1 / 84

UNIT 4 :

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK. Social Arithmetic. Logic Diagrams. UNIT 4 :. Formulae. EXIT. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK. You have chosen to study:. Social Arithmetic. UNIT 4 :. Please choose a question to attempt from the following:. 1. 2. 3. 4. Back to

bing
Download Presentation

UNIT 4 :

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Social Arithmetic UNIT 4 : Please choose a question to attempt from the following: 1 2 3 4 Back to Unit 4 Menu EXIT

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

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

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

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

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

  8. Comments Percentage calculations: 10 100 • Calculate how much of the sales qualify for a commission payment. 10% of £330 = x 330 = 0.10 x 330 etc. • 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. Next Comment Total wage = £180 + £33 = £213 Soc Arith Menu Back to Home

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

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

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

  12. Question 2 = £2812.58 (a) Commission = 2% of £43600 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. = 0.02 x £43600 = £872 Overtime! Gross salary = £1700 + £872 + £240.58 Back to payslip Begin Solution Continue Solution Comments Menu Back to Home

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

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

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

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

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

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

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

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

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

  22. EXIT SOCIAL ARITHMETIC: Question 4 A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts. LP – Loan Protection (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?

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

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

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

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

  27. AMOUNT BORROWED 60 MONTHS 48 MONTHS 24 MONTHS LP Basic LP Basic 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 Additional Comments 4years = 4 x 12 = 48 months (a) They borrow £6000 over 4 years and decide to make basic repayments.How much do they actually pay back? So required monthly repayment = £150.03

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

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

  30. AMOUNT BORROWED 60 MONTHS 48 MONTHS 24 MONTHS LP Basic LP Basic 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 (b) How much extra would they repay if they had opted for a 5 year repayment period with loan protection? Monthly repayment = £127.38

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

  32. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Formulae UNIT 4 : Please choose a question to attempt from the following: 1 2 3 4 5 Back to Unit 4 Menu EXIT

  33. d x x w FORMULAE: Question 1 The surface area of a triangular prism, S cm2, is given by the formula S = x2 + 2dx + dw Answer where all distances are in cm. Full solution Comments Menu (a) Find S when x = 10, w = 14 & d = 30. (b) Find w when x = 5, d = 20 & S = 365. EXIT

  34. d x x w FORMULAE: Question 1 The surface area of a triangular prism, S cm2, is given by the formula S = x2 + 2dx + dw What would you like to do now? where all distances are in cm. Full solution Comments Menu (a) Find S when x = 10, w = 14 & d = 30. = 1120 cm2 w = 7 (b) Find w when x = 5, d = 20 & S = 365. EXIT

  35. Question 1 • Substitute known values into given • formula: S = x2 + 2dx + dw (a) S =x2 + 2dx + dw • Find S when x = 10, • w = 14 & d = 30. = (10 x 10) + (2 x 30 x 10) + (30 x 14) = 100 + 600 + 420 = 1120 Area is 1120cm2 Begin Solution Continue Solution Comments Menu Back to Home

  36. Question 1 • Substitute known values into given • formula: S = x2 + 2dx + dw (b) S =x2 + 2dx + dw • Find w when x = 5, • d = 20 & S = 365. 365 = (5 x 5) + (2 x 20 x 5) + (20 x w) 2. Tidy up then solve equation for target letter: 20w + 200 + 25 = 365 20w = 140 w = 7 Begin Solution Continue Solution Width is 7cm Comments What would you like to do now? Menu Back to Home

  37. Comments When evaluating formulae: (a) S =x2 + 2dx + dw 1. Write formula = (10 x 10) + (2 x 30 x 10) + (30 x 14) 2. Replace known values = 100 + 600 + 420 3. Evaluate = 1120 Area is 1120cm2 Next Comment Menu Back to Home

  38. Comments When evaluating formulae: • Substitute known values into given • formula: (b) S =x2 + 2dx + dw 1. Write formula 365 = (5 x 5) + (2 x 20 x 5) + (20 x w) 2. Replace known values 2. Tidy up then solve equation for target letter: 3. Evaluate 20w + 200 + 25 = 365 20w = 140 4. Solve resulting equation w = 7 Next Comment Width is 7cm Menu What would you like to do now? Back to Home

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

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

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

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

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

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

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

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

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

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

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

  50. Question 3 (b) ( ) ( ) T = 2  T = 2  L L L L L L L 10 10 10 10 10 10 10 ( ) ( ) ( ) 6.28 x = 18.84 2 x 3.14 x = 18.84 = 3 = 32 = 9 ( 6.28) • Find T when L = 40m. • Find L when • T = 18.84secs. Square both Sides ! Take  = 3.14. Begin Solution L = 9 x 10 = 90 Continue Solution Length is 90m when time is 18.84secs Comments Menu What would you like to do now? Back to Home

More Related