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Siva

· started a discussion

· 1 Months ago

Doubt (Question: 108415.0 )

Sir , I have doubt in 60th question.please explain briefly

Question:
DIRECTIONS: In the given question two equations numbered I and II are given. You have to solve both the equations and give answer.


I. 225x2 – 4 = 0

II. \(\sqrt[]{225y}\) + 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:

Ans: (e)

I. 225x2 = 4

⇒ x2 = \(\cfrac{4}{225}\)  ⇒ x = ± \(\cfrac{2}{15}\) =+0.133

II. \(\sqrt[]{225y}\) + 2 = 0

⇒ \(\sqrt[]{225y}\) = -2

Squaring on both sides,

225y = 4

⇒ y = \(\cfrac{4}{225}\) = 0.017

  Hence, No relationship can be estblished.

kanhaiya jha

· commented

· 1 Months ago

Ans: (e)
I. 225x^2 = 4
⇒ x^2 = 4/(225) ⇒ x = ± (2/15) =±0.133
II. √(225)y + 2 = 0
⇒ √(225)y = -2
Squaring on both sides,
225y = 4
⇒ y = 4/(225) = 0.017
Now correlate x and y,
+0.133>0.017 means x>y
-0.133<0.017 means x<y
Hence, No relationship can be established.
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