Margo borrows $1800, agreeing to pay it back with 3% annual interest after 14 months. How much interest will she pay? Round your answer to the nearest cent, if necessary.

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Margo borrows $1800, agreeing to pay it back with 3% annual interest after 14 months. How much interest will she pay? Round your answer to the nearest cent, if necessary.

Mathematics
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Simple interest over time formula: I=(Po) (r) (t)
Yes, that's correct — but this problem has a wrinkle! The time is expressed in months, but you need to find annual interest. Can you express 14 months as a number of years?
Wouldn't it be 1.2?

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1.2 = 1 and 1/5, right? Is 1/5 of a year 2 months?
No
I left out a step: 1.2 = 1 and 2/10 = 1 and 1/5 ...
So we probably won't be within the bounds of acceptable error if we don't use the right amount of time. We know that we have to pay 3% interest per year. How much is that per month, if there are 12 months in a year?
.25
Okay, \[3\%/\text{yr} = \frac{3\%}{12\text{ months}} = 0.25\%/\text{month}\] What do you get for an answer if you use that interest rate and 14 months for the time in your formula?
6,300
Are you sure about that answer? After 1 year, she has to pay 3% of the amount borrowed in interest, or $3 for every $100 borrowed. She's only borrowing $1800. Our answer shouldn't be much larger than 1 year's interest, as we are only doing 1/6 of a year more...
I think you perhaps forgot that \[x\% = \frac{x}{100}\]
Would it be 63?
I don't know why I'm not understanding myself, I was very strong in this unit
Here is what I understand
I understand why we got .25
Yes, $63 is what I get for the answer. For 1 year's interest (12 months), she pays 3% of the $1800, or $54. Then for the additional 2 month, she pays \[\frac{2\text{ months}}{12\text{ months}}*3\%*\$1800 = \$9\]And of course, if you add those two together, you get $63. You could also do it like this: \[I = P_0 * i * t = \$1800 * \frac{3\%}{1\text{ year}}*14 \text{ months}*\frac{1\text{ year}}{12\text{ months}}\]\[=\$1800 * \frac{3\%}{1\cancel{\text{ year}}}*14 \text{ months}*\frac{1\cancel{\text{ year}}}{12\text{ months}}\]\[=\$1800 * \frac{3\%}{1\cancel{\text{ year}}}*14 \cancel{\text{ months}}*\frac{1\cancel{\text{ year}}}{12\cancel{\text{ months}}}\]\[=\$1800*3\%*\frac{14}{12} = \$1800*\frac{3}{100}*\frac{14}{12}\] and if you punch that all in, you'll get $63 again.
notice how all of the units cancelled each other out? This serves as an error check that we have converted between units correctly, instead of accidentally dividing where we should have multiplied or vice versa. If the units didn't cancel out, and instead we had some weird unit like \[\frac{\text{months}^2}{\text{years}^2}\]we would know that we messed up somewhere along the line.
But why are we changing both the time and rate into ratios over 100?
we are changing the percentage into a decimal — that's the division by 100. We are changing months into years: if I do just that portion alone: \[14 \text{ months} * \frac{1\text{ year}}{12\text{ months}} = 14 \cancel{\text{ months}} * \frac{1\text{ year}}{12\cancel{\text{ months}}} = \frac{14}{12}\text{ years} \approx 1.16667 \text{ years}\] \[I = \$1800*\frac{3}{100}*1.16667 = $63.0002 \]
I see it now, makes so much more sense. Thank You so much for clearing this up for me
Glad I could help!
Now, just a word of warning: this is what is called "simple interest", not to be confused with "compound interest". We only compute the interest on the amount borrowed, and there is no interest on the interest, so to speak. In real life, most of the time you have compound interest, where after each compounding period (often a month for things like credit cards, but can be anything), you compute the interest for just that month, and add the interest to the existing balance to become the balance. That difference can lead to a substantially higher interest bill over time! It's also the same mechanism that makes your savings balance grow.

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