TAFLES 
By equation (33) 
€ = 12 — 7.44 + 1.60 = 6.16 years. 
Example. — What is the expectancy of a similar article 1 5 
years old, which has survived its probable term of usefulness 
but is still in service and in good condition? 
Here n = 12 and m = 13. 
By equation (34) 
e = 8.64 — 5.25 = 3.39 years. 
Example. — What is the expectancy of an article 40 years 
old, whose probable life new was 60 years and which is in good 
condition? 
Here n = 60 and m = 4o. 
By equation (33) 
e = 60 — 37.20 + 8.00 = 30.80 years. 
To find the remaining value of an original investment of $100 
in any article or the accrued depreciation when the expectancy 
of the article is known and the probable life new is different from 
any covered by the tables here published, proceed as follows: 
Find the annuity which in # years will amount to $100. From 
tables, such as Tables 26 and 28 in this volume, or by calculation, 
find the amount of this annuity for (# — e) years. This amount 
will represent the accrued depreciation and $100 less this amount 
will be the remaining value of an original investment. 
Example. — What is the remaining value (6 per cent interest) 
of an article which cost $100, whose probable life new was 12 
years, which is 8 years old and which is still in fair condition? 
The expectancy of this article as shown in a preceding example 
is about 6.2 years. The annuity which in 12 years at 6 per cent 
will amount to $100 is $5.93; this annuity in (12 — 6.2) years 
amounts to $40.44, the accrued depreciation. The present worth 
will be $100 — $40.44 = $359.56. 
By use of the ten- and fifteen-year subdivisions of the table 
and interpolation the procedure would be as follows: 
£2 435