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).
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
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
If you have any feedback about our math content, please mail us :
v4formath@gmail.com
We always appreciate your feedback.
You can also visit the 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
Trigonometry 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
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
Recent Articles
-
Angles and Parallel Lines Worksheet
Oct 26, 21 06:29 AM
Angles and Parallel Lines Worksheet
-
Algebra and Angle Measures
Oct 26, 21 06:28 AM
Algebra and Angle Measures - Examples
-
Algebra and Angle Measures Worksheet
Oct 26, 21 06:23 AM
Algebra and Angle Measures Worksheet
