Arithmetic II Solutions and Exam Questions
Explore solutions and exam-style questions in arithmetic, including calculations for investments, interest rates, tax deductions, and savings accounts. Enhance your understanding of financial calculations with these detailed examples.
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CHAPTER 09 Arithmetic II Solutions: Revision Section A
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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. (i) If 12,000 is invested in this account, how much will it amount to at the end of two years? 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
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A Calculate Jose s annual net income. (i) 3. Jose earns 27,500 per annum and has a tax credit of 1,120. Gross tax = 27% of 22000 + 38% of 5500 = 0 27 22000 + 0 38 5500 The standard rate cut off point is 22,000. = 5940 + 2090 = 8030 The standard rate of tax was 27% and the higher rate was 38%. Tax paid = Gross tax Tax credit = 8030 1120 = 6910 Net income = Gross salary Tax paid = 27500 6910 = 20590
09 Revision and Exam Style Questions: Section A (ii) By how much would Jose s net income be increased, if the 27% rate of tax is reduced to 25%? 3. 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
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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 2325 96=2000(1+r)3 1 163=(1+r)3 1 163 =1+r 1 052=1+r 3 0 052 5 2% = = r r
09 Revision and Exam Style Questions: Section A 5. A house valued at 750,000 in 2006 has depreciated by a total of 20% in 6 years. (i) Find the value of the house in 2012. Value = 750,000 0 8 Value = 600 000
09 Revision and Exam Style Questions: Section A 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. (ii) Find the value of the contents in 2012. Give your answer to the nearest euro. 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
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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. (i) Calculate the tax paid by Frank on his income. 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
09 Revision and Exam Style Questions: Section A 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 = 0 03 46800 = 1404 Frank also has to pay a 3% income levy on his gross income, 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
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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 startof 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
09 Revision and Exam Style Questions: Section A 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 12 = 1 $1 = 0 89285 ( 170) $170 = 151 79 = 152
09 Revision and Exam Style Questions: Section A 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. (iii) 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? Balance + Purchase = New Balance 1359 + 152 = 1511 This value exceeds the credit Limit of 1,500.
09 Revision and Exam Style Questions: Section A 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. (iv) Find the annual percentage rate (APR) charged on Pat s credit card, to two decimal places. 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%
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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
09 Revision and Exam Style Questions: Section A 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.
09 Revision and Exam Style Questions: Section A 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. (iii) 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. 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
09 Revision and Exam Style Questions: Section A 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. Thurs at 3 am: 12cm (ii) Thurs at 10 pm: Fri at 8 am: 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 F = 12(0 92)14 = 3 73 cm F = 3 73(0 95)10 = 2 24 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.
