# Math Word Problem — Ship and Anchor

Remember those word problems in high school or freshman college math class? Maybe you, like a lot of people, hated them. Maybe you liked them. If you are in the latter group, here’s one for you:

A ship is twice as old as the anchor was when the ship was as old as the anchor is. The sum of their two present ages is 42 years. How old is each?

You can email me for the answer, or you can solve it and post your answer and solution in the comments. I haven’t looked but I wouldn’t be surprised that you could find the answer and the solution somewhere on the internet. But that would be cheating.

#### Math Word Problem — Ship and Anchor — 6 Comments

1. Let:

S = age ship is
Ai = age anchor is
Aw = age anchor was when ship was age anchor is

From the story:

S + Ai = 42
-> S = 42 – Ai [1]

S = 2 x Aw [2]

Aw = Ai – (S – Ai) [3]

From [3]:

Aw = Ai – S + Ai
-> Aw = 2 x Ai – S [4]

Sub [4] into [2]:

S = 2 x (2 x Ai – S)
-> S = 4 x Ai – 2S
-> 3 x S = 4 x Ai [5]

Sub [1] into [5]:

3 x (42 – Ai) = 4 x Ai
-> 126 – 3 x Ai = 4 x Ai
-> 7 x Ai = 126
-> Ai = 126 / 7
=> Ai = 18

So, from [1], S = 42 – 18 = 24

Thus, the ship is currently 24 and the anchor is 18

END

• Brilliant! Your immediate recognition that the present age of the ship must be twice the age the anchor was, led you down the right path.

2. Yes, that’s how I solved the problem; I believe the objective was to solve the problem.

I used the “iterative approach”, a valid and accepted method used in scientific endeavors. When the situation requires it, the next step would be to generalizem and (hopefully) develop a formula or set of equations for solving similar problems in the future. I did not see the need for a general solution for the problem, and am content to leave that task for those who are interested, or have a continuing need for such. Cheers, R.

3. The ship is [bleep] years old’ the anchor is [bleep] years old.