A group of cows and horses are randomly divided into two equal rows. (The animals are welltrained and stand very still.) Each animal in one row is directly opposite an animal in the other row. If 75 of the animals are horses and the number of cows opposite cows is 10 more than the number of horses opposite horses, determine the total number of animals in the group.
Solution 1

Let $T$ be the total number of animals.

Suppose that, in the first line, there are $C$ cows and $H$ horses.

Suppose that, in the second line, there are $w$ cows and $x$ horses that are standing opposite the cows in the first line, and that there are $y$ cows and $z$ horses that are standing opposite the horses in the first line.

Since there are 75 horses, then $H+x+z=75$.

Since there are 10 more cows opposite cows than horses opposite horses, then $w=z+10$.

The total number of animals is $T=C+H+w+x+y+z$.

But $y+z=H$ and $C=w+x$ so $T=(w+x)+H+w+x+H=2(H+w+x)$.

Now $w=z+10$ so $T=2(H+z+10+x)=20+2(H+x+z)=20+2(75)=170$, and so the total number of animals is 170 .

## Solution 2

Let $h$ be the number of pairs consisting of a horse opposite a horse.

Let $c$ be the number of pairs consisting of a cow opposite a cow.

Let $x$ be the number of pairs consisting of a horse opposite a cow.

Since the total number of horses is 75 , then $75=2 h+x$. (Each "horse-horse" pair includes two horses.)

Also, the given information tells us that $c=h+10$.

The total number of animals is thus

$$
2 h+2 c+2 x=2 h+2(h+10)+2(75-2 h)=2 h+2 h+20+150-4 h=170
$$

ANSWER: 170