A E and D worked on project numbered 5 B and C alone can complete whole project numbered 5 in 20 and...........

Quantitative Aptitude ( CCS) Data Table

Directions : The following table shows the number of days for which 4 individuals A, B, C and D worked on 5 different projects numbered 1 to 5. It also shows the part of respective project that could not be completed by individuals in time. Some values are missing which are denoted by symbol (-). With the help of information in questions and table below answer the questions that follow.

Project No.	No. Of days for which individuals worked on different projects				Part of project uncompleted after all worked
Project No.	A	B	C	D	Part of project uncompleted after all worked
1	6	6	2	3	1/3
2		3	4
3
4	5	3	2		1/6
5	2	2	3	5	1/12

A, E and D worked on project numbered 5. B and C alone can complete whole project numbered 5 in 20 and 30 days respectively. E who is 3/2 times efficient than B and C together replaces both of them and worked for same number of days for which B and C had to work. A completed 1/12^th of the work. Find in how many days all A, B, C, and D can complete the project 5 together.

3 days

6 days

8 days

11 days

9 days

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Explanation:

B and C together can complete work in 12 days [1/20 + 1/30 = 5/60 — 12 days]
Efficiency E : (B+C) = 3/2 : 1 = 3 : 2
So, ratio of number days = 2 : 3
3 == 12
1 == 4
So E can complete whole work in 2 == 8 days
Now E worked for (2+3) = 5 days – total days for which B and C worked.
So E completed 5/8 of work, A completed 1/12 of work. 1/12 is uncompleted work.

Let x is no. of days in which D can complete whole work. So
A’s work + E’s work + D’s work = 1 – uncompleted work
1/12 + 5/8 + 5/x = 1 – 1/12
Solve, x = 24 = no. of days in which D can alone complete project 5.
A completed 1/12^th work in 2 days, so he can complete whole project in 2*12 = 24 days
A = 24, B = 20, C = 30, D = 24
So together they can complete in – 1/24 + 1/20 + 1/30 + 1/24 = 1/6 — 6 days

Hence, option 2 is the correct answer.