## 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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