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If 1/3 of a barrel's contents is emptied with each turn of the wheel,

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If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 11 Jan 2018, 22:40
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Question Stats:

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If 1/3 of a barrel's contents is emptied with each turn of the wheel, what fraction of a full barrel remains after 2 turns of the wheel?

A. 1/9
B. 4/7
C. 4/9
D. 2/3
E. 8/9
[Reveal] Spoiler: OA

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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 12 Jan 2018, 01:16
one turn of wheel empty 1/3 of barrel
two turns of wheel empty 2/3 of barrel

After that remaining = 1-2/3 = 1/3

Bunuel: Can you please check the answer options or my explanation and please correct me If I am wrong.
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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 12 Jan 2018, 02:49
mandeey wrote:
one turn of wheel empty 1/3 of barrel
two turns of wheel empty 2/3 of barrel

After that remaining = 1-2/3 = 1/3

Bunuel: Can you please check the answer options or my explanation and please correct me If I am wrong.


No. The answer is not correct. Let's wait for others to reply.
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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 12 Jan 2018, 02:54
If 1/3 of a barrel's contents is emptied with each turn of the wheel, what fraction of a full barrel remains after 2 turns of the wheel?

A. 1/9
B. 4/7
C. 4/9
D. 2/3
E. 8/9

Let 9 be the vol. of the barrel.

After 1 turn of wheel vol. left =9-3=6;
after 2nd turn vol left 6-2=4
Hence req fraction is 4/9
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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 12 Jan 2018, 04:52
Volume of Barrel left after 1st turn= 1-\(\frac{1}{3}\)= \(\frac{2}{3}\)

Volume of Barrel emptied in 2nd turn= \(\frac{2}{3}\)*\(\frac{1}{3}\)= \(\frac{2}{9}\)

Volume of Barrel left after 2nd turn= \(\frac{2}{3}\)-\(\frac{2}{9}\)= \(\frac{6-2}{9}\)

= \(\frac{4}{9}\)

Answer: C.
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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 12 Jan 2018, 05:24
4/9

Best way to this kind of questions is variable substitution

Let say x=90

Aftr frst round it will have 90-90/3=60 value remain

Aftr scnd round 60-60/3=40

So answer will be 40/90

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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 12 Jan 2018, 05:35
Volume after first turn is 2/3. 1/3of remaining (2/3) is emptied in next turn so remaining 4/9

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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 13 Jan 2018, 19:03
mandeey
The problem with your solution is you left it halfway.
Consider barrel has capacity X.
After one turn capacity of barrel will be = x-1/3x= 2/3x
After one more turn the capacity will be= 2/3x-1/3(2/3x)=4/9x------------------You missed to read question correctly. It says 1/3rd of remaining.
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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 13 Jan 2018, 23:36
Bunuel wrote:
If 1/3 of a barrel's contents is emptied with each turn of the wheel, what fraction of a full barrel remains after 2 turns of the wheel?

A. 1/9
B. 4/7
C. 4/9
D. 2/3
E. 8/9

Let the original capacity of the barrel be = 9

1st turn : Emptied = 3 units ; Left = 6 Units
2nd turn : Emptied = 2 unit ; Left = 4 Units

Thus, after 2 turns of the wheel 4/9 fraction of a full barrel remains in the barrel, answer will be (C)

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Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel, [#permalink]

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New post 17 Jan 2018, 09:57
Full * whats left1 * whats left2 = whats left

-> 1 * (2/3) * (2*3) = 4/9
Re: If 1/3 of a barrel's contents is emptied with each turn of the wheel,   [#permalink] 17 Jan 2018, 09:57
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