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;
Y2
– 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)
X2
+ x – 2x – 2 = x2 + 3x – x – 3
x
= 1/3
(y
+ 3)(y – 2) = (y +1)(y + 2)
Y2
–2y + 3y – 6 = y2 + 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
