 9

7. I. x² = 49

II. y² - 9y = 0

Correct Answer: (c) x = y or No relation
Solution:

From I:
x² = 49
x = -7, 7

From II:
y² - 9y = 0
y(y - 9) = 0
y = 0, 9

8. I. x³ = 512

II. y² - 20y + 96 = 0

Correct Answer: (b) x ≤ y
Solution:

From I:
x³ = 512
x = 8

From II:
y² - 20y + 96 = 0
y² - 12y - 8y + 96 = 0
y(y - 12) - 8(y - 12) = 0
(y - 12)(y - 8) = 0
y = 8, 12

9. I. x² - 19x = - 90

II. y² + 7y - 120 = 0

Correct Answer: (d) x > y
Solution:

From I:
x² - 19x = - 90
x² - 19x + 90 = 0
x² - 9x - 10x + 90 = 0
x(x - 9) - 10(x - 9) = 0
(x - 9)(x - 10) = 0
x = 9, 10

From II:
y² + 7y - 120 = 0
y² + 15y - 8y - 120 = 0
y(y + 15) - 8(y + 15) = 0
(y + 15)(y - 8) = 0
y = -15, 8

10. I. 5x² - 48x + 64 = 0

II. 3y² = 192

Correct Answer: (c) x = y or No relation
Solution:

From I:
5x² - 48x + 64 = 0
5x² - 40x - 8x + 64 = 0
5x(x - 8) - 8(x - 8) = 0
(5x - 8)(x - 8) = 0
x = 8, 8/5

From II:
3y² = 192
y² = 64
y = -8, 8