solve this if you can.
#1
05-24-2008, 10:12 PM
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
#5
05-25-2008, 12:23 AM
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
#9
06-12-2008, 12:24 PM
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!
#10
06-12-2008, 09:05 PM
Join Date: May 2008
Location: Riverview, FL, USA, Terra Firma, Milky Way
Posts: 244
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.
#12
06-12-2008, 09:12 PM
Join Date: May 2008
Location: Riverview, FL, USA, Terra Firma, Milky Way
Posts: 244
#13
06-12-2008, 09:17 PM
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...
#14
06-12-2008, 09:21 PM
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.
#16
06-12-2008, 10:02 PM
It doesn't bug me that you said it so much, it could easily be a typo, just the fact that 15 other people said it without realizing what it was bugs me...
#17
06-12-2008, 10:09 PM
Join Date: May 2008
Location: Riverview, FL, USA, Terra Firma, Milky Way
Posts: 244
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.
#18
06-17-2008, 04:00 PM
8 nickles = 40 cents
2 dimes = 20 cents
50 coins = 1 dollar
#19
06-17-2008, 04:06 PM
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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