\fdO'^ - Old Forge Coal Mines

ARITHMETIC. 45
Explanation.—There are 12 inches in 1 foot; hence, in
5,722 inches there are as many
feet as 12 is contained times
in 5,722 inches, or 476 ft. and 10 inches remaining. Write
these 10 inches as a remainder. There are 3 feet in 1 yard :
as 3 is contained
hence, in 476 feet there are as many yards
times in 476 feet, or 158 yards and 2 feet remaining. There
are 5-^ yards in one rod ; hence, in 158 yards
there are 28 rods
and 4 yards remaining. Then, in 5,722 inches there are
28 rd. 4 yd. 2 ft. 10 in.
(92) 5 weeks 3.5 days.
X
J3
5 days in 5 weeks.
+ 3.5
3 8.5 days.
Then, we find how many seconds there are in 38. 5 days.
3 8.5 days
X 2 4 hours in one day.
1540
770
92 4.0 hours in 38.5 days.
X 6 minutes in one hour.
5 5 4 4 minutes in 38.5 days.
X 6 seconds in one minute.
3 3 2 6 40 seconds in 38. 5 days. Ans.
(93) Since there are 24 gr. in 1 pwt., in 13,750 gr. there
are as many pennyweights as 24 is contained times in
13,750, or 572 pwt. and 22 gr. remaining. Since there are
20 pwt. in 1 oz., in 572 pwt. there are as many ounces as
20 is contained times in 572, or 28 oz, and 12 pwt. remaining.
Since there are 12 oz. in 1 lb. (Troy), in 28 oz. there are
as many pounds as 12 is contained times in 28, or 2 lb. and
4 oz. remaining. We now have the pounds
required by the problem; therefore, in 13,750 gr.
2 lb. 4 oz. 12 pwt. 22 gr.
and ounces
there are