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
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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
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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
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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 = 7x9 = 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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