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# Quant Quiz On Quadratic Equation Questions Day 4 Bag

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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