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

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Math Expert
Joined: 02 Sep 2009
Posts: 52431
One-third of a barrel's contents is emptied with each turn of the whee  [#permalink]

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11 Jan 2018, 20:54
00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:17) correct 20% (01:37) wrong based on 14 sessions

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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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12 Jan 2018, 10:22
1
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?
$$\frac{EndUnits}{OriginalUnits}=\frac{4}{9}$$

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

*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 &nbs [#permalink] 12 Jan 2018, 10:22
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