$A$ could've done a job in 6 hours longer, $B$ in 15 hours longer and $C$ in twice the time needed if they work together. How long together?

Question

I am trying to solve this question:

A, B and C do a piece of work together. A could've done it in 6 hours longer, B in 15 hours longer and C in twice the time. How long did it take all three to do the work together?

My attempt:

Let the time for A, B and C to do it be $x$. Then:

• A takes $x+6$ hours.
• B takes $x+15$ hours.
• C takes $2x$ hours.

LCM of $x+6$ and $x+15$ and $2x$ is $2x(x+6)(x+15)$. In this time:

• A can do $2x(x+15)$ of work.
• B can do $2x(x+6)$ of work.
• C can do $(x+6)(x+15)$ of work.

So in total they do $2x(x+15) + 2x(x+6) + (x+6)(x+15)$ work.

$= 5x^2 + 63x + 90$.

I'm not sure what to do next, or if I'm even on the right method. Am I on the right track? Is there a better method? What do I do next?

Answer 1

Express their rates of work as the amount they do in 1 hour:

• A does $1/(x+6)$

• B does $1/(x+15)$

• C does $1/2x$

So in 1 hour they do $1/(x+6) + 1/(x+15) + 1/2x$ amount of work.

So in x hours they will complete the job; in other words:

$x(1/(x+6) + 1/(x+15) + 1/2x)$ = 1.

Solve this, and we get $x^2 + 7x - 30 = 0$, so $(x+10)(x-3)=0$.

Since x cannot obviously be negative, x = 3.

So 3 hours.

Answer 2

What is your reasoning for using the LCM? This is the part of your approach that does not make sense.

Convert the time it takes for each person to complete the job into a fraction of the job performed per unit of time (in this case, jobs per hour). That is to say, $A$ has a work rate of $1/(x+6)$ of the job per hour; $B$ works at a rate of $1/(x+15)$ of the job per hour; and $C$ works at a rate of $1/(2x)$ of the job per hour. Therefore, when they all work together, their rate of progress is $$\frac{1}{x+6} + \frac{1}{x+15} + \frac{1}{2x}$$ of the job per hour, and since it takes $x$ hours for them to complete the job together, this means $x$ times the above expression must equal $1$ full job completed; i.e., $$x \left( \frac{1}{x+6} + \frac{1}{x+15} + \frac{1}{2x} \right) = 1.$$ Now can you solve this equation for $x$?

Answer 3

I'm not exactly sure what you're doing. You start out correctly, but I don't understand what you mean when you say, "A can do ...".

It takes A $x+6$ hours to do the job, so the fraction he does in $x$ hours is $\frac x{x+6}$. Similar calculations hold for B and C, and we get $$\frac x{x+6}+\frac x{x+15}+\frac x{2x}=1\tag1$$ or $$\frac x{x+6}+\frac x{x+15}=\frac12\tag2$$ The terms on the left-hand side of $(1)$ represent the fraction each worker does, and together, they do the whole job.

Now you need to clear denominators in $(2)$, and solve the resulting quadratic.

Answer 4

What you have written down is correct, but I feel like you do not fully understand how you are manipulating the expressions to get the answer. Here's my answer building on what you have done with explanations.

It would make more sense to say that in $\text{LCM}(x+6, x+15, 2x) = (x+6)(x+15)(2x)$ hours:

• A can do $(x+15)(2x)$ jobs
• B can do $(x+6)(2x)$ jobs
• C can do $(x+6)(x+15)$ jobs

so when $A,B,C$ work together, they can do:

$$(x+15)(2x) + (x+6)(2x) + (x+6)(x+15)$$ $$= 5x^2 + 63x + 90$$

jobs in the given time, as you have said.

Since it is assumed that each person works at a constant rate, then the $\text{(number of jobs)}/\text{(time)}$ is always constant. But $A,B,C$ together must do $1$ job in $x$ hours according to your first assumption, so this means that:

$$\frac{5x^2+63x+90 \text{ jobs}}{(x+6)(x+15)(2x) \text{ hours}} = \frac{1\text{ jobs}}{x \text{ hours}}$$

and now your job is to find $x$.