How to resolve this proportion/equivalence calculation? [simple one]

Question

Let's suppose I have one cat and when buying food for him I have to take into account this:

1 cat eats 2kg of food each 20 days

How can I get a formula to know how many days my food will last based on how many cats I have and how many food I've bought?

Example: 2 cats, I've bought 10kg of food
5 cats, I've bought 6kg of food

How many days the food will last? I need a formula so I can solve any input(cats, food kg bought) and get a output(days food will last)

Thanks!!!!!

Answer 1

From the condition we have that a cat eats $0.1$ kg per day. So therefore if we have $x$ cats and $y$ kgs of food. Then those $x$ cats will eat $x\cdot(0.1)$ kgs of food in one day. Divide $y$ by this and you will get wanted value. In other words

$$\text{Days} = \frac{10 \cdot \text{Kilos of Food}}{\text{No. of Cats}}$$

Answer 2

I take the first example: 2 cats, I've bought 10kg of food.

The basic are the 20 days. $x=20 d\times \ldots $

Then you built one fraction of $2$ cats and $5$ cats and another fraction of $2$ kg food and $6$ kg food.

The more cats you have the shorter the food last. Thus the fraction has to be smaller than $1$: $\frac{2}{5}$

$x=20 d\times \frac{2}{5} \times \ldots$

Now you go on with the food. THe more food you buy the longer the food last. Thus the fraction has to be bigger than $1$: $\frac{10}{6}$

In total we have $x=20 d\times \frac{2}{5} \times \frac{10}{6}=20d\cdot \frac{2}{3}=13\frac{1}{3}d$