Time and work part2

Examples on “LCM method for time and work problems”

Example 1 :
A can do a piece of work in 8 days. B can do the same in 14 days. In how many days can the work be completed if A and B work together? 
Solution :
Let us find LCM for the given no. of days “8” and “14”.
L.C.M of (8, 14) = 56
Therefore, total work = 56 units
A can do =  56 / 8 =  7 units/day
B can do = 56 / 14 =  4 units/day
(A + B) can do = 11 units per day 
No. of days taken by (A+B) to complete  the same work
= 56 / 11 days
Let us look at the next example on “LCM method for time and work problems”
Example 2 :
A and B together can do a piece of work in 12 days and A alone can complete  the work in 21 days. How long will B alone to complete  the same work? 
Solution :
Let us find LCM for the given no. of days “12” and “21”.
L.C.M of (12, 21) =  84
Therefore, total work = 84 units.
A can do =  84 / 21 =  4 units/day
(A+B) can do = 84 / 12 =  7 units/day
B can do = (A+B) – A = 7 – 4 = 3 units/day  
No. of days taken by B alone  to complete  the same work
= 84 / 3
= 28 days
Let us look at the next example on “LCM method for time and work problems”
Example 3 :
A and B together can do a piece of work in 110 days. B and C can do it in 99 days. C and A can do the same  work in 90 days. How  long would each take to complete  the work ?
Solution :
Let us find LCM for the given no. of days “110”, “99” and “90”.
L.C.M of (110, 99, 90) =  990
Therefore, total work = 990 units.
(A + B) = 990/110 = 9 units/day  ———>(1)
(B + C) = 990/99 = 10 units/day  ———>(2)
(A + C) = 990/90 = 11 units/day  ———>(3)
By adding (1), (2) & (3), we get,
2A + 2B + 2C = 30 units/day
2(A + B + C) = 30 units/day
A + B + C = 15 units/day ———>(4)
(4) – (1) ====> (A+B+C) – (A+B) = 15 – 9 = 6 units
C can do = 6 units/day ,  C will take = 990/6 = 165 days
(4) – (2)====> (A+B+C) – (B+C) = 15 – 10 = 5 units
A can do = 5 units/day,   A will take = 990/5 = 198 days     
(4) – (3) ====> (A+B+C) – (A+C) = 15 – 11 = 4 units
B can do = 4 units/day,   B will take = 990/4 = 247.5 days
Let us look at the next example on “LCM method for time and work problems”
Example 4 :
A and B can do a work in 15 days. B and C can do it in 30 days. C and A can do the same  work in 18 days. They all work together for 9 days and then A left. In how many days can B and C finish remaining work?
Solution :
Let us find LCM for the given no. of days “15”, “30” and “18”.
L.C.M of (15, 30, 18) = 90 units 
Therefore, total work = 90 units.
(A + B) = 90/15 = 6 units/day  ———>(1)
(B + C) = 90/30 = 3 units/day  ———>(2)
(A + C) = 90/18 = 5 units/day  ———>(3)
By adding (1), (2) & (3), we get,
2A + 2B + 2C = 14 units/day
2(A + B + C) = 14 units/day
A + B + C = 7 units/day ———>(4)
A, B and C all work together for 9 days. 
No. of units completed in these 9 days = 7×9 = 63 units
Remaining work to be completed by B and C = 90 – 63 = 27 units
B and C will take = 27/3 = 9 days   [ Because (B+C) = 3 units/day ] 
Hence B and C will take 9 days to complete the remaining work.
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