Transcript Percentage Multiplier - Milford Haven School

Percentages
Percentage Multiplier
October 2006
©RSH
Introduction
Percentages
• Percentages are used all over the place
Sale Now On
12% off
Everything !
Pay rises
by 4%
October 2006
House Prices
go up 12%
a year !
VAT 17.5 %
Save with us
and get 4.5%
a year!
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Inflation up
to 3.5%
Introduction
Percent – a reminder
• 5% means 5 out of 100 or
5
100
• To work out 5% of 60, do this
5% of 60
5
 60
=
100
= 0.05 x 60
=3
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Notes
Percentage Increase or Decrease
• John’s wage is £15000 a year. He gets a 5%
pay rise. What is his new wage ?
Answer
5% of 15000
New wage
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5
 15000
=
100
= 750
= 15000 + 750 = £15750
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Notes
Percentage Increase or Decrease
• My car cost £18500 last year. It dropped in
value by 12%. What is it worth now ?
Answer
12
 18500
12% of 18500 =
100
New wage
October 2006
= 2220
= 18500 - 2220 = £16280
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Exercise
Page 49 Ex 4c
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Notes
Percentage Multipliers
Example
• Increase £60 by 5%
Method 1
• 5% of £60
New value
Percentage Multiplier
New value = original + increase
= 100% + 5%
= 105%
= 1.05
5
 60
=
100
=3
= 60 + 3 = £63
Method 2
• New value is 105% of £60
105% of £60 = 1.05 x 60
= £63
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Notes
Percentage Multipliers
Example
• Increase £800 by 6%
Percentage Multiplier
New value = original + increase
= 100% + 6%
= 106%
= 1.06
Answer
New value is 106% of £800
106% of £800 = 1.06 x 800
= £848
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Notes
Percentage Multipliers
Example
• Decrease £800 by 6%
Percentage Multiplier
New value = original - decrease
= 100% - 6%
= 94%
= 0.94
Answer
New value is 94% of £800
94% of £800 = 0.94 x 800
= £752
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Exercise
Use the multiplier method
a)
b)
c)
d)
Increase 200 by 4%
Increase 150 by 12%
Decrease 300 by 15%
Decrease 1200 by 4%
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a)
b)
c)
d)
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1.04
1.12
0.85
0.96
x
x
x
x
200 = 208
150 = 168
300 = 255
1200 = 1152
Notes
Repeated % change
• The multiplier is useful for working out repeated % change problems.
Example
• £1000 is put into a savings account paying 5% p.a., what will be in
the account after
a)
1 year ?
b)
2 years ?
Answer
The multiplier is 1.05.
a) After 1 year, 1.05 x 1000 = £1050
b) After 2 years, 1.05 x 1050 = £1102.50
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Check this
1.05
2
x 1000
Notes
Repeated % change
Example
• A car worth £15000 reduces in value by 15% each year. What will
the car be worth in
a)
1 year ?
b)
2 years ?
c)
3 years ?
Answer
The multiplier is 0.85.
a) After 1 year, 0.85 x 15000 = £12750
b) After 2 years, 0.85 x 12750 = £10837.50
c) After 3 years, 0.85 x 10837.50 = £9211.88
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Check this
0.85
3
x 15000
Notes
Repeated % change
To work out the value after a number of years
(multiplier)years x original value
Example
In 2005, Sara earned £14000. She gets a 5% pay rise every
year. What will she earn in 2010 ?
Answer
The multiplier is 1.05.
After 5 years, she will earn 1.055 x 14000 = £17867.94
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Exercise
1) A computer operator is paid £12000 a year. Assuming her
pay is increased by 6% each year, what will her salary be in
4 years time ?
1.064 x 12000 = £15149.72
2) The price of a small flat was £78000 in 1995. At the end of
each year, the price has increased by 8%. What was the flat
worth in 2005 ?
1.0810 x 78000 = £168396
3) A new bike is valued at £8000. At the end of each year, its
value is reduced by 15% of its value at the start of the year.
What will it be worth after 4 years?
0.854 x 8000 = £4176.05
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Notes
Compound Interest
• Banks and Building Societies pay interest on the amount in a saving
account.
• If you leave the interest in the account, you will get more interest
next year.
• This is called Compound Interest
Example
In 2006, Paul put £4000 in a saving account paying 4% compound
interest every year. What will be in this account in 2016 ?
Answer
The multiplier is 1.04.
After 10 years, Paul will have 1.0410 x 4000 = £5920.98
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Exercise
1) A bank pays interest of 14% on money in deposit accounts.
Mrs. Wells puts £2000 in the bank. How much does she have
after
a) 1.14 x 2000 = £2280
a) 1 year
b) 3 years
b) 1.143 x 2000 = £2963.09
2) £20000 is invested for ten years at 12% compound interest.
What will the investment be worth then ?
a) £62116.96
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