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Advanced Math/Algebra Age Problems


Bob is 6 years years more than 3 times as old as his daughter. Eight years ago, he was 2 times less than 9 times as old as his daughter was then. Find each of their ages.

Let Bob's age be B and his daughter's age be D. Then the first statement says that

B = 6 + 3D.

Let their ages 8 years ago be B' = B - 8 and D' = D - 8. Now, I'm not quite sure how to interpret the 2nd statement "2 times less than 9 times as old", but I think it means that, 8 years ago, Bob was one half 9 times his daughter's age. In which case

B' = (1/2)(9D')     <- notice the primes

B - 8 = (1/2)・9・(D-8) = (9/2)D - 9・8/2

B = (9/2)D - 28.

Using the first eqn to substitute for B gives

3D + 6 = (9/2)D - 28

or D = (2/3)・34 = 22.7 = daughter's age        (sorry, you'll have to do the intermediate algebra)

Which gives B = 6 + 3(22.7) = 72 = Bob's age.

Let me know if this makes sense.


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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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