Math formula question

[Hardbopjazz]

There are 105 animals in a room with a total of 360 legs. The animals are rabbits and chickens.

The question is how many are chickens. I know the answer is 50 chickens and 55 rabbits.

50 * 2 = 100

65 * 4 = 260

---

360.

Now do any of you know a easier way to explain this to a 11 year old?

Edited September 23, 2011 by Hardbopjazz

[jazztrain]

You appear to have 115 animals in the room, not 105.

The answer is 30 chickens and 75 rabbits.

Does the 11 year old do basic algebra?

Edited September 23, 2011 by jazztrain

[Hardbopjazz]

You appear to have 115 animals in the room, not 105.

The answer is 30 chickens and 75 rabbits.

Does the the 11 year old do basic algebra?

Yes at her school they are doing algebra.

[GA Russell]

You beat me jazz train! Here are my calculations.

4x+2y=360

x+y=105

2x+y=180

2x=180-y

-[x=105-y]

x=75

y=105-x

y=105-75

y=30

[jazztrain]

Ok.

Let C = the number of chickens

Let R = the number of rabbits

Number of animals in room = C + R = 105

Number of legs in room = 2C + 4R = 360 (assuming each animal has a normal complement of limbs)

substitute C = 105 - R in the second equation and solve for R.

you get R = 75, so C = 30.

You appear to have 115 animals in the room, not 105.

The answer is 30 chickens and 75 rabbits.

Does the the 11 year old do basic algebra?

Yes at her school they are doing algebra.

And you typed faster than I did!

You beat me jazz train! Here are my calculations.

4x+2y=360

x+y=105

2x+y=180

2x=180-y

-[x=105-y]

x=75

y=105-x

y=105-75

y=30

[bertrand]

(assuming each animal has a normal complement of limbs)

[Jazzmoose]

Quick, lock this thread before the amputee animal porn starts!

[Free For All]

I never was too good at cypherin'. Probably why I went into music.

[JSngry]

http://www.youtube.com/watch?v=Bfq5kju627c

[Hardbopjazz]

That was great Jim

[Leeway]

Hated math in school. Got queasy just reading Hardbopjazz's question. Yet I still scored much better on the SAT (1970 version!) than the current average. Current situation not encouraging, but that's material for another thread probably.

[T.D.]

We could also solve the problem without all the algebra and symbols. Here's an approach:

Start out with a blind guess of 55 rabbits and 50 chickens. Rabbits have 4 legs, and chickens 2, so the total # of legs is 55*4 + 50*2 = 320.

That's 40 legs short of our required 360, so we need to switch some chickens to rabbits to gain 40 total legs.

Now, for every chicken we switch to a rabbit we gain 2 legs (4 - 2). So to gain 40 legs we need to switch 40/2 = 20 chickens to rabbits.

That gives an answer of 55+20=75 rabbits and 50-20=30 chickens.

Check: 75*4 + 30*2 = 300+60 = 360.

But if the kid's taking algebra in school, she's probably supposed to work through the system of 2 equations in 2 unknowns...

Edited September 23, 2011 by T.D.

[JETman]

We could also solve the problem without all the algebra and symbols. Here's an approach:

Start out with a blind guess of 55 rabbits and 50 chickens. Rabbits have 4 legs, and chickens 2, so the total # of legs is 55*4 + 50*2 = 320.

That's 40 legs short of our required 360, so we need to switch some chickens to rabbits to gain 40 total legs.

Now, for every chicken we switch to a rabbit we gain 2 legs (4 - 2). So to gain 40 legs we need to switch 40/2 = 20 chickens to rabbits.

That gives an answer of 55+20=75 rabbits and 50-20=30 chickens.

Check: 75*4 + 30*2 = 300+60 = 360.

But if the kid's taking algebra in school, she's probably supposed to work through the system of 2 equations in 2 unknowns...

That, and the algebraic approach is much faster!