# Women, sacks and cats algebra

1. Mar 24, 2016

### Natasha1

1. The problem statement, all variables and given/known data
In West Cornwall, I met w women each carrying s sacks, each containing c cats. However, a third of the women then lost half their sacks, and the remaining women lost half their cats from a third of their sacks. How many cats remained, it terms of w, s and c?

2. The attempt at a solution

w.s.c - (1/3w.1/2s) - (2/3w.1/2c.1/3s)

. = multiplication?

2. Mar 24, 2016

### Staff: Mentor

Almost. Consider the case w=3, s=2, c=100, and ignore the cat losses for now (so just 1/3 of the women loses 1/2 their sacks). Can w.s.c - (1/3w.1/2s) be true?

* is the usual (ascii) multiplication sign.

3. Mar 24, 2016

### Natasha1

Hallsoflvy why is (w/3)(s/2) sacks so wsc/6 cats. I do not understand this part...

(w/3)(s/2) = 2w3s/6

4. Mar 24, 2016

### Ray Vickson

The correct result is as HallsofIvy says:
$$\frac{w}{3} \frac{s}{2} = \frac{ws}{6}$$
You multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator. Why would you think otherwise?

Last edited: Mar 24, 2016
5. Mar 24, 2016

### Natasha1

(w/3)(s/2) = (ws/6)

then

(2/3w)(s/3)(c/2) = wsc/9

6. Mar 24, 2016

### Natasha1

How did Hallsoflvy get to 11 wsc/6

I get (w/3)(s/2)(c/1) = wsc/6
and
(2/3w)(s/3)(c/2) = wsc/9

Hence
wsc/6 + wsc/9 = 5wsc/18

Last edited: Mar 24, 2016
7. Mar 25, 2016

### Staff: Mentor

Subtract that from the total, and you get the same result as me.

8. Mar 25, 2016

### Natasha1

swc - 5wsc/18 = 13wsc/18 (is this correct?)

9. Mar 25, 2016

### PeroK

You need to get into the habit of testing answers with some real numbers. This sort of problem should be easy to check. Try, for example:

w = 3, s = 6, c = 10

It's even a good idea to work out the problem with these actual numbers first, to establish the pattern of the solution. Then, extend your specific solution using 3, 6, 10 to a general one using w, s, c.