Word problem for math nerds

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I'm cleaning out old papers, and scanning the keepers. Found this old puzzle, likely from the early '80s. Give it a try. The difficulty, as is typical of word problems, is setting up the two formulas. Once you've got them, the algebra is straightforward.

Len and Pam math puzzle 001.jpg
 
Nice job, VP & CR94. It took me ages decades ago, and was no easier this time around.

Wurlitzer, let P be Pam's present age, and L be Len's present age.

1st equation:

"Pam is twice as old as Len was when Pam was as old as Len is (now)"

P = 2 * (L - [P - L])
P = 2 * (L - P + L)
P = 2 * (2L - P)
P = 4L - 2P
3P = 4L
P = 4/3 L

2nd equation:

"Len is half as old as Pam will be when Len is three years older than Pam is now."

L = 1/2 * (P + [P - L] + 3)
2L = P + P - L + 3
2L = 2P - L + 3
3L = 2P + 3

Now replace P with 4/3 L in the 2nd equation (because we know that P = 4/3 L).

3L = 2(4/3L) +3
3L = 8/3L + 3
9/3L = 8/3L + 3
1/3L = 3
L = 9

Len is 9.

Replace L with 9 in the first equation:

P = 4/3(9) = 12

Pam is 12.

Now check by working out the word problem with Pam's age = 12, and Len's age = 9.

Pam (12) is twice as old as Len (now 9) was when Pam (now 12) was as old as Len is now (9). We know that Len is 3 years younger than Pam, so when Pam was 9, Len must have been 6. Sure enough, Pam now, at 12, is twice as old as Len was then (6).

Does the 2nd part hold as well?

Len (9) is half as old as Pam will be when Len is three years older than Pam is now (12). So when Len is three years older than Pam is now, he'll be 15. At that point Pam, being three years older than Len, will be 18. Pam's age then (18) is twice as old as Len (9) is now.

Phew!
crazy.gif
 
I did it with 4 simpler equations with 4 unknowns, but basically the same route Number_35 illustrated. Unfortunately, with my late start and algebraic slowness, Vitus_Probi beat me to the answer.
 
Originally Posted by CR94
I did it with 4 simpler equations with 4 unknowns, but basically the same route Number_35 illustrated. Unfortunately, with my late start and algebraic slowness, Vitus_Probi beat me to the answer.

True confession time ... I knew I'd done this years and years ago, but was getting nowhere, so I plugged in some guesstimate numbers which helped me come up with the correct answer. That then helped me to reverse-engineer the formulas.

Give how quickly you (CR94) and VP responded, you both solved this a lot more quickly than I did. Well done!
 
Decades ago I would have solved this with matrices.

I never had the motivation to do stuff like this just for amusement after I finished college.

Today you could probably set this up in a spreadsheet and have Excel solve it.
 
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Originally Posted by Nyogtha
Decades ago I would have solved this with matrices.

I never had the motivation to do stuff like this just for amusement after I finished college.

Today you could probably set this up in a spreadsheet and have Excel solve it.

I learned about matrices in linear algebra in '75, and don't remember using them since. I should look into them.

I never thought of using a spreadsheet - good one.

I find stuff like this is fun, and it likely keeps the grey cells active.
 
The approach Number_35 gave is the cleanest, I actually introduced another variable X=P-L to help wrap my head around it.
I'm pretty sure I tossed the envelope I was scribbling on to solve this (and, yes, it was on the back), but if it's still sitting by the home computer I'll reproduce the equations later.
 
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