1. Social Arithmetic Logic Diagrams Formulae
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 4 :
Social
Arithmetic
Logic
Diagrams
Formulae
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2. 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
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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
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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
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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
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6. 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
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7. 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
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8. 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
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9. 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
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10. SOCIAL ARITHMETIC: Question 2
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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 / KU34521D H May
£ £240.58
£ £487.25
Answer
Full solution
Comments
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(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. SOCIAL ARITHMETIC: Question 2
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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 / KU34521D H May
£ £240.58
£ £487.25
Full solution
Comments
= £182.82
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What would you like to do now?
= £
(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 (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.
= x £43600
= £872
Overtime!
Gross salary = £ £872 + £240.58
= £
Back to payslip
Begin Solution
Continue Solution
Comments
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13. Question 2 (b) Pension = 6.5% of £2812.58 = 0.065 x £2812.58
Found in part (a)
(b)
Pension = 6.5% of £
(b)
Kerry’s monthly pension contributions are 6.5% of her gross salary.
Find this and hence find her net salary for May.
= x £
= £ to nearest penny
Total deductions
= £ £ £182.82
(see payslip)
Back to payslip
= £855.70
Begin Solution
Net salary = £ £855.70
Continue Solution
Comments
= £
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14. (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
= x £43600
= x £43600
= £872
Overtime!
Gross Pay =
Basic + Commission + Overtime
Gross salary = £ £872 + £240.58
= £
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15. Comments
(b)
Pension = 6.5% of £
Net pay =
Gross Pay
- (Nat. Ins. + Inc Tax + Pension)
= x £
= £ to nearest penny
TOTAL DEDUCTIONS
Total deductions
= £ £ £182.82
(see payslip)
= £855.70
What would you like to do now?
Net salary = £ £855.70
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= £
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16. SOCIAL ARITHMETIC: Question 3
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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 / WA12311F L 37
£265.65
£ £ £45.83
£ £58.70
Answer
Full solution
Comments
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(a) Find her gross wage for this particular week.
(b) If she works a 38 hour basic week then find her hourly rate.

17. SOCIAL ARITHMETIC: Question 3
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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 / WA12311F L 37
£265.65
£ £ £45.83
£ £58.70
= £398.73
Full solution
Comments
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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

18. Question 3 (a) Total deductions (a) = £26.32 + £60.93 + £45.83
= £ £ £45.83
(a)
Find her gross wage for
this particular week.
= £133.08
Work backwards!
Gross wage = £ £265.65
Back to payslip
= £398.73
Begin Solution
Continue Solution
Comments
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19. 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
= £ £ £41.35
Gross from part (a)
Back to payslip
= £298.68
Hourly rate = £  38
Begin Solution
= £7.86
Continue Solution
Comments
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20. (a) Total deductions = £26.32 + £60.93 + £45.83 = £133.08
Comments
Working Backwards
(a)
Total deductions
= £ £ £45.83
Gross pay
= Net pay + Deductions
= £133.08
Work backwards!
Gross wage = £ £265.65
= £398.73
Next Comment
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21. Comments
Working Backwards
Basic wage is before overtime and bonus.
Basic wage
= £ £ £41.35
Gross from part (a)
Hourly rate = Basic wage
Hours worked
= £298.68
Hourly rate = £  38
= £7.86
What would you like to do now?
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22. SOCIAL ARITHMETIC: Question 4
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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?

23. SOCIAL ARITHMETIC: Question 4
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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?
= £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 = £ x 48
= £
Back to table
Begin Solution
Continue Solution
Comments
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25. (b) Monthly repayment = £127.38
Question 4
Answer from (a)
= £
(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 = £ x 60
= £
Extra paid = £ £
Back to table
= £441.36
Begin Solution
What would you like to do now?
Continue Solution
Comments
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26. 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 = £ x 48
= £
Next Comment
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27. Additional Comments
4years = 4 x 12 = 48 months
AMOUNT BORROWED
60 MONTHS
48 MONTHS
24 MONTHS
LP Basic £ £ £ £
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

28. (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 = £ x 48
= £
Next Comment
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29. 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)
= £
(b) Monthly repayment = £127.38
Total repayment = £ x 60
= £
Extra paid = £ £
= £441.36
Next Comment
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30. 5 years = 60 months
AMOUNT BORROWED
60 MONTHS
48 MONTHS
24 MONTHS
LP Basic £ £ £ £
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

31. (b) Monthly repayment = £127.38
Comments
Answer from (a)
= £
(b) Monthly repayment = £127.38
Extra Repaid =
Cost under option 1
- Cost under option 2
Total repayment = £ x 60
= £
Extra paid = £ £
= £441.36
Next Comment
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32. 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
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33. 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 = x dx + dw
Answer
where all distances are in cm. x d w
Full solution
Comments
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(a) Find S when x = 10, w = 14 & d = 30.
(b) Find w when x = 5, d = 20 & S = 365.
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34. 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 = x dx + dw
What would you like to do now?
where all distances are in cm. x d w
Full solution
Comments
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(a) Find S when x = 10, w = 14 & d = 30.
= cm2
(b) Find w when x = 5, d = 20 & S = 365.
w = 7
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35. Question 1 S = x2 + 2dx + dw (a) S = x2 + 2dx + dw = 100 + 600 + 420
Substitute known values into given
formula:
S = x dx + dw
(a)
S = x dx + dw
Find S when x = 10,
w = 14 & d = 30.
= (10 x 10) + (2 x 30 x 10) + (30 x 14) =
= 1120
Area is 1120cm2
Begin Solution
Continue Solution
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36. Question 1 S = x2 + 2dx + dw (b) S = x2 + 2dx + dw
Substitute known values into given
formula:
S = x dx + dw
(b)
S = x dx + 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 = 365
20w = 140
w = 7
Begin Solution
Continue Solution
Width is 7cm
Comments
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37. (a) S = x2 + 2dx + dw = 100 + 600 + 420 = 1120 Area is 1120cm2
Comments
When evaluating formulae:
(a)
S = x dx + dw
1. Write formula
= (10 x 10) + (2 x 30 x 10) + (30 x 14)
2. Replace known values =
3. Evaluate
= 1120
Area is 1120cm2
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38. (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 = x dx + 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 = 365
3. Evaluate
20w = 140
4. Solve resulting equation
w = 7
Next Comment
Width is 7cm
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39. C = 5/9(F - 32) FORMULAE: Question 2
To convert temperatures from °F into °C we use the formula
C = 5/9(F )
Answer
Full solution
Comments
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Change 302°F into °C .
Change -40°F into °C , and comment on your answer.
Change 10°C into °F .
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40. C = 5/9(F - 32) FORMULAE: Question 2
To convert temperatures from °F into °C we use the formula
C = 5/9(F )
What would you like to do now?
Full solution
Comments
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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

41. C = 5/9(F - 32) Question 2 (a) C = 5/9(F - 32) C = 5/9(302 - 32)
BODMAS
(a) C = 5/9(F )
C = 5/9(F )
C = 5/9( )
Change 302°F into °C
Change -40°F into °C
Change 10°C into °F .
C = 5/9 of 270
C =  9 x 5 = 150
302°F = 150°C
Begin Solution
Continue Solution
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42. C = 5/9(F - 32) Question 2 (b) C = 5/9(F - 32) C = 5/9(-40 - 32)
BODMAS
(b) C = 5/9(F )
C = 5/9(F )
C = 5/9( )
Change 302°F into °C
Change -40°F into °C
Change 10°C into °F .
C = 5/9 of -72
C =  9 x 5 = -40
-40°F = -40°C
Begin Solution
Continue Solution
Value same in each scale!!
Comments
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43. 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 )
90 = 5F
5F =
5F = 250
F = 50
Begin Solution
10°C = 50°F
Continue Solution
Comments
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44. (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 )
1. Write formula
2. Replace known values
C = 5/9( )
C = 5/9 of 270
3. Evaluate ( BODMAS)
C =  9 x 5 = 150
B O ÷ / x
+ / -
Brackets Of
÷ / x
302°F = 150°C
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45. Value same in each scale!!
Comments
When evaluating formulae:
BODMAS
(b) C = 5/9(F )
1. Write formula
C = 5/9( )
2. Replace known values
C = 5/9 of -72
3. Evaluate ( BODMAS)
C =  9 x 5 = -40
B O ÷ / x
+ / -
Brackets Of
÷ / x
-40°F = -40°C
Value same in each scale!!
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46. (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 )
1. Write formula
10 = 5/9(F ) (x9)
2. Replace known values
90 = 5(F )
3. Evaluate ( BODMAS)
90 = 5F
B O ÷ / x
+ / -
Brackets Of
÷ / x
5F =
5F = 250
4. Solve resulting equation
F = 50
10°C = 50°F
Next Comment
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47. ( ) 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.
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Find T when L = 40m.
Find L when T = 18.84secs.
Take  =
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48. ( ) 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 =
L = 90
Take  =
EXIT

49. ( ) ( ) ( ) 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 =
Find T when L = 40m.
Find L when
T = 18.84secs.
Time is 12.56secs when length is 40m
Take  =
Begin Solution
Continue Solution
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50. ( ) ( ) ( ) ( ) ( ) Question 3 T = 2  (b) T = 2 
( ) L 10
(b)
T = 2 
( ) L 10
2 x 3.14 x =
( ) L 10
6.28 x =
( ) 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  =
= 9 L 10
Begin Solution
L = 9 x 10 = 90
Continue Solution
Length is 90m when time is 18.84secs
Comments
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51. ( ) ( ) 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 =
B O ÷ / x
+ / -
Brackets Of
÷ / x
Time is 12.56secs when length is 40m
Next Comment
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52. ( ) ( ) ( ) ( ) 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 =
( ) L 10
2. Replace known values
6.28 x =
( ) 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
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53. 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
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54. 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
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55.  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
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Comments
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56.   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 =
p2 = k + m 3
EXAMPLE
x2 = 3
Isolate p
p = m + k 3 
x = 3 
EXAMPLE
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57. 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
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58. 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
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EXIT

59. 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
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60. 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
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61. 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
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62. 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.
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Full solution
EXIT
Answer
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63. 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
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Full solution
EXIT
Comments

64. 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
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cycle
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65. bus Combinations are ... taxi 1. Bus-bus bus 2. Bus-taxi 3. Train-bus
4. Train-taxi
bus
5. Train-cycle
taxi
cycle
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66. 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) =
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67. 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.

68. 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.

69. 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 = x £3850
Find the annual interest for
a tax-payer with £3850.
= £169.40
Begin Solution
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70. 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 = x £3850
= £169.40
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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
Find the annual interest for a tax-payer with £3850.

72. 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
= x £3850
4.4% of £3850 = x £3850
etc.
= £169.40
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73. 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.

74. 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.

75. 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 = 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
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76. 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 = x £6700
= £381.90
Investor is exempt from tax so int
= 1.28 x £381.90
= £488.83
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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
Find the annual interest for a non tax-payer with £6700.

78. 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
= x £6700
5.7% of £6700 = x £6700
etc.
= £381.90
Investor is exempt from tax so int
= 1.28 x £381.90
= £488.83
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79. 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

80. 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

81. 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.
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82. 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
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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

83. 2 3 1 4 5 13 6 7 10 8 11 9 12 Comments Test for Traversing a network PD 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
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84. 2 3 1 4 5 13 6 7 10 8 11 9 12 Comments Test for Traversing a network PD 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
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End of Unit 4