\fdO'^ - Old Forge Coal Mines

56 ARITHMETIC.
Since there are 24 gr. in 1 pwt., in 33,606 gr. there are as
many pwt. as 24 is contained times in 33,606, or 1,400 pwt.
and 6 gr. remaining. This gives us the number of grains
in the answer. "We now reduce 1,400 pwt. to higher denom-
inations. Since there are 20 pwt. in 1 oz., in 1,400 pwt.
there are as many ounces as 20 is contained times in 1,400,
or 70 oz. and pwt. remaining; therefore, there are pwt.
in the answer. We reduce 70 oz. to higher denominations.
Since there are 12 oz. in 1 lb., in 70 oz. there are as
many pounds as 12 is contained times in and 10 oz. remaining. We can not reduce
70, or 5 lb.
5 lb. to any
higher denominations.
6gr.
Therefore, our answer is 5 lb. 10 oz.
Another but more complicated way of working this
problem
is as follows:
To get rid of the decimal
in the pounds, reduce .5 of a
to ounces. Since
pound
1 lb. = 12 oz., .5 of a pound
equals .5 lb. X 12 = 6 oz.
6 oz. +a5 oz. = 21 oz. We
now have 4 lb. 21 oz. 19. 5 pwt.
and 18 gr., but we still have a
decimal in the column of pwt., so we reduce .5 pwt. to grains
to get rid of it. Since 1 pwt. =»-24 gr., .5 pwt. = .5 pwt.
X 24= 12 gr. 12 gr. + 18 gr. = 30 gr. We now have 4
lb. 21 oz. 19 pwt. and 30 gr. Since there are 24 gr. in 1 pwt.,
in 30 gr. there is 1 pwt. and 6 gr. remaining. Place 6 gr.
under the column of grains and add 1 pwt. to the pwt.
column. Adding 1 pwt., we have 19 + 1 = 20 pwt. Since
there are 20 pwt. in 1 oz., we have 1 oz. and pwt. remain-
ing. Write the pwt. under the pwt. column, and reserve
the 1 oz. to the oz. column. 21 oz. + 1 oz. = 22 oz. Since
there are 12 oz. in 1 lb., in 22 oz. there is 1 lb. and 10 oz.
remaining. Write the 10 oz. under the ounce column, and
reserve the 1 lb. to add to the lb. column. 4 lb. + 1 lb.
(reserved) = 5 lb. Hence, the answer equals 5 lb. 10 oz.
6gr.
lb.