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A boat loses r% of its value each year. After 3 years the value of the boat is half its original value. Find the value of r.​

Sagot :

The equation to set up would be

(1-r/100)^3 = 1/2

The r/100 refers to the r% portion. Subtract that from 1 to indicate how much is leftover after the particular year. Cubing that expression lets us see how much remains after 3 years.

We set that expression equal to 1/2 to indicate it has half its value by this point in time.

If you use graphing technology to solve that equation you should get approximately r = 20.6299

Therefore, the boat loses about 20.6299% of its value each year.

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Example: Let's say the boat starts off at $1000

  • End of year 1 it would be worth around 1000*(1-20.6299/100) = 793.70 dollars
  • End of year 2 it would be worth around 793.70*(1-20.6299/100) = 629.96 dollars
  • End of year 3 it would be worth around 629.96*(1-20.6299/100) = 499.99988196 dollars which rounds to 500 when rounding to the nearest hundredth.

There's a bit of rounding error but we get close enough.

A shortcut would be to compute 1000*(1-20.6299/100)^3 = 500.0008958345 which rounds to 500 when rounding to the nearest penny. Interestingly we get different rounding error happening here. This time we're a bit too large compared to the 500 dollar marker.

Whichever method we use, we have shown that the boat reaches half its original value by the end of year 3 (since 500 is half of 1000).

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