Problems on Work Done in a Given Period of Time

Learn how to solve problems on work done in a given period of time when a person A completes 1/n th part of work in one day, then A will take n days to complete the work.

Now we will apply these in working out various word problems to understand the concept to find work done in a given period of time.


Solved problems on work done in a given period of time:

1. Aden and Brad can do a piece of work in 18 days. Brad can Cody can do it in 24 days and Cody and Aden can do it in 36 days. If Aden, Brad, Cody work together, in how many days will they finish the work?

Solution:           

(Aden + Brad)’s 1 day’s work = 1/18                       

(Brad + Cody)’s 1 day’s work = 1/24

(Cody + Aden)’s 1 day’s work = 1/36

Therefore, 2 (Aden + Brad + Cody)’s 1 day’s work = 1/18 + 1/24 + 1/36

                                                                   = (4 + 3 + 2)/72

                                                                   = 9/72

                                                                   = 1/8

(Aden + Brad + Cody)’s 1 day’s work = 1/2 × 8 = 1/16

Therefore, Aden, Brad, Cody together will complete this work in 16 days.

 

2. Jack and Max can do a piece of work in 10 days. Max and Troy can do the same work in 12 days and Troy and Jack can do it in 15 days. Find the number of days taken by each to finish it working alone.

Solution:

(Jack + Max)’s 1 day’s work = 1/10                          

(Max + Troy)’s 1 day’s work = 1/12

(Troy + Jack)’s 1 day’s work = 1/15          

Therefore, 2(Jack + Max + Troy)’s 1 day’s work = 1/10 + 1/12 + 1/15

                                                                = (6 + 5 + 4)/60

                                                                = 15/60

                                                                = 1/4

Therefore, (Jack + Max + Troy)’s 1 day’s work = 1/2 × 1/4

                                                               = 1/8

Now Jack’s 1 day’s work = (Jack + Max + Troy)’s 1 day’s work – (Max + Troy)’s 1 day’s work.

                                 = 1/8 – 1/12

                                 = (3 – 2)/24

                                 = 1/24

Therefore, Jack alone can complete the work in 25 days.

Now Max’s 1 day’s work = (Jack + Max + Troy)’s 1 day’s work – (Jack + Troy)’s 1 day’s work

                                 = 1/8 – 1/15

                                 = (15 – 8)/120

                                 = 7/120

Therefore, Max alone can complete the work in 120/7 days. = 17.1 days.

Now Troy’s 1 day’s work = (Jack + Max + Troy)’s 1 day’s work – (Jack + Max)’s 1 day’s work

                                 = 1/8 – 1/10

                                 = (5 – 4)/40

                                 = 1/40

Therefore, Troy alone can complete the work in 40 days.

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