SOLUTION: A woman is asked her weight. She, appropriately, replies (you should never ask a woman her weight – unless you are really asking for a slap in the face or a divorce) that she weigh

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Question 8876: A woman is asked her weight. She, appropriately, replies (you should never ask a woman her weight – unless you are really asking for a slap in the face or a divorce) that she weighs 60 lbs plus half her weight.
What is her actual weight (in lbs)?

Found 2 solutions by rapaljer, prince_abubu:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is based upon the fact that "Her weight is equal to 60 lbs. plus half her weight."

Srart by letting x = thw weight of the woman (if she is like most women I know, in this case, x is REALLY the unknown!)

"Her weight is equal to 60 lbs. plus half her weight."
+x+++++=++++60+%2B+%281%2F2%29%2A+x+

You probably don't like that fraction, so just multiply both sides of the equation by 2 and get rid of it.
2x+=+120+%2B+2%2A%281%2F2%29%2Ax
2x+=+120+%2B+x

Subtract x from each side, and you get:
2x+-+x+=+120+%2B+x+-+x
x+=+120

Check it out: Her weight is 120, which equals 60 plus half of her weight, which is another 60.

You might figure out a way to do this intuitively (like in your head), but I like equations. They are reliable!

R^2 at SCC

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
Say that her weight is that mysterious W. If she says that her weight is 60 pounds plus 1/2 of her weight, that's

+W+=+60+%2B+0.5W+ <-------------- Start here

+W+-+0.5W+=+60+ <---- subtract 0.5W from both sides

+0.5W+=+60+ <----- combine like terms

+W+=+120+ <----- divide both sides by 0.5. She weighs 120 pounds.