A company takes out a loan of $£200,000$ from a bank, which it pays off as follows. At the end of each year, it pays back interest on the amount it still owes, plus $£8000$ repayment. Initially, the interest rate is $6\%$. a) At the end of each year, how much does the company pay to the bank? b) At the end of one year, the company pays the bank $£16640$. i) What was the amount the company still owed at the end of that year? ii) How many years after taking out the loan was this? c) After 10 years, the interest rate increases. At the end of the eleventh year, the company still owes $£120,000$, it pays the bank $£18200$. What was the new interest rate?

After attempting this question I have got the following answers and want to check that they are correct: a) At the end of the $n^{th}$ year the company pays back $20000-480(n-1)$ b)i) Company owes $£136,000$ ii) After $8$ years c) $8.5$%

Thank you very much for your help!

  • $\begingroup$ a doesn't ask for how much the company owes. it asks for how much the company pays back. $\endgroup$ – user451844 Jul 19 '17 at 22:08
  • $\begingroup$ Sorry meant pays back, still same answer as above. $\endgroup$ – rani Jul 19 '17 at 22:09
  • $\begingroup$ These answers can be checked quite simply with a table and calculator. An Excel spreadsheet would be a faster alternative. $\endgroup$ – Shuri2060 Jul 19 '17 at 22:16
  • $\begingroup$ It wasn't the calculations that I wanted to check per se, but the interpretation of the question that lead to the calculations. $\endgroup$ – rani Jul 19 '17 at 22:18
  • $\begingroup$ well your answer based on my interpretation of b)i) maybe wrong assuming I did the math correct. I got 144000 left. because the interest paid is 8640, now I didn't compensate for APR versus APY, so that may be the difference. $\endgroup$ – user451844 Jul 19 '17 at 22:28

Let $S=200,000$ the loan, $C=8,000$ and $i=6\%$. The schema is the repayment of a loan with constant principal in $n=S/C=25$ years. So we have for $k=1,\ldots,n$ the remaining debt is $$D_k=S-kC=(n-k)C$$ the interest payed is $$I_k=iD_{k-1}=i(n-k+1)C$$ and each installment is $$P_k=I_k+C=[i(n-k+1)+1]C$$ If $P_k=16,640$ we have $I_k=P_k-C=8,640=i(n-k+1)C$ and then $k=8$ and the debt is $$D_8=(n-k)C=(25-8)8,000=136,000$$.

At year $11$ we have $D_{11}=120,000$ and $P_{11}=18,200=[i(25-11+1)+1]\times 8,000$ that is the new interest rate is $i=8.5\%$.


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