# When will total interest be 2500?

Asked by

dotlin (

422)
October 31st, 2010

we have £5000 in a bank with 8% interest

year 1 £400

year 2 £430

work out when total interest will be 2500

The other questions in this book dealt with working out answers using graphs, I know there are other ways to work this out but can you work this out using the line rule y=mx+b and other similar rules?

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## 7 Answers

I don’t necessarily want the answer but an explanation of how to work it out.

I am a little confused by the statement of the problem. The interest the first year would be 400 pounds, as stated. That would give a total of 5400 pounds in the bank. The interest for the second year would be .08 * 5400 = 432 pounds, not 430 pounds.

The key to working with this type of problem is to see that the total amount of money at the end of the first year is 1.08 times greater, once we add interest to principal. The total amount after two years is 1.08*1.08*5000. The total after n years is (1.08)^n*5000. To find the part that is interest, just subtract the original deposit of 5000, so you get (1.08)^n-5000=2500. You can solve this problem by working with logarithms.

If this is compound interest

A = P*(1+r/q)^(nq)

A is the total amount earned, including interest, P is the starting principal, r is the interest rate, n is the # of years, and q is the number of times that the interest is added each year. Just plug in 7500 for A (the total interest plus the strarting 5000) and all of the other numbers, and solve for n. I know that this is not the sort of equation that you are looking for, but the actual equation is extremely complicated. It would be something like y=(log_(1+r/q) (x/P))/q, where you would solve for x = 7500.

Err will continue where everyone else has left off to show you the solution when working with logarithms just in case you didn’t know. As mentioned, the formula which you will be using is:

5000*(1.08)^n = 7500

(5000 is the original deposit, multiplied by 1.08^n where n represents the number of years passed. 7500 is the target amount you want).

so you take the log of both sides:

log 5000(1.08)^n = log 7500.

then we can rearrange this using the rules that *log x^y = y log x* and *log xy = log x + log y* to get:

log 5000 + n log 1.08 = log 7500

we subtract log 5000 from both sides to get:

n log 1.08 = log 7500 – log 5000

this second half of the equation can be rewritten again (using the same idea as the *log xy = log x + log y* rule mentioned earlier, but swapping the multiplication for division and the addition for subtraction) to get:

n log 1.08 = log 1.5

we then finish by dividing by log 1.08 to get:

n = log 1.5 / log 1.08

then we can solve using a calculator to find the solution of 5.27.

Finally, don’t forget to round up to find your answer :D.

Having messed up in giving the solution, let me see if I can redeem myself by showing how you can check your answer by using non-compounded interest. For a low number like 1.08 and a short period of time like 5.7 years, the non-compounded solution should be fairly close. 8% interest on 5000 pounds is 400 pounds. 2500/400= 6.25, or about half a year longer than compounding, which seems reasonable.

don’t invest. its a scheme… legit banks don’t give 8% interest anymore

Get a financial calculator like the TI-BA2+

It does amortization in a heartbeat.

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