[Speed Time Work] Three men can finish a work in x days, share in wages & other special cases
- Prologue
- Case: three men start the work simultaneously.
- Case: Three Men Start Working One After Another
- Case: Share in Daily Wage
- Case: Share in money after work finished?
- Case: Time equations: set of two people working
- Case: Time Ratios are given
Prologue
Speed Time work questions routinely appears in almost every aptitude exam
A can complete a piece of work in 10 days, B can do it in 12 days, C can to it in 15 days. Now A starts the work and continues for 2 days, then B joins and then C joins ….etc.etc. how much time will it take?
- Sometimes, Instead of character names, they’ll give your three pipes. Pipe A can fill the tank in X minutes, Pipe B can do in ….But the approach remains the same. (if pipe is emptying then use minus sign in the speed.)
- Such questions take very little time. Just LCM, fill in the table, and use STD formula and done.
Case: three men start the work simultaneously.
Aiyyar can paint Gokuldham society in 10 days, Bhide Master can do it in 12 days and Champaklal can do it in 15 days. If all three of them start working simultaneously from day 1, how many days will it take for them to finish the painting work?
It is similar to the pipes and cisterns problem.so make a table
Aiyyar | Bhide | Champak | Together | |
Speed | ||||
Time | 10 | 12 | 15 | |
Distance |
The “time” for each individual character, is already given in the problem.
Now take the LCM of Time, means that factorize number into its prime numbers (2,3,5,7,11,13,17,19 etc)
10=2×5
12=2^{2}×3
15=3×5
Now take each prime number in its highest power and multiply them together.
LCM=2^{2}×3^{1}×5^{1}=60
- So LCM (10,12,15)=60. This is the total distance covered, and in this case it means total work done. i.e. entire surface area of walls of Gokuldham is 60 square meters.
- Whether these characters paint the society individually or together, the area of the walls will remain one and same 60.
Aiyyar | Bhide | Champak | Together | |
Speed | ||||
Time | 10 | 12 | 15 | |
Distance | 60 | 60 | 60 | 60 |
Not only empty box is of speed.
Simply write the number required to get 60 by multiplication.
Example Aiyyer’s time is 10 days. So to get 60 from ten, you’ve to multiply 10 with 6.
(Alternatively) for Aiyyer’s case
Speed x time = distance
Speed x 10 days = 60 ; because it is already given in the problem that Aiyyar takes 10 days.
Speed = 60/10 = 6.
Write 6, in Aiyyar’s speed box.
Similarly for Bhide and Champak you’ll get 5 and 4 respectively.
Aiyyar | Bhide | Champak | Together | |
Speed | 6 | 5 | 4 | 6+5+4=15 |
Time | 10 | 12 | 15 | |
Distance | 60 | 60 | 60 | 60 |
When these three characters work together, their speed will increase: 6+5+4=15.
Now we have to find out the time required,
Again STD formula
Speed x time = distance
15 x time = 60
Time = 60/15= 4 days.
Final answer: if these three characters work together, they can paint the society in only four days.
Case: Three Men Start Working One After Another
Same problem, but with little modification
Aiyyar can paint Gokuldham society in 10 days, Bhide Master can do it in 12 days and Champaklal can do it in 15 days. Aiyyar starts painting the society and Bhide joins him after 2 days. Then Both of paint the society together for one day and then Champaklal joins the gang. In this manner, how many total days will be required to finish the paint work?
Part I: Aiyaar starts the work
The table remains seem up to this part
Characters | Work | |||
Aiyyar | Bhide | Champak | Aiyyar starts | |
Speed | 6 | 5 | 4 | 6 |
Time | 10 | 12 | 15 | x2 |
Distance | 60 | 60 | 60 | =12 |
It is given that For the first two days, Aiyyar works alone.