LCM Method for Time and Work :

In this section, we 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.

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.

**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).

Already tank is two-fifth full.

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

Widget is loading comments...

You can also visit our following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Time and work word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**