Math Is Fun Forum

kkkkk

Guest

i need help with this math problem

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

Re: i need help with this math problem

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.

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"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..."

Dionysus

Member

Registered: 2006-03-06

Posts: 9

Re: i need help with this math problem

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)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: i need help with this math problem

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.

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"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..."

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 52,604

Re: i need help with this math problem

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!

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: i need help with this math problem

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.

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Why did the vector cross the road?
It wanted to be normal.

damathgirl

Member

Registered: 2006-03-11

Posts: 3

Re: i need help with this math problem

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)

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: i need help with this math problem

That works!

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

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: i need help with this math problem

Problems that involve martians are always funner.

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A logarithm is just a misspelled algorithm.

ashwil

Member

Registered: 2006-02-27

Posts: 121

Re: i need help with this math problem

I took the question literally:

14 people (humans) = 28 eyes
2 martians = 6 eyes