Some years back I was travelling by a cargo ship from New Zealand to Tahiti. I was curious to look around the ship one day and in the boiler room I asked a man how old the ship was. He smiled and replied to me in this way: 'The ship is twice as old as its boiler was when the ship was as old as the boiler is now. And the combined age of the ship and the boiler is thirty years.' Can you figure out what is the age of the ship and of the boiler?
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2 Answers
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Age of the ship (S) = twice (the boiler's age (B) minus (the ship's age minus the boiler's age)):
S = 2(B-(S-B))
= 2(2B-S)
= 4B-2S | +2S
3S = 4B | :3
S = 4B/3
Combined age is 30:
S+B = 30 | -S
B = 30-S
Solve S:
S = 4(30-S)/3 | *3
3S = 120-4S | +4S
7S = 120
S = 17 1/7
B = 30-S
= 12 6/7
The ship is a bit over 17 years old and the boiler is almost 13 years old.
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First, some definitions:
$\begin{array}{ll} S_n & \text{Age of ship now} \\ S_t & \text{Age of ship then} \\ B_n & \text{Age of boiler now} \\ B_t & \text{Age of boiler then} \\ t & \text{Time between then and now} \end{array}$
This leaves us with the following equations to solve:
$\begin{align} S_n & = S_t + t \tag{a} \\ B_n & = B_t + t \tag{b} \\ S_n & = 2 \cdot B_t \tag{c} \\ S_t & = B_n \tag{d} \\ S_n + B_n &= 30 \tag{e} \\ \end{align}$
Combining (a) and (d) yields $S_n = B_n + t$ or alternatively $t = S_n - B_n$ which is more to the point. Combining this with (b) and (c) gives us
$$\begin{align} S_n & = 2 \cdot B_t \\ S_n & = 2 \cdot (B_n - t) \\ S_n & = 2 \cdot (B_n - (S_n - B_n)) \\ S_n & = 2 \cdot (2 \cdot B_n - S_n) \\ S_n & = 4 \cdot B_n - 2 \cdot S_n \\ 3 \cdot S_n & = 4 \cdot B_n \end{align}$$ Combining this with (e), we can solve it: $$\begin{align} 3 \cdot S_n & = 4 \cdot B_n \\ 3 \cdot S_n & = 4 \cdot (30 - S_n) \\ 3 \cdot S_n & = 120 - 4 \cdot S_n \\ 7 \cdot S_n & = 120 \\ S_n & = \frac{120}{7} \end{align}$$ Remembering that they total to 30 years, we get $$\begin{align} S_n & = \frac{120}{7} & = 17\tfrac{1}{7} \text{years} \\ B_n & = \frac{90}{7} & = 12\tfrac{6}{7} \text{years} \end{align}$$
