UNIT 4 :
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
SocialArithmetic
Logic Diagrams
Formulae
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EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
SocialArithmetic
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2 3 4
SOCIAL ARITHMETIC: Question 1
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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?
SOCIAL ARITHMETIC: Question 1
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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.
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SOCIAL ARITHMETIC: Question 1
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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
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Question 1
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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
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Question 1
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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
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Percentage calculations:
10% of £330 = x 330
= 0.10 x 330
etc.
10100
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
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X 10 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
(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
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(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
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= £2812.58
= £182.82
= £1956.88What would you like to do now?
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Question 2
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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
= £2812.58Back to payslip
Overtime!
(b)
Kerry’s monthly pension contributions are 6.5% of her gross salary.
Find this and hence find her net salary for May.
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Question 2
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(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)
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Percentage Calculations
2% of £43600 = x £43600
= 0.02 x £43600
2100
(a)
Commission = 2% of £43600
= 0.02 x £43600
= £872
Gross salary = £1700 + £872 + £240.58
= £2812.58
Overtime!Gross Pay =
Basic + Commission + Overtime
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(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
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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
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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.
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
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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.
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= £398.73
= £7.86
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Question 3
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(a)
Find her gross wage for
this particular week.
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(a)Total deductions
= £26.32 + £60.93 + £45.83
= £133.08
Gross wage = £133.08 + £265.65
= £398.73
Work backwards!
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Question 3
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(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.
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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!
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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
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Amount 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
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
Amount 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
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
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Question 4
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(a) They borrow £6000 over
4 years and decide to make
basic repayments.
How much do they actually
pay back?
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(a)
Monthly repayment = £150.03
Total repayment = £150.03 x 48
= £7201.44
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Question 4
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(b)
How much extra would they
repay if they had opted for
a 5 year repayment period
with loan protection?
= £7201.44Answer from (a)
= £7642.80
= £441.36
(b) Monthly repayment = £127.38
Total repayment = £127.38 x 60
Extra paid = £7642.80 - £7201.44
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(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 period3. Loan Protection
Additional Comments
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
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?
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(a)
Monthly repayment = £150.03
Total repayment = £150.03 x 48
= £7201.44
Total Repayment =
monthly instalment
x no. of instalments
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= £7201.44Answer 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 period3. Loan Protection
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
Monthly repayment = £127.38
(b) How much extra would they repay if they had opted for a 5 year repayment period with loan protection?
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= £7201.44Answer 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
UNIT 4 :
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
Formulae
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FORMULAE: Question 1
The surface area of a triangular prism, S cm2, is given by the formula
S = x2 + 2dx + dw
where all distances are in cm.
x x
d
w
(a) Find S when x = 10, w = 14 & d = 30.
(b) Find w when x = 5, d = 20 & S = 365.
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FORMULAE: Question 1
The surface area of a triangular prism, S cm2, is given by the formula
S = x2 + 2dx + dw
where all distances are in cm.
x x
d
w
(a) Find S when x = 10, w = 14 & d = 30.
(b) Find w when x = 5, d = 20 & S = 365.
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w = 7
= 1120 cm2
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(a)
S = x2 + 2dx + dw
= (10 x 10) + (2 x 30 x 10) + (30 x 14)
= 100 + 600 + 420
= 1120
Area is 1120cm2
1. Substitute known values into given formula:
S = x2 + 2dx + dw
(a)Find S when x = 10,
w = 14 & d = 30.
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1. Substitute known values into given formula:
S = x2 + 2dx + dw
(b)Find w when x = 5,
d = 20 & S = 365.
(b)
S = x2 + 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:
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1. Write formula
2. Replace known values
3. Evaluate
(a)
S = x2 + 2dx + dw
= (10 x 10) + (2 x 30 x 10) + (30 x 14)
= 100 + 600 + 420
= 1120
Area is 1120cm2
When evaluating formulae:
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1. Substitute known values into given formula:
(b)
S = x2 + 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
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FORMULAE: Question 2
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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 .
FORMULAE: Question 2
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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
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(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
BODMASC = 5/9(F - 32)
(a) Change 302°F into °C
(b) Change -40°F into °C
(c) Change 10°C into °F .
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Question 2
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BODMASC = 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!!
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Question 2
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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
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(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
BODMAS
1. Write formula
2. Replace known values
3. Evaluate ( BODMAS)
When evaluating formulae:
BO÷ / x+ / -
BracketsOf÷ / x+ / -
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BODMAS(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)
BO÷ / x+ / -
BracketsOf÷ / x+ / -
When evaluating formulae:
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(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)
BO÷ / x+ / -
BracketsOf÷ / x+ / -
When evaluating formulae:
4. Solve resulting equation
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FORMULAE: Question 3
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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
10L
where L is the length of the pendulum in metres.
Take = 3.14.
FORMULAE: Question 3
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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
10L
where L is the length of the pendulum in metres.
Take = 3.14.
T = 12.56
L = 90
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(a) Find T when L = 40m.
(b) Find L when
T = 18.84secs.
T = 2 ( )L
10 T = 2 x 3.14 x ( )4010
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.
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Question 3
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(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
= 32L10
= 9L10
L = 9 x 10 = 90
Length is 90m when time is 18.84secs
Square bothSides !
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T = 2 x 3.14 x ( )4010
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:
BO÷ / x+ / -
BracketsOf÷ / x+ / -
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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
= 32L10
= 9L10
L = 9 x 10 = 90
Length is 90m when time is 18.84secs
Square bothSides !
1. Write formula
2. Replace known values
3. Evaluate ( BODMAS)
BO÷ / x+ / -
BracketsOf÷ / x+ / -
When evaluating formulae:
4. Solve resulting equation
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FORMULAE: Question 4
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Change the subject of the formula m = 3p2 – k to p.
FORMULAE: Question 4
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Change the subject of the formula m = 3p2 – k to p.
p = m + k3
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Change the subject of
the formula
m = 3p2 – k to p.
m = 3p2 – k
3p2 – k = m
3p2 = m + k
p2 = k + m3
Swap sides.
Isolate 3p2
Isolate p2
Isolate p
p = m + k3
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10 = 3x2 – 4
3x2 – 4 = 10
3x2 = 10 + 4
x = 10 + 4 3
Apply the same rules as in a simple equation.
x2 = 10 + 4 3
m = 3p2 – k
3p2 – k = m
3p2 = m + k
p2 = k + m3
Swap sides.
Isolate 3p2
Isolate p2
Isolate p
p = m + k3
EXAMPLE
EXAMPLE
EXAMPLE
EXAMPLE
EXAMPLE
FORMULAE: Question 5
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Change the subject of the formula Q = 2kn – 1 to n . 3
FORMULAE: Question 5
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Change the subject of the formula Q = 2kn – 1 to n . 3
n = 3Q + 3 2k
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Question 5
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Change the subject of the
formula
Q = 2kn – 1 to n . 3
3Q = 2kn - 3
2kn – 3 = 3Q
2kn = 3Q + 3
Q = 2kn – 13
X 3 to eliminate fraction.
Swap sides.
Isolate 2kn .
Isolate n .
n = 3Q + 3 2k
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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 – 13
n = (3 x 10) + 3 2 x 5
3Q = 2kn - 3
2kn – 3 = 3Q
2kn = 3Q + 3
Q = 2kn – 13
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
EXAMPLE
EXAMPLE
EXAMPLE
EXAMPLE
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UNIT 4 :
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EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
Logic Diagrams
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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
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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
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Combinations are ...
1. Bus-bus 2. Bus-taxi 3. Train-bus
4. Train-taxi 5. Train-cycle
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Question 1
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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
taxicycle
bus
train
bus
taxi
bus
taxicycle
Combinations are ...
1. Bus-bus
2. Bus-taxi
3. Train-bus
4. Train-taxi
5. Train-cycle
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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 travelledby train then bus?
Probability (uses bus) = 35
What is the probability that he Will use a bus at some point in his journey?
Probability (train,bus) = 15
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.
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
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Question 2
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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
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Amount is under £5000
so rate = 4.4%
4.4% of £3850 = 0.044 x £3850
= £169.40
Take care to ensure thatyou follow correct “flow” of
diagram by answering each question carefully.
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.
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Percentage Calculations
4.4% of £3850 = x £3850
= 0.044 x £3850
4.4100
etc.
Amount is under £5000
so rate = 4.4%
4.4% of £3850 = 0.044 x £3850
= £169.40
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.
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
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Question 3
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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
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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 thatyou follow correct “flow” of
diagram by answering each question carefully.
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.
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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
5.7100
etc.
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
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
23
45
67
89
1011
12
13
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Question 4
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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.
(b) One possible route around the network is
P
GF
E
D
C
B
A
1
P G
2
F
3
E
4
D
5
P
6
C
7
D
8
B
9
A
10
C
11
B
12
A
13
P
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P
GF
E
D
C
B
A
1
2
3
4
5
67
89
10
11
12
13
Test for Traversing a network
Network must have no more than two odd vertices.
2 odd1 even traversible
Note: must start on an odd vertex
3
43
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P
GF
E
D
C
B
A
1
2
3
4
5
67
89
10
11
12
13
Test for Traversing a network
Network must have no more than two odd vertices.
NOTtraversible
3
33
2
3
4 odd1 even
End of Unit 4