Bunuel wrote:

A circular hole is cut in a circular disc of diameter 10/3 m so that the weight of the disc is reduced to 1/4 of its original weight. The of the hole is

A. \(3*\sqrt 3\) m

B. 4.5 m

C. 5.4 m

D. \(5\sqrt 3\) m

E. \(2*\sqrt 4\) m

Dear

Bunuel,

I guess answer choice

D has a typo error.D should be \(\frac{5}{\sqrt{3}}\)m,

then Correct answer is DLet, radius of the hole =r

So we can write, \(\pi\)\((\frac{5}{3})^2\)-\(\pi\)\(r^2\)=\(\frac{1}{4}\)\(\pi\)\((\frac{5}{3})^2\),or \((\frac{5}{3})^2\)-\(r^2\)=\(\frac{1}{4}\)\((\frac{5}{3})^2\),or \(\frac{3}{4}\)\((\frac{5}{3})^2\)=\(r^2\)

So, Diameter of the hole==2*\(\sqrt{\frac{3}{4}(\frac{5}{3})^2}\)=2(\(\sqrt{3}\)/2)*\(\frac{5}{3}\)=\(\frac{5}{\sqrt{3}}\)

Correct Answer is \(\frac{5}{\sqrt{3}}\)
_________________

Md. Abdur Rakib

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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges