Nat 5 Money Compound Interest Appreciation & Depreciation Working Backwards Exam Type Questions.

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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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Monday, April 10, 2023Monday, April 10, 2023 created by Mr. lafferty @ created by Mr. lafferty @ www.mathsrevision.comwww.mathsrevision.com 33

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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Easier MethodEasier Method


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.


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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


n

100

%1IV

n = 3

± = increase 1+0.07=1.07I =400


Nat 5

Now try Ex 10.2Ch2 (page 43)

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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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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



Nat 5

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

Monday, April 10, 2023 created by Mr. lafferty @ www.mathsrevision.com

12

Quicker MethodEasier

% £100 64995179 05.116341£64995

100

179

Nat 5


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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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


Nat 5

Now try N5 TJ Ex 2.5

Ch2 (page 26)

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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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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.



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.



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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Nat 5 Money Compound Interest Appreciation & Depreciation Working Backwards Exam Type Questions.

Documents

house value depreciation

new value

old value depreciation

compound formula

value important

car value

initial value v

terms appreciation