LIC AAO | QUANT QUIZ - 20 | S W A G

Directions (Q. 1-10): In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer:

1). I. x² – 8x + 15 = 0                II. y² – 4y + 3 = 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

2). I. x² + 12x + 35 = 0              II. y² + 7y + 10 = 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

3). I. (59 / √x) – (97 / √x) = 19√x                  II. (47√y / 23) – (√y / 23) = (1 / √y)

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

4). I. (2/x) + (3/y) = 13              II. (5/x) – (4/y) = -2

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

5). I. [(x+3) / (x+2)] = [(3x-7) / (2x-3)]           II. (4/y) – 3 = (5 / 2y+3)

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

6). I. x² + 2x – 15 = 0                II. y² + 6y – 7 = 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

7). I. 2x² + x – 6 = 0                 II. 2y² – 11y + 14 = 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

8). 14x² – 17x + 15 = 0             II. 4y² + 4y – 3= 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

9). 8x² + 22x + 15 = 0               II. 4y² + 16y + 15 = 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

10). I. X² + 10x + 21 = 0           II. y² + 2y – 8 = 0

a)   If x > y

b)   If x > y

c)   If x < y

d)   If x < y

e)   If x = y or the relationship cannot be established

#Answers:

1. I. x² – 8x + 15 = 0

Or, x² – 5x – 3x + 15 = 0

Or, x(x – 5) – 3(x – 5) = 0

Or, (x – 5) ( x – 3) = 0

:. X = 5, 3

II. y² – 4y + 3 = 0

Or, y² – 3y – y + 3 = 0

Or, y(y – 3) – 1(y – 3) = 0

Or, (y – 3) ( y – 1) = 0

:. Y = 1, 3

Hence x > y

Answer: b)

2. I. x² + 12x + 35 = 0

Or, x² + 7x + 5x + 35 = 0

Or, x(x + 7) + (x + 7) = 0

Or, (x + 7) ( x + 5) = 0

:. X = -7, -5

II. y² + 7y + 10 = 0

Or, y² + 5x + 2x + 10 = 0

Or, y( y + 5) + 2(y + 5) = 0

Or, ( y + 5) ( y+ 2) = 0

:. Y = -5, -2

Hence x < y

Answer: d)

3. I. (59 / √x) – (97 / √x) = 19√x

Or, - (38 / √x) = 19 √x

Or, - (38 / 19) = x

:. X = -2

II. (47√y / 23) – (√y / 23) = (1 / √y)

Or, [(46√y) / 23] = (1 /√y)

Or, 2√y = (1/√y)

Or, 2y = 1

:. Y = (1/2)

Hence x < y

Answer: c)

4. Let (1/x) = u

And (1/y) = v

Now, 2u + 3v = 13   …(i)

And 5u – 4u = -2     …(ii)

Solving equation (i) × 4 and adding equation (ii) × 3, we get

8u + 12u = 52

[(15u – 12u = -6) / (23u = 46)]

:. U = 2 and v = 3

So, (1/x) = 2

:. X = (1/2)

Again, (1/y) = 3

:. Y = (1/3)

Hence x > y

Answer: a)

5. I. [(x+3) / (x+2)] = [(3x-7) / (2x-3)]

Or, (x + 3) (2x – 3) = (x + 2) (3x – 7)

Or, x² – 4x – 5 = 0

Or, x² – 5x + x – 5 = 0

Or, (x – 5) + 1(x – 5) = 0

Or, (x -5) (x + 1) = 0

:. X = 5, -1

II. (4/y) – 3 = (5 / 2y+3)

Or, [(4 – 3y)/ y] = [5 / (2y + 3)

Or, 6y² + 6y – 12 = 0

Or, y² + y – 2 = 0

Or, y² + 2y – y – 2 = 0

Or, y(y + 2) – 1(y = 2) = 0

Or, (y + 2) (y – 1) = 0

Or, y = -2, 1

Hence relationship can’t be established

Answer: e)

6. I. x² + 2x – 15 = 0

Or, x² + 5x – 3x – 15 = 0

Or, x(x + 5) – 3(x + 5) = 0

Or, (x – 3) (x + 5) = 0

:. X = 3, -5  

II. y² + 6y – 7 = 0

Or, y² + 7y – y – 7 = 0

Or, y(y + 7) – 1(y + 7) = 0

Or, (y -1) (y + 7) = 0

Or, y = 1, -7

Hence relationship can’t be established

Answer: e)

7. I. 2x² + x – 6 = 0

Or, 2x² + 4x – 3x – 6 = 0

Or, 2x (x + 2) – 3(x + 2) = 0

Or, (2x – 3) (x + 2) = 0

:. X = (3/2), -2

II. 2y² – 11y + 14 = 0

Or, 2y² – 11y + 14 = 0

Or, y(2y – 7) – 2(2y – 7) = 0

Or, (y – 2) (2y – 7) = 0

:. Y = 2, (7/2)

Hence x < y

Answer: c)

8. I. 14x² – 17x + 15 = 0

Or, 14x² – 10x – 7x + 5 = 0

Or, 2x(7x – 5) – 1(7 – 5) = 0

Or, (2x – 1) (7x – 5) = 0

:. X = (1/2), (5/7)            

II. 4y² + 4y – 3= 0

Or, 4y² + 6y – 2y – 3 = 0

Or, 2y(2y + 3) – 1(2y + 3) = 0

Or, (2y – 1) (2y + 3) = 0

:. Y = (1/2), -(3/2)

Hence x > y

Answer: b)

9. I. 8x² + 22x + 15 = 0

Or, 8x² + 10x + 12x + 15 = 0

Or, 2x(4x + 5) + 3(4x + 5) = 0

Or, (2x + 3) (4x + 5) = 0

:. X = -(3/2), -(5/4)

II. 4y² + 16y + 15 = 0

Or, 4y² + 10y + 6y + 15 = 0

Or, (2y + 5) + 3(2y + 5) = 0

Or, (2y + 5) (2y + 3) = 0

:. Y = -(5/2), -(3/2)

Hence x > y

Answer: b)

10. I. X² + 10x + 21 = 0

Or, x² 7x – 3x + 21 = 0

Or, x(x + 7) + x(x + 7) = 0

Or, (x + 7) (x + 3) = 0

:. X = -7, -3           

II. y² + 2y – 8 = 0

Or, y² + 4y – 2y – 8 = 0

Or, y(y + 4) – 2(y + 4) = 0

Or, (y – 2) (y + 4) = 0

Or, (y -2) (y + 4) = 0

:. Y = 2, -4

Hence relationship can’t be established.

Answer: e)