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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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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

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

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Re: An empty barrel is filled with oil at a constant rate that takes 10 ho [#permalink]
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]
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Reworking on the equation of universal work we have,

The amount of work to be done = 1 - 6/7 =1/7 and we have to find T2.

T1/W1 =T2/W2

=> 10/(6/7) = T2/ (1/7)

=>T2 = 5/3 hours = 1 hour 40 min

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