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F = P (1 + i ) t

1. A sum of €12,000 is invested in a ten year government bond with an annual equivalent rate (AER) of 1·5%. Find the value of the investment, when it matures in ten years. P = €12,000 , i = 1·5%, t = 10. F = P (1 + i ) t. F = 12000(1 + 0·015) 10. F = 12000(1·015) 10. = €13926·49.

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F = P (1 + i ) t

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  1. 1. A sum of €12,000 is invested in a ten year government bond with an annual equivalent rate (AER) of 1·5%. Find the value of the investment, when it matures in ten years. P = €12,000, i= 1·5%, t = 10 F = P(1 + i)t F = 12000(1 + 0·015)10 F = 12000(1·015)10 = €13926·49

  2. 2. A deposit account will earn 3% interest in the first year and 6% interest in the second year. The interest is added to the account at the end of each year. If €12,000 is invested in this account, how much will it amount to at the end of two years? (i) Year 1: P = €12,000, i = 3%, t = 1 F = 12000(1 + 0·03)1 F =12000(1·03) F = €12360 Year 2: P = €12,360, i = 6%, t = 1 F = 12360(1 + 0·06)1 F = 12360(1·06) F = €13101·60

  3. 2. Show that, to the nearest euro, the same amount of interest is earned by investing the money for two years in an account that pays compound interest at 4·49% AER. (ii) P = €12,000, i = 4·49%, t = 2 F = 12000(1 + 0·0449)2 F = 12000(1·0449)2 = €13101·79212 = €13102

  4. 3. Calculate Jose’s annual net income. (i) Jose earns €27,500 per annum and has a tax credit of €1,120. Gross tax = 27% of €22000 + 38% of €5500 The standard rate cut off point is €22,000. = 0·27 × 22000 + 0·38 × 5500 = 5940 + 2090 The standard rate of tax was 27% and the higher rate was 38%. = €8030 Tax paid = Gross tax – Tax credit = €8030 – €1120 = €6910 Net income = Gross salary – Tax paid = €27500 – €6910 = €20590

  5. 3. By how much would Jose’s net income be increased, if the 27% rate of tax is reduced to 25%? (ii) Jose earns €27,500 per annum and has a tax credit of €1,120. The standard rate cut off point is €22,000. Gross tax = 25% of €22000 + 38% of €5500 = 5500 + 2090 The standard rate of tax was 27% and the higher rate was 38%. = €7590 Difference in gross tax = €8030 – €7590 = €440 increase

  6. 4. A savings account gives 4% at the end of the first year, 5% at the end of the second year and 6·5% at the end of the third year. Find: the final value of €2,000, invested for three full years (i) Year 1: P = €2000, i = 4%, t = 1 F = 2000(1 + 0·04)1 = 2000(1·04) = €2080 Year 2: P = €2080, i = 5%, t = 1 F = 2080(1 + 0·05)1 = 2080(1·05) = €2184 Year 3: P = €2184, i = 6·5%, t = 1 F = 2184(1 + 0·065)1 = 2184(1·065) = €2325·96 Therefore, the €2000 has a final value of €2325·96

  7. 4. A savings account gives 4% at the end of the first year, 5% at the end of the second year and 6·5% at the end of the third year. Find: the final value of 2,000, invested for three full years (ii) F = €2325·96, P = €2000, t = 3 F = P(1 – i)t

  8. 5. A house valued at €750,000 in 2006 has depreciated by a total of 20% in 6 years. Find the value of the house in 2012. (i) Value = €750,000 × 0·8 Value = €600 000

  9. 5. The contents of the house (furniture, electrical goods, valuables) were valued at €30,000 in 2006. The contents depreciated by 6% every year, by the reducing balance method. Find the value of the contents in 2012. Give your answer to the nearest euro. (ii) P = €30,000, i = 6%, t = 3 F = P(1 − i)t F = 30000(1 – 0·06)6 F = 30000(0·94)6 = €20696·09 = €20,696

  10. 5. House insurance is charged at a rate of 0·1% of the value of the house and its contents. Find the difference in cost of the insurance policy, to the nearest euro, for house insurance in 2006 and 2012. (iii) In 2006: (€750 000 + €30 000) × 0·001 = €780 In 2012: (€600 000 + €20696) × 0·001 = €620·70 Difference  €159·30

  11. 6. Frank is married and earned €46,800 last year. The standard rate cut-off point for a married person was €35,300. The standard rate of income tax was 21% and the higher rate was 42%. Frank has tax credits of €3,600. Calculate the tax paid by Frank on his income. (i) Gross tax = 21% of €35300 + 42% of €11500 = 0·21 × 35300 + 0·42 × 11500 = 7413 + 4830 = €12,243 Tax paid = Gross tax – Tax credit = €12243 – €3600 = €8,643

  12. 6. Frank is married and earned €46,800 last year. The standard rate cut-off point for a married person was €35,300. The standard rate of income tax was 21% and the higher rate was 42%. Frank has tax credits of €3,600. Levy = 3% of €46800 Frank also has to pay a 3% income levy on his gross income, = 0·03 × 46800 = €1404 USC of 2% on the first €12,500 of his salary, 4% on the next €6,200 and 6% on all remaining income. USC:  2% of €12500 = 250 4% of €6200 = 248 6% of €28100 = 1686 Frank also pays a monthly health insurance payment of €90, €2184 a weekly pension contribution of €85 Health ins: €90 × 12 = €1080 Pension: €85 × 52 = €4420 and a weekly trade union subscription of €8. Union: €8 × 52 = €416

  13. 6. Frank is married and earned €46,800 last year. The standard rate cut-off point for a married person was €35,300. The standard rate of income tax was 21% and the higher rate was 42%. Frank has tax credits of €3,600. Calculate Frank’s annual net income after all deductions have been made. (ii) Total deductions: Tax: €8643 Levy: €1404 USC: €2184 Health ins: €1080 Pension: €4420 Union: €416 €18147 Net income = Gross salary – Deductions = €46800 – €18147 = €28653

  14. 7. Pat has a credit card with a €1,500 limit. Interest is charged on the balance due at the end of each month, at a rate of 1·8% per month. At the start of April, Pat owes €1,200 on the card. Given that he makes no payments off the card, find the total amount he owes at the start of November. (i) Start of April to the start of November = 7 months P = €1200, i= 1·8%, t = 7 F = P(1 + i)t F = 1200(1 + 0·018)7 F = 1200(1·018)7 F = €1359·61

  15. 7. Pat has a credit card with a €1,500 limit. Interest is charged on the balance due at the end of each month, at a rate of 1·8% per month. During November, Pat purchases an item online, costing $170. Given the exchange rate is €1 = $1·12, find to the nearest euro, the sum which will be charged to Pat’s credit card. (ii) $1·12 = €1 ( 1·12) $1 = €0·89285 (× 170) $170 = €151·79 = €152

  16. 7. Pat has a credit card with a €1,500 limit. Interest is charged on the balance due at the end of each month, at a rate of 1·8% per month. The credit card company reject this payment and Pat’s order is cancelled. Give a reason why this might be the case. What advice would you give Pat? (iii) Balance + Purchase = New Balance €1359 + €152 = €1511 This value exceeds the credit Limit of €1,500.

  17. 7. Pat has a credit card with a €1,500 limit. Interest is charged on the balance due at the end of each month, at a rate of 1·8% per month. Find the annual percentage rate (APR) charged on Pat’s credit card, to two decimal places. (iv) Based on €1 for one year: P = €1, i= 1·8%, t = 12 F = P(1 + i)t F = 1(1 + 0·018)12 F = 1(1·018)12 F = €1·2387 Annual rate: = 1·2387 – 1 = 0·2387 × 100 = 23·87%

  18. 8. €8,500 was invested for 2 years at compound interest. The rate of interest for the first year was 4%. Find the amount of the investment at the end of the first year. (i) P = €8500, i= 4%, t = 1 F = P(1 + i)t F = 8500(1 + 0·04)1 F = 8500(1·04) = €8840 F = €8840

  19. 8. €8,500 was invested for 2 years at compound interest. The amount of the investment at the end of the second year was €9,237·80 . Find the rate of interest for the second year, to one decimal place. (ii) F = €9237·80, P = €8840, t = 1 F = P(1 + i)t 9237·8 = 8840(1 + r)1 1·045 = 1 + r 0·045 = r 4·5% = r

  20. 9. On a Wednesday night in winter, 12 cm of snow fell in a town. On Thursday morning, from 8 am onwards, the snow started to melt at a steady rate of 8% per hour. Find how many centimetres of snow remained at 3 pm? Give your answer to two decimal places. (i) P = 12, i= 8%, t = 7 (8am to 3pm is 7 hours) F = P(1 − i)t F = 12(1 – 0·08)7 F = 12(0·92)7 F = 6·69 cm

  21. 9. On a Wednesday night in winter, 12 cm of snow fell in a town. Thursday morning, from 8 am onwards, the snow started to melt at a steady rate of 8% per hour. Show that by 10 pm, over half of the snow had melted. (ii) P = 12, i= 8%, t = 14 (8am to 10pm is 14 hours) F = P(1 − i)t F = 12(1 – 0·08)14 F = 12(0·92)14 F = 3·73 cm 3·73 cm is remaining, so 8·27 cm has melted, which is more than half of the original 12 cm.

  22. 9. On a Wednesday night in winter, 12 cm of snow fell in a town. Thursday morning, from 8 am onwards, the snow started to melt at a steady rate of 8% per hour. From 10 pm Thursday night until 8 am on Friday morning, the melting rate reduced to 5% per hour. Find the depth of the snow at 8 am the following morning. Give your answer to once place of decimal. (iii) P = 3·73, i= 5%, t = 10 (10pm to 8am is 10 hours) F = P(1 − i)t F = 3·73(1 – 0·05)10 F = 3·73(0·95)10 F = 2·23 cm = 2·2 cm

  23. 9. On a Wednesday night in winter, 12 cm of snow fell in a town. Thursday morning, from 8 am onwards, the snow started to melt at a steady rate of 8% per hour. If the rate of snow melting continues in this way (8% per hour from 8 am to 10 pm and 5% per hour from 10 pm to 8 am), use trial and improvement to estimate the day and time when there will be 0·5cm of snow remaining. (ii) Thurs at 3 am: 12cm Thurs at 10 pm: F = 12(0·92)14 = 3·73 cm Fri at 8 am: F = 3·73(0·95)10 = 2·24 cm Fri at 10 pm: F = 2·24(0·92)14 = 0·696 cm Sat at 8 am: F = 0·696(0·95)10 = 0·42 cm Between Friday at 10 pm and Saturday 8 am, the depth drops to below 0·5cm Sat at 4 am: F = 0·696(0·95)6 = 0·51 cm Sat at 5 am: F = 0·696(0·95)7 = 0·49 cm There will be 0·5 cm of snow at approximately 4·30 am on Saturday morning.

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