UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.
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Transcript of UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.
![Page 1: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/1.jpg)
UNIT 4 :
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
SocialArithmetic
Logic Diagrams
Formulae
![Page 2: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/2.jpg)
UNIT 4 :
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Please choose a question to attempt from the following:
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
SocialArithmetic
1
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2 3 4
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SOCIAL ARITHMETIC: Question 1
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EXIT
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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?
![Page 4: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/4.jpg)
SOCIAL ARITHMETIC: Question 1
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Reveal answer only
EXIT
Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200.
(a) How much does she earn in a week when her sales are £530?
(b) What do her sales need to be so her wages are £250?
1. Calculate how much of the sales qualify for a commission
payment.
2. Apply given percentage to this
sum to find Commission due.
3. Remember total Wage includes
basic
4. In (b) use reverse percentage to work
out sales required to give the amount above
basic.
5. In (b) Remember sales must be £200 before commission earned.
What would you like to do now?
![Page 5: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/5.jpg)
SOCIAL ARITHMETIC: Question 1
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Go to Social Arithmetic MenuEXIT
Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200.
(a) How much does she earn in a week when her sales are £530?
(b) What do her sales need to be so her wages are £250? = £330
= £900
What would you like to do now?
![Page 6: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/6.jpg)
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Question 1
Soc Arith Menu
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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
![Page 7: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/7.jpg)
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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
What would you like to do now?
![Page 8: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/8.jpg)
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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
![Page 9: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/9.jpg)
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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
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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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EXIT
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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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EXIT
= £2812.58
= £182.82
= £1956.88What would you like to do now?
![Page 12: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/12.jpg)
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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!
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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.
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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
What would you like to do now?
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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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EXIT
Diana Marr works in an electronics factory. Her partially completed payslip is shown below.
(a) Find her gross wage for this particular week.
(b) If she works a 38 hour basic week then find her hourly rate.
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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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EXIT
Diana Marr works in an electronics factory. Her partially completed payslip is shown below.
(a) Find her gross wage for this particular week.
(b) If she works a 38 hour basic week then find her hourly rate.
What would you like to do now?
= £398.73
= £7.86
![Page 18: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/18.jpg)
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Question 3
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(a)
Find her gross wage for
this particular week.
Back to payslip
(a)Total deductions
= £26.32 + £60.93 + £45.83
= £133.08
Gross wage = £133.08 + £265.65
= £398.73
Work backwards!
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Question 3
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Back to payslip
(b)
If she works a 38 hour
basic week then find her
hourly rate.
(b)Hourly rate = Basic wage
Hours worked
Basic wage
= £398.73 - £58.70 - £41.35
= £298.68
Hourly rate = £298.68 38
= £7.86
Gross from part (a)
Basic wage is before overtime and bonus.
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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!
![Page 21: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/21.jpg)
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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
What would you like to do now?
![Page 22: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/22.jpg)
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
![Page 23: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/23.jpg)
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
![Page 24: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/24.jpg)
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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?
Back to table
(a)
Monthly repayment = £150.03
Total repayment = £150.03 x 48
= £7201.44
![Page 25: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/25.jpg)
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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
What would you like to do now?
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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
![Page 27: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/27.jpg)
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?
![Page 28: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/28.jpg)
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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
![Page 29: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/29.jpg)
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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
![Page 30: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/30.jpg)
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?
![Page 31: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/31.jpg)
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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
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UNIT 4 :
You have chosen to study:
Please choose a question to attempt from the following:
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
Formulae
1
Back toUnit 4 Menu
2 3 4 5
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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.
EXIT
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![Page 34: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/34.jpg)
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.
EXIT
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w = 7
= 1120 cm2
![Page 35: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/35.jpg)
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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.
![Page 36: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/36.jpg)
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Question 1
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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:
![Page 38: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/38.jpg)
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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 .
![Page 40: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/40.jpg)
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 .
![Page 42: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/42.jpg)
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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!!
![Page 43: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/43.jpg)
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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+ / -
![Page 45: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/45.jpg)
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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:
![Page 46: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/46.jpg)
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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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![Page 47: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/47.jpg)
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.
![Page 48: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/48.jpg)
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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Question 3
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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.
![Page 50: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/50.jpg)
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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+ / -
![Page 52: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/52.jpg)
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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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![Page 53: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/53.jpg)
FORMULAE: Question 4
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Answer
Change the subject of the formula m = 3p2 – k to p.
![Page 54: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/54.jpg)
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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Question 4
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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
![Page 57: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/57.jpg)
FORMULAE: Question 5
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Change the subject of the formula Q = 2kn – 1 to n . 3
![Page 58: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/58.jpg)
FORMULAE: Question 5
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Change the subject of the formula Q = 2kn – 1 to n . 3
n = 3Q + 3 2k
![Page 59: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/59.jpg)
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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
![Page 60: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/60.jpg)
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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 :
You have chosen to study:
Please choose a question to attempt from the following:
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
Logic Diagrams
1
Back toUnit 4 Menu
2 3 4
![Page 62: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/62.jpg)
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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![Page 63: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/63.jpg)
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
![Page 64: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/64.jpg)
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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
![Page 65: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/65.jpg)
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
![Page 67: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/67.jpg)
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.
![Page 68: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/68.jpg)
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
![Page 69: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/69.jpg)
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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
![Page 70: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/70.jpg)
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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.
![Page 71: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/71.jpg)
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.
![Page 72: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/72.jpg)
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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
![Page 73: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/73.jpg)
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.
![Page 74: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/74.jpg)
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
![Page 75: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/75.jpg)
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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.
![Page 77: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/77.jpg)
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.
![Page 78: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/78.jpg)
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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.
![Page 79: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/79.jpg)
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
![Page 80: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/80.jpg)
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
![Page 81: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/81.jpg)
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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.
![Page 82: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/82.jpg)
(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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![Page 83: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/83.jpg)
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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
![Page 84: UNIT 4 : EXIT INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Social Arithmetic Logic Diagrams Formulae.](https://reader035.fdocuments.net/reader035/viewer/2022062309/56649ddc5503460f94ad3bdd/html5/thumbnails/84.jpg)
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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