# LCM METHOD FOR TIME AND WORK PROBLEMS

In this section, you will learn how to solve time and work problems using least common multiple.

Let us look at the steps involved in solving time and work problems using least common multiple.

Step 1 :

Find the least common multiple for all the given days / hours / minutes.

Step 2 :

Least common multiple found in step 1 is considered as total amount of work to be completed.

Step 3 :

Use the formula given below to solve the problem. Further process from step 3 will be depending upon the situation given in the problem.

It has been explained clearly in the examples given below.

## Examples

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) is 56.

Therefore, the total work is 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

So, no. of days taken by (A + B) to complete  the same work is

=  56 / 11 days

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) is 84.

Therefore, the total work is 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

So, no. of days taken by B alone to complete the same work is

=  84/3

=  28 days

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) is 990.

Therefore, the total work is 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

Therefore B can do 4 units per day.

So, the number of days taken by B to complete the work is

=  990/4

=  247.5 days

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) is 90 units.

Therefore, the total work is 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 is

=  7 ⋅ 9

=  63 units

Remaining work to be completed by B and C is

=  90 - 63

=  27 units

So, no. of days taken by B and C to complete the work is

=  27/3

=  9 days

[Because (B + C)  =  3 units/day]

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) is 60 units.

Therefore, the total work is 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 is

=  6 ⋅ 3

=  18 units

Then the work done by (A + B) is

=  60 - 18

=  42 units

So, initially no. of days worked by A and B together is

=  42/7

=  6 days

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 and B working capability ratio is 3 : 1.

Then, A and B time taken ratio is 1 : 3.

From the ratio,

Time taken by A  =  k

Time taken by B  =  3k

Given : A takes 30 days less than B.

Then, we have

3k - k  =  30

2k  =  30

k  =  15

Therefore,

Time taken by A  =  15 days

Time taken by B   =  3 ⋅ 15  =  45 days

LCM (15, 45) is 45.

Total work is 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

So, no. of days taken by (A + B) to complete the same work is

=  45/4

=  11 1/4 days

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) is 40.

Total work is 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 is

=  9 units -----(1)

Total units completed in the 3rd day and 4th day is

=  9 units -----(2)

Total units completed in the 5th day and 6th day is

=  9 units -----(3)

Total units completed in the 7th day and 8th day is

=  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 is

=  40 - 36

=  4 units

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

So, the work will be completed in 9 days.

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) is 80.

Total work is 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

So, no. of minutes taken by (A + B) to fill the tank is

=  80/9

=  8 8/9 minutes

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) is 60.

Total work is 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 is

=  60/11

= 5 5/11 minutes

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) is 30.

Total work is 30 units (to fill the empty tank).

So, the work completed already is

=  (2/5) ⋅ 30

=  12 units

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

A can fill  =  30/10  =  3 units/min

B can empty  =  30/6  =  5 units/min

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

=  5 - 3

=  2 units/minute

The tank is already two-fifth full (12 units).

So, no. of minutes taken to empty the tank is

=  12/2

=  6 minutes Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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