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David and Michael together can finish a job in 4 days 19

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David and Michael together can finish a job in 4 days 19  [#permalink]

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New post 15 Oct 2019, 04:35
1
7
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

45% (02:47) correct 55% (03:26) wrong based on 60 sessions

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David and Michael together can finish a job in 4 days 19 hours 12 minutes. If David works at two-thirds Michel's speed, how long does it make Micheal alone to finish the same job?

a. 6 days
b. 8 days
c. 10 days
d. 12 days
e. 9 days 14 hours 24 minutes

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David and Michael together can finish a job in 4 days 19  [#permalink]

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New post 15 Oct 2019, 06:09
jackfr2 wrote:
David and Michael together can finish a job in 4 days 19 hours 12 minutes. If David works at two-thirds Michel's speed, how long does it make Micheal alone to finish the same job?

a. 6 days
b. 8 days
c. 10 days
d. 12 days
e. 9 days 14 hours 24 minutes

Posted from my mobile device



D and M working together take 4 days, 19 hours, and 12 minutes. Converting time into minutes,
it will be \(4*24*60 + 19*60 + 12 = 5760 + 1140 + 12 = 6912\) minutes.

We are told that David works at \(\frac{2}{3}\)rd of Michael's speed. If we assume the total units of work
to be 69120 units, together they do 10 unit in a minute. If Michael does 6 units in a minute,
David will do \(\frac{2}{3}*6 = 4\) units of work in a minute.

Therefore, the time taken by Michael to complete the work is \(\frac{69120}{6*24*60}\) = 8 days(Option B)
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Re: David and Michael together can finish a job in 4 days 19  [#permalink]

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New post 15 Oct 2019, 12:34
1
jackfr2 wrote:
David and Michael together can finish a job in 4 days 19 hours 12 minutes. If David works at two-thirds Michel's speed, how long does it make Micheal alone to finish the same job?

a. 6 days
b. 8 days
c. 10 days
d. 12 days
e. 9 days 14 hours 24 minutes

Posted from my mobile device


Instead of converting to minutes, convert to days-> 12 minutes = \(\frac{1}{5}\)hr.
19\(\frac{1}{5}\) hrs= \(\frac{96}{5}\) hrs
\(\frac{96}{5}\)hrs = \(\frac{96}{24*5}\) Days =.8 days, so 19\(\frac{1}{5}\) hrs = .8 days So total days 4.8

Final equation :
\(\frac{2}{3}\frac{1}{x}\)+\(\frac{1}{x}\) = \(\frac{1}{4.8}\)
\(x\)= 8
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Re: David and Michael together can finish a job in 4 days 19  [#permalink]

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New post 08 Nov 2019, 15:46
If we were to look at this critically:

David is two-thirds Michael's speed: \(D = \frac{2M}{3}\)
While working TOGETHER with Michael, the total speed is \(\frac{2M}{3} + M = \frac{5M}{3}\) (which is a little less than 2 Michaels working together)

So if it takes a little less than 2 Michaels 4 days 19 hours 12 minutes to complete the task, then logically it would take little less than double the time for 1 Michael.
With (E) being double the time, so a little less than (E) would be the answer (B)
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Re: David and Michael together can finish a job in 4 days 19   [#permalink] 08 Nov 2019, 15:46
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David and Michael together can finish a job in 4 days 19

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