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Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days

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Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days



To solve question 4: Question 4

To read the article: 3 Deadly Mistakes You Must Avoid in Time and Work Questions

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Originally posted by EgmatQuantExpert on 30 May 2018, 03:36.
Last edited by EgmatQuantExpert on 13 Aug 2018, 00:19, edited 3 times in total.
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 30 May 2018, 06:16
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EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days


Given data: Leo, Shelly,and Raj can complete a certain job in 10,12,and 15 days.

We assume the work that needs to be done as LCM(10,12,15) = 60 units.
Individual rates are as follows: Leo - 6 units/day | Shelly - 5 units/day | Raj - 4 units/day

If they work together, they will complete 6+5+4 = 15 units in a day. When Raj leaves 2 days before,
Leo & Shelly do 11 units/day. When Leo also leaves work 1 day before, Shelly does 5 units in a day.

Let x be the number of days.
\(15(x-2) + 11 + 5 = 60\) -> \(15x - 30 + 16 = 60\) -> \(15x - 14 = 60\) -> \(x = \frac{74}{15}\) days

Therefore, Leo, Shelly and Raj complete the job in \(\frac{74}{15}\) days (Option E)
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 30 May 2018, 11:10
EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days


shouldn't the answer be A?
According to the question my soln comes 89/15. where am i going wrong?
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 30 May 2018, 11:36
EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days


Let's total unit of work LCM[10,12,15] = 60 units of work
Leo does 60 /10 = 6 units of work per day
shelly does 60/112 = 5 units of work per day
Raj does 60 / 15 = 4 units of work per day

now Raj leaves 2 before the completion day and leo leaves 1 day before the completion
Assume the work completed on x days
Therefore,
(6 + 4 +5)(x - 2) + (6 + 5) + 5= 60
==>15x - 30 = 60 - 16
==> 15x = 44 + 30 = 74

Hence x = 74 / 15
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 30 May 2018, 12:51
EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days


let d=days to complete job
d(15/60)-14/60=1
d=74/15 days
E
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 31 May 2018, 06:55
EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days



To solve question 4: Question 4

To read the article: 3 Deadly Mistakes You Must Avoid in Time and Work Questions


Responding to a pm:

On the last day of the job, only Shelly worked so she could have done only 1/12 of the work.

So 11/12 of the work was done before the completion day.

On the day before last, only Leo and Shelly worked so they would have done 1/10 + 1/12 = 11/60 of the work

So 11/12 - 11/60 = 11/15 of the work was done before by all three.
Combined rate of all three = 1/10 + 1/12 + 1/15 = 1/4
For this, time taken = (11/15) / (1/4) = 44/15

Total time taken = 44/15 + 2 = 74/15

Note: Usually in questions with fractional number of days, the fractional day is the last day. It is certainly unsettling that as per this question, the work was started in the middle of the day some time.
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 31 May 2018, 10:48
pushpitkc wrote:
EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days


Given data: Leo, Shelly,and Raj can complete a certain job in 10,12,and 15 days.

We assume the work that needs to be done as LCM(10,12,15) = 60 units.
Individual rates are as follows: Leo - 6 units/day | Shelly - 5 units/day | Raj - 4 units/day

If they work together, they will complete 6+5+4 = 15 units in a day. When Raj leaves 2 days before,
Leo & Shelly do 11 units/day. When Leo also leaves work 1 day before, Shelly does 5 units in a day.

Let x be the number of days.
\(15(x-2) + 11 + 6 = 60\) -> \(15x - 30 + 16 = 60\) -> \(15x - 14 = 60\) -> \(x = \frac{74}{15}\) days

Therefore, Leo, Shelly and Raj complete the job in \(\frac{74}{15}\) days (Option E)


There's a typo in your first equation - it should be \(15(x-2) + 11 + 5\)
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 31 May 2018, 11:01
By the way, how do I understand that the completion day does not vary, i.e. it is fixed?

I've spent ~40 freaking mins trying to solve this, because I thought that the completion day changes once a person leaves the team, which is logical. That is, when Raj left, I changed the Work value to \(1/2\) and Rate value to \(11/60\). Using these values I calculated the time needed to finish the work - \(30/11\). Then I figured out that Leo leaves at moment \(30/11 - 11/11 = 19/11\). I calculated the amount of work done by both Leo and Shelly, which is \(11/60 * 19/11 = 19/60\).

Now, knowing that the amount of Work left is \(1 - (30/60) - (19/60) = (11/60)\), I calculated the time needed for Shelly to finish the work alone, which is \((11/60) : (5/60) = (11/5)\).

Finally, I added up all the time values, i.e. \(2 days + (19/11) days + (11/5) days\), and obtained a result \((326/55)\).

This did not match any of the answers provided here, and I tried to find my mistake for 40 FREAKING MINUTES. As you can see, my logic was not wrong. So, how should have I figured out what I was supposed to calculate?
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 31 May 2018, 11:26
Krotishka1 wrote:
pushpitkc wrote:
EgmatQuantExpert wrote:
3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 3

Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days respectively. They are assigned to work together to complete the job. All of them started the job together. Leo left 1 day before the completion day and Raj left 2 days before the completion day of the job. In how many days the job gets completed?

    A. \(\frac{34}{5}\) days

    B. \(\frac{48}{5}\) days

    C. \(\frac{57}{12}\) days

    D. \(\frac{66}{15}\) days

    E. \(\frac{74}{15}\) days


Given data: Leo, Shelly,and Raj can complete a certain job in 10,12,and 15 days.

We assume the work that needs to be done as LCM(10,12,15) = 60 units.
Individual rates are as follows: Leo - 6 units/day | Shelly - 5 units/day | Raj - 4 units/day

If they work together, they will complete 6+5+4 = 15 units in a day. When Raj leaves 2 days before,
Leo & Shelly do 11 units/day. When Leo also leaves work 1 day before, Shelly does 5 units in a day.

Let x be the number of days.
\(15(x-2) + 11 + 6 = 60\) -> \(15x - 30 + 16 = 60\) -> \(15x - 14 = 60\) -> \(x = \frac{74}{15}\) days

Therefore, Leo, Shelly and Raj complete the job in \(\frac{74}{15}\) days (Option E)


There's a typo in your first equation - it should be \(15(x-2) + 11 + 5\)


Thanks for notifying Krotishka1 - Made the necessary change
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 31 May 2018, 23:54
Krotishka1 wrote:
By the way, how do I understand that the completion day does not vary, i.e. it is fixed?

I've spent ~40 freaking mins trying to solve this, because I thought that the completion day changes once a person leaves the team, which is logical. That is, when Raj left, I changed the Work value to \(1/2\) and Rate value to \(11/60\). Using these values I calculated the time needed to finish the work - \(30/11\). Then I figured out that Leo leaves at moment \(30/11 - 11/11 = 19/11\). I calculated the amount of work done by both Leo and Shelly, which is \(11/60 * 19/11 = 19/60\).

Now, knowing that the amount of Work left is \(1 - (30/60) - (19/60) = (11/60)\), I calculated the time needed for Shelly to finish the work alone, which is \((11/60) : (5/60) = (11/5)\).

Finally, I added up all the time values, i.e. \(2 days + (19/11) days + (11/5) days\), and obtained a result \((326/55)\).

This did not match any of the answers provided here, and I tried to find my mistake for 40 FREAKING MINUTES. As you can see, my logic was not wrong. So, how should have I figured out what I was supposed to calculate?


Hey Krotishka1

The mistake you are making is that you have calculated the time for Leo and Shelly to complete
the work, but the work doesn't get completed. When you are adding up the individual times you
have added the time that Leo and Shelly would have taken to complete the work :)

Whenever you want to solve this kind of problem using this method you need to go backward from
the last day to the first day. The solution provided by VeritasPrepKarishma should make it clear.

Hope this helps you.
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days  [#permalink]

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New post 02 Jun 2018, 12:02

Solution



Given:
    • Leo, Shelly, and Raj can complete a certain job in 10, 12, and 15 days respectively
    • All of them started working on the job together
    • Leo left the job 1 day before the completion day
    • Raj left the job 2 days before the completion day

To find:
    • In how many days the job gets completed

Approach and Working:
If we assume the total job to be LCM (10, 12, 15) = 60 units, then
    • 1-day work of Leo = \(\frac{60}{10}\) units = 6 units
    • 1-day work of Shelly = \(\frac{60}{12}\) units = 5 units
    • 1-day work of Raj = \(\frac{60}{15}\) units = 4 units

If the total job takes d days to complete, then1
    • Shelly worked for d days, and completed \(5 * d = 5d\) units of work
    • Leo worked for (d – 1) days, and completed \(6 * (d – 1) = 6d – 6\) units of work
    • Raj worked for (d – 2) days, and completed \(4 * (d – 2) = 4d – 8\) units of work

As the total work is 60 units, we can say
    • 5d + 6d – 6 + 4d – 8 = 60
    Or, 15d = 60 + 14
    Or, 15d = 74
    Or, d = \(\frac{74}{15}\) days

Hence, the correct answer is option E.

Answer: E

Important Observation Related to the Article


• When the question mentions points like Leo left 1 day before completion or Raj left 2 days before completion, the calculation is done keeping in mind that the finishing day remains same.
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Re: Leo, Shelly and Raj can complete a certain job in 10, 12, and 15 days &nbs [#permalink] 02 Jun 2018, 12:02
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