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An empty barrel is filled with oil at a constant rate that takes 10 ho

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An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]

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An empty barrel is filled with oil at a constant rate that takes 10 hours to fill 6/7 of its capacity. How much more time will it take to finish filling the barrel?

A. 1 hr. 50 mins.
B. 1 hr. 40 mins.
C. 1 hr. 30 mins.
D. 1 hr. 20 mins.
E. 1 hr.
[Reveal] Spoiler: OA

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Re: An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]

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New post 08 Aug 2017, 01:42
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Bunuel wrote:
An empty barrel is filled with oil at a constant rate that takes 10 hours to fill 6/7 of its capacity. How much more time will it take to finish filling the barrel?

A. 1 hr. 50 mins.
B. 1 hr. 40 mins.
C. 1 hr. 30 mins.
D. 1 hr. 20 mins.
E. 1 hr.


10 hrs = 6/7 units

6/70 units = 1 hr
1 unit = 70/6 hrs
1/7 units = 70/6 * 1/7 = 10/6 = 1 hr 40 mins
B
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Re: An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]

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New post 08 Aug 2017, 01:45
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Bunuel wrote:
An empty barrel is filled with oil at a constant rate that takes 10 hours to fill 6/7 of its capacity. How much more time will it take to finish filling the barrel?

A. 1 hr. 50 mins.
B. 1 hr. 40 mins.
C. 1 hr. 30 mins.
D. 1 hr. 20 mins.
E. 1 hr.


\(\frac{6}{7}\) capacity of empty barrel is filled with oil in \(10\) hours.

Left capacity to be filled \(= 1 - \frac{6}{7} = \frac{7-6}{7} = \frac{1}{7}\)

Therefore we need to find how much time it will take to fill \(\frac{1}{7}\) of capacity of the barrel.

Let the unknown time be \(= x\)

\(\frac{10}{(6/7)} = \frac{x}{(1/7)}\)

\(\frac{70}{6} = 7x\)

\(x = \frac{10}{6} = \frac{5}{3} = 1\) hour \(40\) mins.


Answer (B)...

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An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]

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New post 08 Aug 2017, 05:52
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Bunuel wrote:
An empty barrel is filled with oil at a constant rate that takes 10 hours to fill 6/7 of its capacity. How much more time will it take to finish filling the barrel?

A. 1 hr. 50 mins.
B. 1 hr. 40 mins.
C. 1 hr. 30 mins.
D. 1 hr. 20 mins.
E. 1 hr.

At what rate is the barrel filled? \(\frac{W}{t}=r\)

\(\frac{(\frac{6}{7})}{10}\) = \(\frac{6}{70}\) = \(\frac{3}{35}\) = r

Work remaining: \(\frac{1}{7}\)

Remaining time to finish, \(\frac{W}{r}=t\):

(1/7)/(3/35) = \(\frac{1}{7}\) * \(\frac{35}{3}\) = \(\frac{5}{3}\)

= 1\(\frac{2}{3}\) hrs or 1 hr 40 minutes

Answer B

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Re: An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]

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New post 10 Aug 2017, 10:46
Bunuel wrote:
An empty barrel is filled with oil at a constant rate that takes 10 hours to fill 6/7 of its capacity. How much more time will it take to finish filling the barrel?

A. 1 hr. 50 mins.
B. 1 hr. 40 mins.
C. 1 hr. 30 mins.
D. 1 hr. 20 mins.
E. 1 hr.


The remaining capacity is 1/7 of the barrel. We can use a proportion to determine how much time it will take to finish filling the barrel.

10/(6/7) = x/(1/7)

70/6 = 7x

70 = 42x

70/42 = x

10/6 = x

5/3 = x

So, it takes 1⅔ hours or 1 hour and 40 minutes more to finish filling the barrel.

Answer: B
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Re: An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]

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New post 14 Aug 2017, 00:20
6/7th part takes 10 hours.

Therefore 1 part(whole) takes =10*7/6 hours= 35/3 hours= 11 hours 40 mins.

Therefore it takes 1 hour 40 minutes more to fill the barrel.
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Re: An empty barrel is filled with oil at a constant rate that takes 10 ho   [#permalink] 14 Aug 2017, 00:20
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