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- Question 1 of 10
##### 1. Question

Directions:1-5)

**In each of these questions, two equations I and II are given. You have to solve both the equations and given answer**I)a

^{7}– {(45)^{15/2 }/√a} =0II) 44/ √b + √b =89 / √b

Correct**I)**a^{7}– {(45)^{15/2 }/ √a} =0

a

^{15/2 }=(45)^{15/2}a=45

**II)**(44+b) / √b = 89/√b

44+b=89

b=89-44=45

Incorrect**I)**a^{7}– {(45)^{15/2 }/ √a} =0

a

^{15/2 }=(45)^{15/2}a=45

**II)**(44+b) / √b = 89/√b

44+b=89

b=89-44=45

- Question 2 of 10
##### 2. Question

- (a+189)
^{1/2}+(5184)^{1/2 }=86 - b
^{2}-11b+30 =0

Correct- (a+189)
^{1/2}+(5184)^{1/2}=86

Or,(a+189)

^{1/2}=86-72=14Now, squaring both sides, we get

a+189=196

Or, a=196-189=7

II.b

^{2}-11b+30=0Or,b

^{2}-6b-5b+30=0Or, b(b-6)-5(b-6)=0

Or,(b-5)(b-6)=0

∴ b= 5, 6

Hence a>b

Incorrect- (a+189)
^{1/2}+(5184)^{1/2}=86

Or,(a+189)

^{1/2}=86-72=14Now, squaring both sides, we get

a+189=196

Or, a=196-189=7

II.b

^{2}-11b+30=0Or,b

^{2}-6b-5b+30=0Or, b(b-6)-5(b-6)=0

Or,(b-5)(b-6)=0

∴ b= 5, 6

Hence a>b

- (a+189)
- Question 3 of 10
##### 3. Question

- a=
^{3}√512 - b=√289

Correcta =

^{3}√512a = 8

b = √289 = 17

a<b

Incorrecta =

^{3}√512a = 8

b = √289 = 17

a<b

- a=
- Question 4 of 10
##### 4. Question

I.15a

^{2-}8a+1=0II.18b

^{2}-11b+1=0Correct**I.**15a^{2}-8a+1 =0Or,15a

^{2}-5a-3a+1=0Or,5a(3a-1)-1(3a-1) =0

Or,(5a-1) (3a-1) =0

∴a=1/5,1/3

II.18b

^{2}-11b+1=0or,18b

^{2}-9b-2b+1=0or,9b(2b-1)-1(2b-1) =0

or,(9b-1) (2b-1) =0

∴b =1/9,1/2

Hence, no relation can be established between a and b.

Incorrect**I.**15a^{2}-8a+1 =0Or,15a

^{2}-5a-3a+1=0Or,5a(3a-1)-1(3a-1) =0

Or,(5a-1) (3a-1) =0

∴a=1/5,1/3

II.18b

^{2}-11b+1=0or,18b

^{2}-9b-2b+1=0or,9b(2b-1)-1(2b-1) =0

or,(9b-1) (2b-1) =0

∴b =1/9,1/2

Hence, no relation can be established between a and b.

- Question 5 of 10
##### 5. Question

- a
^{2}-9a=0 - 3a
^{2}+5a+2 =0

Correct**I.**a^{2}-9a =0Or,a

^{2}=9a∴a=9

II.3a

^{2}+5a+2 =0Or,3a

^{2}+3a+2a+2=0Or,3a(a+1)+2(a+1) =0

Or,(3a+2)(a+1) =0

∴a= -2/3, -1

Hence a>b

Incorrect**I.**a^{2}-9a =0Or,a

^{2}=9a∴a=9

II.3a

^{2}+5a+2 =0Or,3a

^{2}+3a+2a+2=0Or,3a(a+1)+2(a+1) =0

Or,(3a+2)(a+1) =0

∴a= -2/3, -1

Hence a>b

- a
- Question 6 of 10
##### 6. Question

Directions:6-10)

**In the following questions, two equations I and II are given. You have to solve both the equations and g****ive Answer as,**- 4x
^{2}– 15x + 9 = 0 - 2y
^{2}– 13y + 21 = 0

CorrectI.4x

^{2}– 15x + 9 = 0=> 4x

^{2}– 12x – 3x+ 9 = 0=> 4x(x – 3) – 3(x – 3) = 0

=> (4x – 3) (x – 3) = 0

=> x = 3, 4/3

II. 2y

^{2}– 13y + 21 = 0=> 2y

^{2}– 7y – 6y + 21 = 0=> y(2y – 7) – 3(2y – 7) = 0

=> y= 3, 7/2

Y≥X

IncorrectI.4x

^{2}– 15x + 9 = 0=> 4x

^{2}– 12x – 3x+ 9 = 0=> 4x(x – 3) – 3(x – 3) = 0

=> (4x – 3) (x – 3) = 0

=> x = 3, 4/3

II. 2y

^{2}– 13y + 21 = 0=> 2y

^{2}– 7y – 6y + 21 = 0=> y(2y – 7) – 3(2y – 7) = 0

=> y= 3, 7/2

Y≥X

- 4x
- Question 7 of 10
##### 7. Question

- 3x
^{2}– 17x + 20 = 0 - 3y
^{2}+ y – 10 = 0

CorrectI.3x

^{2}– 17x + 20 = 0=> 3x

^{2}– 12x – 5x + 20 = 0=> 3x(x – 4) – 5(x – 4) = 0

=> (3x – 5)(x – 4) = 0

X = 5/3, 4

II. 3y

^{2}+ y – 10 = 0=> 3y

^{2}+ 6y – 5y – 10 = 0=> 3y( y + 2) – 5( y + 2) = 0

=> (y + 2)(3y – 5) = 0

Y = -2, 5/3

X≥Y

IncorrectI.3x

^{2}– 17x + 20 = 0=> 3x

^{2}– 12x – 5x + 20 = 0=> 3x(x – 4) – 5(x – 4) = 0

=> (3x – 5)(x – 4) = 0

X = 5/3, 4

II. 3y

^{2}+ y – 10 = 0=> 3y

^{2}+ 6y – 5y – 10 = 0=> 3y( y + 2) – 5( y + 2) = 0

=> (y + 2)(3y – 5) = 0

Y = -2, 5/3

X≥Y

- 3x
- Question 8 of 10
##### 8. Question

- 3x
^{2}– 4x – 32 = 0 - 3y
^{2}+ 14y + 16 = 0

CorrectI. 3x

^{2}– 4x – 32 = 0=> 3x

^{2}– 12x + 8x – 32 = 0=> 3x(x – 4) + 8(x – 4) = 0

=> (x – 4) (3x + 8) = 0

X = 4, -8/3

II. 3y

^{2}+ 14y + 16 = 0=> 3y

^{2}+ 8y + 6y + 16 = 0=> Y(3y + 8) + 2(3y + 8) = 0

=> (y + 2)(3y + 8) = 0

Y= -8/3, -2

X=Y or No relation

IncorrectI. 3x

^{2}– 4x – 32 = 0=> 3x

^{2}– 12x + 8x – 32 = 0=> 3x(x – 4) + 8(x – 4) = 0

=> (x – 4) (3x + 8) = 0

X = 4, -8/3

II. 3y

^{2}+ 14y + 16 = 0=> 3y

^{2}+ 8y + 6y + 16 = 0=> Y(3y + 8) + 2(3y + 8) = 0

=> (y + 2)(3y + 8) = 0

Y= -8/3, -2

X=Y or No relation

- 3x
- Question 9 of 10
##### 9. Question

- 4x
^{2}+ 4x – 3 = 0 - 2y
^{2 }– 13y + 21 = 0

CorrectI.4x

^{2}+ 4x – 3 = 0=> 4x

^{2}+ 6x – 2x – 3 = 0=> 2x(2x + 3) – 1(2x + 3) = 0

=> (2x – 1)(2x + 3) = 0

X = ½ , – 3/2

II. 2y

^{2 }– 13y + 21 = 0=> 2y

^{2 }– 7y – 6y + 21 = 0=> Y(2y– 7) – 3(2y – 7) = 0

=> (y – 3)(2y – 7)= 0

Y = 3, 7/2

Y>X

IncorrectI.4x

^{2}+ 4x – 3 = 0=> 4x

^{2}+ 6x – 2x – 3 = 0=> 2x(2x + 3) – 1(2x + 3) = 0

=> (2x – 1)(2x + 3) = 0

X = ½ , – 3/2

II. 2y

^{2 }– 13y + 21 = 0=> 2y

^{2 }– 7y – 6y + 21 = 0=> Y(2y– 7) – 3(2y – 7) = 0

=> (y – 3)(2y – 7)= 0

Y = 3, 7/2

Y>X

- 4x
- Question 10 of 10
##### 10. Question

- 3x
^{2}– 7x – 20 = 0 - 3y
^{2}+ 20y + 25 = 0

CorrectI. 3x

^{2}– 7x – 20 = 0=> 3x

^{2}– 12x + 5x – 20 = 0=> 3x(x – 4) + 5(x – 4) = 0

=> (x – 4)(3x + 5)= 0

X= 4, -5/3

II. 3y

^{2}+ 20y + 25 = 0=> 3y

^{2}+ 15y + 5y + 25 = 0=> 3y(y + 5) + 5(y + 5) = 0

=> (y + 5)(3y + 5) = 0

Y = – 5, -5/3

X≥Y

IncorrectI. 3x

^{2}– 7x – 20 = 0=> 3x

^{2}– 12x + 5x – 20 = 0=> 3x(x – 4) + 5(x – 4) = 0

=> (x – 4)(3x + 5)= 0

X= 4, -5/3

II. 3y

^{2}+ 20y + 25 = 0=> 3y

^{2}+ 15y + 5y + 25 = 0=> 3y(y + 5) + 5(y + 5) = 0

=> (y + 5)(3y + 5) = 0

Y = – 5, -5/3

X≥Y

- 3x