An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How

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An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How

by BTGmoderatorDC » Wed Nov 03, 2021 6:42 am

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An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How much more time will it take to fill the bucket to full capacity?

A. 7/18
B. 9/18
C. 2
D. 18/7
E. 18/5

OA D

Source: Manhattan Prep

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Re: An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How

by [email protected] » Fri Nov 05, 2021 7:46 am
BTGmoderatorDC wrote:
Wed Nov 03, 2021 6:42 am
An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How much more time will it take to fill the bucket to full capacity?

A. 7/18
B. 9/18
C. 2
D. 18/7
E. 18/5

OA D

Source: Manhattan Prep
We can create the proportion:

6/(7/10) = n/(3/10)

60/7 = 10n/3

180 = 70n

180/70 = n

18/7 = n

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Re: An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How

by swerve » Sat Nov 06, 2021 6:38 pm
BTGmoderatorDC wrote:
Wed Nov 03, 2021 6:42 am
An empty bucket being filled with paint at a constant rate takes 6 minutes to be filled to 7/10 of its capacity. How much more time will it take to fill the bucket to full capacity?

A. 7/18
B. 9/18
C. 2
D. 18/7
E. 18/5

OA D

Source: Manhattan Prep
Suppose, capacity $$= 70 .$$
So, to fill $$\dfrac{7}{10}$$ of $$70$$ i.e. $$49,$$ it took $$6$$ minutes.

Hence, to fill $$70$$ it will take $$= 10\ast \dfrac{6}{7} = \dfrac{60}{7}$$ minutes.

So, additional time $$= \dfrac{60}{7} - 6 = \dfrac{18}{7} \Longrightarrow$$ D

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