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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 diameter of the hole is

Re: A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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31 Jul 2016, 05:07

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 diameter 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

area of disc =pi*r^2=pi*25/9 let cutout disc=Pi*r^2 so pi*25/9-Pi*r^2=(1/4)*pi*25/9

but i mthinking the dia of cutout disc must be smaller than dia of bigger one which is 10/3 ~3m but none of option is there less than 3???

Plz explain anyone

Thanks

Last edited by rohit8865 on 31 Jul 2016, 05:44, edited 1 time in total.

Re: A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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31 Jul 2016, 05:34

Top Contributor

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 D

Let, 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 \(\frac{5}{\sqrt{3}}\) _________________

Re: A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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31 Jul 2016, 05:43

Top Contributor

rohit8865 wrote:

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 diameter 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

area of disc =pi*r^2=pi*25/9 let cutout disc=Pi*r^2 so pi*25/9-Pi*r^2=(1/4)*pi*25/9 solving 2*r(dia) = 4.33~4.5m Ans B

but i mthinking the dia of cutout disc must be smaller than dia of bigger one which is 10/3 ~3m but none of option is there less than 3???

Plz explain anyone

Thanks

Your approach is similiar to me,but you've made a calculation error on the highlighted portion above

Re: A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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31 Jul 2016, 05:46

AbdurRakib wrote:

rohit8865 wrote:

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 diameter 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

area of disc =pi*r^2=pi*25/9 let cutout disc=Pi*r^2 so pi*25/9-Pi*r^2=(1/4)*pi*25/9 solving 2*r(dia) = 4.33~4.5m Ans B

but i mthinking the dia of cutout disc must be smaller than dia of bigger one which is 10/3 ~3m but none of option is there less than 3???

Plz explain anyone

Thanks

Your approach is similiar to me,but you've made a calculation error on the highlighted portion above

2r=\(\frac{5}{\sqrt{3}}\)

might be unnecessary i pulled myself towards any choices above but yes if any choices is 5/sqrt3 then i have blindly hit that as correct one...

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