### Brain Teasers

# Animal Fair

Long long ago, in a village there used to be held an annual animal fair where only elephants, horses and sheep used to be sold. During one year the price-list in the fair was as follows:-

One Elephant --- $5

One Horse --- $1

Twenty Sheep --- $1

A merchant sent his servant with $100 to the fair and instructed him to buy 100 animals with the money....but the condition was that there had to be at least one animal of each kind in the lot and there must be no money left over.

How did the servant go about carrying out his task ?

One Elephant --- $5

One Horse --- $1

Twenty Sheep --- $1

A merchant sent his servant with $100 to the fair and instructed him to buy 100 animals with the money....but the condition was that there had to be at least one animal of each kind in the lot and there must be no money left over.

How did the servant go about carrying out his task ?

### Answer

The servant bought 19 elephants, 1 horse and 80 sheep.Total number of animals = 19 + 1 + 80

= 100

Total cost = (19 x 5)+(1x1)+(80 x 1/20)

= 95 + 1 + 4

= 100

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## Comments

You know, there are other possiblities too.

He could've bought 10 elephants, 49 horses, and 20 sheep; 9 elephants, 50 horses, and 100 sheep; 1 elephant, 1 horse, and 1920 sheep....There are WAY too many possible answers. BAD TEASER. It should be something like "How many animals does he have to buy so the number of each type of animal is closest."

It says that the total animals bought should equal to 100, and I don't think there's another combination that could work ... correct me if I'm wrong

e + h + s = 100

5e + h + s/20 = 100

e + h + s = 5e + h + s/20

Then use trial and error to try and balance the equation. It only helps a bit though.

5e + h + s/20 = 100

e + h + s = 5e + h + s/20

Then use trial and error to try and balance the equation. It only helps a bit though.

Sane's last equation is shaky because it equates dollars to animals.

Knowing you must buy sheep in multiples of twenty to avoid remnants of a dollar means $4 or less will be spent on sheep and the number elephants plus the number of horses will be even.

There is no way to spend $97 or $99 on horses and elephants. If you first buy an odd number of horses, no matter how often you trade five horses for one elephant, the total number of animals is odd.

Happily, this cuts calculations to figuring if we can get 60 horses and elephants for $98 or 20 fo $96.

Knowing you must buy sheep in multiples of twenty to avoid remnants of a dollar means $4 or less will be spent on sheep and the number elephants plus the number of horses will be even.

There is no way to spend $97 or $99 on horses and elephants. If you first buy an odd number of horses, no matter how often you trade five horses for one elephant, the total number of animals is odd.

Happily, this cuts calculations to figuring if we can get 60 horses and elephants for $98 or 20 fo $96.

Let there be x elephants, y horses and 100 â€“ (x + y) sheep

The cost is 5x + y + [100 â€“ (x + y)]/20 = 100

100x + 20y + 100 â€“ x â€“ y = 2000

99x + 19y = 1900

19y = 1900 â€“ 99x

y = 100 â€“ 99x/19 and y must be an integer

Since 19 is a prime number and does not divide 99, x is a multiple of 19.

Hence x = 19 (Elephants) , y = 1 (Horse) and there are 80 sheep.

Note if x > 19 then y becomes < 1 which is impossible.

The cost is 5x + y + [100 â€“ (x + y)]/20 = 100

100x + 20y + 100 â€“ x â€“ y = 2000

99x + 19y = 1900

19y = 1900 â€“ 99x

y = 100 â€“ 99x/19 and y must be an integer

Since 19 is a prime number and does not divide 99, x is a multiple of 19.

Hence x = 19 (Elephants) , y = 1 (Horse) and there are 80 sheep.

Note if x > 19 then y becomes < 1 which is impossible.

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