|
Consumer
arithmetic
Clement
Radcliffe, Contributor
 |
| A
student of the Rusea's High School in Lucea, Hanover, raises a number of issues,
which she said needs to be examined as it relates to tourism, globalisation and
rural development during the 2006 'Dialogue for Development Forum on Globalisation,
Rural Development and Tourism'. The event, held last Thursday, was sponsored by
the Planning Institute of Jamaica. - Photo by Noel Thompson |
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.
The cash price of a dining-room suite consisting of a table and six identical
chairs is $880. If the price of the table is $250, what is the price for each
chair? Solution
Since
the table and six chairs value $880 and the table values $250, then the six chairs
value $880 - $250 = $630. | ie.
the price of each chair is | $630 | =
$105. | | | 6 | |
Answer
= $105. 2.
The dining-room suite may be bought on hire purchase for a deposit of $216 plus
monthly payments of $35 for a period of two years. Calculate (i)
The total hire purchase price of the suite (ii)
The extra cost of buying on hire purchase as a percentage of the cash price. (CXC
January 2004, 2000, 1b, c) Solution
(i)
The suite is bought on hire purchase for a deposit of $216. Monthly
payments of $35 for two years = $35 x 24 = $840. Total
hire purchase price is $840 + $216 = $1056.
(ii)Since
the hire purchase price is $1056 and the original cash price is $880
Therefore,
the extra cost price is $1056 - $880 = $176. The
extra cost price as a percentage of the cash price is:
Answer
= 20%. Let
us continue with another topic under consumer arithmetic. The
values of equipment, for example, motor cars, normally reduce over time. This
illustrates depreciation which is literally the percentage decrease in the cost
price. Example
Mrs.
White bought a computer on 1 January 1996, at a cost of Bds$4260. Given that the
value of the computer depreciated by 20% each year, calculate, in Barbados dollars,
the value of the computer at the end of 1997. Solution
Since
the cost of the computer at 1 January 1996 is Bds$4,260 and the value of the computer
depreciates by 20% each year: Then
cost at the end of the first year is | Bds$4,260
x | 80 | =
Bds $3,408 | | | 100 | |
ie.
the value at the end of 1997, one year later, is: | Bds
$3,408 x | 80 | =
Bds $2,726.40 | | | 100 | |
Answer
= Bds$3,408. 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 | |
Therefore,
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 four years is $8,100. Calculate the rate per cent
per annum. Solution:
| Simple
Interest = | Principal
x Time x Rate | | | | 100 | |
| Therefore,
Rate = | S.I
x 100 | | | | P
x T | |
=
8,100 x 100 | =
27/2 % or 13.5% | 15,000
x 4 | | |
As
usual, I will close with your homework. 1.
Mr. Mitchell deposited $40,000 in a bank and earned simple interest at 7% per
annum for two years. Calculate
the amount he will receive at the end of the two-year period.
2.
Mr. Williams bought a plot of land for $40,000. The value of the land appreciated
by 7% per each year.
Calculate
the value of the land after a period of two years. (CXC,June
2004) | 
