# Ratio. Number of sheep and chickens

At a farm, the ratio of the number of chickens to the number of sheep was 5:2. After the farmer sold 15 chickens, there was an equal number of chickens and sheep. How many chickens and sheep were there at the farm in the end?

My work:

Number of chickens = C Number of Sheep = S

*We know (C/S) = (5/2)

*We know that the farmer sold 15 chickens and then had equal number of both animals. So C - 15 = S

Then I plugged C - 15 = S into (C/S) = (5/2)

So I got C = 25. (So I'm guessing there where 25 chickens before?)

Since we know that there were 25 chickens before, I pluged C = 25 into C - 15 = S

So 25 - 15 = S

So, we know that there are 10 Chickens and 10 Sheep. So the answer should be 20 total?

If I'm correct, is there a better way to do this?

• ... or you could do $\frac CS = \frac 52$ so $\frac {25}S = \frac 52$ so $S = 10$. Commented Oct 4, 2018 at 5:52

Initially you have $$5$$ units of chickens and $$2$$ units of sheep.

After selling, we have $$2$$ units of chickens and $$2$$ units of sheep.

So $$3$$ units is equal to $$15$$ animals and each unit represents $$5$$ animals.

In the end we have $$4 \times 5 = 20$$ animals. $$10$$ chickens and $$10$$ sheep.

• This is a simple and powerful method for solving many problems of this type. If it seems too weird to say things like "$5$ units of chickens," you can say there were $5x$ chickens and $2x$ sheep at the beginning. That's just saying that one "unit of chickens" is $x$ chickens. Commented Oct 4, 2018 at 20:39

The answer is right, the calculations look correct, and it is a very reasonable way to reach the answer.

Alternatively, you could've noted that the $$15$$ chickens sold were three fifths of all the chickens, and gone from there. It might have saved you a few seconds of calculation, in exchange for having to spend a few seconds writing down an explanation for why this is true.

You can check your answer by plugging your solution into the initial equation (think of systems of equations back in algebra).

We know without doubt, like you said, that $$\frac{C}{S}=\frac52$$ represents the ratio of chickens to sheep, and that $$C-15=S$$ represents 15 chickens being sold and only then becoming equal to $$S$$, therefore,

$$\$$ We plug $$S=C-15$$ into $$\frac{C}{S}$$ as follows:

$$\frac{C}{C-15}=\frac52$$ evaluate as follows $$(C-15)\frac{C}{C-15}=\frac52(C-15)$$ $$C=\frac{5c-75}{2}$$ $$2(C)=(\frac{5c-75}{2})2$$ once evaluated, you get $$2C=5C-75$$

reword $$C-15=S$$ as $$\color{red}C\color{red}=\color{red}1\color{red}5\color{red}+\color{red}S$$ then plug $$\color{red}C$$ into the above equation to solve for $$\color{blue}S$$ $$2(\color{red}1\color{red}5\color{red}+\color{red}S)=5(\color{red}1\color{red}5\color{red}+\color{red}S)-75$$ $$30+3S=75+5S-75$$ $$30+2S=5S$$ $$30=3S$$ $$\color{blue}{S=10\text{ sheep}}$$

Finally plug this $$S$$ into the original equality, $$\frac{\color{red}C}{\color{blue}S}=\frac52$$

$$\frac{\color{red}C}{\color{blue}1\color{blue}0}=\frac52$$ $$2C=\color{blue}5\color{blue}0$$ $$\color{red}{C=25\text{ chickens prior to selling}}$$

Therefore, you are correct to say 25 chickens PRIOR to selling and 10 sheep in the end, and because the problem says there is an equal amount of animals 10 sheep implies there must be 10 chickens as well.

$$C$$ number of chickens, $$S$$ number of sheep.

1) $$C/S=5/2;$$

2) $$C-15 = S$$;

1) $$S= (2/5)C$$

Combining with 2):

$$C-15=(2/5)C;$$

$$5C -75 =2C;$$

$$3C = 75$$; or $$C=25.$$

$$C=25$$ is the number of chickens before the transaction .

After:

$$C_f= 25-15 =10$$, the number of chickens after transaction .

$$C_f=S=S_f=10.$$

• Barry Cipra. Thanks. Commented Oct 4, 2018 at 20:08

... or you could do $$\frac CS = \frac 52$$ so $$\frac {25}S = \frac 52$$ so $$S = 10$$.

Another way is $$\frac CS = \frac 52 = \frac {5a}{2a}$$ for some $$a$$ where $$C = 5a$$ and $$S = 2a$$.

Then $$C - 15 = S$$ means $$5a - 15 = 2a$$ so $$3a = 15$$ and $$a =5$$. So there were $$25$$ chickens and $$10$$ cows originally.

I'm not sure if that second way is better but captures the spirit of proportions more that the straight algebra.

You can actually do this by inspection in your head and by using a multiplication factor to get the ratio into actual # of animals.

The ratio originally was $$5:2$$ (Chickens : Sheep) but then when $$15$$ chickens were sold, the ratio changed to $$2:2$$ (keeping the original ":2" of the sheep). Why didn't we make it $$5:5$$ you ask? Because we are taking chickens away so it is more clear to lower the first part of the ratio from $$5:$$ to $$2:$$.

By taking $$15$$ chickens away and changing the ratio from $$5:2$$ to $$2:2$$, we see that the $$5$$ changed to a $$2$$ so that we multiply the ratio by $$5$$ to get the actual number of animals because we see that $$15$$ less chickens changed the first part of the ratio from $$5$$ to $$2$$ which is a difference of only $$3$$, not $$15$$ and not something else. So for ratio $$5:2$$, multiply by $$5$$ and we get $$25:10$$ animals initially, then when the ratio drops to $$2:2$$, multiply both by $$5$$ and get $$10:10$$ animals afterwards.

Note that this answers differs from many others here in that this one you can just do in your head, you don't need paper and pencil. One other person had a very similar answer which I upvoted but I didn't see it until I had already posted my answer.