how doesn't know simple math
b(bottle) + c(cork) = $1.05
b= c+$1.00
c+$1.00 + c = $1.05
c+c= $1.05 -$1.00
2c = $.05
c = $.025
or cork = 2 1/2 cents
bottle = $1.025 or 1 dollar and 2 1/2 cents
if the cork was 5 cents and bottle was $1 more than the cork than bottle and cork would add up to $1.10 ( $.05 + $1.05)
2006-06-22 06:19:31
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answer #1
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answered by dch921 3
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Either 5 cents or 2.5. I know there aren't any half pennys but depending on the problem, that might not matter. I say 2.5 cents because if the bottle cost a dollar more then the cork, the bottle is $1 plus whatever the cork cost. So if the cork cost 2.5 cents then you would have 2.5 + $1 and 2.5 cents, which equals 1.05.
2006-06-22 13:11:01
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answer #2
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answered by Asterisk_Love♥ 4
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If the cork costs X, then bottle cost 1 + X,
So, X + ( 1 + X) = 1.05
=> 2X = 0.05
=> X = 0.025
So, the cork is of $0.025.
2006-06-22 13:46:00
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answer #3
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answered by Sachin A 2
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It could be one dollar and the cork could be 5 cents. Then again, the bottle could be 1.01, 1.02, 1.03 or 1.04 and therefore the cork would be .04, .03, .02 or .01 respectively. I'm probably completely wrong and it's one of those questions where you'll kick yourself when you get the answer!
2006-06-22 13:09:44
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answer #4
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answered by crazyoldben 2
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$ 0.025
set x = Cork than
x + (x +1) = 1.05
2x + 1 = 1.05
2x = 0.05
x = 0.025
2006-06-22 13:07:26
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answer #5
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answered by dhaval70 2
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This is a simple algebra question. First, translate this into 2 equations:
B+C = 1.05
B=1+C.
Now substitute:
(1+C)+C = 1.05
1+2C = 1.05
2C = .05
C=0.025 (that's a pretty cheap cork!)
2006-06-22 13:19:09
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answer #6
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answered by vishalarul 2
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bottle costs $1.025 and cork costs $0.025 (assuming you can use mills to price the goods)
2006-06-22 17:40:56
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answer #7
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answered by jimbob 6
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Maximum $0.025, so I guess 1 or 2 cent, or free.
2006-06-22 13:08:05
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answer #8
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answered by User1 2
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if x=cork then
x+$1.00=$1.05
x=$0.05
x=5cents
2006-06-22 13:19:15
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answer #9
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answered by *ღ♥۩ THEMIS ۩♥ღ* 6
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Twenty -Five pennies
2006-06-22 13:16:21
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answer #10
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answered by Ms Phil 3
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