If the value of a $2000 machine depreciates by 20% at the end of each year, what is its value at the end of 12 years? I just need help with the ratio, is -0.2 correct? Thank you.
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What do you mean by "Is $-0.2$ correct"? Correct how? – 5xum Nov 02 '17 at 11:35
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I meant, is it the right ratio? – shhh Nov 02 '17 at 11:36
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Ratio between what and what? – 5xum Nov 02 '17 at 11:36
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This seems like a homework problem. Those generally aren’t received well. Maybe include your own attempt in full? – gen-ℤ ready to perish Nov 02 '17 at 12:51
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nope the ratio you are looking fore is 0.8 because the value in 1 year is $previous\ value- previous\ value\times 0.2=(1-0.2) \times previous\ value=0.8\times \times previous\ value$
J. Sadek
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Whatever the ratio is, it ought be the ratio that takes us from $\$2000$ to $80\%$ of $\$2000$, which is $\$1600$. What ratio could that possibly be? Let's call it $r$ and find out: $$ \$1600 = r\cdot \$2000\\ \frac{\$1600}{\$2000} = r $$ Cancelling the $\$$, a couple of zeroes and a factor of $4$, we're left with $r = \frac{4}{5} = 0.8$.
Alternative approach: Decrease by $20\%$ means decrease to $80\%$, from $100\%$. Since $\%$ just means $\frac1{100}$, this is the same as saying decrease to $0.8$ from $1$. What ratio could possibly do that? It's $0.8$.
Note that your potential answer $-0.2$ is partway to the correct answer. You just forgot to take the whole into account: $1-0.2 = 0.8$.
Arthur
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This may seem dumb, but I quite don't understand how we got 80% instead of 20% when it says in the problem that it depreciates by 20%. Thank you for your time! – shhh Nov 02 '17 at 11:41
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@m18 Because when it decreases by $20%$, then that means it decreases to $80%$ of its original value. The ratio, the way I guess you intend to use it, takes you from old value to new value, so when you calculate the ratio, you have to use the old value and the new value, and not the old value and the depreciation. – Arthur Nov 02 '17 at 11:42