How Do You Calculate the Age Ratio Between a Ship and a Boiler?

[DaveC426913]

The ship is twice as old as the boiler was when the ship was as old as the boiler is.

What is the ratio of their ages?


I do believe I have solved this (without Googling) and have uploaded my answer online with a timestamp of 10:35PM 4/27/2011 (i.e. before I posted this).

Let me know what you think.

 

[Jimmy Snyder]

I got:

Spoiler

4 / 3
Let n be a positive integer. If the boiler was 2n when the ship was 3n, then the boiler is 3n and the ship is 4n (twice what the boiler was when the ship was 3n). 4n / 3n = 4 / 3.

 

[Filip Larsen]

Nice one. I had to make a timeline diagram to get it sorted out.

 

[DaveC426913]

I tried to solve it formulaically but quickly got mired. So I solved it geometrically.

http://www.davesbrain.ca/science/ships-boiler.png

 

[Jimmy Snyder]

I solved it algebraicly:

Spoiler

SW = ship was; SI = ship is; BW = boiler was; BI = boiler is:
"ship was as old as the boiler is"
SW = BI
"The ship is twice as old as the boiler was"
SI = 2BW
And the fact that both age the same:
SI - SW = BI - BW
This is three equations in 4 unknowns, but only a ratio is required so there is sufficient input.

 

[BobG]

I went algebraically as well:

Spoiler

x = age boiler was
2x = age of ship
y = years since boiler was x
x+y = current age of boiler
x+y = age of ship when boiler was x

ship age = (x+y)+y= x+2y
ship age = 2x
therefore:
2x = x+2y
(1/2)x=y

current age of ship = x + 2(1/2)x = 2x
current age of boiler = x + (1/2)x = (3/2)x

ratio of ship to boiler
2x:(3/2)x or 4:3

I do like Dave's geometric solution, though.

 

[DaveC426913]

Nice one. I had to make a timeline diagram to get it sorted out.

I'd like to see it.

The guy who gave me the puzzle it turns out also was working on a geometric solution and it looks very much like mine.

 

[Borek]

This is not much different from https://www.physicsforums.com/showthread.php?t=245602

 

[Filip Larsen]

I'd like to see it.

Well, my head was spinning from getting the past and present tense sorted out, so it probably ended up being a bit more elaborate than necessary.

Spoiler

Make made two time lines representing the "life" of the ship and boiler:

|---------------S₀----->S₁
|..d..|---------B₀----->B₁

where the timeline for the boiler starts (i.e. has zero) a time d later than for the ship. The following equations can then be extracted: S₁ = 2B₀, B₁-B₀ = d, B₁ = S₀, and S₁-S₀ = d. Solving these for S₁/B₁ yields 4/3.

 

[regor60]

I found this one challenging:

https://www.physicsforums.com/showthread.php?t=360070


but once you come up with the method, all of these are easy