\fdO'^ - Old Forge Coal Mines

46 ARITHMETIC.
24 ) 13750 gr.
20 ) 5 7 2 pwt. + 22 gr.
12 oz.
)_2^ +12 pwt.
2 lb. + 4 oz.
Ans. = 2 lb. 4 oz. 12 pwt. 22 gr.
(94) 100 ) 47 6 325 4 li.
Ans. = 595 mi. 32 ch. 54 li.
80 ) 47632 + 54 li."
5 9 5 + 32 ch.
Explanation.—There ar« 100 li. in one chain; hence, in
4,763,254 li. there are as many chains as 100 is contained
times in 4,763,254 li., or 47,632 ch. and 54 li. remaining.
Write the 54 li. as a remainder. There are 80 ch. in one
mile ; hence, in 47,632 ch. there are as many miles as 80 is con-
tained times in 47,632 ch., or 595 miles and 32 ch. remaining.
Then, in 4,763,254
(96) 1728 ) 764325 cu. in.
li. there are 595 mi. 32 ch. 54 li.
27 ) 442 + 549 cu. in.
1 6 cu. yd. + 10 cu. ft.
Ans. = 16 cu. yd. 10 cu. ft. 549 cu. in.
Explanation.—There are 1,728 cu. in. in one cubic foot;
hence, in 764,325 cu. in. there are as many cubic feet as
1,728 is contained times in 764,325, or 442 cu. ft. and
549 cu. in. remaining. Write the 549 cu. in. as a remainder.
There are 27 cu. ft. in one cubic yard; hence, in 442 cu. ft.
there are as many cubic yards
as 27 is contained times in
442 cu. ft., or 16 cu. yd. and 10 cu. ft. remaining. Then, in
764,325 cu. in. there are 16 cu. yd.
10 cu. ft. 549 cu. in.
(96) We must arrange the different terms in columns,
taking care to have like denominations in the same column,
rd. yd. ft. in.
2 2 2 3
4 19
2 7
3 2i 7
or 3 2 2 1 Ans.