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# Quant Quiz On Quadratic Equation Questions Day 4 Bag

130+ SSC CGL Previous Papers With Solution Free >>

# Quant Quiz On Quadratic Equation Questions Day 4 Bag

In the following questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x2 +23x +120 = 0
II. 15y2 +14y -49 = 0
Change Language
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
Solution :A✔️
From equation I:
x2 +23x +120 = (x + 15)(x + 8)= 0
=> x = -15, -8
From equation II:
15y2 +14y -49 = (3y + 7)(5y -7) = 0
=> y = -7/3, 7/5
X = -15 X = -8
Y = -7/3 x < y x < y
Y = 7/5 x < y x < y
So, x < y
I. x2 -35x +300 = 0
II. y2 -14y +45 = 0
Change Language
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
B✔️
Solution :
From equation I:
x2 -35x +300 = (x -15)(x -20)= 0
=> x = 15, 20
From equation II:
y2 -14y +45 = (y -5)(y -9) = 0
=> y = 5, 9
X = 15 X = 20
Y = 5 x > y x > y
Y = 9 x > y x > y
So, x > y
I. x2 -12x -45 = 0
II. y2 +37y +340 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
B✔️
Solution :
From equation I:
x2 -12x -45 = (x -15)(x + 3)= 0
=> x = 15, -3
From equation II:
y2 +37y +340 = (y + 20)(y + 17) = 0
=> y = -20, -17
X = 15 X = -3
Y = -20 x > y x > y
Y = -17 x > y x > y
So, x > y
I. x2 +15x -34 = 0
II. 10y2 +7y -45 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
E✔️✔️
From equation I:
x2 +15x -34 = (x + 17)(x -2)= 0
=> x = -17, 2
From equation II:
10y2 +7y -45 = (5y -9)(2y + 5) = 0
=> y = 9/5, -5/2
X = -17 X = 2
Y = 9/5 x < y x > y
Y = -5/2 x < y x > y
So, relationship cannot be established between x and y
In the following questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 10×2 -41x +40 = 0
II. 10y2 -7y -45 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
E✔️
From equation I:
10×2 -41x +40 = (5x -8)(2x -5)= 0
=> x = 8/5, 5/2
From equation II:
10y2 -7y -45 = (2y -5)(5y + 9) = 0
=> y = 5/2, -9/5
X = 8/5 X = 5/2
Y = 5/2 x < y x = y
Y = -9/5 x > y x > y
So, relationship cannot be established between x and y
I. x2 -20x +100 = 0
II. y2 -38y +357 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
A✔️✔️
From equation I:
x2 -20x +100 = (x -10)(x -10)= 0
=> x = 10, 10
From equation II:
y2 -38y +357 = (y -21)(y -17) = 0
=> y = 21, 17
X = 10 X = 10
Y = 21 x < y x < y
Y = 17 x < y x < y
So, x < y
I. x2 +25x +46 = 0
II. y2 +47y +552 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
D✔️
From equation I:
x2 +25x +46 = (x + 2)(x + 23)= 0
=> x = -2, -23
From equation II:
y2 +47y +552 = (y + 24)(y + 23) = 0
=> y = -24, -23
X = -2 X = -23
Y = -24 x > y x > y
Y = -23 x > y x = y
So, x ≥ y
I. 10×2 -37x +30 = 0
II. y2 -31y +184 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
Solution :A✔️✔️✔️
From equation I:
10×2 -37x +30 = (2x -5)(5x -6)= 0
=> x = 5/2, 6/5
From equation II:
y2 -31y +184 = (y -23)(y -8) = 0
=> y = 23, 8
X = 5/2 X = 6/5
Y = 23 x < y x < y
Y = 8 x < y x < y
So, x < y
In the following questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 361^1/2 + x^3/2 = 12^2
B
II. 10y^1/2 + 14 = 16^3/2
a) x < y
b) x ≥ y
c) x > y
d) x ≤ y
e) x = y or the relationship between X and Y cannot be determined
E✔️
From I,
19 + x3/2 = 144
=> x3/2 = 125
=> x = 25
From II,
10y1/2 + 14 = 64
=> 10y1/2 = 50
=> y1/2 = 5
=> y = 25
Hence x=y
I. x2 – 2.9x + 2.08 = 0
II. y2 – 4.3y + 4.42 = 0
a) x < y
b) x ≥ y
c) x > y
d) x ≤ y
e) x = y or the relationship between X and Y cannot be determined
Solution :A✔️
From I,
(x-1.3)(x-1.6) = 0
=> x = 1.3, 1.6
From II,
(y-1.7)(y-2.6) = 0
=> y = 1.7, 2.6
Hence x < y
I. 12x + 5y = 391
II. 10x – 3y = 161
a) x<y
b) x>y
c) x≤y
d) x≥y
e) x=y
E✔️
I x 3 + II x 5 => 36x + 15y + 50x – 15y = 1173 + 805
=> 86x = 1978
=> x = 23
Putting x=23 in II, 230 – 3y = 161
=> 3y = 69 or y=23
Hence x=y
x2 -15x -34 = 0
II. y2 -13y -230 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
E✔️
From equation I:
x2 -15x -34 = (x -17)(x + 2)= 0
=> x = 17, -2
From equation II:
y2 -13y -230 = (y -23)(y + 10) = 0
=> y = 23, -10
X = 17 X = -2
Y = 23 x < y x < y
Y = -10 x > y x > y
So, relationship cannot be established between x and y
I. x2 -34x +285 = 0
II. 15y2 -64y +64 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
Solution :B✔️✔️
From equation I:
x2 -34x +285 = (x -19)(x -15)= 0
=> x = 19, 15
From equation II:
15y2 -64y +64 = (5y -8)(3y -8) = 0
=> y = 8/5, 8/3
X = 19 X = 15
Y = 8/5 x > y x > y
Y = 8/3 x > y x > y
So, x > y
I. 9×2 -18x +5 = 0
II. 15y2 -14y -8 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
E✔️
From equation I:
9×2 -18x +5 = (3x -1)(3x -5)= 0
=> x = 1/3, 5/3
From equation II:
15y2 -14y -8 = (5y + 2)(3y -4) = 0
=> y = -2/5, 4/3
X = 1/3 X = 5/3
Y = -2/5 x > y x > y
Y = 4/3 x < y x > y
So, relationship cannot be established between x and y
I. 4×2 -26x +42 = 0
II. 4y2 -32y +63 = 0
a) x < y
b) x > y
c) x ≤ y
d) x ≥ y
e) x = y or relationship cannot be established
C✔️
From equation I:
4×2 -26x +42 = (2x -6)(2x -7)= 0
=> x = 6/2, 7/2
From equation II:
4y2 -32y +63 = (2y -9)(2y -7) = 0
=> y = 9/2, 7/2
X = 6/2 X = 7/2
Y = 9/2 x < y x < y
Y = 7/2 x < y x = y
So, x ≤ y