__Quantitative Aptitude Quadratic Equations Questions (14 – 09 – 2017)__**Directions (1-5): In each of these questions, two or three equations are given. You have to solve both the equations and give answer**

**1. 6x + 9y = 23; 4x + 7y = 42; x + z = 42**

a) x < y = z

b) x < y < z

c) x > y > z

d) x ≤ y < z

**3. √(x + 5) + √1024 = √1369; √y(y-5) = √1764 – √1296**

a) x > y

b) x < y

c) x ≥ y

d) x ≤ y

e) x = y or relation cannot be established

**4. (3x – 2)/y = (3x + 6)/(y + 16)**

**(x + 2)/(y + 4) = (x + 5)/(y + 10)**

a) x > y

b) x < y

c) x ≥ y

d) x ≤ y

e) x = y or relation cannot be
established

**5. (x – 2)(x + 1) = (x – 1)(x + 3)**

**(y + 3)(y – 2) = (y + 1)(y + 2)**

a) x > y

b) x < y

c) x ≥ y

d) x ≤ y

e) x = y or relation cannot be
established

**Directions (6 – 15): In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer**

a) If X > Y

b) If X ≥ Y

c) If X< Y

d) If X≤ Y

e) X=Y or Relationship cannot be
established.

**Solutions:**

**1. B)**Solving 1st and 2nd equations, we get x = 217/6 ≈ -36; y = 80/3 ≈ 26

substituting
in 3rd eqn, z = 78

X
< Y < Z

**3. A)**√(x + 5) + 32 = 37; √(x + 5) = 5;

Squaring
on both sides, (x + 5) = 25

X
= 20;

√y(y-5)
= 42 – 36; √y(y-5) = 6;

Squaring
on both sides, y(y-5) = 36;

Y

^{2}– 5y – 36 = 0; y = +9, -4;
X
> Y

**4. B)**(3x – 2)/y = (3x + 6)/(y + 16)

y
+ 16(3x – 2) = y (3x + 16)

48x
– 8y = 32

6x
– y = 4 —(1)

(x
+ 2)/(y + 4) = (x + 5)/(y + 10)

(x
+ 2)(y + 10) = (x + 5)(y + 4)

2x
= y —(2)

Substitute
(2) in (1),

x
= 1, y = 2

**5. A)**(x – 2)(x + 1) = (x – 1)(x + 3)

X

^{2}+ x – 2x – 2 = x^{2}+ 3x – x – 3
x
= 1/3

(y
+ 3)(y – 2) = (y +1)(y + 2)

Y

^{2}–2y + 3y – 6 = y^{2}+ 2y
y
= – 4

**6. C)**I. (257)

^{1/4}X + (217)

^{1/3}=0

Or, 4.004 X+6.01 = 0

Or, 4.004X= - 6.01

X = - 1.5

II.
√1100 Y + √1295 = 0

Or,
33.17 Y+ 35.99 = 0

Or, Y=
- 1.1

**11. B)**x = -3/4, -1/2; y = -3, -4/5 Therefore x > y

**12. D)**x = -2/3, -1/3; y = -7/4, -2/3 Therefore x ≥ y

**13. D)**x = 1/5, ½; y = 1/7, 1/5 Therefore x ≥ y

**14. B)**I. x – 7√2 x + 24 = 0 => (√x - 3√2)( √x - 4√2) = 0;

If
√x = 3√2 => x = 18 and if √x = 4√2 => x = 32

II.
y – 5√2 y + 12 = 0 => (√y - 2√2)( √y - 3√2) = 0;

If
√y = 2√2 => y = 8 and if √y = 3√2 => y = 18

Therefore
x ≥ y