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If water is leaking from a certain tank at a constant rate of 1,200 mi

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If water is leaking from a certain tank at a constant rate of 1,200 mi  [#permalink]

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New post 21 Jul 2018, 07:50
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Question Stats:

73% (00:57) correct 27% (00:58) wrong based on 22 sessions

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If water is leaking from a certain tank at a constant rate of 1,200 milliliters per hour, how many seconds does it take for 1 milliliter of water to leak from the tank?

A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. 2
D. 3
E. 20

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Re: If water is leaking from a certain tank at a constant rate of 1,200 mi  [#permalink]

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New post 21 Jul 2018, 08:04
ayush98 wrote:
If water is leaking from a certain tank at a constant rate of 1,200 milliliters per hour, how many seconds does it take for 1 milliliter of water to leak from the tank?

A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. 2
D. 3
E. 20


OA: D
Leakage rate =\(\frac{{1200 milliliters}}{{1hr}}\)
Time taken for 1200 milliliters of water to leak = \(1\)hr
Time taken for 1 milliliters of water to leak = \(\frac{{1*60*60}}{{1200}}\) sec = \(3\)sec
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If water is leaking from a certain tank at a constant rate of 1,200 mi  [#permalink]

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New post 21 Jul 2018, 08:53
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ayush98 wrote:
If water is leaking from a certain tank at a constant rate of 1,200 milliliters per hour, how many seconds does it take for 1 milliliter of water to leak from the tank?

A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. 2
D. 3
E. 20

How many seconds does it take for 1 milliliter of water to leak from the tank?

1 hour = 3,600 seconds

Rate conversion (see below for all the conversion ratios):
\(\frac{1,200mL}{1hr}=\frac{1,200mL}{3,600secs}=\frac{1mL}{3secs}\)

One mL of water takes 3 seconds to leak from the tank

Answer D

Conversion

This conversion problem is not hard. Others will be. One way to tackle conversion problems takes very little time to learn and saves a lot of headaches when complex conversions are involved.

One hour = 3,600 seconds?

Convert with an equation set up so that units cancel (treat units to be canceled as if they were numbers in a long set of multiplied ratios - see reference material cited below)

\(\frac{1,200mL}{1hr}*\frac{1hr}{60mins}*\frac{1min}{60secs}=\frac{1,200mL}{3,600secs}\)

Cross off units that show up opposite one another (numerator/denominator and vice versa).

In this case, hours and minutes cancel and we are left with units we need, namely, milliliters per second.

(To get the answer to this question, we just simplify the result of the conversion equation above: \(\frac{1,200mL}{3,600secs}=\frac{1mL}{3secs}\))

For more on how to convert and cancel units, see Setting up conversion problems, here
And
Really good examples of the method, here

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Re: If water is leaking from a certain tank at a constant rate of 1,200 mi  [#permalink]

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New post 21 Jul 2018, 09:05
Answer is D
quite easy if you can just convert b/w the hour and sec.
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Re: If water is leaking from a certain tank at a constant rate of 1,200 mi &nbs [#permalink] 21 Jul 2018, 09:05
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