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Simple Interest / Compound
Calculating PercentagesCalculating PercentagesInt 2
Compound Interest
Appreciation / Depreciation
Inflation / Working back
StarterStarter
3
(a) 3.5647 (b) 4.55732 (c) 3.98111
Q2. Write in standard form.
(a) 12300 (b) 0.0034 (c) 0.0005
(a) 2.8 x 10 (b) 1.19 x
Q1. Round the following to 2 decimals places.
Q3. Write out in full.
- 2 - 1 10 (c) 4.5 x 10
Int 2
Int 2
Calculating PercentagesCalculating Percentages
Just Calculating Percentages
Simple Interest
Compound
Interest
Appreciation
More
Depreciation
Less
InflationRising Prices
Workingbackwards
Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. To know the meaning of To know the meaning of the term simple interest.the term simple interest.
1. To understand theterm simple interest and compound interest.
2.2. To know the meaning of To know the meaning of the term compound the term compound interestinterest..
Int 2
Calculating PercentagesCalculating Percentages
3.3. Know the difference Know the difference between simple and between simple and compound interest.compound interest.
Just working out
percentages
Int 2
Calculating PercentagesCalculating Percentages
Simple InterestSimple Interest
I have £400 in the Bank. At the end of each year I receive 7% of £400 in interest. How much interestdo I receive after 3 years. How much do I now have?
I nterest = 7 ÷ 100 × 400 = £28
After 3 years interest is 3 x £28 = £84.
Total in bank is = £400 + £84 = £484
Int 2
Calculating PercentagesCalculating Percentages
Compound InterestCompound Interest
Interest calculated on
new value every year
Real life Interest is not a fixed quantity year after Real life Interest is not a fixed quantity year after year. One year’s interest becomes part of the year. One year’s interest becomes part of the next year’s amount. Each year’s interest is next year’s amount. Each year’s interest is calculated on the amount at the start of the year.calculated on the amount at the start of the year.
Example
Daniel has £400 in the bank. He leaves it in the Daniel has £400 in the bank. He leaves it in the bank for 3 years. The bank for 3 years. The interest is 7%interest is 7% each year. each year. Calculate the Calculate the compound interestcompound interest and the and the amount he has in the bank after 3 years.amount he has in the bank after 3 years.
Principal value
Interest calculated on
new value every year
Int 2
Calculating PercentagesCalculating PercentagesCompound InterestCompound Interest
Daniel has £400 in the bank. He leaves it in the Daniel has £400 in the bank. He leaves it in the bank for 3 years. The bank for 3 years. The interest is 7%interest is 7% each year. each year. Calculate the Calculate the compound interestcompound interest and the and the amount he has in the bank after 3 years.amount he has in the bank after 3 years.
Year 1 : Interest = 7% of £400 = £28 Amount = £400 + £28 = £428
Year 2 : Interest = 7% of £428 = £29.96
Amount = £428 + £29.96 = £457.96
Year 3 : Interest = 7% of £457.96 = £32.06
Amount = £457.96 + £32.06 = £490.02
Compound interest is £490.02 - £400 = £90.02
StarterStarter
(a) 4.66647 (b) 8.75732 (c) 7.49111
Q2. Write in standard f orm.
(a) 300 (b) 0.04 (c) 0.0105
Q1. Round the f ollowing to 2 decimals places.
Q3. Find the compound interest on £200
over 2 years at 10% interest.
Int 2
Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. To calculate compound To calculate compound interest using interest using calculator..calculator..
1. To understand how to use the calculator to calculate compound interest easier. 2.2. Show appropriate Show appropriate
workingworkingwhen solving problems.when solving problems.
Int 2
Calculating PercentagesCalculating Percentages
This is called the multiplier.
Int 2
Calculating PercentagesCalculating Percentages
Using calculator Using calculator to calculate Compound
Calculate the compound interest on £400 over 3 Calculate the compound interest on £400 over 3 years if interest rate is 7%.years if interest rate is 7%.
Year 1 : Total = 107% of £400 = 1.07 x £400
Year 2 : it is worth 107% of (1.07 x £400)
= 1.07 x 1.07 x £400 = (1.07)2 x £400
Year 3 : it is worth 107% of (1.072) x £400
= 1.07 x 1.07 x 1.07 x £400 = (1.07)3 x £400 = £490.02
StarterStarter
3 -1(a) 8.567 ×10 (b) 2.6×10
Q2. Write in standard f orm.
(a) 100 (b) 0.035
Q1. Write out in f ull.
Q3. Find the compound interest on £ 600
over 2 years at 5% interest.
Int 2
Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. To know the terms To know the terms appreciation and appreciation and depreciation.depreciation.
1. To understand the terms appreciation and depreciation.
2.2. Show appropriate Show appropriate workingworkingwhen solving problems when solving problems containing appreciation containing appreciation and depreciation.and depreciation.
Int 2
Calculating PercentagesCalculating Percentages
Int 2
Calculating PercentagesCalculating Percentages
Appreciation / DepreciationAppreciation / Depreciation
Appreciation : Going up in value e.g. House value
Depreciation : Going down in value e.g. car value
Average house prices in Ayr have appreciated by 79% Average house prices in Ayr have appreciated by 79% over the past 10 years.over the past 10 years.
If you bought a house for £64995 ten years ago, how If you bought a house for £64995 ten years ago, how much would the house be worth now ?much would the house be worth now ?
Appreciation Appreciation = 79% x £ 64995= 79% x £ 64995= 0.79 x £64995= 0.79 x £64995
= = £ 51346.05£ 51346.05
New value New value = Old Value + Appreciation= Old Value + Appreciation= £64995 + £51346.05= £64995 + £51346.05
= = £ 116341.05£ 116341.05
Just working out
percentages
A Mini Cooper cost £14 625 in 2002A Mini Cooper cost £14 625 in 2002
At the end 2003 it At the end 2003 it depreciateddepreciated by 23% by 23%
At the end 2004 it will depreciate by a further 16%At the end 2004 it will depreciate by a further 16%
What will the mini cooper worth at end 2004?What will the mini cooper worth at end 2004?
End 2003 End 2003
Depreciation Depreciation = 23% x £14625= 23% x £14625
= 0.23 x £14625= 0.23 x £14625
= £3363.75= £3363.75
New valueNew value = Old value - Depreciation= Old value - Depreciation
= £14625 - £3363.75= £14625 - £3363.75
= = £11261.25£11261.25
Int 2
Calculating PercentagesCalculating Percentages
Int 2
Calculating PercentagesCalculating Percentages
End 2003 End 2003
Depreciation Depreciation = 23% x £14625= 23% x £14625
= 0.23 x £14625= 0.23 x £14625
= £3363.75= £3363.75
New valueNew value = Old value - Depreciation= Old value - Depreciation
= £14625 - £3363.75= £14625 - £3363.75
= £11261.25= £11261.25
End 2004 End 2004
Depreciation Depreciation = 16% x £11261.25= 16% x £11261.25
= 0.16 x £11261.25= 0.16 x £11261.25
= £1801.80= £1801.80
New Value New Value = £11261.25 - £1801.80= £11261.25 - £1801.80
= £9459.45= £9459.45
StarterStarter
Q1. Write these perecentages as decimals.
(a) 50% (b) 25% (c) 60%
Q2. Find 40% of £ 400.
Q3. Find the area of the shape.
Int 2
6cm
5cm
Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. Know the term inflation.Know the term inflation.1. Understand term inflationand work out associated real-life problems.
2.2. Work out real-life Work out real-life problems involving problems involving inflation.inflation.
Int 2
Calculating PercentagesCalculating Percentages
Int 2
Calculating PercentagesCalculating Percentages
Inflation Inflation
Measure of how much prices rise each year.
Inflation is normally given in percentage form and is normally in the range 0 – 10%
Example 1
In 2009 a worker received a wage of £300 per week. If inflation is 2% in 2009, what should be his wage be in 2010.
2009 inflation = 2%
2% of £300 = £6
His wage should be £300 + £6 = £306
Int 2
Calculating PercentagesCalculating Percentages
Inflation Inflation
Measure of how much prices rise each year.
Inflation is normally given in percentage form and is normally in the range 0 – 10%
Example 2
In 2002 a CD cost £8. The cost increases in line with inflation.What is the price in 2003 if inflation is 1.5%.
2002 inflation = 1.5%
1.5% of £8 = £0.12
Price is £8 + £0.12 = £8.12
Int 2
Calculating PercentagesCalculating Percentages
Example 1
After a 10% increase the price of a house is £88 000. What was the price before the increase.
Reversing the change
Price before is 100% : £800 x 100 = £80 000
£88 000 110 = £8001 % :
100 % + 10 % = £88 000Deduce from question :
110 % = £88 000We have :
Int 2
Calculating PercentagesCalculating Percentages
Example 2
The value of a car depreciated by 15%. It is now valuedat £2550. What was it’s original price.
Reversing the change
Price before is 100% : £30 x 100 = £3 000
£2 550 85 = £301 % :
100 % - 15 % = 85%Deduce from question :
85 % = £2 550We have :