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LOGARITHMS. 87
can he save in 5 years in this way ? (Note.—\% per year =
2% per 6 months, or per period of time, and #25 a month = #150
every 6 months, or per period of time. The interest is computed
semi-annually ; therefore 5 years = 10 periods of time).
Solution by formula
:
,
v _ , 100 6 (rn — 1)
a = p X rn
+ ^ ’—
50 X 1.0210
-f-
100 X 150 X (102 10 — 1)
a = 50 X 1.219
100 X 150 X 0.219
a = 60.95 + 1642.50
a = #1703.45 = Amount.
The original sum of #50 has increased to #60.95, and the
monthly payments amounted to #1500. The last six payments
did not draw any interest, as they were deposited in the last
six months of the fifth year and would commence to draw inter-
est at the beginning of the sixth year if the amount had not
been withdrawn.
Example.
A man has #800 invested at 0%. How much must he save
and invest at the same interest every year in order to increase it
to #3000 in five years ? Interest is computed annually.
Solution by formula
:
^ __ a y —p y x >n
lOO rn — 100
3000 X 5 — 800 X 5 X 1.055
b =
6 =
b =
100 X 1.055 — 100
15000 — 1.2763 X 4000
100 X 1.2763 — 100
15000 — 5105.2
127.63
9894.8
100
27.63
b = 358.118 = #358.12 to be paid in each year.
The total payments will be :
800 + 5 X 358.12 = #800 + #1790.60 = #2590.60.
The rest of the amount is accumulated interest. The last pay-
ment is made at the end of the fifth year ; therefore this money
does not draw interest.
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