Calculate Savings Account Value and Interest

2. 1. A savings account will earn 2% interest in the first year, 3·5% interest in the second year and 5% in the third year. The interest is added to the account at the end of each year. If a person invested €8,000 in this account, how much will they have in the account at the end of the three years? Give your answer to the nearest cent. (i) Year 1: P = €8000, r = 2%, t = 1 F = 8000(1 + 0·02)1 = 8000(1·02) = €8160 Year 2: P = €8160, r = 3·5%, t = 1 F = 8160(1 + 0·035)1 = 8160(1·035) = €8445·60 Year 3: P = €8445·60, r = 5%, t = 1 F = 8445·6(1 + 0·05) = 8445·6(1·05) = €8867·88 There will be €8867·88 in the account at the end of the three years.

3. 1. A savings account will earn 2% interest in the first year, 3·5% interest in the second year and 5% in the third year. The interest is added to the account at the end of each year. Show that, to the nearest euro, the same amount of interest is earned by investing the money for three years in an account that pays compound interest at 3·49% (AER) to two decimal places. (ii) P = €8000, r = 3·49%, t = 3 F = P(1 + i)t F = 8000(1 + 0·0349)3 F = 8000(1·0349)3 F = €8867·17 F = €8867

4. 2. Calculate Alex’s net income. (i) The standard rate of income tax is 20% and the higher rate is 41%. Gross tax = 20% of €36500 + 41% of €11000 The standard rate cut-off point is €36,500. = 0·2 × 36500 + 0·41 × 11000 = 7300 + 4510 = €11810 Alex has a gross income of €47,500 and total tax credits of €1,830. Tax paid = Gross tax – Tax credit = €11810 – €1830 = €9980 Net income = Gross salary – Tax paid = €47500 – €9980 = €37520

5. 2. The following year Alex’s gross income increases. The tax rates, cut-off point and tax credits remain unchanged. His net tax now amounts to €15,105. What is his new gross income? (ii) The standard rate of income tax is 20% and the higher rate is 41%. The standard rate cut-off point is €36,500. Tax paid = Gross tax – Tax credit Alex has a gross income of €47,500 and total tax credits of €1,830. €15105 = Gross tax – €1830 €16935 = Gross tax Gross tax = 20% of 36500 + 41% of €x €16935 = €7300 + 0·41x €9635 = 0·41x €23500 = x €36500 + €23500 = €60000 Gross income.

6. 3. A government bond gives 25% interest after 12 years. Find the AER for this bond, correct to two decimal places. Based on an investment of €1: F = €1·25 , P = €1, t = 12 F = P(1 + i)t 1·25 = 1(1 + r)12 1·25 = (1 + r)12 1·01877 = 1 + r 0·01877 = r 1·877% = r 1·88% = r

7. 4. What sum of money invested at 6% per annum compound interest will amount to €4,000 in 5 years? Give your answer correct to the nearest euro. F = €4000 , i = 6%, t = 5 F = P(1 + i)t 4000 = P(1 + 0·06)5 4000 = P(1·06)5 €2989·03 = P

8. 5. In city A, house prices have increased by 3% each year for the last three years. If a house cost €180,000 three years ago, calculate, to the nearest euro, its value today. (i) P = €180000 , i = 3%, t = 3 F = P(1 + i)t F = 180000(1 + 0·03)3 F = 180000(1·03)3= 180 000 × 1·092727 F = €196690·86 F = €196,691

9. 5. In city B, a house worth €100,000 three years ago is now valued at €119,102. Calculate the yearly percentage increase in the value of this house. (ii) F = €119102, P = €100000, t = 3 F = P(1 + i)t 119102 = 100000(1 + r)3 1·19 = (1 + r)3 1·06 = 1 + r 0·060 = r 6% = r

10. 6. David bought a car for €10,000. The car depreciations as a constant annual rate. If after nine and a half years the value of the car is halved, find the rate of depreciation, correct to the nearest percent. F = €5000, P = €10000, t = 9·5 F = P(1 + i)t 5000 = 10000(1 – r)9·5 0·5 = (1 – r)9·5 0·9296 = 1 – r r = 1 – 0·9296 r = 0·0704 r = 7%

11. 7. At the start of each academic year, Jane took out a student loan of €P at an APR of 8% per annum. At the start of year 1, Jane borrows €P for 3 years at 8% per annum compound interest. Find, in terms of P, the value of this loan at the end of the 3 years. (i) P = P, i = 8%, t = 3 F = P(1 + i)t F = P(1 + 0·08)3 F = P(1·08)3 F = 1·2597P

12. 7. At the start of each academic year, Jane took out a student loan of €P at an APR of 8% per annum. At the start of year 2, Jane borrows €P for 2 years at 8% per annum compound interest. Find, in terms of P, the value of this loan at the end of the 2 years. (ii) P = P, i = 8%, t = 2 F = P(1 + i)t F = P(1 + 0·08)2 F = P(1·08)2 F = 1·1664P

13. 7. At the start of each academic year, Jane took out a student loan of €P at an APR of 8% per annum. At the start of year 3, Jane borrow €P for 1 year at 8% per annum compound interest. Find, in terms of P, the value of this loan at the end of the 1 year. (iii) P = P, i = 8%, t = 1 F = P(1 + i)t F = P(1 + 0·08)1 F = 1·08P

14. 7. At the start of each academic year, Jane took out a student loan of €P at an APR of 8% per annum. At the end of the three years, Jane owes a total of €8765·28, calculate P. (iv) Total due = 1·2597P + 1·1664P + 1·08P €8765·28 = 3·5061P €2,500 = P

15. 8. A scientist is running an experiment. Under certain conditions, the number of bacteria in a particular culture increase by 40% every hour. If there are 500 bacteria in the sample at the start of the experiment, find: the number of bacteria present after three hours. (i) P = 500, i = 40%, t = 3 F = P(1 + i)t F = 500(1 + 0·4)3 F = 500(1·4)3 = 1372 bacteria after three hours

16. 8. A scientist is running an experiment. Under certain conditions, the number of bacteria in a particular culture increase by 40% every hour. If there are 500 bacteria in the sample at the start of the experiment, find: the number of bacteria present after five hours. (ii) P = 500, i = 40%, t = 5 F = P(1 + i)t F = 500(1 + 0·4)5 F = 500(1·4)5 = 2689·12 = 2689 bacteria after five hours

17. 8. A scientist is running an experiment. Under certain conditions, the number of bacteria in a particular culture increase by 40% every hour. If there are 500 bacteria in the sample at the start of the experiment, find: After five hours of testing, the scientist introduces a chemical which reduces the rate of growth of the bacteria from 40% to a new value. Seven hours later, there are 7,153 bacteria present in the sample. Find the new rate of growth, after the chemical has been added. (iii) F = 7153, P = 2689, t = 7 F = P(1 + i)t 7153 = 2689(1 + r)7 2·66 = (1 + r)7 1·15 = 1 + r 0·15 = r Therefore, r = 15%

18. 9. €600 is invested at r % per annum. The interest earned is subject to tax at 35%. After one year, tax of €8·40 was paid on the interest earned. Find the total amount of interest earned. (i) €8·40 = 35% of the interest earned (35) €0·24 = 1% (× 100) €24 = 100% of interest earned.

19. 9. €600 is invested at r % per annum. The interest earned is subject to tax at 35%. After one year, tax of €8·40 was paid on the interest earned. Hence, calculate r. (ii) F = €624, P = €600, t = 1 F = P(1 + i)t 624 = 600(1 + r)1 1·04 = 1 + r 0·04 = r 4% = r

20. 10. A savings bond earns an interest rate of 1·25% for each year the bond is kept. The interest is compounded annually. The interest earned is taxed at 20% when the bond is redeemed. €15,000 is invested in this bond, for 4 years. What is the net amount payable (after tax is deducted) at the end of the 4 years? (i) P = €15000, i = 1·25%, t = 4 F = P(1 + i)t F = 15000(1 + 0·0125)4 F = 15000(1·0125)4 F = 15764·18 €15764·18 – €15000 = €764·18 interest €764·18 × 0·8 = €611·34 net interest after tax Net value of bond = €15,000 + €611·34 = €15,611·34

21. 10. A savings bond earns an interest rate of 1·25% for each year the bond is kept. The interest is compounded annually. The interest earned is taxed at 20% when the bond is redeemed. P = €10000, i = 1·25%, t = 8 If the investor keeps the bond for 8 years, a bonus of 15% is paid on the interest earned, before tax is deducted. What is the amount payable on an investment of €10,000, invested in this bond for 8 years? (ii) F = 10000(1 + 0·0125)8 =10000 × 1·01258 = €11,044·86 Interest = €11,044·86 – €10,000 = €1044·86 15% Bonus = 0·15 × €1044·86 = €156·73 Total interest = €1044·86 + €156·73 = €1201·59 Tax of 20% deducted: €1201·59 × 0·8 = €961·27 : Net interest Net value of bond = €10,000 + €961·27 = €10,961·27