nat 5 money compound interest appreciation & depreciation working backwards exam type questions
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Nat 5
MoneyMoneyw
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Compound Interest
Appreciation & Depreciation
Working Backwards
Exam Type Questions
Nat 5
Monday 10 April 2023Monday 10 April 2023
Starter QuestionsStarter Questions
10, 4, 10, 1, 5
Q1. Find the standard deviation for the data below
Q2. Find the coordinateswhere the line
and curve meet.
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Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. To know when to use To know when to use compound formula.compound formula.
1. We are learning how to use the compound formula for appropriate problems.
2.2. Solve problems Solve problems involving compound involving compound formula.formula.
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Compound InterestCompound Interest
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Compound InterestCompound InterestInterest
calculated on new value every year
Real life Interest is not a fixed quantity year after year. One year’s interest becomes part of the next year’s amount. Each year’s interest is calculated on the amount at the start of the year.
Example
Daniel has £400 in the bank. He leaves it in the bank for 3 years. The interest is 7% each year. Calculate the simply interest and then the compound interest after 3 years.
Initial value
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Daniel has £400 in the bank. He leaves it in the bank for 3 years. The interest is 7% each year. Calculate the compound interest and the amount he has in the bank after 3 years.
Y1 : Interest = 7% of £400 = £28
Amount = £400 + £28 = £428
Y 2 : Interest = 7% of £428 = £29.96
Amount = £428 + £29.96 = £457.96
Y 3 : Interest = 7% of £457.96 = £32.06Amount = £457.06 + £32.06 = £490.02
Compound is £490.02 - £400 = £90.02Simple Interest is only £84
Interest = 7% of £400 = £28
3 x 28 = £84
Simple Interest
Interest calculated on
new value every year
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Compound InterestCompound Interest
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Easier MethodEasier Method
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n
100
%1IV
n = period of timeDays, months years
± = increase or decrease
I = initial value
V = Value
IMPORTANT
Can only use this when percentage
is fixed
This is called the multiplier.
Compound InterestCompound Interest
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Calculate the money in the bank after 3 years if the Calculate the money in the bank after 3 years if the compound interest rate is 7% and the initial value is compound interest rate is 7% and the initial value is £400.£400.
V= 400 x (1.07)3 =
£490.02
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n
100
%1IV
n = 3
± = increase 1+0.07=1.07I =400
Compound InterestCompound Interest
Nat 5
Now try Ex 10.2Ch2 (page 43)
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Compound InterestCompound Interest
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Monday 10 April 2023Monday 10 April 2023
Starter QuestionsStarter Questions
4 0
2 9
x y
x y
Q1. Solve the equations
Q2. Find the coordinateswhere the line
and curve meet.
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Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. To know the terms To know the terms appreciation and appreciation and depreciation.depreciation.
1. We are learning about 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.
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Appreciation & DepreciationAppreciation & Depreciation
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Appreciation : Going up in value e.g. House value
Depreciation : Going down in value e.g. car value
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Appreciation & DepreciationAppreciation & Depreciation
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Average house price in Ayr has appreciated by 79% over past 10 years.
If you bought the house for £64995 ten years ago in 1994 how much would the house be worth now ?
Appreciation = 79% x £ 64995= 0.79 x £64995
= £ 51346.05New value = Old Value + Appreciation
= £64995 + £51346.05= £ 116341.05
Just working out
percentages
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Quicker MethodEasier
% £100 64995179 05.116341£64995
100
179
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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
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Appreciation & DepreciationAppreciation & Depreciation
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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
Appreciation & DepreciationAppreciation & Depreciation
Nat 5
Now try N5 TJ Ex 2.5
Ch2 (page 26)
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Appreciation & DepreciationAppreciation & Depreciation
Nat 5
Monday 10 April 2023Monday 10 April 2023
Starter QuestionsStarter Questions
2 3 6
3 3 19
x y
x y
Q1. Solve the equations
Q2. Solve the coordinateswhere the line
and curve meet.
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Learning IntentionLearning Intention Success CriteriaSuccess Criteria
1.1. To understand the process To understand the process of work backwards.of work backwards.
1. We are learning how to work backwards to find the original/Initial value.
2.2. Solve problems using Solve problems using backwards process.backwards process.
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Work BackwardsWork Backwards
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What was the price before the increase.
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Work BackwardsWork Backwards
Method% £
110 88 000 100
Are we expecting more or
less
80000£88000110
100
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Example 2
The value of a car depreciated by 15%. It is now valuedat £2550. What was it’s original price.
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Work BackwardsWork Backwards
Method% £100 2550
85
Are we expecting more or
less
3000£255085
100
Nat 5
Now try N5 TJEx 2.6
Ch2 (page 27)
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Work BackwardsWork Backwards