Version 1 Question 1.
A ,B and C can do a piece of work in 24 days, 36 days and 54 days respectively. They started working together but after 4 days A left. B and C worked for the next 5 days and then B left. C worked alone for further 9 days and then he called D to complete the work and the work was completed in the next 9 days. If D was assigned to do the work alone, in how many days could he alone have furnished the work?
1. 90 days 2. 80 days 3. 70 days 4.100 days 5.120 days
A,B and C does 1/24th ,1/36th and 1/54 of the work in one day respectively.Together they do =>(1/24+1/36+1/54) = 18/216 th of work.
They work for 4 days =>(18/216) × 4 = 1/3
Remaining 2/3 work is left B and C work for 5 days =>(1/36+1/54 ) x 5 =>25/108
Work remaining 2/3 -25/108 =47/108.
C alone worked 9 days. (1/54)×9=1/6.
remaining work =47/108 – 1/6=29/108
The remaining work is completed by C and D. Let us assume D took x days to
complete the work =>9(1/54+1/x)=29/108 =>x=90 days.(approximately)
Let us assume that the total work done comprises 216 units
.LCM of 24,36,54 days. Let us calculate the number of units completed by each of
them per day
A =>216/24 =>8 units per day.
B =>216/36=>6 units per day
C =>216/54=>4 units per day.
Number of units completed in one day =>8+6+4=18 units.
All of them worked 4 days. Units completed in 4 days=>18×4=72 units
Remaining units 216-72=144 units
B and C worked for together for 5 days = 5 x 10 =50 (In one day B & C would have completed 6+4=10 units)
Remaining units left 144-50=94 units
C worked for 9 days=>9×4=36 units.
Remaining units left 94-36=58.
Let us assume D does x units of work per day and both of them together completed the work.
C and D worked for 9 days =9(4+x)=58. x=22/9.
If D works alone and he should complete 216 units of work. In 1 day he completes
22/9 units.Hence he takes 216/(22/9) =>90 days(approximately)