Three countrymen met at a cattle market.

"Look here," said Hodge to Jakes, "I'll give you six of my pigs for one of your horses, and then you'll have twice as many animals here as I've got."

"If that's your way of doing business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a horse, and then you'll have three times as many animals as I."

"Well, I'll go better than that," said Jakes to Durrant; "I'll give you four cows for a horse, and then you'll have six times as many animals as I've got here."

How many animals Jakes, Hodge, and Durrant must have taken to the cattle market?

AMUSEMENTS IN MATHEMATICS by HENRY ERNEST DUDENEY

### Solution:

- Jakes must have taken 7 animals to market
- Hodge must have taken 11
- Durrant must have taken 21
- There were thus 39 animals altogether.

### Explanation:

Let's say the number of animals taken to the market by Hodge, Jakes and Durrant are p, q and r respectively.
We learned that, for Hodge and Jakes:

(p - 6 + 1 ) x 2 = q - 1 + 6

2p - q = 15

2p - q - 15 = 0

For Durrant and Hodge:

(r - 14 + 1) x 3 = p -1 + 14

3r - p = 52

p - 3r + 52 = 0

For Jakes and Durrant:

(q - 4 + 1) x 6 = r + 4 - 1

6q - r = 21

6q - r - 21 = 0

Solve the 3 simultaneous equations:

2p - q - 15 = 0

p - 3r + 52 = 0

6q - r - 21 = 0

p = 11, q = 7, r = 21

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