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# A circular hole is cut in a circular disc of diameter 10/3 m so that

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

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31 Jul 2016, 03:14
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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

A. $$3*\sqrt 3$$ m

B. 4.5 m

C. 5.4 m

D. $$\frac{5}{\sqrt 3}$$ m

E. $$2*\sqrt 4$$ m
[Reveal] Spoiler: OA

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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.
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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 $$\frac{5}{\sqrt{3}}$$
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Last edited by AbdurRakib on 31 Jul 2016, 05:53, edited 1 time in total.
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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

2r=$$\frac{5}{\sqrt{3}}$$
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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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Joined: 15 Mar 2015
Posts: 113
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, 09:54
Great question.

I am not sure if I would have enough time to do the calculations in time at the test, however here is my solution.

area of the disc: 25*pi/9
hole: r^2*pi
area left of the disc after erasing the hole = 1/4.

(25*pi/9-r^2*pi)/(25*pi/9)=1/4.
25*pi/9-r^2*pi=25*pi/36
r^2=25/9-25/36
r^2=100/36-25/36
r^2=75/36
r^2= (5*5*3)/(2*3*2*3) = (5*5)/(2*2*3)
r=5/(2*sqrt(3))
d=2r=5/sqrt(3).

As noted before, this does not match any of the answer choices.
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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, 10:27
I am also getting the answer as 5/sqrt(3).

Please confirm if I am missing anything or there is a problem with the choices.
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Re: A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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24 Aug 2016, 01:28
I am confused as to what is 'weight' referring to in this context?
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Re: A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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27 Aug 2016, 02:12
ameyaprabhu wrote:
I am confused as to what is 'weight' referring to in this context?

Weight is simply the weight of the disc. Say 10grams, 20 grams, etc.
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A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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30 Sep 2017, 10:34
abhimahna wrote:
I am also getting the answer as 5/sqrt(3).

Please confirm if I am missing anything or there is a problem with the choices.

Hi

abhimahna Can you please explain this.
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A circular hole is cut in a circular disc of diameter 10/3 m so that [#permalink]

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01 Oct 2017, 00:28
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udit harnal wrote:
Hi

abhimahna Can you please explain this.

Hi udit harnal ,

Weight will reduce to $$1/4$$ means the area will be reduced to $$1/4$$ the original area

Or I can say:

Original Area - Removed area = 1/4 Original area --- (1)

Original area = pi $$(5/3)^2$$

Removed area = pi $$r^2$$

Substitute these values in (1) above, you will get the answer.

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

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01 Oct 2017, 00:48
abhimahna Yes got it

Thank You
Re: A circular hole is cut in a circular disc of diameter 10/3 m so that   [#permalink] 01 Oct 2017, 00:48
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