A friend of mine posted this on LinkedIn. Quite interesting problems. It should be solvable at 8-10th grade levels.

## Solutions

### 1. Number of Apes

Let A be the number of apes. It then follows:

A = \left(A/8\right)^2 + 12

\implies A^2 - 64A + 12.64 = 0

\implies A = (64 \pm \sqrt{64^2 - 48.64})/2

\implies A = (64 \pm 8.\sqrt{16})/2

\implies A = (64 \pm 32)/2

\implies A = 48 or 16

It is interesting to note there are two solutions to this problem, 48 and 16. I was searching for the maximum or minimum number of apes in a pack, but it was not conclusive. So we leave it at this.

### 2. Horses, Camels, Mules and Oxen

It is easy to see that it forms a system of simultaneous equations. However, we don’t know how much fortune, so we assume 1 and go from there.

Let a, b, c and d be the cost of a Horse, Camel, Mule and Ox respectively.

5a+2b+8c+7d = 1

3a+7b+2c+d = 1

6a+4b+c+2d=1

8a+b+3c+d=1

I am not in the mood to solve this by hand, so I have plugged it in Wolfram Alpha and got the answer.

Since we don’t know the actual value of the treasure, the answer is a = 85x, b = 76x, c = 31x, and d = 4x, where x is some multiple.

### 3. Peacock and Snake

We assume the peacock glides in a straight line. Let x be the distance from the pole. Hence,

(27-x)^2 = 81 + x^2

\implies 3^6 + x^2 -54x = 81 + x ^ 2

\implies 54x = 3^6 - 81

\implies x = \frac {3^6 - 3^4} {54} = \frac{3^4 (9 - 1)}{3^3.2}

\implies x = 12 C