Venn Diagram Problem

**latinolp85** · October 19th 2007, 05:32 AM

Ok i'm having an issue with this problem here:

In a group of cows and chickens i count 48 heads and 122 feet. How many cows and chickens do i have?

i tried drawing it out on the venn diagram and came out with a completely wrong looking one, please if someone could help me with the answer to this one along with a picture of the venn diagram?

**ThePerfectHacker** · October 19th 2007, 06:31 AM

Originally Posted by latinolp85

Ok i'm having an issue with this problem here:

In a group of cows and chickens i count 48 heads and 122 feet. How many cows and chickens do i have?

i tried drawing it out on the venn diagram and came out with a completely wrong looking one, please if someone could help me with the answer to this one along with a picture of the venn diagram?

Let $x$ be cow $y$ for the other one. So $x+y=48$ and $4x+2y=122$ . Does that make sense? (And before to think about it before saying "No").

**Mark@Work** · October 19th 2007, 08:14 AM

I don't realy see that your problem suits a Venn diagram,
The best I could come up with was a kind of donut showing animals with at least 2 legs and animals with more than two legs.
Which is not very helpful.

as ThePerfectHacker suggests, the problem is an absolutely classic simultaneous equation.

Are you sure that you understood your assignment correctly and that your teacher definately specified Venn diagrams?

Mark.

**Soroban** · October 19th 2007, 10:42 AM

Hello, latinolp85!

Who assigned this problem?
A Venn diagram is the worst way to solve this problem!

In a group of cows and chickens i count 48 heads and 122 feet.
How many cows and fowl do i have?

I spent several minutes trying to create a Venn diagram for this problem.

Let one circle be the set of cows.
Everything outside the circle is a fowl.

Code:

      * - - - - - - - - - - - - - - - *
      |                               |
      |                               |
      |     * - - - - - - *           |
      |     |             |           |
      |     |             |           |
      |     |     Cows    |           |
      |     |             |           |
      |     |             |           | 
      |     * - - - - - - *           |
      |                               |
      |                               | 
      |                 Fowl          |
      |                               |
      |                               |
      * - - - - - - - - - - - - - - - *

Let a second circle be the sets of Heads.
Everything outside the circle is a foot.

Code:

      * - - - - - - - - - - - - - - - *
      |                               |
      |                               |
      |      Feet                     |
      |                               |
      |                               |
      |           * - - - - - - *     |
      |           |             |     |
      |           |             |     | 
      |           |    Heads    |     |
      |           |             |     |
      |           |             |     | 
      |           * - - - - - - *     |
      |                               |
      |                               |
      * - - - - - - - - - - - - - - - *

Then our Venn diagram looks like this.

Code:

      * - - - - - - - - - - - - - - - *
      |                               |
      |                               |
      |     * - - - - - - *           |
      |     |  Cow        |           |
      |     |  feet       |           |
      |     |     * - - - + - - *     |
      |     |     |  Cow  |     |     |
      |     |     | heads |     |     | 
      |     * - - + - - - *     |     |
      |           |     Fowl    |     |
      |           |     heads   |     |
      |           * - - - - - - *     |
      |    Fowl                       |
      |    feet                       |
      * - - - - - - - - - - - - - - - *

Then I spent several more minutes trying to solve the problem
. . using this Venn diagram.
I failed abysmally. .I was forced to rely on Algebra.
. . This meant all the above work was a total waste of time.

Let $C$ = number of cows .and $F$ = number of fowl.

Since each cow has one head and four feet: . $\begin{array}{ccc}\text{Cow heads} & = & C \\ \text{Cow feet} & = & 4C\end{array}$

Since each fowl has one head and two feet: . $\begin{array}{ccc}\text{Fowl heads} & = & F \\ \text{Fowl feet} & = & 2F\end{array}$

We are told that: . $\begin{array}{ccc}\text{[Cow heads] + [Fowl heads]} & = & 48 \\<br /> \text{[Cow feet] + [Fowl feet]} & = & 122 \end{array}$

This becomes the system: . $\begin{array}{ccc}C + F & = & 48 \\ 4C + 2F & = & 122\end{array}$

. . which has the solution: . $C\,+\,13,\;F\,=\,35$

Like I said, the worst way to solve it.

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