LCM METHOD TO SOLVE TIME AND WORK PROBLEMS

About "LCM method to solve time and work problems"

On this web page, we are going to see LCM method to solve time and work problems.

Let us look at the steps involved in solving time and work problems using LCM method.

Further process from step 3 will be depending upon the situation given in the problem. 

It has been explained clearly in the example problems given below.

Let us look at some example problems on "LCM method to solve time and work problems".

Examples on "LCM method to solve 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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Example 5 :

A and B each working alone can do a work in 20 days and 15 days respectively. They started the work together, but B left after sometime and A finished the remaining work in 6 days. After how many days from the start, did B leave?

Solution :

Let us find LCM for the given no. of days "20" and "15".

L.C.M of (20, 15) = 60 units 

Therefore, total work = 60 units.

A can do = 60/20 = 3 units/day 

B can do = 60/15 = 4 units/day 

(A + B) can do = 7 units/day

The work done by A alone in 6 days = 6x3 = 18 units

Then the work done by (A+B) = 60 - 18  = 42 units

Initially, no. of days worked by A and B together 

= 42/7 = 6 days

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Example 6 :

A is 3 times as fast as B and is able to complete the work in 30 days less than B. Find the time in which they can complete  the work together.

Solution :

A & B working capability ratio = 3 : 1

A & B time taken ratio = 1 : 3

From the ratio, time taken by A = k and time taken by B = 3k

From "A takes 30 days less than B", we have 

3k - k = 30

2k = 30 ===> k = 15

Time (A) = 15 days, Time (B) = 3x15 = 45 days

LCM (15, 45) = 45

Total work = 45 units

A can do =  45 / 15 =  3 units/day

B can do = 45 / 45 =  1 unit/day

(A + B) can do = 4 units per day 

No. of days taken by (A+B) to complete  the same work

= 45 / 4  = 11 1/4 days

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Example 7 :

A and B working separately can do a piece of work in 10 and 8 days  respectively. They work on alternate days starting with A on the first day. In how many days will the work be completed ?

Solution :

Let us find LCM of the given no. of days "10" and "8"

LCM of (10, 8) = 40

Total work = 40 units

A can do = 40/10 = 4 units/day

B can do = 40/8 = 5 units/day

On the first two days,

A can do 4 units on the first day and B can do 5 units on the second day. (Because they are working on alternate days)

Total units completed in the 1st day and 2nd day = 9 units ----(1)

Total units completed in the 3rd day and 4th day = 9 units ----(2)

Total units completed in the 5th day and 6th day = 9 units ----(3)

Total units completed in the 7th day and 8th day = 9 units ----(4)

By adding (1),(2),(3) & (4), we get 36 units. 

That is, in 8 days 36 units of the work completed. 

Remaining work = 40 - 36 = 4 units

These units will be completed by A on the 9th day.

Hence the work will be completed in 9 days.

Let us look at the next example on "LCM method to solve time and work problems"

Example 8 :

Two pipes A and B can fill a tank in 16 minutes and 20 minutes respectively. If both the pipes are opened simultaneously, how long will it take to complete  fill the tank ?

Solution :

Let us find LCM of the given no. of minutes "16" and "20"

LCM of (16, 20) = 80

Total work = 80 units

A can fill = 80/16 = 5 units/min

B can fill = 80/20 = 4 units/min

(A+B) can fill = 9 units/min

No. of minutes taken by (A+B) to fill the tank

= 80/9 = 8 8/9 minutes

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Example 9 :

Pipe A can fill a tank in 10 minutes. Pipe B can fill the same tank in 6 minutes. Pipe C can empty the tank in 12 minutes. If all of them work together, find the time taken to fill the empty tank.

Solution :

Let us find LCM of the given no. of minutes "10", "6" and "12"

LCM of (10,6, 12) = 60

Total work = 60 units

A can fill = 60/10 = 6 units/min

B can fill = 60/6 = 10 units/min

(A+B) can fill = 16 units/min

C can empty = 60/12 = 5 units/min

If all of them work together,  

(6 + 10 - 5)  =  11 units/min will be filled

If all of them work together, time taken to fill the empty tank

= 60/11 = 5 5/11 minutes

Let us look at the next example on "LCM method to solve time and work problems"

Example 10 :

A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

Solution :

Let us find LCM of the given no. of minutes "10" and "6".

LCM of (10, 6) = 30

Total work = 30 units (to fill the empty tank)

Already tank is two-fifth full.

So, the work completed already = (2/5)x30 = 12 units

Out of 30 units, now the tank is 12 units full.

A can fill = 30/10 = 3 units/min (filling)

B can empty  = 30/6 = 5 units/min (emptying)

If both the pipes are open, the tank will be emptied

2 units/minute

No of minutes taken to empty the tank (already two-fifth filled)

= 12/2 =  6 minutes 

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