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**kkkkk****Guest**

3 eyed martians met with earthlings (each with 2 eyes) in the first ever intergalactic. observers noticed that there were only 14 people, yet there were 34 eyes. how many of the people are earthlings?

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Here's the way I did it, a very non-matimatical way.

Let's assume that everyone is human. Then there are 28 eyes. For every human we replace with a martian, the net gain is 1 eye.

We need to gain 34-28 = 6 eyes, meaning there are 6 martians. Thus, there are 8 humans.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Dionysus****Member**- Registered: 2006-03-06
- Posts: 9

Or you could do it the mathmatical way:

3x + 2y = 34 (3 eyes for each martian, 2 for each human, is equal to 34)

x + y = 14 (martians and humans combine to 14 people total)

we need to solve for y, ill choose the second equation and get:

y=-x + 14

Now substitue that in for y in the first equation and simplify:

3x + 2(-x + 14) = 34

3x - 2x + 28 = 34

x = 6

Now since X stands for martians, we know we have 6 martians, leaving us with 8 humans.

If you plug those in for the first equation you will find 3(6) + 2(8) = 34, thus there are:

6 Martians (18 eyes) and 8 (16 eyes) Humans.

*Last edited by Dionysus (2006-03-08 17:10:25)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

My way took about 5 seconds

But I also had a feeling kkkkk didn't know how to solve systems of equations yet. But I see no other mathimatical way to do it.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,495

3m+2e=34

m+e=14--->2m+2e=28

Therefore, m=6, e=8.

No. of maritans=6, no. of earthlings=8.

(Time taken = about 15 seconds)

This is how I teach my students to approach problems involving Simultaneous equations. Most examinations give about half a minute or less to solve such problems, and I think this is the best way it can be done!

Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Ricky wrote:

Here's the way I did it, a very non-matimatical way.

Let's assume that everyone is human. Then there are 28 eyes. For every human we replace with a martian, the net gain is 1 eye.

We need to gain 34-28 = 6 eyes, meaning there are 6 martians. Thus, there are 8 humans.

That's not very non-mathematical. It's logical at least, and logic fits in very micely with maths.

It's just a bit unconventional. It's how I'd have done it, though.

Why did the vector cross the road?

It wanted to be normal.

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**damathgirl****Member**- Registered: 2006-03-11
- Posts: 3

I got 6 martians and 8 earthlings.

Everyone had 2 eyes at least so that would be 14*2=28 take 28 from 34 and you know that you need 6 martains to equal 34 eyes!? or

*Last edited by damathgirl (2006-03-12 05:50:41)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

That works!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Problems that involve martians are always funner.

A logarithm is just a misspelled algorithm.

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**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

I took the question literally:

14 people (humans) = 28 eyes

2 martians = 6 eyes

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