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Reviewing
consumer arithmetic Part II
Clement
Radcliffe, Contributor
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| Students
of Mico Teachers' College recite
a prayer to heal the nation. The
occasion was a general assembly,
held on the grounds of the college
on April 12, 2005. - Junior Dowie
Photo |
LAST
WEEK we began to review CONSUMER ARITHMETIC
with particular emphasis on Percentage
and problems based on it. We will continue
to review the topic by first going through
the homework.
1.
A discount of 20% is given on the
cost of a pair of shoes which was
originally priced at $1,000. What
is the selling price of the shoes?
SOLUTION
Since
a discount of 20% is given, then the
selling price is 80% of the cost price.
As
the cost price is $1000
ie.
the selling price is 80% of $1000
=
80/100 x $1000 = $800
2.
A company sells its printers to customers
in order to make a profit of 25%.
Calculate
(i)
The price a customer pays for a printer
which the company bought for $1,700
.
(ii)
The price the company paid for a printer
which was sold to a customer for $2,500.
Solution
(i)
The cost price of the printer is $1,700.
As the profit is 25%
ie.
the selling price is represented by
125%
ie.
the customer pays 125/100
x $1,700 = $2,125.
Answer
is $2,125.
(ii)
Since the customer paid $2,500 representing
a profit of 25%
ie.
125% is $2,500
ie.
100% is $2,500 x 100/125 = $2,000.
ie.
the customer paid $2,000 for the printer.
The
answer is $2,000.
Let's
look at a problem which deals with
depreciation.
Some
years ago, the US$1.00 was equivalent
to J$50.00. After devaluation, the
J$1.00 was worth 70% of its original
value. Calculate the new rate of exchange
for US$1.00 and hence calculate the
amount of Jamaican dollars that would
be equivalent to US$2,400.
Solution:
Since
US$1.00 was equivalent to J$50.00
ie. J$1.00
= US$ 1/500
Given
that the Jamaican Dollar had a 70%
devaluation:
ie.
Its worth after devaluation is:
J$1.00
= US$ 1/50 x 70/100
=US$7/500
ie.
The new rate of exchange is
J$1.00 = US$ 7/500
ie.
US$1.00 = J$1.00 x 500/7
=
J$71.43
ie.
US$2,400 = J$71.43 x 2,400 = J171,432.
We
will now take a look at Simple Interest.
This fairly popular topic is based
on the formula:
| Simple
Interest = |
Principal
x Time x Rate |
| |
100
|
Example
1
The
simple interest on $400 for 3 years
at 10% per annum is:
(a)
$1.20 (b) $12 (c) $120 (d) $0.12
Solution:
| Simple
Interest = |
Principal
x Time x Rate |
| |
100
|
| |
=
$400 x 3 x 10 = $120 |
| |
100
|
ie.
The answer is (c)
As
you can see, problems involving simple
interest are quite straightforward.
Once the formula is known, then you
are simply required to perform the
algebraic operations of Transposition
and Substitution in order to find
the Principal, Time or Rate. For example,
if you are asked to find the Principal,
the equation is
| Principal
= |
Simple
Interest x 100 |
| |
Time
x Rate
|
Example
2
The
simple interest on $15,000 for 4 years
is $8100. Calculate the rate per cent
per annum.
Solution:
| Simple
Interest = |
Principal
x Time x Rate |
| |
100
|
|
ie.
Rate =
|
S.I.
x 100
|
| |
P
x T
|
|
=
8100 x 100
|
=
27/2 % or 13.5% |
|
15000
x 4
|
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As
usual, I will close with your homework.
1.
A woman bought a stove for $12,800.
After using it for 2 years she decided
to trade in the stove. The company
estimated a depreciation of 15% for
the first year of its use and a further
15 % on its reduced value, for the
second year.
(i)
Calculate the value of the stove after
the two years.
(ii)
Express the value of the stove after
two years as a percentage of the original
value. (CXC June, 1988)
2.
Calculate the simple interest if $4,000
is placed in a bank at 8% per annum
for 6 months.
3.
The simple interest on $12,000 invested
at 8% per annum is $6,720. Calculate
the number of years for which the
sum was invested.
*
Clement Radcliffe is principal
of Glenmuir High School in Clarendon.
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