1a_ch3(1). 1a_ch3(2) 3.1simple problems involving percentages a using percentage to find a number b...
TRANSCRIPT
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1A_Ch3(1)
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1A_Ch3(2)
3.1 Simple Problems Involving Percentages
A Using Percentage to Find a
Number
B Finding the Percentage
C Finding the Original Number
from a Given Percentage
Index
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1A_Ch3(3)
3.2 Percentage Change
A Percentage Increase
B Percentage Decrease
C Percentage Change
Index
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1A_Ch3(4)
3.3 Profit and Loss
A Profit
B Loss
Index
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Using Percentage to Find a Number
y% of a number A = A × y%
Index
A)
1A_Ch3(5)3.1 Simple Problems Involving Percentages
Example
Index 3.1
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Find the value of each of the following.
(a) 850 × 25%
Index
1A_Ch3(6)3.1 Simple Problems Involving Percentages
(a) 25% of 850 (b) 10% of 12.8
= 850 × 0.25
= 212.5
(b) 12.8 × 10% = 12.8 × 0.1
= 1.28
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Index
1A_Ch3(7)3.1 Simple Problems Involving Percentages
There are 1 400 staff in a company. It is
known that 88% of the staff in the company
are university graduates. How many staff in
that company are university graduates?
Number of staff who are university graduates
= 1 400 × 88%
10088
= 1 400 ×
= 1 232
Fulfill Exercise Objective
Use percentage to find a number.
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Index
1A_Ch3(8)3.1 Simple Problems Involving Percentages
In 2002, the population of Hong Kong was 6.8 million and
16% of them were aged under 15. How many people were
15 or above?
Number of people aged under 15 = 6.8 × 16% million
= 6.8 × 0.16 million
= 1.088 million
∴ Number of people aged 15 or above
= (6.8 – 1.088) million
= 5.712 million
Fulfill Exercise Objective
Use percentage to find a number.
Key Concept 3.1.1
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Finding the Percentage
Index
B)
1A_Ch3(9)3.1 Simple Problems Involving Percentages
1. To find out what percentage of a is b,
we write .ba× 100%
2. To find out what percentage of b is a,
we write .ab× 100%
Example
Index 3.1
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(a) What percentage of 45 is 36?
(b) What percentage of 36 is 45?
Index
1A_Ch3(10)
3.1 Simple Problems Involving Percentages
(a) The required percentage = %1004536
= 80%
(b) The required percentage = %1003645
= 125%
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Index
1A_Ch3(11)
3.1 Simple Problems Involving Percentages
Among the 800 people in the election
team, 256 of the team members vote
for candidate A and the rest vote for
candidate B.
(a) What percentage of the votes has gone to
candidate A?
(b) What percentage of the number
of votes for A is that for B?
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Index
1A_Ch3(12)
3.1 Simple Problems Involving Percentages
(a) The required percentage = %100800256
= 32%
(b) The number of votes that candidate B gets = 800 – 256
= 544
The required percentage = %100256544
= 212.5%
Fulfill Exercise Objective
Find the required percentage.
Back to Question
Key Concept 3.1.2
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Finding the Original Number from a Given Percentage
Index
C)
1A_Ch3(13)
3.1 Simple Problems Involving Percentages
Given : x% of the original number A = y
Theny
x%A =
Example
Index 3.1
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Find the unknown in each of the following.
Index
1A_Ch3(14)
3.1 Simple Problems Involving Percentages
(a) $24 is 60% of $m. (b) 180 kg is 125% of n kg.
m × 0.6= 24
m = 24 ÷ 0.6
= 40
(a) $m × 60% = $24
n × 1.25 = 180
n = 180 ÷ 1.25
= 144
(b) n kg × 125%= 180 kg
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Index
1A_Ch3(15)
3.1 Simple Problems Involving Percentages
A football team won 85% of its games
last year. If they won 34 games
altogether, how many games did the team
play?Let n be the total number of games the team played.
Then 85% of n is 34.
i.e. 85% × n = 34
n × 0.85= 34
n = 34 ÷ 0.85
= 40 ∴ The total number of games the team played was 40.
Fulfill Exercise Objective
Find the original number.
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Index
1A_Ch3(16)
3.1 Simple Problems Involving Percentages
In the academic year 1999 – 2000, 55%
of the students in the University of Hong
Kong were male. If the number of female
students in that year was 6 300, then what
was the total number of students?
Percentage of female students
= 100% – 55%
= 45%
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Index
1A_Ch3(17)
3.1 Simple Problems Involving Percentages
Let n be the total number of students.
Then 45% of n is 6 300.
i.e. 45% × n= 6 300
n × 0.45= 6 300
n = 6 300 ÷ 0.45
= 14 000
∴ The total number of students was 14 000.
Fulfill Exercise Objective
Find the original number.
Back to Question
Key Concept 3.1.3
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Percentage Increase
Index
A)
1A_Ch3(18)
3.2 Percentage Change
Example
Index 3.2
1. Increase = New value – Original value
2. Percentage increase = Increase
Original value× 100%
3. Increase = Original value × Percentage increase
4. New value = Original value × (1 + Percentage increase)
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Find the percentage increase in each of the following.
Index
1A_Ch3(19)
3.2 Percentage Change
(a) Increase = 156 – 120
= 36
∴ Percentage increase
= %10012036
= 30%
(a) An increase from 120 to 156.
(b) An increase of 10.5 mL from 25 mL.
(b) Increase = 10.5 mL
∴ Percentage increase
= %100mL 25mL 5.10
= 42%
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Increase each of the following quantities by the given percentage:
Index
1A_Ch3(20)
3.2 Percentage Change
(a) New value = 82 × (1 + 15%)
= 82 × 1.15
= 94.3
(a) 82 by 15% (b) $144 by 35%
(b) New value = $144 × (1 + 35%)
= $144 × 1.35
= $194.4
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Index
1A_Ch3(21)
3.2 Percentage Change
The price increase of a movie recorded on DVD and
on video CD is both $5. If the original prices of a
DVD and a video CD are $125 and $40 respectively,
find the percentage increase in the price of
(a) a DVD,
(b) a video CD.
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Index
1A_Ch3(22)
3.2 Percentage Change
(a) Percentage increase in the price of a DVD
= %100125$5$
= 4%
(b) Percentage increase in the price of a video CD
= %10040$5$
= 12.5%
Fulfill Exercise Objective
Find the percentage increase.
Back to Question
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Index
1A_Ch3(23)
3.2 Percentage Change
To celebrate the New Year, a certain
brand of chocolate beans added 10% to
the weight of each pack for free.
If the original weight of each pack of the chocolate beans
was 50 g, what was the weight of each pack after the
increase?
The new weight = 50 × (1 + 10%) g
= 50 × 1.1 g
= 55 gFulfill Exercise Objective
Find the new value. Key Concept 3.2.1
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Percentage Decrease
Index
B)
1A_Ch3(24)
3.2 Percentage Change
Example
Index 3.2
1. Decrease = Original value – New value
2. Percentage decrease = Decrease
Original value× 100%
3. Decrease = Original value × Percentage decrease
4. New value = Original value × (1 – Percentage
decrease)
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Find the percentage decrease in each of the following.
Index
1A_Ch3(25)
3.2 Percentage Change
(a) Decrease = 300 – 48
= 252
∴ Percentage decrease
= %100300252
= 84%
(a) An decrease from 300 to 48.
(b) An decrease of 3 g from 12 g.
(b) Decrease = 3 g
∴ Percentage decrease
= %100g 12g 3
= 25%
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Decrease each of the following quantities by the given percentage:
Index
1A_Ch3(26)
3.2 Percentage Change
(a) New value = 150 × (1 – 27%)
= 150 × 0.73
= 109.5
(a) 150 by 27% (b) 855 cm by 60%
(b) New value = 855 × (1 – 60%) cm
= 855 × 0.4 cm
= 342 cm
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Index
1A_Ch3(27)
3.2 Percentage Change
The worldwide population of the black r
hinoceroes has dropped from 65 000 in t
he early 1970s to 2 405 in the late 1990s.
What is the percentage decrease in the n
umber of black rhinoceroes?
The decrease = 65 000 – 2 405
= 62 595
The percentage decrease = %100000 65595 62
= 96.3%
Fulfill Exercise Objective
Find the percentage decrease.
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Index
1A_Ch3(28)
3.2 Percentage Change
Kitty scored 80 marks in Mathematics
last term. If her score is decreased by
5% this term, what is her score in this
term?
The score of this term = 80 × (1 – 5%)
= 80 × 0.95
= 76Fulfill Exercise Objective
Find the new value. Key Concept 3.2.2
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Percentage Change
Index
C)
1A_Ch3(29)
3.2 Percentage Change
Example
Index 3.2
‧ Percentage change = New value – Original value
Original value× 100%
New value > Original value
New value < Original value
Change Sign of percentage change
Increase
Decrease
+
–
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Index
1A_Ch3(30)
3.2 Percentage Change
(a) If 100 is changed to 110, then
percentage change = %100100
100110
= +10%
(b) If 50 is changed to 40, then
percentage change = %10050
5040
= –20%
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Index
1A_Ch3(31)
3.2 Percentage Change
The price of a digital camera was
$4 000 last year. This year the price
becomes $3 400. Find the percentage
change in price.
Percentage change
Fulfill Exercise Objective
Find the percentage change.
= %100000 4
000 4400 3
= %100000 4600
= –15% Key Concept 3.2.3
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Profit
Index
A)
1A_Ch3(32)
3.3 Profit and Loss
1. When a merchant pays to buy goods, the amount he pays is called the cost price.
2. When the merchant sells goods at a price, the amount he receives is called the selling price.
3. If selling price > cost price, the merchant will make a profit.
4. If we compare the profit with the cost price of the goods and express the result as a percentage, the percentage is called the profit per cent (written as profit %).
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Profit
Index
A)
1A_Ch3(33)
3.3 Profit and Loss
Example
Index 3.3
5. If selling price > cost price,
ii. Profit % = Profit
Cost price× 100%
iii. Profit = Cost price × Profit %
iv. Selling price = Cost price × (1 + Profit %)
i. Profit = Selling price – Cost price
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The selling price of a toy car is $225, and the
cost price is $180.
Index
1A_Ch3(34)
3.3 Profit and Loss
(a) Find the profit of the toy car.
(b) Find the profit per cent of the toy car.
(a) Profit = $(225 – 180) = $45
(b) Profit % = %100180$45$
= 25%
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Index
1A_Ch3(35)
Mr Wong bought a pair of jeans for $80
and sold them for $130. Mr Cheung bou
ght a T-shirt for $40 and sold it for $90.
Who made a greater profit %?
Profit made by Mr Wong
= %10080$50$
3.3 Profit and Loss
= $(130 – 80)
= $50
∴ Profit %
= 62.5%
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Index
1A_Ch3(36)
Profit made by Mr Cheung
= %10040$50$
3.3 Profit and Loss
= $(90 – 40)
= $50
∴ Profit %
= 125%
∴Mr Cheung made a greater profit %.
Fulfill Exercise Objective
Find the profit %.
Back to Question
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Index
1A_Ch3(37)
A car was bought for $350 000 and
was sold at a profit of 20%. How
much was it sold for?
Selling price
Key Concept 3.3.1
3.3 Profit and Loss
= $350 000 × (1 + 20%)
= $420 000
= $350 000 × 1.2
Fulfill Exercise Objective
Find the selling price.
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Loss
Index
B)
1A_Ch3(38)
3.3 Profit and Loss
1. If selling price < cost price, there will be a loss.
2. We can express loss as a percentage of the cost pric
e. The result is called the loss per cent (written as l
oss %).
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Loss
Index
B)
1A_Ch3(39)
3.3 Profit and Loss
Example
3. If selling price < cost price,
ii. Loss % = Loss
Cost price× 100%
iii. Loss = Cost price × Loss %
iv. Selling price = Cost price × (1 – Loss %)
i. Loss = Cost price – Selling price
Index 3.3
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Index
1A_Ch3(40)
If a bicycle is bought at $1 000 and sold for $800, then
loss = $(1 000 – 800)
= $200
and loss % = %100000 1$
200$
= 20%
3.3 Profit and Loss
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Index
1A_Ch3(41)
A box of 100 coloured pencils costs $250. In
a sale, each pencil is sold at $2 a piece.What i
s the loss per cent after all of the pencils have
been sold?
Total cost price
Key Concept 3.3.2
3.3 Profit and Loss
= $250
Total selling price= $2 × 100 = $200
Total loss= $(250 – 200) = $50
∴ Loss % = %100250$50$
= 20%
Fulfill Exercise Objective
Find the loss %.
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Discount
Index
1A_Ch3(42)
3.4 Discount
1. The difference between the marked price and the
selling price is called the discount.
2. We can express the discount as a percentage of the
marked price, which is called the discount per cent
(written as discount %).
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Discount
Index
1A_Ch3(43)
3.4 Discount
Example
3. i. Discount = Marked price – Selling price
ii. Discount % = Discount
Marked price× 100%
iii. Discount = Marked price × Discount %
iv. Selling price = Marked price × (1 – Discount %)
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A sofa is marked at $2 800 and sold at $2 100 in
a sale. What is the discount and discount %?
Index
1A_Ch3(44)
Discount= $(2 800 – 2 100)
= $700
= 25%
3.4 Discount
Discount % %100800 2$
700$ =
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Index
1A_Ch3(45)
A pack of E-Power Battery marked at $24 is now sold
for $21 only. A pack of D-cell Battery, on the other
hand, marked at $20 is now sold for $17.9.
Which pack of Battery is sold at a larger discount per
cent?
3.4 Discount
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Index
1A_Ch3(46)
For E-Power Battery,
3.4 Discount
= $(24 – 21)
discount % = %10024$3$ = 12.5%
For D-cell Battery,
= $(20 – 17.9)
discount % = %10020$
1.2$ = 10.5%
= $3
= $2.1
∴ E-Power Battery is sold at a larger discount per cent.
Fulfill Exercise Objective
Find the discount %.
discount
discount
Back to Question
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Index
1A_Ch3(47)
After a 44% discount, a radio is now
sold for $112. What was the original
marked price of the radio?
3.4 Discount
44% off
Now $112 only
Let $P be the original marked price of the radio.
Then112 = P × (1 – 44%)
112 = P × 0.56
P = 112 ÷ 0.56
= 200
∴ The original marked price was $200.
Fulfill Exercise Objective
Find the marked price.
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Index
1A_Ch3(48)
A dictionary marked at $260 is sold at
a discount of 25%.
3.4 Discount
(a) Find the selling price.
(b) If the cost price of the dictionary is
$200, find the loss %.
Dictionary
25% off
(a) The selling price = $260 × (1 – 25%)
= $260 × 75%
= 10075
260$
= $195
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Index
1A_Ch3(49)
Key Concept 3.4.1
3.4 Discount
(b) Loss = cost price – selling price
= $(200 – 195)
= $5
= %100200$
5$
= 2.5%
Loss %
Fulfill Exercise Objective
Miscellaneous problems.
Back to Question
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Simple Interest
Index
1A_Ch3(50)
3.5 Simple Interest
1. If the interest in each period is calculated on the
same principal, then the interest obtained is called
simple interest.
2. The calculation of interest is based on a percentage of
the principal and that percentage is called the interest
rate.
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Simple Interest
Index
1A_Ch3(51)
3.5 Simple Interest
3. Let $I stand for the simple interest, $P stand for the
principal, R% stand for the interest rate per annum,
T years stand for the period of time, $A stand for
the amount, then
Examplei.100
TRPI
ii. )100
1(RT
PIPA Example
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Index
1A_Ch3(52)
If $1 000 is deposited in a bank at an interest rate of 6%
p.a., find the interest received a year later.
3.5 Simple Interest
Interest received = $1 000 × 6%
= $60
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Index
1A_Ch3(53)
Deposit $6 000 in a bank at an interest rate of 8% p.a.
for 2 years. What is the simple interest received?
3.5 Simple Interest
Simple interest received
=100
28000 6$
= $960
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Index
1A_Ch3(54)
Simon wants to deposit $30 000 in a bank
for 2.5 years. Bank A offers an interest
rate of 8% p.a. and Bank B offers 7.5%
p.a. How much more simple interest will
Bank A give to Simon than Bank B?
Simple interest given by Bank A
3.5 Simple Interest
=100
5.28000 30$
= $6 000
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Index
1A_Ch3(55)
Simple interest given by Bank B
3.5 Simple Interest
=100
5.25.7000 30$
= $5 625
The difference between the simple interests
∴ Bank A will give $375 more simple interest to Simon
than Bank B.
Fulfill Exercise Objective
Find the simple interest.
= $(6 000 – 5 625)
= $375
Back to Question
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Index
1A_Ch3(56)
Kimmy borrows some money from a bank to start her o
wn business. The interest rate is 5% p.a.
Find the amount of money she borrows
from the bank if she has to pay a total
of $30 000 as simple interest
after 2 years.
3.5 Simple Interest
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Index
1A_Ch3(57)
Let $P be the amount of money Kimmy borrows, then
3.5 Simple Interest
30 000 = 100
25P
30 000 = 10P
P = 30 000 × 10
= 300 000
∴ The amount of money Kimmy borrows is $300 000.
Fulfill Exercise Objective
Find the principal.
Back to Question
Key Concept 3.5.1
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Index
1A_Ch3(58)
A bank pays simple interest for money deposited at an
interest rate of 10% p.a. If $5 000 is deposited in the bank,
find the amount received after 3 years.
3.5 Simple Interest
Amount received after 3 years
=
100310
1000 5$
= $6 500
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Index
1A_Ch3(59)
A bank pays simple interest at an
interest rate of 8% p.a. If $25 000 is
deposited in the bank, find
3.5 Simple Interest
(a) the amount received after 10 months, correct to the nearest $100,
(b) the time required to get the amount $27 500.
(a) Amount received after 10 months
=
100
81000 25$
1012
= $26 700 , cor. to the nearest 100
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Index
1A_Ch3(60)
3.5 Simple Interest
(b) Let T years be the time required, then
27 500 =
1008
125000T
1008
1T =
000 25500 27
1008 T
= 0.1
T =8
10
= 1.25
∴ The time required is 1.25 years, i.e. 1 year and 3 months.
Fulfill Exercise Objective
Find the amount. Find the period of time.
Back to Question
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Index
1A_Ch3(61)
John deposits $1 000 000 in a bank where simple interest
is calculated at a rate of R% p.a.
3.5 Simple Interest
(a) If the amount received by John doubles the principal
after 10 years, find the value of R.
(b) Suppose R = 8.
(i) Find the amount received by John after 20
years.
(ii) Peter borrows a sum of money from John
and the interest rate is 8% p.a. If the amount that
Peter has to pay back John after 2.5 years is
$120 000, how much does Peter borrow?
Soln
Soln
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Index
1A_Ch3(62)
3.5 Simple Interest
(a) The principal is $1 000 000.
The amount received by John after 10 years
Hence 2 000 000 =
10010
1000 000 1R
2 = 1 + 0.1R
1 = 0.1R
∴ R = 10
= $1 000 000 × 2
= $2 000 000
Back to Question
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Index
1A_Ch3(63)
3.5 Simple Interest
(b) (i) Amount received after 20 years=
100208
1000 000 1$
= $2 600 000
(ii) If $P is the sum of money that Peter borrows from John,
then120 000 =
1005.28
1P
120 000 = P(1 + 0.2)
P =2.1000 120
= 100 000
∴ Peter borrowed $100 000 from John.
Fulfill Exercise Objective
Find the interest rate. Find the principal.
Key Concept 3.5.1
Back to Question