Question

Can someone please help?

The number of bags of grass seed, n, needed to reseed a yard varies directly with the area, a, to be seeded and inversely with the weight, w, of a bag of seed. If it takes two 3-lb bas to seed an area of 3000ft2, how many 3-lb bas will seed 9000ft2?

Answer 1

3lb/3000ft2 = ?lb/9000ft^2

Answer 2

The question is NOT telling you this:
\[n \space = \space \frac{a}{w}\]
Rather, it is telling you this:
\[n \space \alpha \space \frac{a}{w}\]
That is, as a increases, n increases, and as w increases, n decreases. The question tells you the relationship between these variables, but not their "equivalence". You can't just express any two variables as a ratio the way you have, because then there's a third variable you aren't accounting for (unless that variable is held constant).

To make side one equal to the other, there needs to be a constant of proportionality (k):
\[n \space = \space k \times\frac{a}{w}\]
For instance, R is the constant of proportionality in the gas law, pi is the constant of proportionality in calculating the circumference of a circle, and so on.

In the last sentence, we are given information that allows us to solve for k. However, because k is a CONSTANT, we can use it set that situation (1) equal to the one we're trying to solve (2):

\[\frac{n_{1} \times w_{1}}{a_{1}}=k \space \space \space and \space \space \space \frac{n_{2} \times w_{2}}{a_{2}}=k\]
so:
\[\frac{n_{1} \times w_{1}}{a_{1}}= \frac{n_{2} \times w_{2}}{a_{2}}\]
\[\frac{2 \times 3}{3000}= \frac{n_{2} \times 3}{9000}\]
\[n_{2}=6\]

Answer 3

If you want to do this more easily, you can do a simple ratio, but you need to use the variables that change. In this case, the weight isn't changing (it's 3 lbs in both cases), so it's not useful putting that in the ratio. Instead, find the ratio of n/a:

\[\frac{n_{1}}{a_{1}}=\frac{n_{2}}{a_{2}}\]
\[\frac{2}{3000}=\frac{n_{2}}{9000}\]
\[n_{2}=6\]