In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

$I . 4 x^{2} = 16$
$I I . 2 y^{2} - 20 y + 50 = 0$

x = y

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x > y

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x < y

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Solution

The correct option is C x < y
$F r o m I, 4 x^{2} = 16 x^{2} = 4 x = 2 o r x = - 2 F r o m I I, 2 y^{2} - 20 y + 50 = 0 y^{2} - 5 y - 5 y + 25 = 0 (y - 5) (y - 5) = 0 y = 5 \dots (2)$
From (1) and (2), we have y > x.

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In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option: I. 4x2 = 16 II. 2y2−20y+50 = 0

The correct option is C x < yFrom I,4x2=16x2=4x=2 or x=−2From II,2y2−20y+50=0y2−5y−5y+25=0(y−5)(y−5)=0y=5⋯(2) From (1) and (2), we have y > x.

In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

$I . 4 x^{2} = 16$
$I I . 2 y^{2} - 20 y + 50 = 0$

The correct option is C x < y
$F r o m I, 4 x^{2} = 16 x^{2} = 4 x = 2 o r x = - 2 F r o m I I, 2 y^{2} - 20 y + 50 = 0 y^{2} - 5 y - 5 y + 25 = 0 (y - 5) (y - 5) = 0 y = 5 \dots (2)$
From (1) and (2), we have y > x.