Example 1: A worker earns 15
Baht in 6 hours. How much does he earn in 4 hours?We have solved earlier on tasks in which corresponding quantities were proportional. We can now solve the same problems by application of proportions, by the functional method and graphically.
Example 1: A worker earns 15
Baht in 6 hours. How much does he earn in 4 hours?
a) Solution by direct calculation: In 6 hours, the worker earns 15 Baht, that is, in 1 hour 15/6 Baht and in 4 hours 15·4 /6= 10 Baht.
b) Solution by proportion: The worker's wage (y) is proportional to his working hours (x):
y/15 = 4/6, that is y = 4·15/6 = 10.
c) The functional equation
is y = mx. Substitution of x = 6, y =
15 yields 15 = m· 6 or m = 15/6, whence follows the
equation
y = x·15/6; for x = 4, it yields y
= x·15/6 = 10.
d) Graphical solution: The equation y = mx yields a line through the origin. A second point A on it is given by x=6 and y=15. Hence, for x = 4, the ordinate is y = 10. The ordinate for x = 1 is his hourly wage - 2.50 Baht.
All
tasks involving direct proportions can be solved by
a) calculation, b) proportions, c) the functional method and
d) graphically by means of a straight line through the origin.
Example 2: 1.6 kg of a substance cost 4 Baht. What is the price of 2.4 kg?
a) Solution by computation: 0.8 kg cost 2 Baht, 2.4 kg cost three times that much, that is 2·3 = 6 Baht.
b) Solution by proportions: The price (y) is proportional to the amount of substance (x):
y/4 = 2.4/1.6, whence y = 4·2.4/1.6 = 6.
c) Solution by functional method: For x = 1.6 and y =
4 , the equation y = mx yields 4 =
m ·1.6, whence m = 4/1.6, that is y = 4/1.6·x.
Thus for x = 2.4
y = 4·2.4/1.6 = 6.
d) Graphical solution: The point A has the coordinates x = 1.6, y = 4. For x = 2.4, y = 6, whence 2.4 kg cost 6 Baht. The ordinate for x = 1 gives for 1 kg the price 2.50 Baht.
Frequently, direct proportions are confined to a certain range. 1 postcard costs 5 Satang; 6 post cards 25 Satang; 1 pencil costs 20 Satang, 1 dozen 10 Baht; 1 pound of sugar costs 30 Satang, 100 pounds less than 100 times that price. One of the splendid examples with persistent proportionality is the percent calculus.
A proportionality applies frequently only within certain limits.
The validity of a proportional relation frequently extends algebraically over a range larger than the restricted range where the task has a meaning.