solve this if you can.
solve this if you can.
solve this problem
you 100$ you need to buy 100 chickens
roosters= $5.00
hens= $1.00
peps= .05
you have to buy atleast one of each and you need to spend exactly 100$
i cant get it see how many of you can
you 100$ you need to buy 100 chickens
roosters= $5.00
hens= $1.00
peps= .05
you have to buy atleast one of each and you need to spend exactly 100$
i cant get it see how many of you can
I come close but not quite there. I guess I got 2 chicken for free. LOL.
09 roosters x $5.00 each = $45.00
53 hens x $1.00 each = $53.00
09 roosters x $5.00 each = $45.00
53 hens x $1.00 each = $53.00
40 peps x $0.05 eahc = $02.00
102 chicken for $100.00
102 chicken for $100.00
Im done with school so screw this. lol
Member
Joined: Jan 2008
Posts: 554
From: Orange County
let r = rooster, h=hens, p=peps
Two conditions that you've set:
5r + 1h+ .05p = 100
r + h + p = 100
Now for the substitutions:
p = 100 - h - r
5r + 1h +.5*(100 - h - r) = 100
4.5r + .5h = 50 (then times it by 2 to make it whole numbers)
9r +h = 100
h = 100 - 9r
p = 100 - 100 - 9r -r = -10r
So we have
r+ 100-9r + -10r = 100
hmm...looks like no solution but let me check. shorter way of telling is to subtract first equation from second. here's the easier way. I messed up with .05 vs .5 somewhere up there.
5r + 1h+ .05p = 100
r + h + p = 100
_________________
4r +0 + -.95p =0
r = .95/4 *p
.95/4 p + h + p = 100
h = 100 - 1.2375p
.2375p + 100 - 1.2375p + p = 100
uhh... .2375p + 1.2375p + 0 = 0 ... no solution because you have to have at least one pep.
Then there's the matrix form for the mathematicians:
[5, 1, .05] 100
[1, 1, 1] 100
[0 0 0] 0
As you can see if you subtract the second row with first row, there's no way of getting 1 in the middle column. Therefore, there's no solution because coeficient matrix cannot be transformed into identity matrix. Muhahaha!
Actually Roosters are not chickens, and it can be argued if a peps are even a real thing (maybe it is suppose to be peeps?) so the answer would be 100 Chickens and $1.00 each.
Member
Joined: Aug 2007
Posts: 153
From: Florida
Hmm... linear algebra.
let r = rooster, h=hens, p=peps
Two conditions that you've set:
5r + 1h+ .05p = 100
r + h + p = 100
Now for the substitutions:
p = 100 - h - r
5r + 1h +.5*(100 - h - r) = 100
4.5r + .5h = 50 (then times it by 2 to make it whole numbers)
9r +h = 100
h = 100 - 9r
p = 100 - 100 - 9r -r = -10r
So we have
r+ 100-9r + -10r = 100
hmm...looks like no solution but let me check. shorter way of telling is to subtract first equation from second. here's the easier way. I messed up with .05 vs .5 somewhere up there.
5r + 1h+ .05p = 100
r + h + p = 100
_________________
4r +0 + -.95p =0
r = .95/4 *p
.95/4 p + h + p = 100
h = 100 - 1.2375p
.2375p + 100 - 1.2375p + p = 100
uhh... .2375p + 1.2375p + 0 = 0 ... no solution because you have to have at least one pep.
Then there's the matrix form for the mathematicians:
[5, 1, .05] 100
[1, 1, 1] 100
[0 0 0] 0
As you can see if you subtract the second row with first row, there's no way of getting 1 in the middle column. Therefore, there's no solution because coeficient matrix cannot be transformed into identity matrix. Muhahaha!
let r = rooster, h=hens, p=peps
Two conditions that you've set:
5r + 1h+ .05p = 100
r + h + p = 100
Now for the substitutions:
p = 100 - h - r
5r + 1h +.5*(100 - h - r) = 100
4.5r + .5h = 50 (then times it by 2 to make it whole numbers)
9r +h = 100
h = 100 - 9r
p = 100 - 100 - 9r -r = -10r
So we have
r+ 100-9r + -10r = 100
hmm...looks like no solution but let me check. shorter way of telling is to subtract first equation from second. here's the easier way. I messed up with .05 vs .5 somewhere up there.
5r + 1h+ .05p = 100
r + h + p = 100
_________________
4r +0 + -.95p =0
r = .95/4 *p
.95/4 p + h + p = 100
h = 100 - 1.2375p
.2375p + 100 - 1.2375p + p = 100
uhh... .2375p + 1.2375p + 0 = 0 ... no solution because you have to have at least one pep.
Then there's the matrix form for the mathematicians:
[5, 1, .05] 100
[1, 1, 1] 100
[0 0 0] 0
As you can see if you subtract the second row with first row, there's no way of getting 1 in the middle column. Therefore, there's no solution because coeficient matrix cannot be transformed into identity matrix. Muhahaha!
Well Im going to go shoot myself for actually reading through this breakdown...
Member
Joined: Dec 2007
Posts: 111
From: Kansas City, MO
THANK YOU! What the **** is a pep?? I'm assuming it's supposed to be peep, but as far as I know, they're called chicks.... Dictionary.com < go see what a pep is, I dare you.
I was just messing with you. I though you must have meant Chick. Just probably a language thing. Although a Rooster really is not a chicken. My answer is correct given the literal meaning of your post.
Member
Joined: Sep 2006
Posts: 97
From: Texas City/Huntsville
8 nickles = 40 cents
2 dimes = 20 cents
50 coins = 1 dollar
Master FitFaker. CHEA!
Joined: Mar 2008
Posts: 5,317
From: Marble Falls, TX
Hmm... linear algebra.
let r = rooster, h=hens, p=peps
Two conditions that you've set:
5r + 1h+ .05p = 100
r + h + p = 100
Now for the substitutions:
p = 100 - h - r
5r + 1h +.5*(100 - h - r) = 100
4.5r + .5h = 50 (then times it by 2 to make it whole numbers)
9r +h = 100
h = 100 - 9r
p = 100 - 100 - 9r -r = -10r
So we have
r+ 100-9r + -10r = 100
hmm...looks like no solution but let me check. shorter way of telling is to subtract first equation from second. here's the easier way. I messed up with .05 vs .5 somewhere up there.
5r + 1h+ .05p = 100
r + h + p = 100
_________________
4r +0 + -.95p =0
r = .95/4 *p
.95/4 p + h + p = 100
h = 100 - 1.2375p
.2375p + 100 - 1.2375p + p = 100
uhh... .2375p + 1.2375p + 0 = 0 ... no solution because you have to have at least one pep.
Then there's the matrix form for the mathematicians:
[5, 1, .05] 100
[1, 1, 1] 100
[0 0 0] 0
As you can see if you subtract the second row with first row, there's no way of getting 1 in the middle column. Therefore, there's no solution because coeficient matrix cannot be transformed into identity matrix. Muhahaha!
let r = rooster, h=hens, p=peps
Two conditions that you've set:
5r + 1h+ .05p = 100
r + h + p = 100
Now for the substitutions:
p = 100 - h - r
5r + 1h +.5*(100 - h - r) = 100
4.5r + .5h = 50 (then times it by 2 to make it whole numbers)
9r +h = 100
h = 100 - 9r
p = 100 - 100 - 9r -r = -10r
So we have
r+ 100-9r + -10r = 100
hmm...looks like no solution but let me check. shorter way of telling is to subtract first equation from second. here's the easier way. I messed up with .05 vs .5 somewhere up there.
5r + 1h+ .05p = 100
r + h + p = 100
_________________
4r +0 + -.95p =0
r = .95/4 *p
.95/4 p + h + p = 100
h = 100 - 1.2375p
.2375p + 100 - 1.2375p + p = 100
uhh... .2375p + 1.2375p + 0 = 0 ... no solution because you have to have at least one pep.
Then there's the matrix form for the mathematicians:
[5, 1, .05] 100
[1, 1, 1] 100
[0 0 0] 0
As you can see if you subtract the second row with first row, there's no way of getting 1 in the middle column. Therefore, there's no solution because coeficient matrix cannot be transformed into identity matrix. Muhahaha!
but it's summer, so i don't care!
school ftl
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