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Topic: Algebra Word Problem - Ages (Read 525 times) |
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TruthlessHero
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Algebra Word Problem - Ages
« on: Apr 21st, 2007, 10:23pm » |
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The ages of Old and Young total 48. Old is twice as old as Young was when Old was half as old as Young will be when Young is three times as old as Old was when Old was three times as old as Young.
Anyone?
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| « Last Edit: Apr 21st, 2007, 10:36pm by TruthlessHero » |
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Eigenray
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Re: Algebra Word Problem - Ages
« Reply #1 on: Apr 22nd, 2007, 10:34am » |
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Break it down into a system of 5 equations in 5 unknowns:
Old is twice is old as Young was T1 years ago.
T1 years ago, Old was half as old as Young will be T2 years from now.
In T2 years, Young will be three times as old as Old was T3 years ago.
T3 years ago, Old was three times as old as Young.
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TruthlessHero
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Re: Algebra Word Problem - Ages
« Reply #2 on: Apr 22nd, 2007, 10:53am » |
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That's a little more complex than I ended up doing it.
O + Y = 48
O = 48 - Y
Y = 48 - O
g = age gap = O - Y = O - (48 - O) = O - 48 + O = 2O - 48
x = number of years since Old was three times as old as Young
>>Old is twice as old as Young was when Old was half as old as Young will be when Young is three times as old as (O-x = 3(Y-x)).
Solve for x, to be used later:
O-x = 3(Y-x)
O-x = 3Y - 3x
48 - y - x = 3Y - 3x
48 + 2x = 4Y
x = 2Y-24
>>Old is twice as old as Young was when Old was half as old as 9(Y-x).
>>Old is twice as old as Young was when Old was (9(Y-x))/2.
>>Old is twice as old as (9(Y-x))/2 -g.
>>Old is 2((9(Y-x))/2 -g).
Replace x with 2Y-24, Y with 48 - O, and g with 2O - 48.
O = 9(Y-x) - 2g
O = 9(Y - 2Y + 24) - 2g = 9(24 - Y) -2g
O = 9(24 - (48 - O)) - 2g = 9(24 - 48 + O) - 2g = 9(O-24) - 2g
O = 9O - 216 - 2g
8O = 216 + 2g = 236 + 2(2O - 48) = 4O + 216 - 96
4O = 120
O = 30
Y = 18
>>Old is 30. Young is 18.
I guess mine is lengthier, but it worked :P.
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| « Last Edit: Apr 22nd, 2007, 10:54am by TruthlessHero » |
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