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One-third of a barrel's contents is emptied with each turn of the whee

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One-third of a barrel's contents is emptied with each turn of the whee [#permalink]

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New post 11 Jan 2018, 21:54
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One-third of a barrel's contents is emptied with each turn of the wheel. If the barrel begins full, what fraction of the 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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One-third of a barrel's contents is emptied with each turn of the whee [#permalink]

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New post 12 Jan 2018, 11:22
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Bunuel wrote:
One-third of a barrel's contents is emptied with each turn of the wheel. If the barrel begins full, what fraction of the barrel remains after 2 turns of the wheel?

A. 1/9

B. 4/7

C. 4/9

D. 2/3

E. 8/9

Assign values
Let the barrel contain 9 units
One wheel turn: the barrel loses \(\frac{1}{3}\) of its units
One turn, number lost: \(\frac{1}{3}\)* 9 = 3 lost
One turn, # remaining: (9-3)= 6 remain
Second turn, number lost: \(\frac{1}{3}\)*6= 2 lost
Second turn, # remaining: (6-2) = 4 remain

After these 2 turns, what fraction remains?
\(\frac{EndUnits}{OriginalUnits}=\frac{4}{9}\)

Answer C

Algebraically
Let x = original number of units in the barrel
First turn, number lost: \(x*\frac{1}{3}=\frac{1}{3}x\)
First turn, number remaining:
\((1x - \frac{1}{3}x)= \frac{2}{3}x\)

Second turn, number lost:
\(\frac{2}{3}x*\frac{1}{3}=\frac{2}{9}x\)
Second turn, number remaining: \(\frac{2}{3}x -\frac{2}{9}x=\frac{4}{9}x\) =
Fraction that remains after 2 turns of the wheel

Answer C

*Strictly defined, fraction that remains is
\(\frac{(\frac{4}{9})x}{x}=\frac{4}{9}x*\frac{1}{x}=\frac{4}{9}\). No need for this step: x = 1 and we're working with fractions. It's for the meticulous folks.
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One-third of a barrel's contents is emptied with each turn of the whee   [#permalink] 12 Jan 2018, 11:22
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