Work & workers word problem

Here's another one of those job / workers word problems (inverse or direct variation). Try and see if you can solve it using the "table" method instead of equations:

A certain job can be done by 18 clerks in 26 days.  How many clerks are needed to perform the job in 12 days?


Again, we can set up a table and reason this out. Initially set it up like this:

jobs | clerks | days
--------------------------
  1  |  18    |  26
--------------------------
  1  |        |  
--------------------------
  1  |   ?    |  12

Then think of the "days" column. We want to "go" from 26 to 12. You could use a proportion here... or first figure out how many clerks are needed to do this job in 2 days, and then from that go to 12 days.

If 18 clerks do it in 26 days, then how many clerks would do it in 2 days... which is 1/13 the amount of time.... so we need 13 times as many clerks.
13 x 18 = 234 clerks are needed.

jobs | clerks | days
--------------------------
  1  |  18    |  26
--------------------------
  1  |  234  |   2  
--------------------------
  1  |   ?    |  12

Now, if 234 clerks do it in 2 days, how many clerks would do it in 12 days? Now, the time increases 6-fold, so we need only 1/6 as many workers.

234 / 6 = 39.

So 39 clerks are needed.

Houses/workers/days problem (direct and inverse variation)

Today I want to feature a great video from a fellow blogger and Youtuber, Dave Marain. He solves this question, which at first can sound intimidating:

If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days?

It sounds like you'd need algebra, proportions, inverse or direct variation, etc. And true, you could use those. But his method completely avoids all that and is based on setting up a simple TABLE, AND using common sense!

|  HOUSES  |  WORKERS  |  DAYS  |
----------------------------------
|    3     |    10     |  60    |
----------------------------------
|    1     |    10     |        |
----------------------------------
|    2     |    10     |        |
----------------------------------
|    5     |           |        |
----------------------------------

See the video below:




And here's a link to his other Youtube videos: www.youtube.com/user/MathNotationsVids.

Solving direct and inverse variations in chart form

Dave Marain recently featured my blog on his, and now it just so happens I get to promote his, because I really liked his post about learning direct and inverse variations.

He uses a beagle problem with interesting numbers:

"Three beagles can dig 4 holes in five days. How many days will it take 6 beagles to dig 8 holes?"

The solution is actually quite easy — just think how there are exactly twice as many beagles and also twice as many holes. Dave shows a chart that can help youngsters grasp the solution.

I want to go one step further with this. Let's try a little more awkward numbers:

"Three beagles can dig 7 holes in eight days. How many days will it take 5 beagles to dig 9 holes?"

I want to show you how a "chart" approach will still work. This situation describes joint variation, because there is both inverse and direct variation involved: the number of days it takes to dig the holes varies inversely with the number of beagles (the more beagles, the less days it takes), and varies directly with the number of holes (the more holes, the more days it takes).

In each step on our chart, we change ONE variable (either the number of beagles, the number of holes, or the number of days), keep ONE variable unchanged, and figure out how the third variable changes. You need to carefully think if that third variable is multiplied or divided — if it is in direct or inverse variation.

For example: if the number of beagles is halved, and there are the same amount of holes, how will the number of days change?
Or: if the number of holes is quadrupled, and there are the same amount of beagles, how will the number of days change?

Let's start with the situation given in the problem.

Beagles       Holes     Days
----------------------------------
3             7         8

We will want to find out how many days it takes ONE beagle to dig ONE whole, and then use that as a "stepping stone" to find how many days it takes 5 beagles to dig 9 holes .

So, we first change the chart so we have only ONE beagle, digging the same amount of holes. This, of course, TRIPLES the amount of days.

Beagles       Holes     Days
----------------------------------
3             7         8
1             7         24

Next, we find out how long it takes this one beagle to dig just ONE hole. Now, the amount of days is divided by 7.

Beagles       Holes     Days
----------------------------------
3             7         8
1             7         24
1             1        24/7 = 3 3/7

Then, to the question. How many days will it take 5 beagles to dig 9 holes? First, let's increase the amount of beagles to 5, digging the one hole. That will slash the amount of days by 5:

Beagles       Holes     Days
----------------------------------
3             7         8
1             7         24
1             1        24/7
5             1       (24/7)/5 = 24/35

Lastly, we increase the number of holes from 1 to 9, so that the amount of days will also increase 9-fold:

Beagles       Holes     Days
----------------------------------
3             7         8
1             7         24
1             1        24/7
5             1        (24/7)/5 = 24/35
5             9       9 * 24/35 ≈ 6.17

You could also go through this in some other order. But the beauty of this chart approach is that it will work for any numbers. So, kids who have hard time with joint variation formulas might be able to use such chart approach successfully for all these types of work problems.

---

Joint Variation: The formula

For completeness sake, I'll also solve this problem of mine using the formula. Let d be the number of days, h be the number of holes, and b be the number of beagles. And, k is the constant of variation. We know that d varies directly with the number of holes (the more holes, the more days it takes). Also, d varies inversely with the number of beagles (the more beagles, the less days). So:

d = k * (h/b)

To solve for k, we plug in the values from the original situation (Three beagles can dig 7 holes in eight days).

8 = k * (7/3) from which k = 24/7.

Then we have our formula ready to use:

d = (24/7) * (h/b)

The question was: how many days will it take 5 beagles to dig 9 holes? So b is 5 and h is 9:

d = (24/7) * (9/5) = 216/35 ≈ 6.17 days.