Question 164035: 1.A man is now 42 years old and his friend is 33 years old. How many years ago was the man twice as old as his friend was then.
2.Paul is 3 years younger than his friend Peter. In seven years the product of their ages will be five more than the product of their ages 5 years ago. How old are they now?
Please answer this... i really need help tomorrow is the submission of my project so answer it quickly if possible. For now, i am working hard to solve the problems...
thx!!!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 1.A man is now 42 years old and his friend is 33 years old. How many years ago was the man twice as old as his friend was then.
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Present ages
M = 42
F = 33
So M = F+9 (always)
Earlier, x years ago:
42-x = 2*(33-x)
42-x = 66-2x
x = 24
24 years ago, M (the Man) was 18 and F (the Friend) was 9.
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2.Paul is 3 years younger than his friend Peter. In seven years the product of their ages will be five more than the product of their ages 5 years ago. How old are they now?
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p1 = Paul's age
p2 = Peter's age
p2 = p1+3 (it says that)
The product 7 years from now is (p1+7)*(p2+7)
The product 5 years ago was (p1-5)*(p2-5)
And, (p1+7)*(p2+7) = (p1-5)*(p2-5) + 5
Sub p1+3 for p2
(p1+7)(p1+10) = (p1-5)(p1-2) + 5
p1^2 + 17p1 + 70 = p1^2 - 7p1 + 15
17p1 + 70 = -7p1 + 15
24p1 = -55
p1 = -55/24 (Paul's age, he's not yet born)
p2 = +17/24 (Peters age)
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Odd ages, but it does check. I suspect a typo somewhere.
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