Two old problems
New Inductive Arithmetics (The Complete Arithmetic, Oral and Written, on the basis of works By Benjamin Greenleaf, published by Leach, Shewell, and Sanborn, in Boston and New York in 1881), contains the following problem (problem 22, page 346).
If 5 horses eat as much as 6 oxen, and 8 horses and 12 oxen eat 12 tons of hay in 40 days, how much hay will 7 horses and 15 oxen eat in 65 days?
Remark.
One ton weighs 2000 pounds. Today we call it a "short ton". One metric ton is heavier; it weighs 1000 kilograms, namely, 2200 pounds.
A modern solution.
Variables and constants:
x weight of hay in pounds that a horse eats in one day,
y weight of hay in pounds that an ox eats in one day,
1 ton = 2000 pounds.
Equations.
5*x = 6*y< 5 horses eat as much as 6 oxen (per day)..
(8*x + 12*y)*40 = 12*2000 8 horses and 12 oxen eat 12 tons of hay in 40 days.
Question.
What is the value of (7*x + 15*y)*65/2000? How much will 7 horses and 15 oxen eat in 65 days? (The answer is in tons.)
Algebraic manipulations.
x = 6/5*y
8*x + 12*y = 600, so
8*(6/5)*y + 12*y= 600
48/5*y + 60/5*y = 600
108/5*y = 600
y = 5/108*600 = 250/9 = 27.7777... lb
x = 6/5*250/9 = 100/3 pounds
Computation.
Use a calculator, or mental arithmetic, to compute the answer.
(7*x + 15*y)*65/2000 =
(7*100/3 + 15*250/9)*65/2000 = 21.125 = 21 1/8 tons.
Another old problem
This is another problem from Greenleaf's Arithmetics. (This was his spelling of arithmetic.)
A cistern has 3 pipes. The first will fill it in 12 hours, the second in 16, and the third in 18 hours. If all run together, in what time will they fill it? (State this example as a proportion, if you can.)
(I cannot state it as a proportion. And there is no need for it.)
The cistern has (unknown) volume V (measured in gallons). But we know the rate at which each pump fills the cistern:
First pump: V/12 gallons per hour.
Second pump: V/16 gallons per hour.
Third pump: V/18 gallons per hour.
When all pumps work together, their rates add up. So the rate of the three of them is,
All pumps: V/12 + V/16 + V/18 gallons per hour.
The total time of filling a cistern is its volume divided by the rate it is filled. Therefore
Time_of_filling = V/(V/12 + V/16 + V/18) = (1/12 + 1/16 +
1/18) -1
Computation on the TI-34 II.
(1/12+1/16+1/18)-1 Use the green key for / to get common fractions.
4_28/29
The answer, "4 and 28/29 of an hour", is expected in a story problem. But if you want to pretend that the problem is real, say, "About 5 hours."