A can do a work in 36 days, B in 18 days and C in 12 days. Every 2nd day B and every 3rd day C, helps A .Then in how many days the work will be completed?
A. 12
B. 14
C. 10
D. 8
Answer: Option B
Solution(By Examveda Team)
Let to work Total Work = 36One day work of A = $$\frac{{36}}{{36}}$$ = 1 unit/day
One day work of B = $$\frac{{36}}{{18}}$$ = 2 unit/day
One day work of C = $$\frac{{36}}{{12}}$$ = 3 unit/day
In 3 days cycle total work done is A, A + B, A + C = 1 + (1 + 2) + (1 + 3) = 8 unit/Cycle
∴ 32 units of the work completed in 4 cycle and reminder 4 units of works in next two days.
1 cycle = 3 days
∴ 4 cycle = 12 days
And remaining 4 unit work done in next two days. In days 13, A will work 1 unit and in day 14, A and B will work 3 units.
Total number the day required to complete the work in the given condition is 14 days
This Question Belongs to Arithmetic Ability >> Time And Work
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Ans is 12 days
Total work 36 units
In 12 days work done=12*1+6*2+4*3
=>36 units
So,required days 12 (ans)
@subbrato kumer. B works 3 days and c works 2 days how it is obtained??
Answer-12days
During the 6th, 12th.....days all three of them would have worked together. So the number of days would be less....???
LCM of 2 and 3=6
in first 6 days A work all the day, B works 3 days and C works for 2 days
A+B+C in first 6 days Done=(6/36+3/18+2/12)=1/2 part
1/2 part can Complete in 6 days
so, 1 part Complete in 6X2=12 Days