Question 724604: Building A was built 20 years before building B and 30 years after building C. In 10 years, building C will be 20 years younger than the combined ages of buildings A and B. What is the present age of each building?
I started to set up the problem and solve it, but my solution doesn't make any sense.
BUILDING A : Now = x-20 -> In 10 yrs = (x-20)+10
BUILDING B : Now = x -> In 10 yrs = x+10
BUILDING C : Now = x+30 -> In 10 yrs = (x+30)+10
BUILDING B
[(x+30)+10]-20=(x-20)+10+x+10
x+20=2x -> x-x+20=2x-x
x=20 <- Doesn’t Make Any Sense for the Age of B Now.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Building A was built 20 years before building B and 30 years after building C. In 10 years, building C will be 20 years younger than the combined ages of buildings A and B. What is the present age of each building?
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Let age of B be "x".
Then age of A is "x+20"
And age of C is "x + 30"
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Equation:
x+30+10 = (x+10)+(x+20+10)-20
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x + 40 = 2x+20
x = 20 age of B
x + 20 age of A
x + 30 age of C
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Cheers,
Stan H.
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