Page 67 - Calculating Agriculture Cover 20191124 STUDENT - A

CH 5] Calculating Agriculture 5-5

Example C: Subtract: 41 gal 2 qt from 48 gal 1 qt

Solution: Set up the problem as 48 gal 1 qt minuend
In subtraction, a
- 41 gal 2 qt subtrahend subtrahend is
subtracted from a
Explanation: Note that 2 quarts being subtracted from 1 quart poses the same minuend to
problem whereby borrowing from the minuend to the left is required; in calculate a
this example 1 gallon is equal to 4 quarts, so you borrow 4 quarts (1 difference. In the
gallon) from the 48 gallons (minuend) and adding this to the 1 quart (1 problem, 9 — 6 = 3,
qt + 4 qt=) 5 quarts and reduces the 48 gallons to (48 gal -1 gal = ) 47 9 is the minuend, 3
gallons. Thus this problem is then restated as: is the subtrahend,
You begin with 48 gal 1 qt minuend and 6 is the
difference.
- 41 gal 2 qt subtrahend

and it becomes 48 gal 1 qt
- 1 gal + 4 qt Borrowed 1 gal adds 4 qts
47 gal 5 qt minuend
- 41 gal 2 qt subtrahend
6 gal 3 qt answer (difference)

The resulting difference then is 6 gallons 3 quart.

In the division of denominate numbers, you will encounter two kinds of problems:
(1) those requiring division by a whole number, that will determine equivalent
portions and (2) those requiring division of one denominate number by another
denominate number, that determine how many times the one dominate number is
divisible by the other.

Example D: (1) Division by a whole number. Divide 22 gallons 2 quarts into
4 equal quantities.

Solution: 5 gal 2 qt 1 pt
4)22 gal 2 qt 5
20 gal
2 gal = 8 qt Parts in Division: The 4
10 qt different parts of a
8 qt division problem are
2 qt = 4 pt dividend, divisor,
4 pt quotient and remainder.

Explanation: Note in this division you work with one denominate number such 19 ÷ 6 (19 divided by 6)
as our starting with gallons and then working our way through to Dividend: the number
the other liquid volumes. being divided. (19)

(2) When dividing one denominate number by another denominate number it is best to Divisor: the number
convert both numbers to the smallest term, unless you want a headache. When both doing the dividing. (6)
numbers are expressed in the same unit the arithmetic is made easier.
Quotient: The answer.
Example E: Divide 50 pounds 5 ounces by 7 pounds 3 ounces. 19 ÷ 6 = 3 + remainder

Solution: Let’s think this through; both the divisor and the dividend are expressed as Remainder: The value
compound dominate numbers with a common measurement, ounces, the remaining when the
smallest unit. There are 16 ounces in every pound and thus both elements arithmetic does not yield
can be expressed as ounces, granted larger numbers than those we start a whole number.
19 ÷ 6 = 3 + r 1
with but their equivalent values are maintained.
Check: 6 x 3 + r 1 =
If both numbers are expressed in ounces then each becomes: Divisor, 7 pounds 3 ounces 18 + r 1 = 19.
becomes ([7 lbs x 16 oz/lb] + 3oz =) 115 ounces and the dividend of 50 pounds 5 ounces
becomes ([50 lbs x 16 oz/lb] + 5 oz =) 805 ounces. Expressed arithmetically for solving:

805 x oz 805 oz 805 NOTE: Any value, (oz),
805 oz ÷ 115 oz = ————— = —— x —— = —— x 1 = 7 divided by itself, (oz),
115 x oz 115 oz 115 always equals 1.

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