Question 127

If $$\ \sqrt{5}\ $$= 2.236, then what is the value of $$\ \frac{\sqrt{5}}{2}+\frac{5}{3\sqrt{5}}-\sqrt{45}$$?

Solution

Given : $$\sqrt5=2.236$$

To find : $$\frac{\sqrt{5}}{2}+\frac{5}{3\sqrt{5}}-\sqrt{45}$$

= $$\frac{\sqrt5}{2}+\frac{\sqrt5}{3}-3\sqrt5$$

= $$\sqrt5(\frac{1}{2}+\frac{1}{3}-3)$$

= $$\sqrt5(\frac{3+2-18}{6})$$

= $$2.236\times\frac{-13}{6}\approx-4.845$$

=> Ans - (B)

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