Photo:
THE MIRROR
You can tell how old the following task is by the small amounts of money involved.
A pet dealer bought a certain number of mice and half as many pairs of budgies.
She bought two dollars per mouse and one dollar per budgie.
She sells the animals at a 10 percent premium on the purchase price.
A few days later, when she counted the budgerigars and mice, she made an interesting observation: The previous income from the sale of the animals corresponds exactly to the expenses for buying the budgies and mice.
Seven animals have not yet been sold.
The income from the sale of the remaining animals, which we hope will soon be realized, therefore corresponds to their profit.
How big is this profit?
The profit is
$13.20
.
We denote by x the number of mice purchased.
The number of budgies is also x.
Seven animals have not yet been sold, of which y mice - and accordingly 7-y budgerigars.
Now we set up an equation: On the left are the expenses for the purchase, on the right the income for all animals except for the seven that have not yet been sold.
2*x + 1*x = 2.2*(x - y) + 1.1*(x - (7 - y))
We multiply both sides by 10 and then rearrange:
30x = 22x - 22y + 11x - 77 + 11y
0 = 3x - 11y - 77
3x = 11y + 77
We now look for all integer solutions of this equation, where y is between 0 and 7.
There are only solutions for y = 2, x = 33 and y = 5, x = 44.
There is one more condition: x must be an even number.
Because half of x corresponds to the number of pairs of budgerigars purchased.
Therefore only x = 44 and y = 5 can be considered as a solution.
So the pet dealer still has five mice and two budgies in the shop.
The sales value and therefore your profit is 5*2.2 + 2*1.1 = $13.20.
I discovered this puzzle in the book "My Best Mathematical and Logical Puzzles" by Martin Gardner.
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