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ujjawal

· started a discussion

· 1 Months ago

wrong answer (Question: 98817.0 )

the right answer is a.

rajeev

· commented

· 1 Months ago

it may hav option a and c both correct..depends the procedure u undergo

Knowledge Expert

· commented

· 1 Months ago

Given answer is correct....x is less than y...

This topic is not avialable

Saurabh Singh

· started a discussion

· 1 Months ago

wrong answer (Question: 98817.0 )

sqrt(x)= -sqrt(400/500) so after squaring x will be approximately 0.8 so y

Question:
DIRECTION: In the given question two equations numbered I and II are given. You have to solve both the equations and find if:


\(I. \sqrt{500}x +\sqrt{402} = 0\)

\(II. \sqrt{360}y + (200)^{\frac{1}{2}} = 0\)

Options:
A) x>y
B) x≥y
C) x<y
D) x≤y
E) x = y or the relationship cannot be established
Solution:
\(I. \sqrt{500}x= -\sqrt{402}\)

\(⇒ x = - \sqrt[]{\cfrac{402}{500}}≈-\sqrt[]{\cfrac{400}{500}}≈-0.9\)

\(II. \sqrt{360}y = -\sqrt{200}\)

\(y =-\sqrt[]{\cfrac{200}{360}}= - 0.74\)

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