\fdO'^ - Old Forge Coal Mines

ARITHMETIC. 47
Explanation.—We begin to add at the right-hand column.
7 + 9 + 3 = 19 in. as 12 in. make one •, foot, 19 in. =
1 ft. and 7 in. Place the 7 in. in the inches column, and
reserve the 1 ft. to add to the next column.
1 (reserved) 4-^ + 1 + ^ = ft. Since 3 ft. make 1 yard,
6 ft. = 2 yd. and ft. remaining. Place the cipher in the
column of feet and reserve the 2 yd. for the next column.
2 (reserved) -f 4 -f 2 = 8 yd.
Since 5- yd. = 1 rod, 8 yd. =
1 rd. and 2— yd. Place 2— yd. in the yards column, and
reserve 1 rd. for the next column ; 1 (reserved) -f- 2 = 3 rd.
Ans. = 3 rd. 2^ yd. ft. 7 in.
or, 3 rd. 2 yd. 1 ft. 13 in.
or, 3 rd. 2 yd. 2 ft. 1 in. Ans.
(97) We write the compound numbers so that the units
of the same denomination shall stand in the same column.
Beginning to add with the lowest denomination, we find that
the sum of the gills is 1 + 2 +
gal.
3 = 6. Since there are 4 gi. in
1 pint, in 6 gi. there are as many
pints as 4 is contained times in
6, or 1 pt. and 2 gi. We place
2 gi. under the gills column
16 gal. 3 qt. Opt. 2 gi. and reserve the 1 pt. for the
pints column; the sum of the
pints is 1 (reserved) -|-5-l-l-|-l = 8. Since there are 2 pt.
in 1 quart, in 8 pt. there are as many quarts as 2 is con-
tained times in 8, or 4 qt. and pt. We place the cipher
under the column of pints and reserve the 4 for the quarts
column. The sum of the quarts is 4 (reserved) -f- 8 -|- 3 = 15.
Since there are 4 qt. in 1 gallon, in 15 qt. there are as many
gallons as 4 is contained times in 15, or 3 gal. and 3 qt. remaining.
We now place the 3 under the quarts column
and reserve the 3 gal. for the gallons column. The sum of
the gallons column is 3 (reserved) + 4 + 6 + 3 = 16 gal.
Since we can not reduce 16 gal. to any higher denomination,
we have 16 gal. 3 qt. pt. and 2 gi. for the answer.