Arithmetic Progression Time and Work Problem

Question

A group of men working at the same rate can finish a job in $45$ hours. However, the men report to work, one at a time, at equal intervals over a period of time. Once on the job, each man stays until the job is finished.
If the first man works five times as many hours as the last man, find :

1) The number of hours the first man works.
2) The total number of men in the group.

I know it is an AP problem but I can't figure out how to solve it. I also know I have to use the formula $n = (l-a)/d + 1$ where $l$ = last term, $a$ = first term, $d$ = common difference, $n$ = no. of terms

Try writing things down explicitly rather than relying on the formula. — Cameron Williams, Mar 10 '19 at 07:22
I am not sure how to proceed like that. Can I get some hints? — RaphX, Mar 10 '19 at 07:56

score 0 · Accepted Answer · edited Jan 15 '21 at 14:21

0

Suppose that there are $M$ men in the group. We are told that it takes $45M$ man-hours to finish the job. Now suppose the first man comes to work at time $0$ and works until time $T,$ and that the other men arrive at intervals of $h$ hours. The second man arrives at time $h$ and works until time $T$ so he works $T-h$ hours. The second man arrives at time $2h,$ so he works $T-2h$ hours and so on.

The total number of man-hours worked is $$ T + (T-h)+(T-2h)+\cdots+(T-(M-1)h))=M\left({2T-(M-1)h\over2}\right)=45M$$ so that $$2T-(M-1)h=90\tag{1}$$

The first man works $T$ hours, the last man works $T-(M-1)h$ hours and we are told that $$T=5(T-(M-1)h)$$ so that $$4T=5(M-1)h\tag{2}$$ Solving $(1)$ and $(2)$ simultaneously gives $$\begin{align} (M-1)h&=60\\ T&=75\end{align}$$

We can say that the first man works $75$ hours (without any rest!) and the last man $15,$ but the number of men is not determined. If $M=2,$ the first man works $75$ hours, the last man works $15$ and the whole job takes $$75+15=90=45M\text{ man-hours.}$$ If $M=3,$ then $h=30,$ the second man works $45$ hours, and the whole job takes $135$ man-hours. Any integer $\ge2$ gives a valid value for $M.$

edited Jan 15 '21 at 14:21

answered Mar 10 '19 at 09:52

saulspatz

53,131

The answer is given as 61 for the 2nd part. – RaphX Mar 11 '19 at 16:25
@RaphX Do you see a mistake in my answer? – saulspatz Mar 11 '19 at 16:28
@RaphX Perhaps here was some information given in the textbook that you didn't transcribe. Did they tell you at what intervals the men arrive, perhaps? – saulspatz Mar 11 '19 at 17:13
Nope, this was an exact copy of the question. The answer for the first part is correct i.e. 75 but for the 2nd one they have given 61 as the answer. – RaphX Mar 12 '19 at 03:19
Well, it appears that is wrong. You can check the examples I gave yourself. – saulspatz Mar 12 '19 at 04:26
Ok, can you please elaborate why each man works until time T? – RaphX Mar 12 '19 at 15:11
@RaphX It says, "Each man stays until the job is finished," and I'm calling the finishing time $T.$ – saulspatz Mar 12 '19 at 15:19
Got it , thanks! – RaphX Mar 12 '19 at 15:23

Arithmetic Progression Time and Work Problem

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