## anonymous one year ago 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. • This Question is Closed 1. anonymous Simple interest over time formula: I=(Po) (r) (t) 2. whpalmer4 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? 3. anonymous Wouldn't it be 1.2? 4. whpalmer4 1.2 = 1 and 1/5, right? Is 1/5 of a year 2 months? 5. anonymous No 6. whpalmer4 I left out a step: 1.2 = 1 and 2/10 = 1 and 1/5 ... 7. whpalmer4 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? 8. anonymous .25 9. whpalmer4 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? 10. anonymous 6,300 11. whpalmer4 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...

12. whpalmer4

I think you perhaps forgot that $x\% = \frac{x}{100}$

13. anonymous

Would it be 63?

14. anonymous

I don't know why I'm not understanding myself, I was very strong in this unit

15. anonymous

Here is what I understand

16. anonymous

I understand why we got .25

17. whpalmer4

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.

18. whpalmer4

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.

19. anonymous

But why are we changing both the time and rate into ratios over 100?

20. whpalmer4

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: $14 \text{ months} * \frac{1\text{ year}}{12\text{ months}} = 14 \cancel{\text{ months}} * \frac{1\text{ year}}{12\cancel{\text{ months}}} = \frac{14}{12}\text{ years} \approx 1.16667 \text{ years}$ $I = \1800*\frac{3}{100}*1.16667 = 63.0002$

21. anonymous

I see it now, makes so much more sense. Thank You so much for clearing this up for me

22. whpalmer4