Heads and Legs!

Algebra Level 1

In a group of buffaloes and ducks the number of legs are 24 more than twice the number of heads. What is the number of buffaloes in the group?

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Image Credit: Flickr Majestic Duck by Daz /The duck daddy

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2 solutions

Tín Phạm Nguyễn
Mar 16, 2015

Let number of buffaloes = $b$ and number of ducks = $d$

Axiom 1: A buffalo has four legs and a duck has two legs

Axiom 2: Each buffalo and duck has only one head

Then:

$4b+2d=2(b+d)+24$

$\Leftrightarrow 2b=24$

$\Leftrightarrow b=12$

Hence, there are $\boxed{12}$ buffaloes.

Quick solution with clear explanation. You know what I find funny about this problem? There must be $12$ buffaloes, but for the number of ducks is irrelevant, within bounds ( $d\in \mathbb{N}$ ). For all we know, there could be one hundred trillion ducks, and $12$ buffaloes, and the conditions would still work.

Caleb Townsend - 6 years, 2 months ago

Thanks, I've fixed it. What a shame!

Tín Phạm Nguyễn - 6 years, 2 months ago

A Former Brilliant Member
Nov 21, 2017

Let $B$ be the number of buffaloes and $D$ be the number of ducks. Note that a buffalo has four feet and a duck has two feet. Then the equation is

$4B+2D=24+2(B+D)$

Expanding and simplifying, we have

$4B+2D=24+2B+2D$

$2B=24$

$B=12$

Heads and Legs!

Image Credit: Flickr Majestic Duck by Daz /The duck daddy

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