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

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

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

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

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

Question 1 Commissionable sales = £530 - £200 = £330 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

Sales on which commission earned = 10 x £70 = £700 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

Commissionable sales = £530 - £200 = £330 Commission = 10% of £330 Comments Percentage calculations: Calculate how much of the sales qualify for a commission payment. 10 100 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

Sales on which commission earned = 10 x £70 = £700 Comments Calculate how much of the sales qualify for a commission payment. Working backwards (reverse %) 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

SOCIAL ARITHMETIC: Question 2 EXIT SOCIAL ARITHMETIC: Question 2 Kerry Owen works for a builders’ supplies merchant. Her partly completed payslip for May is shown below 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 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.

SOCIAL ARITHMETIC: Question 2 EXIT SOCIAL ARITHMETIC: Question 2 Kerry Owen works for a builders’ supplies merchant. Her partly completed payslip for May is shown below 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 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. = £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.

Question 2 (a) Commission = 2% of £43600 = 0.02 x £43600 = £872 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 = £2812.58 Back to payslip Begin Solution Continue Solution Comments Menu Back to Home

Question 2 (b) Pension = 6.5% of £2812.58 = 0.065 x £2812.58 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 = £1956.88 Menu What would you like to do now? Back to Home

(a) Commission = 2% of £43600 = 0.02 x £43600 = £872 Comments Percentage Calculations (a) Commission = 2% of £43600 2 100 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 = £2812.58 Next Comment Menu Back to Home

Comments (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 = £1956.88 Menu Back to Home

SOCIAL ARITHMETIC: Question 3 EXIT SOCIAL ARITHMETIC: Question 3 Diana Marr works in an electronics factory. Her partially completed payslip is shown below. 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 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.

SOCIAL ARITHMETIC: Question 3 EXIT SOCIAL ARITHMETIC: Question 3 Diana Marr works in an electronics factory. Her partially completed payslip is shown below. 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 = £398.73 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. = £7.86

Question 3 (a) Total deductions (a) = £26.32 + £60.93 + £45.83 = £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 = £398.73 Begin Solution Continue Solution Comments Menu

Hourly rate = Basic wage Question 3 (b) Hourly rate = Basic wage Hours worked (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 = £7.86 Continue Solution Comments What would you like to do now? Menu

(a) Total deductions = £26.32 + £60.93 + £45.83 = £133.08 Comments 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 = £398.73 Next Comment Menu

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

SOCIAL ARITHMETIC: Question 4 EXIT SOCIAL ARITHMETIC: Question 4 A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts. Amount 60 months 48 months 24 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 (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?

SOCIAL ARITHMETIC: Question 4 EXIT SOCIAL ARITHMETIC: Question 4 A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts. Amount 60 months 48 months 24 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 (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

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

(b) Monthly repayment = £127.38 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

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

Additional Comments 4years = 4 x 12 = 48 months AMOUNT BORROWED 60 MONTHS 48 MONTHS 24 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 (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

(a) Monthly repayment = £150.03 Total repayment = £150.03 x 48 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

tables that you identify relevant categories. In this case: Be careful when using tables that you identify relevant categories. In this case: Amount Repayment period Loan Protection Comments 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

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

(b) Monthly repayment = £127.38 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

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

FORMULAE: Question 1 S = x2 + 2dx + dw d x w 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. x d w 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

FORMULAE: Question 1 S = x2 + 2dx + dw d x w = 1120 cm2 w = 7 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. x d w Full solution Comments Menu (a) Find S when x = 10, w = 14 & d = 30. = 1120 cm2 (b) Find w when x = 5, d = 20 & S = 365. w = 7 EXIT

Question 1 S = x2 + 2dx + dw (a) S = x2 + 2dx + dw = 100 + 600 + 420 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

Question 1 S = x2 + 2dx + dw (b) S = x2 + 2dx + dw 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

(a) S = x2 + 2dx + dw = 100 + 600 + 420 = 1120 Area is 1120cm2 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

(b) S = x2 + 2dx + dw 20w + 200 + 25 = 365 20w = 140 w = 7 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: 20w + 200 + 25 = 365 3. Evaluate 20w = 140 4. Solve resulting equation w = 7 Next Comment Width is 7cm Menu What would you like to do now?

C = 5/9(F - 32) 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

C = 5/9(F - 32) 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 . 302°F = 150°C -40°F = -40°C 10°C = 50°F EXIT

C = 5/9(F - 32) Question 2 (a) C = 5/9(F - 32) C = 5/9(302 - 32) 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 = 150 302°F = 150°C Begin Solution Continue Solution Comments Menu

C = 5/9(F - 32) Question 2 (b) C = 5/9(F - 32) C = 5/9(-40 - 32) 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 -40°F = -40°C Begin Solution Continue Solution Value same in each scale!! Comments Menu

C = 5/9(F - 32) Question 2 (c) 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

(a) C = 5/9(F - 32) C = 5/9(302 - 32) C = 5/9 of 270 Comments 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 B O ÷ / x + / - Brackets Of ÷ / x 302°F = 150°C Next Comment Menu

Value same in each scale!! Comments 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 B O ÷ / x + / - Brackets Of ÷ / x -40°F = -40°C Value same in each scale!! Next Comment Menu

(c) C = 5/9(F - 32) 10 = 5/9(F - 32) (x9) 90 = 5(F - 32) 90 = 5F - 160 Comments 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 B O ÷ / x + / - Brackets Of ÷ / x 5F = 90 + 160 5F = 250 4. Solve resulting equation F = 50 10°C = 50°F Next Comment Menu What would you like to do now?

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

( ) 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 L T = 2  ( ) L 10 Full solution Comments where L is the length of the pendulum in metres. Menu Find T when L = 40m. Find L when T = 18.84secs. T = 12.56 L = 90 Take  = 3.14. EXIT

( ) ( ) ( ) Question 3 T = 2  (a) T = 2  T = 2 x 3.14 x ( ) L 10 (a) T = 2  ( ) L 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

( ) ( ) ( ) ( ) ( ) Question 3 T = 2  (b) T = 2  ( ) L 10 (b) T = 2  ( ) L 10 2 x 3.14 x = 18.84 ( ) L 10 6.28 x = 18.84 ( ) L 10 ( 6.28) Find T when L = 40m. Find L when T = 18.84secs. = 3 ( ) L 10 Square both Sides ! = 32 L 10 Take  = 3.14. = 9 L 10 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?

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

( ) ( ) ( ) ( ) T = 2  (b) 2 x 3.14 x = 18.84 6.28 x = 18.84 Comments When evaluating formulae: T = 2  ( ) L 10 (b) 1. Write formula 2 x 3.14 x = 18.84 ( ) L 10 2. Replace known values 6.28 x = 18.84 ( ) L 10 ( 6.28) 3. Evaluate ( BODMAS) = 3 ( ) L 10 Square both Sides ! B O ÷ / x + / - Brackets Of ÷ / x = 32 L 10 = 9 L 10 4. Solve resulting equation L = 9 x 10 = 90 Next Comment Length is 90m when time is 18.84secs Menu What would you like to do now?

Change the subject of the formula m = 3p2 – k to p. FORMULAE: Question 4 Change the subject of the formula m = 3p2 – k to p. Answer Full solution Comments Menu EXIT

FORMULAE: Question 4 Change the subject of the formula m = 3p2 – k to p. What would you like to do now? p = m + k 3  Full solution Comments Menu EXIT

 Question 4 Change the subject of the formula m = 3p2 – k to p. Swap sides. m = 3p2 – k 3p2 – k = m Isolate 3p2 3p2 = m + k Isolate p2 p2 = k + m 3 Isolate p p = m + k 3  Begin Solution Continue Solution What would you like to do now? Comments Menu

  Swap sides. m = 3p2 – k 10 = 3x2 – 4 3p2 – k = m Isolate 3p2 Comments Apply the same rules as in a simple equation. Swap sides. EXAMPLE m = 3p2 – k 10 = 3x2 – 4 3p2 – k = m Isolate 3p2 EXAMPLE 3x2 – 4 = 10 3p2 = m + k Isolate p2 EXAMPLE 3x2 = 10 + 4 p2 = k + m 3 EXAMPLE x2 = 10 + 4 3 Isolate p p = m + k 3  x = 10 + 4 3  EXAMPLE Next Comment Menu

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

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

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

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

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

LOGIC DIAGRAMS: Question 1 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. Menu Full solution EXIT Answer Comments

LOGIC DIAGRAMS: Question 1 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. Combinations are ... 1. Bus-bus 2. Bus-taxi 3. Train-bus 4. Train-taxi 5. Train-cycle Menu Full solution EXIT Comments

Question 1 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 taxi bus train bus taxi Begin Solution cycle Continue Solution Comments Menu

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

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

LOGIC DIAGRAMS: Question 2 The following flowchart is used to calculate the annual interest on a building society account. 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 EXIT Find the annual interest for a tax-payer with £3850.

LOGIC DIAGRAMS: Question 2 The following flowchart is used to calculate the annual interest on a building society account. = £169.40 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 EXIT Find the annual interest for a tax-payer with £3850.

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

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

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 Find the annual interest for a tax-payer with £3850.

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

LOGIC DIAGRAMS: Question 3 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 EXIT Find the annual interest for a non tax-payer with £6700.

LOGIC DIAGRAMS: Question 3 The following flowchart is used to calculate the annual interest on a building society account. = £488.83 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 EXIT Find the annual interest for a non tax-payer with £6700.

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

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

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 Find the annual interest for a non tax-payer with £6700.

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

LOGIC DIAGRAMS: Question 4 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. P G F E D C B A Explain why you can cover the above network without retracing any part of your route. List one possible route to illustrate this. EXIT

LOGIC DIAGRAMS: Question 4 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. P G F E D C B A 2 because each vertex/point has an even number of streets meeting at it. 3 1 4 5 13 6 7 10 8 11 9 12 Explain why you can cover the above network without retracing any part of your route. List one possible route to illustrate this. EXIT

Question 4 P G F 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. Begin Solution Continue Solution Comments Menu

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

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

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