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

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!!!!!

• 1 cat eats 0.1 kg / day. Commented Jun 16, 2016 at 0:30

## 2 Answers

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

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

• And how do I transform this into days my food will last? Since I don't have this information. The outcome in this situation: 10kg of food / 2 cats * 0.1) will result in 0.5. This can't be the number of days the food will last? Sry I am bit lost Commented Jun 16, 2016 at 0:48
• @HenriqueM. You divide by 0.1, not multiply by 0.1. So the answer will be 50 days instead. Commented Jun 16, 2016 at 0:50
• Awh! Now i see that. Then please edit your answer because I get a bit confused. Voted as right! Thanks! Commented Jun 16, 2016 at 0:51
• @HenriqueM. The answer is alright, but as dividing by 0.1 is same as multiplying with 10 I can fix that. Commented Jun 16, 2016 at 0:52

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