\fdO'^ - Old Forge Coal Mines

ARITHMETIC.
only 5 tens remain. 3 tens from 5 tens leaves 2 tens. In the hundreds column we
have 3 hundreds from hundreds leaves
G hundreds. We can not subtract 3 thou-
59902
3333
'^'^^^^ ^ns.
sands from thousands, so we take 1 ten-thousand from
5 ten-thousands and add it to the thousands. 1 /en-
tJiousand = 10 tJiausands, and 10 thousands + thousands
= 10 thousands. Subtracting, we have 3 thousands from
10 thousands leaves 7 thousands. We took 1 ten-thousand
from 5 ten-thousands and have 4 ten-thousands remaining.
Since there are no ten-thousands in the subtrahend, the
4 in the ten-thousands column in the minuend is brought
down into the same column in the remainder, because from
4 leaves 4.
{b) 15 339
10001
5 3 3 8 Ans.
(lO) {a) 70968 {b) 100000
32975 98735
3 7993 Ans. 1265 Ans.
(11) We have given the minuend or greater number
(1,004) and the difference or remainder (49). Placing these
1004
in the usual form of subtraction we have in which
49
the dash ( ) represents the number sought. This number
is evidently less than 1,004 by the difference 49, hence,
1,004 — 49 = 955, the smaller number. For the sum of the
10 4 larger
two numbers we then have 9 5 5 smaller
19 5 9 sum. Ans.
Or, this problem may be solved as follows: If the greater
of two numbers is 1,004, and the difference between them is
49, then it is evident that the smaller number must be
equal to the difference between the greater number (1,004)