....

1. Compound interest
MATHSWATCH CLIP
164
GRADE
5
L.O. To be able to
 Calculate compound interest and repeated percentage change
Key Words
interest, compound, repeated, increase, decrease, depreciate,
annum, accrued

2. Compound interest
MATHSWATCH CLIP
164
GRADE
5
COMPOUND INTEREST

3. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Interest
 Interest is paid on amounts deposited in a bank account
 An interest rate is given, as a percentage
 Compound interest is the usual type of interest applied to
bank accounts
 Compound interest is when the interest each year is
applied to the total amount in account at the end of that
year (including previous interest paid)

4. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Interest
 Over time, compound interest gives a better return on your
investment than simple interest (where the interest is just calculated
on the original amount invested each year)
£33.10
£30.00
Total Interest gained
£133.10
£130
Amount after 3years
£121
£120
Amount after 2 years
£110
£110
Amount after 1 year
Compound Interest
Simple Interest
The table below shows how £100 grows over a 3
year period if it is invested at a rate of 10%

5. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Example
Jack puts £500 into a savings account with an annual compound
interest rate of 5%.
How much will he have in the account at the end of 4
years if he doesn’t add or withdraw any money?
At the end of each year interest is added to the total amount in the
account. This means that each year 5% of an ever larger amount is
added to the account.
To increase the amount in the account by 5% we need to multiply
it by 105% or a percentage multiplier of 1.05.
We can do this for each year that the money is in the account.

6. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Example
At the end of year 1 Jack has £500 × 1.05 = £525
At the end of year 2 Jack has £525 × 1.05 = £551.25
At the end of year 3 Jack has £ 551.25 × 1.05 = £578.81
At the end of year 4 Jack has £578.81 × 1.05 = £607.75
(These amounts are written to the nearest penny.)
We can write this in a single calculation as
£500 × 1.05 × 1.05 × 1.05 × 1.05 = £607.75
Or using index notation as
£500 × 1.054 = £607.75

7. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
*
Example
How much would Jack have after 10 years?
How long would it take for the money to double?

8. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
*
Example
How much would Jack have after 10 years?
After 10 years the investment would be worth
£500 × 1.0510 = £814.45 (to the nearest 1p)
How long would it take for the money to double?
£500 × 1.0514 = £989.97 (to the nearest 1p)
£500 × 1.0515 = £1039.46 (to the nearest 1p)
Using trial and improvement,
It would take 15 years for the money to double.

9. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Compound interest
To work out the value of an investment, at a given compound
interest rate, for a given number of years…
 First take your percentage interest rate and write as a
multiplier for an increase by that percentage (x 1. _ _ )
 Use this formula:
Value of investment = original amount × % multiplier no.of years

10. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Example
£ 500 is put into a bank account paying 5%
interest per annum. How much will there be
after 2 years ?
Total amount after 2 years = 500 x 1.052
= £ 551.25

11. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
*
Example
£ 2000 is put into a bank account paying 12%
interest per annum. How much will there be
after 3 years ?

12. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
*
Example
£ 2000 is put into a bank account paying 12%
interest per annum. How much will there be
after 3 years ?
Total amount after 3 years = 2508.80 x 1.123
= £ 2809.86

13. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
**
Task
1) £ 240 at 10 % p.a. for 2 years.
2) £ 800 at 4 % p.a. for 2 years.
3) £ 15000 at 8 ½ % p.a. for 2 years.
4) £ 7000 at 15 % p.a. for 3 years.
5) £ 50000 at 12 ¼ % p.a. for 3 years.

14. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
**
Task
1) £ 240 at 10 % p.a. for 2 years.
2) £ 800 at 4 % p.a. for 2 years.
3) £ 15000 at 8 ½ % p.a. for 2 years.
4) £ 7000 at 15 % p.a. for 3 years.
5) £ 50000 at 12 ¼ % p.a. for 3 years.
£ 290.40
£ 865.28
£ 17658.38
£ 10646.13
£ 70717.85

15. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
***
Task
Q1. £8000 is invested at 7% compound interest for 6 years.
Find: (a) the amount in the account at the end of the period (nearest £) and
(b) the interest accrued (nearest £)
Q2. £1250 is invested at 9% compound interest for 10 years.
Find: (a) the amount in the account at the end of the period (nearest £) and (b)
the interest accrued (nearest £)
Q3 .£3750 is invested at 4.5% compound interest for 8 years.
Find the amount in the account at the end of the period (nearest £100)
Q4. £7500 is invested at 8.5% compound interest for 15 years.
Find the amount in the account at the end of the period (nearest £100)

16. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
***
Solutions
Question 1
£8000 is invested at 7% compound interest for 6 years.
Find: (a) the amount in the account at the end of the period (nearest £) and (b) the
interest accrued (nearest £)
Question 2
£1250 is invested at 9% compound interest for 10 years.
Find: (a) the amount in the account at the end of the period (nearest £) and (b) the
interest accrued (nearest £)
(a) 8000 x 1.076 = £12,006 (b) 12,006 – 8000 = £4,006
(a) 1250 x 1.0910 = £2959 (b) 2959 – 1250 = £1709

17. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
***
Solutions
Question 5
£3750 is invested at 4.5% compound interest for 8 years.
Find the amount in the account at the end of the period (nearest £100)
Question 6
£7500 is invested at 8.5% compound interest for 15 years.
Find the amount in the account at the end of the period (nearest £100)
Answer: 3750 x 1.0458 = £5300
Answer: 7500 x 1.08515 = £25 500

18. Compound interest
MATHSWATCH CLIP
164
GRADE
5
REPEATED PERCENTAGE
CHANGE

19. Compound interest
MATHSWATCH CLIP
164
GRADE
5
We can use the “compound interest” formula to help solve many
problems involving repeated percentage increase and decrease.
The population of a village increases by 2% each year.
If the current population is 2345, what will it be in 5 years?
To increase the population by 2% we multiply it by 1.02.
After 5 years the population will be
2345 × 1.025 = 2589 (to the nearest whole)
What will the population be after 10 years?
After 5 years the population will be
2345 × 1.0210 = 2859 (to the nearest whole)

20. Compound interest
MATHSWATCH CLIP
164
GRADE
5
The car costs £24 000 in 2005.
How much will it be worth in 2013?
To decrease the value by 15% we multiply it by 0.85.
After 8 years the value of the car will be
£24 000 × 0.858 = £6540 (to the nearest pound)
The value of a new car depreciates at a
rate of 15% a year.
There are 8 years between 2005 and 2013.

21. Compound interest
MATHSWATCH CLIP
164
GRADE
5
 Sometimes repeated percentage change involves the
percentage itself differing over time, or an increase being
followed by a decrease
 The next examples involve these types of problems

22. Compound interest
MATHSWATCH CLIP
164
GRADE
5
A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the
price by a further 10%.
What is the total percentage discount?
When a percentage change is followed by another percentage
change do not add the percentages together to find the total
percentage change.
The second percentage change is found on a new amount and not
on the original amount.
It is not 30%!

23. Compound interest
MATHSWATCH CLIP
164
GRADE
5
To find a 10% decrease we multiply by 90% or 0.9.
A 20% discount followed by a 10% discount is equivalent to
multiplying the original price by 0.8 and then by 0.9.
To find a 20% decrease we multiply by 80% or 0.8.
original price × 0.8 × 0.9 = original price × 0.72
A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the
price by a further 10%.
What is the total percentage discount?

24. Compound interest
MATHSWATCH CLIP
164
GRADE
5
This is equivalent to a 28% discount.
The sale price is 72% of the original price.
A 20% discount followed by a 10% discount
A 28% discount
A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the
price by a further 10%.
What is the total percentage discount?

25. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
**
Problem solving!
Jenna invests in some shares.
After one week the value goes up by 10%.
The following week they go down by 10%.
Has Jenna made a loss, a gain or is she back to
her original investment?

26. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Difficulty
**
Solution
To find a 10% increase we multiply by 110% or 1.1.
To find a 10% decrease we multiply by 90% or 0.9.
original amount × 1.1 × 0.9 = original amount × 0.99
Fiona has 99% of her original investment and has therefore made
a 1% loss.

27. Compound interest
MATHSWATCH CLIP
164
GRADE
5
Repeated percentage change
 Work out the percentage multipliers for each change
which is to be applied
 Multiply these decimals together to find the overall
percentage multiplier
 This can be used to:
 calculate the final value, or
 decide what the overall percentage change would be