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Two people together can do a typing job in d hours. If the first person worked alone for a hours, it would then take the second b hours to finish. In how many hours could they each do the work alone? I do not know how to set up the problem.

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closed as off-topic by Hans Lundmark, user91500, kingW3, B. Goddard, Davide Giraudo Sep 12 '17 at 19:37

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    $\begingroup$ What are your attempts on the problem? $\endgroup$ – TomGrubb Sep 12 '17 at 3:10
  • $\begingroup$ Please use MathJax to format your posts. $\endgroup$ – gen-z ready to perish Sep 12 '17 at 3:28
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Let $x$, $y$ be the efficiency of Person 1 and Person 2 respectively. Where efficiency is the reciprocal of the time period to finish the task.

$$d(x+y)=1$$

$$ax+by=1$$

$$x+y=\frac{1}{d}$$

$$x=\frac{1}{d}-y$$

$$a(\frac{1}{d}-y)+by=1$$

$$\frac{a}{d}-ay+by=1$$

$$y(b-a)=1-\frac{a}{d}$$

$$y=\frac{d-a}{d(b-a)}$$

Therefore, it takes Person 2 $\frac{d(b-a)}{d-a}$ hours. You can solve for $x$ to find how long it takes Person 1 to type by himself.

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Here's a general explanation of this sort of problem. You are supposed to assume the following:

  • All work is fungible: Whatever work one person does is interchangeable with whatever work someone else does. The only difference between people is the rate at which they do work.

  • The work is infinitely divisible: One person could do exactly $0.47$ of the work, and someone else could do exactly $0.53$ of the work.

  • Each person works at a constant rate: If someone can do the whole job in $10$ hours, then they can do half the job in $5$ hours, or a fifth of the job in $2$ hours, or one percent of the job in $6$ minutes (that is, $0.1$ hours).

With all that in mind, let the time it takes for the first person to do the job be $t_1$ hours, and the time it takes for the second person to do the job be $t_2$ hours. Another way of saying that is that in one hour, the first person can do $r_1 = \frac{1}{t_1}$ of the job, and the second person can do $r_2 = \frac{1}{t_2}$ of the job. Here, we've designated two new variables $r_1$ and $r_2$ to represent the rate at which the two people work.

The first sentence tells us that the two people working together can do the job in $d$ hours. Another way of saying that is that in one hour, the two people together do $\frac{1}{d}$ of the job. In terms of $r_1$ and $r_2$, we can write

$$ r_1+r_2 = \frac{1}{d} $$

The second sentence tells us that if the first person works for $a$ hours, and the second person works for $b$ hours (both separately), then they would finish the job. In $a$ hours, the first person does $ar_1$ of the job; in $b$ hours, the second person does $br_2$ of the job. The sum of those two is the entire job—that is, the fraction of the job done after these $a+b$ hours is $1$:

$$ ar_1+br_2 = 1 $$

We now have two equations in two unknowns. Solving these for $r_1$ and $r_2$ (in terms of $a$, $b$, and $d$) allows us to obtain $t_1 = \frac{1}{r_1}$ and $t_2 = \frac{1}{r_2}$, and those are the answers to your problem.

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Suppose, Rate of work done by first person $$=k_1 \frac{\text{%work done}}{hours}$$ Rate of work done by second person$$=k_2 \frac{\text{%work done}}{hours}$$

They both completed work in a hours d hours $$\text{work done by }P_1\text{ in d hours}+\text{work done by }P_2 \text{ in d hours}=1$$

$$k_1 d+ k_2 d=1$$

Now if one takes $P_1 $ takes $ a $ hours and then $P_2 $ takes $b $ hours, still work is completed,

$$k_1 a+k_2 b=1$$

You can solve further....

$$k_1+k_2=\frac{1}{d}$$$$k_1=\frac{1}{d}-k_2$$$$$$$$a (\frac{1}{d}-k-2)+b k_2=1$$$$(a-b)k_2=\frac{d-a}{d}$$$$$$$$k_2=\frac{d-a}{d(a-b)}$$$$$$$$k_1=\frac{d-b}{d(b-a)}$$

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