A and B can do a piece of work in 9 days, B and C in 12 days, A and C in 18 days. If all of them work together, then how much time will they take?

Question

This is how I did it.

$A+B=9 \tag 1$
$B+C=12 \tag 2$
$A+C=18 \tag 3$

Adding (1), (2) and (3) we get:

$A+B+B+C+A+C=9+12+18$
$2(A+B+C)=39$
$A+B+C=19.5$

So, they complete the work together in $19.5$ days.

But the book says this answer is wrong and the correct solution is:

Work done by $(A+B)$ in $1$ day = $\frac {1}{9}$
Work done by $(B+C)$ in $1$ day = $\frac {1}{12}$
Work done by $(A+C)$ in $1$ day = $\frac {1}{18}$

$(A+B)+(B+C)+(A+C)= \frac {1}{9} + \frac {1}{12} + \frac {1}{18}$
$A+B+C=\fracBE {1}{8}$
BE

So, all of them take $8$ days to complete the task.

Could anybody please tell me why my approach is wrong and the book's approach is right?

Answer 1

Why are you adding them?

If I can clean a room it $2$ hours and you can clean a room in $3$ does it make sense to thing if we work together it would take us $5$ hours? Why would you helping me make it take $3$ hours longer? Shouldn't your helping me make it go faster?

So why are you adding the time?

So if we don't add what do we do?

Well, how many rooms can I clean in a day? It takes $2$ hours per room and I have $24$ hours... that is $\frac {24}2=12$ rooms. And how many room can you clean in a day? It take you $3$ hours per room you can clean $\frac{24}3=8$ rooms. So together how many room can we do? Well, I can do $12$ any you can do $8$ so together we can to $20$.

Note: YOu don't add how long it takes you! you add how much work you can do in an ammount of time!

And if it takes us $24$ hours to do $20$ rooms how long does it take us to do $1$ room? Well $\frac {24}{20} = 1.2$. So together it takes us $1.2$ hours to do one rooom. Notice that is less time than either of us.

So ... how do we do this if we don't want to just guess?

If I can do a job in $m$ time. Then I can to $\frac 1m$ of the job in one unit of time. And you can do a job in $n$ time then you can do a $\frac 1n$ of the job in one unit of time. Togethere we can do $\frac 1m + \frac n = \frac {n+m}{mn}$ of the job in $1$ unit of time.

So how many units of time does it take to do the whole job? We need $x$ units of time and we can to $\frac {n+m}{nm}\cdot x = 1$ job the those units of time so it takes us $x = \frac {nm}{n+m}$ time to do one job.

Ex: If I take $2$ hours I can do $\frac 12$ a job in an hour. You take $3$ hours so you can do $\frac 13$ a job in an hour. Together we can do $\frac 12 + \frac 13 = \frac 56$ of the job in an hour. After $x$ hours we can to $\frac 56 x$ jobs. To do one job exactly we have $\frac 56 x = 1$ so $x = \frac 65=1.2$ hours.

So your problem. $A$ and $B$ together take $9$ days. so Together $A$ and $B$ can to $\frac 19$ of the job in $1$ day. $B$ and $C$ together can do the job in $12$ days so together they can do $\frac 1{12}$ of the job in $1$ day. And $A$ and $C$ together can do the job in $18$ days$ so together they can do $\frac 1{18}$ of the job in $1$ day.

If you cloned them all and had two copies of them working together then 2$Anne$s and 2$Bernie$s and 2$Claudia$s can do $\frac 1{9} + \frac 1{12} + \frac 1{18} = \frac 4{36} + \frac 3{36} +\frac 2{36} = \frac 9{36} = \frac 14$. So two copies of the three people can do $\frac 14$ of the work in a day.

So one copy of the three people to do $\frac 12\frac 14 = \frac 18$ of the work in a day.

So how many days does it take to do on full job if the can do $\frac 18$ in a day? Then in $x$ days then can do $\frac 18 x = 1$ job. SO $x = 8$ it takes them eight days working together.

Answer 2

Here your variables $A,B,C$ doesn't exactly represent what you want them to. Your first equation is $A+B=9$ . Now this line says that $A$ is that certain amount of work done by the person A in those 9 days. Your third equation is however $A+C=18$ , but here $A$ represents a different amount of work the person has done in 18 days and thus the ambiguity.

First we need to understand that a person's capability of doing a certain work in a given context is fixed per day. So let $W_A$ units be the amount of work A does in 1 day alone, similarly let $W_B , W_C$ be defined.

Now the work they are trying to complete is worth say $N$ units. As we have defined their capabilities in units, so this is consistent.

A and B can together do $W_A+W_B$ units of work in a day, so in 9 days they complete the work. This gives us : $$ 9(W_A+W_B)=N$$ Similarly for others we have, $$ 12(W_B+W_C) = N$$ $$ 18(W_C+W_A)=N$$ Work done by all three in one day is $(W_A+W_B+W_C)$ units , so if $X$ days are required for them to complete the work we must have , $$ X(W_A+W_B+W_C)=N$$ Now it is straightforward from here to find $X$ using basic algebra.

Answer 3

Another solution that I like for this problem goes like so:

Imagine each of $A$, $B$, and $C$ has an identical twin, and all $6$ of the people go to work. In the $6$ people, we have one copy of each pair, $\{AB\}$, $\{AC\}$, and $\{BC\}$.

We will let them work for $36$ days, because $36=$lcm$(9,12,18)$.

In that time, $\{AB\}$ will complete $4$ jobs, $\{AC\}$ will complete $2$ jobs, and $\{BC\}$ will complete $3$ jobs. All together that is $9$ jobs in $36$ days, or $4$ days per job.

But since there are two of each person they can work twice as fast as normal. Thus $A$, $B$, and $C$ working together will complete a job in $8$ days.

Answer 4

When you wrote your three equations you have not clearly thought what you want to say with them. This kind of problem requires a careful choice of the used units and meanings of the unknowns.

Assume that $A$, $B$, and $C$ as single persons can do the fractions $a$, $b$, $c$ of the "piece of work" in one day. (The numbers $a$, $b$, $c$ are $\ll1$.) The first equation then would say $$a+b={1\over 9}\ ,$$ because $A$ and $B$ together need $9$ days to complete it all. Similarly we get $$b+c={1\over12},\qquad a+c={1\over18}\ .$$ Adding these three equations gives $$2(a+b+c)={1\over9}+{1\over12}+{1\over18}={1\over4}\ ,$$ or $\>a+b+c={1\over8}$. This says that all three together can do ${1\over8}$ of the total work in one day. Therefore the three will need $8$ days to do the "piece of work".