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.
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