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

Question

3 Deadly Mistakes You Must Avoid in Time and Work Questions – Practice question 4

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, and Raj left 2 days before the original scheduled 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 1:    Question 1

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

EgmatQuantExpert · Accepted Answer

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 originally scheduled completion day
• Raj left the job 2 days before the originally scheduled 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

Now, if all of them would have worked together, the time they would have taken to complete the job =  \frac{60}{(6 + 5 + 4)}  days =  \frac{60}{15}  days = 4 days

Now, if the total work gets completed in d days, we can say
 • Shelly worked for d days, and completed  5 * d = 5d  units of work
• Leo worked for (4 – 1) = 3 days and completed  6 * 3 = 18  units of work
• Raj worked for (4 – 2) = 2 days and completed  4 * 2 = 8  units of work

As the total work is 60 units, we can say
 • 5d + 18 + 8 = 60
Or, 5d = 60 – 26
Or, 5d = 34
Or, d =  \frac{34}{5}  days

Hence, the correct answer is option A.

Answer: A

Important Observation Related to the Article 
 • The originally scheduled completion day = the number of days all 3 would have taken together, to complete the work

prady2231 · Answer

Leo – 10 days
Shelly – 12 days
Raj – 15 days.

The trick is take the LCM or any comman multiple of the number of days and assume it the work to be done. In this case –
I choose 120 So let say work is 120 unit. If you want you can choose 60 as well which is LCM of 10, 12, 15.
 Now Leo alone will be able to do 120 units in 10 days and per day he will do 12 units of work
Shelly alone will be able to do 120 units in 12 days and per day she will do 10 units of work
Raj alone will be able to do 120 units in 15 days and per day he will do 8 units of work
Now if they start working together – work will be finished in 4day. How? 120/(12+10+8)
Leo left 1 day before the completion that means he was there for 3days. He must have performed 36 units.
Raj left 2 days before the completion that means he was there for 2 days. He must have done 16 units.
Note all three were there for 2days so 60 units of work is done.
For 3rd day only Leo and Shelly work – units of work they will complete together – 22 units ( 10 + 12).
Total work complete – 60 +22 = 82Units.
On fourth day, Shelly will be alone and work left = 38 Units. For this to complete she will need 38/10 days.
Total No. of days = 3 + 38/10 = 34/5 days.

u1983 · Answer

In 1 day together Leo, Shelly and Raj does =  \frac{1}{10}+\frac{1}{12}+\frac{1}{15}  =  \frac{(6+5+4)}{60}  =  \frac{15}{60}  portion of the job.
==> they need 4 days if the were working on the job together.
According to the Q stem : 
1.Leo, Shelly and Raj worked together for 2 days. hence the portion done=  \frac{(15*2)}{60}=\frac{30}{60} 
2. Then Leo & Shelly worked for 1 day . Hence the portion done = \frac{(6+5)}{60}  =  \frac{11}{60} .... Thus the portion remaining=  \frac{30}{60}-\frac{11}{60}=\frac{19}{60} ... This is the portion Shelly worked alone
3. Hence time required for Shelly to complete  \frac{19}{60}  portion of the work =work remaining/Shelly's daily working rate= \frac{19}{60}*\frac{60}{5} = \frac{19}{5}

Hence to tal days required :  2+3+\frac{19}{60}= \frac{34}{5} ..... Hence I would go for  Option A .

failatmath · Answer

let time shelly worked = t
leo = t-1
raj = t-2

+  +  =1
t=74/15

IMO ans should be E.

please help?
 Bunuel

EgmatQuantExpert · Answer

Hey  failatmath ,
The question says "Leo left 1 day before, and Raj left 2 days    before the original scheduled completion day    of the job."

If you assume t as the total completion time, then for Leo and Raj, the working time is not t-1 and t-2. The 1-day less and 2-day less times are with respect to the originally scheduled completion time.

Check the solution above, and let us know if you have any confusion.

premprabhs111 · Answer

I understand the given solution. However, what am I doing wrong when I try the below. Please let me know.

Rate when all of them work together = 1/4

Rate when Leo and Shelly work together = 11/60

Rate when Shelly alone works = 1/12

Leo and Shelly worked for - 1 day

Shelly worked alone for - 1 day

Leo, Shelly and Raj worked together for x days

Now,

(x)1/4 + (1)11/60 + (1)1/12 = 1

therefore, x = 44/15

total days = 1 + 1 + 44/15 = 59/15

I know I am doing something wrong, just want to know what it is.

Thanks in advance.

EgmatQuantExpert · Answer

Hey  premprabhs111 ,

check this line mentioned in the question: "Leo left 1 day before, and Raj left 2 days before the original scheduled completion day of the job."

As per you analysis, if the work finally completed in n days, then Raj worked for n-2 days and Leo worked for n-1 days.

However, as per the question statement, if the work would have been completed in d days(when all of them were working), then Raj worked for d-2 days and Leo worked for d-1 days. The values of n and d are not same.

MV94 · Answer

L = 6/60 per day
S = 5/60 per day
R = 4/60 per day

Together = (6 + 5 + 4) / 60 = 15/60 per day = 4 days

15 units per day together, 60 units needed
R works 2 days
L works 3 days

15units x 2 = cumulative 30 units & 2 days
11units x 1 = cumulative 41 units & 3 days

19 units remain, 19 units over 5 units per day = 19/5 days + 3 earlier (15/5) =  34 / 5 days

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

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