**Directions (1-5):** In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

- I. x = (4)^
^{1/2}/ (9)^^{1/2}

II. 9y^{2}-12y +4 =0x>=yx>yy>=xRelation cannot be established.y>xOption D

I.x = (4)^{1/2}/ (9)^^{1/2}

=>x = 2/3

II.9y^{2}-12y +4 =0

=> (3y – 2)^{2}= 0

=> y = 2/3

Hence, x= y or relation cannot b established. - I. 20x
^{2}â€“ 79x + 77 = 0

II. 4y^{2}+ 9y â€“ 28 = 0x>=yy>xy>=xx>yRelation cannot be established.Option A

I. 20x^{2}â€“ 79x + 77 = 0

=> 5x(4x-7) -11 (4x-7) = 0

=> (4x-7)(5x-11)=0

=> x = 11/5 , 7/4

II. 4y^{2}+ 9y â€“ 28 = 0

=> 4y(y+4) â€“ 7(y+4) = 0

=> (y + 4 ) (4y – 7) = 0

=> y = -4 & 7/4

Hence, x >= y - I. x
^{2}+ 13x + 42 =0

II. y^{2}= 36y>xRelation cannot be establishedy>=xy>xx>yOption C

I. x^{2}+ 13x + 42 =0

=>(x+7)(x+6)=0

=> x = -7 & -6

II. y^{2}= 36

=> y = -6 & +6

Hence, y >=x - I. 3x
^{2}+ 2x -1 =0

II. 2y^{2}+ 7y + 6 =0x>yx>=yy>xy>=xRelation cannot be established .Option A

I. 3x^{2}+ 2x -1 =0

=>(3x-1)(x+1)=0

=> x = 1/3 & -1

II. 2y^{2}+ 7y + 6 =0

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

=> y = -3/2 & -2

Hence, x>y - I. 6x
^{2}+ 5x +1 = 0

II. 20y^{2}+ 9y = -1Relation cannot be established.y>=xy>xx>=yx>yOption C

I. 6x^{2}+ 5x +1 = 0

=>(3x+1)(2y+1)=0

=>x = -1/3 & -1/2

II. 20y^{2}+ 9y = -1

=>(4y+1)(5y+1) =0

=>y = -1/4 & -1/5

Hence, y>x - I. 9m
^{2}+ 45m + 26 =0

II. 7n^{2}â€“ 59n â€“ 36 = 0n>mn>=mm>nm>=nRelation cannot be established .Option A

I. 9m^{2}+ 45m + 26 =0

=> m = – (6/9) & – (39/6)

II. 7n^{2}â€“ 59n â€“ 36 = 0

=> n = -(4/7) & 9

Hence, n > m - I. m â€“ (2401)
^{1/2}= 0

II. n^{2}â€“ 2401 = 0n>=mm>nRelation cannot be established .n>mm>=nOption E

I. m â€“ (2401)^{1/2}= 0

=> m = 49

II. n^{2}â€“ 2401 = 0

=> n = -49 & +49

Hence, m>=n - I. m
^{7}â€“ (28*7)^{7.5}/ (m)^{1/2}= 0

II. (n)^{1/3}= 27m>=nm>nn>mn>=mRelation cannot be established.Option C

I. m^{7}â€“ (28*7)^{7.5}/ (m)^{1/2}= 0

=> m^{7}* (m)^{1/2}â€“ (28*7)^{7.5}= 0

=> m^7.5 â€“ (28*7)^7.5 = 0

=> m = 196

II. (n)^{1/3}= 27

=> n = (27)^{3}

Hence, n>m - I. (24-10m)
^{1/2}= 3 â€“ 4m

II. 6n^{2}â€“ 5n â€“ 25 =0Relation cannot be established .n>mm>=nm>nn>=mOption A

I. (24-10m)^{1/2}= 3 â€“ 4m

=> (24 â€“ 10m) = (3 â€“ 4m)^{2}

=> 16m^{2}â€“ 14m â€“ 15 =0

=> (2m-3)(8m+5)=0

=>m = 3/2 & -5/8

II. 6n^{2}â€“ 5n â€“ 25 =0

=>(3n+5)(2n-5)=0

=>n = -5/3 & 5/2

Relation canâ€™t be established . - I. (625m
^{2})^{1/2}â€“ 375 =0

II. (2116)^{1/2}n â€“ 736 = 0n>mm>nn>=mm>=nRelation cannot be established .Option A

I. (625m^{2})^{1/2}â€“ 375 =0

=> 25m â€“ 375 = 0

=> m = 375/25 = 15

II. (2116)^{1/2}n â€“ 736 = 0

=> 46 n â€“ 736 = 0

=> n = 16

Hence,n>m

**Directions (6-10):** Find out the relationship between m and n and choose a required option.

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