1. UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae

2. UNIT 4 : You have chosen to study: Please choose a question to attempt from the following: EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic 1 Back to Unit 4 Menu 234

3. SOCIAL ARITHMETIC: Question 1 Go to full solution Go to Comments Go to Social Arithmetic Menu Reveal answer only EXIT Get hint Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200. (a)How much does she earn in a week when her sales are £530? (b)What do her sales need to be so her wages are £250?

4. SOCIAL ARITHMETIC: Question 1 Go to full solution Go to Comments Go to Social Arithmetic Menu Reveal answer only EXIT Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200. (a)How much does she earn in a week when her sales are £530? (b)What do her sales need to be so her wages are £250? 1. Calculate how much of the sales qualify for a commission payment. 2. Apply given percentage to this sum to find Commission due. 3. Remember total Wage includes basic 4. In (b) use reverse percentage to work out sales required to give the amount above basic. 5. In (b) Remember sales must be £200 before commission earned. What would you like to do now?

5. SOCIAL ARITHMETIC: Question 1 Go to full solution Go to Comments Go to Social Arithmetic Menu EXIT Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200. (a)How much does she earn in a week when her sales are £530? (b)What do her sales need to be so her wages are £250? = £330 = £900 What would you like to do now?

6. Comments Begin Solution Question 1 Soc Arith Menu Back to Home Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200. (a)How much does she earn in a week when her sales are £530? 1.Calculate how much of the sales qualify for a commission payment. (a)Commissionable sales = £530 - £200 = £330 2. Apply given percentage to this sum to find Commission due. Commission = 10% of £330 = £33 3. Remember total Wage includes basic. Continue Solution Total wage = £180 + £33 = £213

7. Comments Begin Solution Question 1 Soc Arith Menu Back to Home Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200. 1.Calculate how much of the sales qualify for a commission payment. 2. Use reverse percentage to work out sales required to give this figure. 3. Remember sales must be £200 before commission earned. Continue Solution (b) What do her sales need to be so her wages are £250? (b)Total commission = £250 - £180 = £70 Sales on which commission earned = 10 x £70 = £700 Total sales = £700 + £200 = £900 What would you like to do now?

8. Comments Soc Arith Menu Back to Home Next Comment Percentage calculations: 10% of £330 = x 330 = 0.10 x 330 etc. 10 100 1.Calculate how much of the sales qualify for a commission payment. (a)Commissionable sales = £530 - £200 = £330 2. Apply given percentage to this sum to find Commission due. Commission = 10% of £330 = £33 3. Remember total Wage includes basic. Total wage = £180 + £33 = £213

9. Comments Soc Arith Menu Back to Home Next Comment X 10 Working backwards (reverse %) From 10% require 100% 1.Calculate how much of the sales qualify for a commission payment. 2. Use reverse percentage to work out sales required to give this figure. 3. Remember sales must be £200 before commission earned. (b)Total commission = £250 - £180 = £70 Sales on which commission earned = 10 x £70 = £700 Total sales = £700 + £200 = £900 10% = £70 100%= £700

10. (b) Kerry’s monthly pension contributions are 6.5% of her gross salary. Find this and hence find her net salary for May. Kerry Owen works for a builders’ supplies merchant. Her partly completed payslip for May is shown below (a)Kerry’s basic monthly salary is £1700 plus overtime plus commission of 2% on all her sales. Find her gross salary for May when her sales totalled £43600. Name Employee No. N.I. No. Tax Code Month Basic salary Commission Overtime Gross salary Nat.Insurance Income tax Pension Total Deductions Net salary K.Owen 34/09852 KU34521D 498H May £1700 £240.58 £185.63 £487.25 SOCIAL ARITHMETIC: Question 2 Full solution Comments Menu Answer EXIT

11. (b) Kerry’s monthly pension contributions are 6.5% of her gross salary. Find this and hence find her net salary for May. Kerry Owen works for a builders’ supplies merchant. Her partly completed payslip for May is shown below (a)Kerry’s basic monthly salary is £1700 plus overtime plus commission of 2% on all her sales. Find her gross salary for May when her sales totalled £43600. Name Employee No. N.I. No. Tax Code Month Basic salary Commission Overtime Gross salary Nat.Insurance Income tax Pension Total Deductions Net salary K.Owen 34/09852 KU34521D 498H May £1700 £240.58 £185.63 £487.25 SOCIAL ARITHMETIC: Question 2 Full solution Comments Menu EXIT = £2812.58 = £182.82 = £1956.88 What would you like to do now?

12. Comments Begin Solution Continue Solution Question 2 Menu Back to Home Kerry’s basic monthly salary is £1700 plus overtime plus commission of 2% on all her sales. Find her gross salary for May when her sales totalled £43600. (a) Commission = 2% of £43600 = 0.02 x £43600 = £872 Gross salary = £1700 + £872 + £240.58 = £2812.58 Back to payslip Overtime!

13. (b) Kerry’s monthly pension contributions are 6.5% of her gross salary. Find this and hence find her net salary for May. Comments Begin Solution Continue Solution Question 2 Menu Back to Home Back to payslip (b) Pension = 6.5% of £2812.58 = 0.065 x £2812.58 = £182.82 to nearest penny Found in part (a) Total deductions = £185.63 + £487.25 + £182.82 = £855.70 Net salary = £2812.58 - £855.70 = £1956.88 (see payslip) What would you like to do now?

14. Comments Menu Back to Home Next Comment Percentage Calculations 2% of £43600 = x £43600 = 0.02 x £43600 2 100 (a) Commission = 2% of £43600 = 0.02 x £43600 = £872 Gross salary = £1700 + £872 + £240.58 = £2812.58 Overtime! Gross Pay = Basic + Commission + Overtime

15. Comments Menu Back to Home Next Comment (b) Pension = 6.5% of £2812.58 = 0.065 x £2812.58 = £182.82 to nearest penny Total deductions = £185.63 + £487.25 + £182.82 = £855.70 Net salary = £2812.58 - £855.70 = £1956.88 (see payslip) Net pay = Gross Pay - (Nat. Ins. + Inc Tax + Pension) TOTAL DEDUCTIONS What would you like to do now?

16. Name Employee No. N.I. No. Tax Code Week Basic wage Bonus Overtime Gross wage Nat.Insurance Income tax Pension Total Deductions Net wage D.Marr 2001/0789 WA12311F 395L 37 £265.65 £26.32£60.93 £45.83 £41.35 £58.70 SOCIAL ARITHMETIC: Question 3 Full solution Comments Menu Answer EXIT Diana Marr works in an electronics factory. Her partially completed payslip is shown below. (a)Find her gross wage for this particular week. (b)If she works a 38 hour basic week then find her hourly rate.

17. Name Employee No. N.I. No. Tax Code Week Basic wage Bonus Overtime Gross wage Nat.Insurance Income tax Pension Total Deductions Net wage D.Marr 2001/0789 WA12311F 395L 37 £265.65 £26.32£60.93 £45.83 £41.35 £58.70 SOCIAL ARITHMETIC: Question 3 Full solution Comments Menu EXIT Diana Marr works in an electronics factory. Her partially completed payslip is shown below. (a)Find her gross wage for this particular week. (b)If she works a 38 hour basic week then find her hourly rate. What would you like to do now? = £398.73 = £7.86

18. Comments Begin Solution Continue Solution Question 3 Menu Back to Home (a) Find her gross wage for this particular week. Back to payslip (a) Total deductions = £26.32 + £60.93 + £45.83 = £133.08 Gross wage = £133.08 + £265.65 = £398.73 Work backwards!

19. Comments Begin Solution Continue Solution Question 3 Menu Back to Home Back to payslip (b) If she works a 38 hour basic week then find her hourly rate. (b) Hourly rate = Basic wage Hours worked Basic wage = £398.73 - £58.70 - £41.35 = £298.68 Hourly rate = £298.68  38 = £7.86 Gross from part (a) Basic wage is before overtime and bonus. What would you like to do now?

20. Comments Menu Back to Home Next Comment Working Backwards Gross pay = Net pay + Deductions (a) Total deductions = £26.32 + £60.93 + £45.83 = £133.08 Gross wage = £133.08 + £265.65 = £398.73 Work backwards!

21. Comments Menu Back to Home Next Comment Basic wage = £398.73 - £58.70 - £41.35 = £298.68 Hourly rate = £298.68  38 = £7.86 Gross from part (a) Basic wage is before overtime and bonus. Working Backwards Hourly rate = Basic wage Hours worked What would you like to do now?

22. Amount60 months48 months24 months LPBasicLPBasicLPBasic £8000169.83166.51204.04200.03376.23368.86 £6000127.38124.88153.03150.03282.18276.66 £400084.9283.26102.02100.02188.12184.43 £200042.4641.6351.0150.0194.0692.22 SOCIAL ARITHMETIC: Question 4 EXIT A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts. (a)They borrow £6000 over 4 years and decide to make basic repayments. How much do they actually pay back? (b)How much extra would they repay if they had opted for a 5 year repayment period with loan protection? LP – Loan Protection

23. Amount60 months48 months24 months LPBasicLPBasicLPBasic £8000169.83166.51204.04200.03376.23368.86 £6000127.38124.88153.03150.03282.18276.66 £400084.9283.26102.02100.02188.12184.43 £200042.4641.6351.0150.0194.0692.22 SOCIAL ARITHMETIC: Question 4 EXIT A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts. (a)They borrow £6000 over 4 years and decide to make basic repayments. How much do they actually pay back? (b)How much extra would they repay if they had opted for a 5 year repayment period with loan protection? LP – Loan Protection = £7201.44 = £441.36

24. Comments Begin Solution Continue Solution Question 4 Menu Back to Home (a) They borrow £6000 over 4 years and decide to make basic repayments. How much do they actually pay back? Back to table (a) Monthly repayment = £150.03 Total repayment = £150.03 x 48 = £7201.44

25. Comments Begin Solution Continue Solution Question 4 Menu Back to Home Back to table (b) How much extra would they repay if they had opted for a 5 year repayment period with loan protection? = £7201.44 Answer from (a) = £7642.80 = £441.36 (b) Monthly repayment = £127.38 Total repayment = £127.38 x 60 Extra paid = £7642.80 - £7201.44 What would you like to do now?

26. Comments Menu Back to Home Next Comment (a) Monthly repayment = £150.03 Total repayment = £150.03 x 48 = £7201.44 Be careful when using tables that you identify relevant categories. In this case: 1.Amount 2.Repayment period 3.Loan Protection

27. Additional Comments AMOUNT BORROWED 60 MONTHS48 MONTHS24 MONTHS LP Basic £8000 169.83 166.51 204.04 200.03 376.23 368.86 £6000 127.38 124.88 153.03 150.03 282.18 276.66 £4000 84.92 83.26 102.02 100.02 188.12 184.43 £2000 42.46 41.63 51.01 50.01 94.06 92.22 LP - Loan Protection 4years = 4 x 12 = 48 months So required monthly repayment = £150.03 (a)They borrow £6000 over 4 years and decide to make basic repayments. How much do they actually pay back?

28. Comments Menu Back to Home Next Comment (a) Monthly repayment = £150.03 Total repayment = £150.03 x 48 = £7201.44 Total Repayment = monthly instalment x no. of instalments

29. Comments Menu Back to Home Next Comment = £7201.44 Answer from (a) = £7642.80 = £441.36 (b) Monthly repayment = £127.38 Total repayment = £127.38 x 60 Extra paid = £7642.80 - £7201.44 Be careful when using tables that you identify relevant categories. In this case: 1.Amount 2.Repayment period 3.Loan Protection

30. AMOUNT BORROWED 60 MONTHS48 MONTHS24 MONTHS LP Basic £8000 169.83 166.51 204.04 200.03 376.23 368.86 £6000 127.38 124.88 153.03 150.03 282.18 276.66 £4000 84.92 83.26 102.02 100.02 188.12 184.43 £2000 42.46 41.63 51.01 50.01 94.06 92.22 LP - Loan Protection 5 years = 60 months Monthly repayment = £127.38 (b)How much extra would they repay if they had opted for a 5 year repayment period with loan protection?

31. Comments Menu Back to Home Next Comment = £7201.44 Answer from (a) = £7642.80 = £441.36 (b) Monthly repayment = £127.38 Total repayment = £127.38 x 60 Extra paid = £7642.80 - £7201.44 Extra Repaid = Cost under option 1 - Cost under option 2

32. UNIT 4 : You have chosen to study: Please choose a question to attempt from the following: EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Formulae 1 Back to Unit 4 Menu 2345

33. FORMULAE: Question 1 The surface area of a triangular prism, S cm 2, is given by the formula S = x 2 + 2dx + dw where all distances are in cm. xx d w (a) Find S when x = 10, w = 14 & d = 30. (b) Find w when x = 5, d = 20 & S = 365. EXIT Full solution Comments Menu Answer

34. FORMULAE: Question 1 The surface area of a triangular prism, S cm 2, is given by the formula S = x 2 + 2dx + dw where all distances are in cm. xx d w (a) Find S when x = 10, w = 14 & d = 30. (b) Find w when x = 5, d = 20 & S = 365. EXIT Full solution Comments Menu What would you like to do now? w = 7 = 1120 cm 2

35. Comments Begin Solution Continue Solution Question 1 Menu Back to Home (a) S = x 2 + 2dx + dw = (10 x 10) + (2 x 30 x 10) + (30 x 14) = 100 + 600 + 420 = 1120 Area is 1120cm 2 1.Substitute known values into given formula: S = x 2 + 2dx + dw (a)Find S when x = 10, w = 14 & d = 30.

36. Comments Begin Solution Continue Solution Question 1 Menu Back to Home 1.Substitute known values into given formula: S = x 2 + 2dx + dw (b)Find w when x = 5, d = 20 & S = 365. (b) S = x 2 + 2dx + dw 365 = (5 x 5) + (2 x 20 x 5) + (20 x w) 20w + 200 + 25 = 365 20w = 140 w = 7 Width is 7cm 2. Tidy up then solve equation for target letter: What would you like to do now?

37. Comments Menu Back to Home Next Comment 1. Write formula 2. Replace known values 3. Evaluate (a) S = x 2 + 2dx + dw = (10 x 10) + (2 x 30 x 10) + (30 x 14) = 100 + 600 + 420 = 1120 Area is 1120cm 2 When evaluating formulae:

38. Comments Menu Back to Home Next Comment 1.Substitute known values into given formula: (b) S = x 2 + 2dx + dw 365 = (5 x 5) + (2 x 20 x 5) + (20 x w) 20w + 200 + 25 = 365 20w = 140 w = 7 Width is 7cm 2. Tidy up then solve equation for target letter: 1. Write formula 2. Replace known values 3. Evaluate When evaluating formulae: 4. Solve resulting equation What would you like to do now?

39. FORMULAE: Question 2 EXIT Full solution Comments Menu Answer To convert temperatures from °F into °C we use the formula C = 5 / 9 (F - 32) (a)Change 302°F into °C. (b)Change -40°F into °C, and comment on your answer. (c)Change 10°C into °F.

40. FORMULAE: Question 2 EXIT Full solution Comments Menu To convert temperatures from °F into °C we use the formula C = 5 / 9 (F - 32) (a)Change 302°F into °C. (b)Change -40°F into °C, and comment on your answer. (c)Change 10°C into °F. 302°F = 150°C -40°F = -40°C 10°C = 50°F What would you like to do now?

41. Comments Begin Solution Continue Solution Question 2 Menu Back to Home (a)C = 5 / 9 (F - 32) C = 5 / 9 (302 - 32) C = 5 / 9 of 270 C = 270  9 x 5 = 150 302°F = 150°C BODMA S C = 5 / 9 (F - 32) (a)Change 302°F into °C (b)Change -40°F into °C (c)Change 10°C into °F.

42. Comments Begin Solution Continue Solution Question 2 Menu Back to Home BODMA S C = 5 / 9 (F - 32) (a)Change 302°F into °C (b)Change -40°F into °C (c)Change 10°C into °F. (b)C = 5 / 9 (F - 32) C = 5 / 9 (-40 - 32) C = 5 / 9 of -72 C = -72  9 x 5 = -40 -40°F = -40°C Value same in each scale!!

43. Comments Begin Solution Continue Solution Question 2 Menu Back to Home C = 5 / 9 (F - 32) (a)Change 302°F into °C (b)Change -40°F into °C (c)Change 10°C into °F. (c)C = 5 / 9 (F - 32) 10 = 5 / 9 (F - 32)(x9) 90 = 5(F - 32) 90 = 5F - 160 5F = 90 + 160 5F = 250 F = 50 10°C = 50°F What would you like to do now?

44. Comments Menu Back to Home Next Comment (a)C = 5 / 9 (F - 32) C = 5 / 9 (302 - 32) C = 5 / 9 of 270 C = 270  9 x 5 = 150 302°F = 150°C BODMA S 1. Write formula 2. Replace known values 3. Evaluate ( BODMAS) When evaluating formulae: B O ÷ / x + / - Brackets Of ÷ / x + / -

45. Comments Menu Back to Home Next Comment BODMA S (b)C = 5 / 9 (F - 32) C = 5 / 9 (-40 - 32) C = 5 / 9 of -72 C = -72  9 x 5 = -40 -40°F = -40°C Value same in each scale!! 1. Write formula 2. Replace known values 3. Evaluate ( BODMAS) B O ÷ / x + / - Brackets Of ÷ / x + / - When evaluating formulae:

46. Comments Menu Back to Home Next Comment (c)C = 5 / 9 (F - 32) 10 = 5 / 9 (F - 32)(x9) 90 = 5(F - 32) 90 = 5F - 160 5F = 90 + 160 5F = 250 F = 50 10°C = 50°F 1. Write formula 2. Replace known values 3. Evaluate ( BODMAS) B O ÷ / x + / - Brackets Of ÷ / x + / - When evaluating formulae: 4. Solve resulting equation What would you like to do now?

47. FORMULAE: Question 3 EXIT Full solution Comments Menu Answer The time, T secs, for a pendulum to swing to & fro is calculated by the formula (a)Find T when L = 40m. (b)Find L when T = 18.84secs. T = 2   ( ) L 10 L where L is the length of the pendulum in metres. Take  = 3.14.

48. FORMULAE: Question 3 EXIT Full solution Comments Menu The time, T secs, for a pendulum to swing to & fro is calculated by the formula (a)Find T when L = 40m. (b)Find L when T = 18.84secs. T = 2   ( ) L 10 L where L is the length of the pendulum in metres. Take  = 3.14. T = 12.56 L = 90 What would you like to do now?

49. Comments Begin Solution Continue Solution Question 3 Menu Back to Home (a)Find T when L = 40m. (b)Find L when T = 18.84secs. T = 2   ( ) L 10 T = 2 x 3.14 x  ( ) 40 10 T = 2 x 3.14 x  4 T = 12.56 Time is 12.56secs when length is 40m T = 2   ( ) L 10 (a) Take  = 3.14.

50. Comments Begin Solution Continue Solution Question 3 Menu Back to Home (a)Find T when L = 40m. (b)Find L when T = 18.84secs. T = 2   ( ) L 10 Take  = 3.14. T = 2   ( ) L 10 (b) 2 x 3.14 x = 18.84  ( ) L 10 6.28 x = 18.84  ( ) L 10 (  6.28) = 3  ( ) L 10 = 3 2 L 10 = 9 L 10 L = 9 x 10 = 90 Length is 90m when time is 18.84secs Square both Sides ! What would you like to do now?

51. Comments Menu Back to Home Next Comment T = 2 x 3.14 x  ( ) 40 10 T = 2 x 3.14 x  4 T = 12.56 Time is 12.56secs when length is 40m T = 2   ( ) L 10 (a) 1. Write formula 2. Replace known values 3. Evaluate ( BODMAS) When evaluating formulae: B O ÷ / x + / - Brackets Of ÷ / x + / -

52. Comments Menu Back to Home Next Comment T = 2   ( ) L 10 (b) 2 x 3.14 x = 18.84  ( ) L 10 6.28 x = 18.84  ( ) L 10 (  6.28) = 3  ( ) L 10 = 3 2 L 10 = 9 L 10 L = 9 x 10 = 90 Length is 90m when time is 18.84secs Square both Sides ! 1. Write formula 2. Replace known values 3. Evaluate ( BODMAS) B O ÷ / x + / - Brackets Of ÷ / x + / - When evaluating formulae: 4. Solve resulting equation What would you like to do now?

53. FORMULAE: Question 4 EXIT Full solution Comments Menu Answer Change the subject of the formula m = 3p 2 – k to p.

54. FORMULAE: Question 4 EXIT Full solution Comments Menu Change the subject of the formula m = 3p 2 – k to p. p = m + k 3  What would you like to do now?

55. Comments Begin Solution Continue Solution Question 4 Menu Back to Home Change the subject of the formula m = 3p 2 – k to p. m = 3p 2 – k 3p 2 – k = m 3p 2 = m + k p 2 = k + m 3 Swap sides. Isolate 3p 2 Isolate p 2 Isolate p p = m + k 3  What would you like to do now?

56. Comments Menu Back to Home Next Comment 10 = 3x 2 – 4 3x 2 – 4 = 10 3x 2 = 10 + 4 x = 10 + 4 3  Apply the same rules as in a simple equation. x 2 = 10 + 4 3 m = 3p 2 – k 3p 2 – k = m 3p 2 = m + k p 2 = k + m 3 Swap sides. Isolate 3p 2 Isolate p 2 Isolate p p = m + k 3  EXAMPLE

57. FORMULAE: Question 5 EXIT Full solution Comments Menu Answer Change the subject of the formula Q = 2kn – 1 to n. 3

58. FORMULAE: Question 5 EXIT Full solution Comments Menu Change the subject of the formula Q = 2kn – 1 to n. 3 n = 3Q + 3 2k

59. Comments Begin Solution Continue Solution Question 5 Menu Back to Home Change the subject of the formula Q = 2kn – 1 to n. 3 3Q = 2kn - 3 2kn – 3 = 3Q 2kn = 3Q + 3 Q = 2kn – 1 3 X 3 to eliminate fraction. Swap sides. Isolate 2kn. Isolate n. n = 3Q + 3 2k

60. Comments Menu Back to Home Next Comment 3 x 10 = 2 x 5 x n - 3 2 x 5 x n – 3 = 3 x 10 2 x 5 x n = (3 x 10) + 3 10 = 2 x 5 x n – 1 3 n = (3 x 10) + 3 2 x 5 3Q = 2kn - 3 2kn – 3 = 3Q 2kn = 3Q + 3 Q = 2kn – 1 3 X 3 to eliminate fraction. Swap sides. Isolate 2kn. Isolate n. n = 3Q + 3 2k Apply the same rules as in a simple equation. EXAMPLE What would you like to do now?

61. UNIT 4 : You have chosen to study: Please choose a question to attempt from the following: EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Logic Diagrams 1 Back to Unit 4 Menu 234

62. Bryan lives in the country and works in town. His journey to work is split into two parts. He can get from home into town by either bus or train. When he arrives in town he can then get to his office by bus or taxi or he can cycle there provided the first part of his journey was made by train. List all the possible ways he can make his way to work. LOGIC DIAGRAMS: Question 1 EXIT Full solution Comments Menu Answer

63. Bryan lives in the country and works in town. His journey to work is split into two parts. He can get from home into town by either bus or train. When he arrives in town he can then get to his office by bus or taxi or he can cycle there provided the first part of his journey was made by train. List all the possible ways he can make his way to work. LOGIC DIAGRAMS: Question 1 EXIT Full solution Comments Menu Combinations are... 1.Bus-bus 2.Bus-taxi3.Train-bus 4.Train-taxi5.Train-cycle

64. Comments Begin Solution Continue Solution Question 1 Menu Back to Home Bryan’s journey to work is split into two parts. He can get from home into town by either bus or train. He can then get to his office by bus or taxi or he can cycle there provided the first part of his journey was made by train. bus train bus taxi bus taxi cycle

65. bus train bus taxi bus taxi cycle Combinations are... 1.Bus-bus 2.Bus-taxi 3.Train-bus 4.Train-taxi 5.Train-cycle Comments Menu Back to Home What would you like to do now?

66. Comments Menu Back to Home Next Comment Combinations are... 1.Bus-bus 2.Bus-taxi 3.Train-bus 4.Train-taxi 5.Train-cycle This question may be combined with probability. If each journey is equally likely what is the probability he travelled by train then bus? Probability (uses bus) = 3535 What is the probability that he Will use a bus at some point in his journey? Probability (train,bus) = 1515

67. LOGIC DIAGRAMS: Question 2 EXIT MULT INT BY 1.18 STOP IS THERE TAX EXEMPTION? START IS AMOUNT > £10000? IS AMOUNT > £5000? INT = 5.3% OF AMOUNT INT = 4.4% OF AMOUNT INT = 6.2% OF AMOUNT Yes No Yes No Yes No The following flowchart is used to calculate the annual interest on a building society account. Find the annual interest for a tax-payer with £3850.

68. LOGIC DIAGRAMS: Question 2 EXIT MULT INT BY 1.18 STOP IS THERE TAX EXEMPTION? START IS AMOUNT > £10000? IS AMOUNT > £5000? INT = 5.3% OF AMOUNT INT = 4.4% OF AMOUNT INT = 6.2% OF AMOUNT Yes No Yes No Yes No The following flowchart is used to calculate the annual interest on a building society account. Find the annual interest for a tax-payer with £3850. = £169.40

69. Comments Begin Solution Continue Solution Question 2 Menu Back to Home Find the annual interest for a tax-payer with £3850. Back to flowchart Amount is under £5000 so rate = 4.4% 4.4% of £3850 = 0.044 x £3850 = £169.40

70. Comments Menu Back to Home Next Comment Amount is under £5000 so rate = 4.4% 4.4% of £3850 = 0.044 x £3850 = £169.40 Take care to ensure that you follow correct “flow” of diagram by answering each question carefully.

71. LOGIC DIAGRAMS: Question 2 MULT INT BY 1.18 STOP IS THERE TAX EXEMPTION? START IS AMOUNT > £10000? IS AMOUNT > £5000? INT = 5.3% OF AMOUNT INT = 4.4% OF AMOUNT INT = 6.2% OF AMOUNT Yes No Yes No Yes No Find the annual interest for a tax-payer with £3850.

72. Comments Menu Back to Home Next Comment Percentage Calculations 4.4% of £3850 = x £3850 = 0.044 x £3850 4.4 100 etc. Amount is under £5000 so rate = 4.4% 4.4% of £3850 = 0.044 x £3850 = £169.40

73. LOGIC DIAGRAMS: Question 3 EXIT The following flowchart is used to calculate the annual interest on a building society account. MULT INT BY 1.28 STOP IS THERE TAX EXEMPTION? START IS AMOUNT > £10000? IS AMOUNT > £5000? INT = 5.7% OF AMOUNT INT = 4.5% OF AMOUNT INT = 6.8% OF AMOUNT Yes No Yes No Yes No Find the annual interest for a non tax-payer with £6700.

74. LOGIC DIAGRAMS: Question 3 EXIT The following flowchart is used to calculate the annual interest on a building society account. MULT INT BY 1.28 STOP IS THERE TAX EXEMPTION? START IS AMOUNT > £10000? IS AMOUNT > £5000? INT = 5.7% OF AMOUNT INT = 4.5% OF AMOUNT INT = 6.8% OF AMOUNT Yes No Yes No Yes No Find the annual interest for a non tax-payer with £6700. = £488.83

75. Comments Begin Solution Continue Solution Question 3 Menu Back to Home Find the annual interest for a non tax-payer with £6700. Back to flowchart Amount is between £5000 & £10000 so rate = 5.7% 5.7% of £6700 = 0.057 x £6700 Investor is exempt from tax so int = 1.28 x £381.90 = £488.83 = £381.90 What would you like to do now?

76. Comments Menu Back to Home Next Comment Amount is between £5000 & £10000 so rate = 5.7% 5.7% of £6700 = 0.057 x £6700 Investor is exempt from tax so int = 1.28 x £381.90 = £488.83 = £381.90 Take care to ensure that you follow correct “flow” of diagram by answering each question carefully.

77. LOGIC DIAGRAMS: Question 3 MULT INT BY 1.28 STOP IS THERE TAX EXEMPTION? START IS AMOUNT > £10000? IS AMOUNT > £5000? INT = 5.7% OF AMOUNT INT = 4.5% OF AMOUNT INT = 6.8% OF AMOUNT Yes No Yes No Yes No Find the annual interest for a non tax-payer with £6700.

78. Comments Menu Back to Home Next Comment Amount is between £5000 & £10000 so rate = 5.7% 5.7% of £6700 = 0.057 x £6700 Investor is exempt from tax so int = 1.28 x £381.90 = £488.83 = £381.90 Percentage Calculations 5.7% of £6700 = x £6700 = 0.057 x £6700 5.7 100 etc.

79. LOGIC DIAGRAMS: Question 4 EXIT The diagram below shows a network of streets near a post office (P). When making deliveries the postman/woman tries to cover all streets without going along the same street more than once if possible. (a)Explain why you can cover the above network without retracing any part of your route. (b)List one possible route to illustrate this. P GF E D C B A

80. LOGIC DIAGRAMS: Question 4 EXIT The diagram below shows a network of streets near a post office (P). When making deliveries the postman/woman tries to cover all streets without going along the same street more than once if possible. (a)Explain why you can cover the above network without retracing any part of your route. (b)List one possible route to illustrate this. (a)because each vertex/point has an even number of streets meeting at it. P GF E D C B A 1 2 3 4 5 6 7 8 9 10 11 12 13

81. Comments Begin Solution Continue Solution Question 4 Menu Back to Home P GF E D C B A (a) The network can be traversed without repeating any route because each vertex/point has an even number of streets meeting at it.

82. (b)One possible route around the network is P GF E D C B A 1 PG 2 F 3 E 4 D 5 P 6 C 7 D 8 B 9 A 10 C 11 B 12 A 13 P Comments Menu Back to Home What would you like to do now?

83. Comments Menu Back to Home Next Comment P GF E D C B A 1 2 3 4 5 6 7 8 9 10 11 12 13 Test for Traversing a network Network must have no more than two odd vertices. 2 odd 1 even traversible Note: must start on an odd vertex 3 4 3

84. Comments Menu Back to Home Next Comment P GF E D C B A 1 2 3 4 5 6 7 8 9 10 11 12 13 Test for Traversing a network Network must have no more than two odd vertices. NOT traversible 3 3 3 2 3 4 odd 1 even End of Unit 4