RRB Junior Engineer Stage 1 Model Papers 8

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Question : 9 of 100

Marks: +1, -0

The third proportional to

(x/y + y/x)

and

√{x^2+y^2}

is

xy
$√{xy}$
$√^3{xy}$
$√^4{xy}$

Solution:

Third proportional of a and b

a : b : c

a/b=b/c

c=b^2/a={(√{x^2+y^2})^2}/{x/y+y/x}={x^2+y^2}/{{x^2+y^2}/{xy}}=xy

Alternate Method :

\First \Proportional × \Third \Proportional = (\Mid\Proportional)^2

∴\text "Third Proportional "={\text "(Mid Proportional)"^2}/{\text "First Proportional"}

Required value

={(√{x^2+y^2})^2}/{x/y+y/x}=x^2+y^2×{xy}/{x^2+y^2}=xy

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