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Bunuel
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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 11 Jan 2018, 23:40
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C
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25% (medium)

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77% (01:14) correct 23% (01:22) wrong based on 118 sessions

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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
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C
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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 12 Jan 2018, 02:16
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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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 12 Jan 2018, 03:49
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mandeey
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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No. The answer is not correct. Let's wait for others to reply.
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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 12 Jan 2018, 03:54
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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

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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 12 Jan 2018, 05:52
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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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 12 Jan 2018, 06:24
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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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 12 Jan 2018, 06:35
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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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 13 Jan 2018, 20:03
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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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 14 Jan 2018, 00:36
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Bunuel
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
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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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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 17 Jan 2018, 10:57
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Full * whats left1 * whats left2 = whats left

-> 1 * (2/3) * (2*3) = 4/9
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If 1/3 of a barrel's contents is emptied with each turn of the wheel, 30 May 2025, 12:35
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1 - 1/3 = 2/3 remain after 1 turn of wheel
2/3 * 1/3 = 2/9 is removed after second turn of wheel
2/3 - 2/9 = 6/9 - 2/9 = 4/9 remain after 2 turns of wheel
mandeey
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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