let the weight of the clown be x and the trapeze be y
x=60+y
2/3x=y
1/3x=60
x=180
y=120
2006-09-02 06:48:13
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answer #1
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answered by raj 7
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120. I know this isn't the correct way to solve the problem, but sometimes, if it is multiple choice, I take the answers and work backwards, if I don't know how to solve the problem. For instance, start with 125 for the weight of the trapeze artist. If the clown weighs 60 lbs more, he would weigh 185. Since the trapeze artist weighs 2/3 the weight of the clown, multiply 180 by 2/3 -- you'll get a little over 123. 123 is not equal to 125, so 125 isn't the correct answer. Go on to 120. Add 60 to get the weight of the clown -- 180. Multiply the 180 by 2/3 and you get 120.
2006-09-02 07:00:10
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answer #2
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answered by kycheerchick 1
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The trapeze artist weighs 120 pounds, the clown weighs 180.
You can do the proof:
C=T+60
T= 2/3C
The math works if 120 is the weight of the trapeze artist - 120 + 60 is 180, 2/3rds of 180 is 120 (180 X 2 = 360, 360 divided by 3 is 120).
2006-09-02 06:54:19
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answer #3
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answered by MCB 2
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Let the clown weigh x lb and the trapeze artist weigh y lb.
y+60=x---(1)
2/3x=y---(2)
From (1),
y+60=x
x=y+60---(3)
Sub (3) into (2)
2/3(y+60)=y
2y+120=3y
y=120
Sub y=120 into (1)
120+60=x
x=180
So the clown weighs 180 lbs and the trapeze artist weighs 120 lbs.
2006-09-02 08:41:36
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answer #4
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answered by Mysterious 3
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Math-A clown weighs 180 and the trapeze artist weighs 120
2006-09-02 06:54:20
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answer #5
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answered by BANG P 7
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Trapeze artist = 120 lbs
Clown = 180 Lbs.
Dubya (an even bigger clown) = 220 Lbs.
2006-09-02 06:46:20
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answer #6
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answered by F. Frederick Skitty 7
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120 for the trapeze artist, 180 for the clown
2006-09-02 06:45:54
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answer #7
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answered by Anonymous
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clown=trapeze + 60 which is c=t+60
Trapeze = 2/3 clown which is t=2/3 c
now rewrite the first equation to be t=c-60 (you subtract 60 from each side)
Subtract the second equation from the new first equation.
t=c-60
t=2/3c
0=1/3 c-60
Add 60 to each side
60=1/3c
now multiply by 3
180=c
2006-09-02 06:49:17
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answer #8
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answered by Anonymous
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shouldn't you be doing your own homework. I won't tell you the answer but i will tell you that math.com can really help with math homework.
2006-09-02 06:46:16
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answer #9
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answered by alba g 1
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