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I have this statement:

It was calculated that $750$ meters of a ditch could be excavated in $10$ days. If $7$ workers made $350$ meters and later with $5$ assistants they finished the work in the fixed term, how many days did the assistants work?

My attempt was:

Workers | Meters | Days
   7       350m    x
   12      400m   10-x
  • $1)$ More workers, less days: indirect proportion
  • $2)$ More meters, more days: direct proportion

From $1)$ $\frac{x}{10-x}= \frac{350}{400}$,

From $2)$ $7x=12(10-x) = \frac{x}{(10-x)} = \frac{12}{7}$

But, I cannot get $x$ from $1)$ or $2)$ independently, if not that, I must gather both information in a single equation. So my doubt is,

What is the logic behind, to join the information from $1), 2)$ and get the result of $x$?

ESCM
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2 Answers2

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Presumably the assistants are as effective as the workers, so $7x$ worker-days did $350$ and $12(10-x)$ worker-days did $400$. Your equation should then be $$\frac {350}{7x}=\frac {400}{12(10-x)}$$

Ross Millikan
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  • I don't understand what $7x$ represent – ESCM Aug 04 '19 at 23:58
  • It is $7$ workers for $x$ days. Similarly the $12(10-x)$ is $12$ workers for $10-x$ days. We assume that each worker or assistant digs a certain length each day. – Ross Millikan Aug 04 '19 at 23:59
  • But, why ´$7$ workers for $x$ days´ is represented by $7x$?, $7x$ is equal to $28$, what is $28$ in this problem? – ESCM Aug 05 '19 at 00:03
  • The general idea is that the work done is proportional to the number of workers times the number of days. $7x$ is the number of worker-days that got $350$ done. $\frac {350}{7x}$ is the amount of ditch one worker can dig in one day. Similarly for the right side. No, $x \neq 4$ – Ross Millikan Aug 05 '19 at 00:13
  • The $\frac{350}{x}$ was useful, thanks a lot. – ESCM Aug 05 '19 at 00:25
2

It may be easier to work with the two setup equations below,

$$7(10-n)r=350$$ $$(7+5)nr=750-350$$

where $r$ is the rate at which each works. The first equation accounts for the work done in the first $10-n$ days by the 7 workers and the second for the rest $n$ days together with the 5 assistants.

Then, one can eliminate $r$ in the joint equations and get

$$\frac{70-7n}{12n}=\frac{350}{400}$$

with the solution

$$n=4$$

Quanto
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