INTERPRETATION OF RESULTS 430. When the roots obtained by solving an equation satisfy the equation, the solution has been properly performed; but the results found in solving a problem may sometimes be at variance with some condition of the problem. Consequently, the interpretation of results becomes important. POSITIVE RESULTS 431. Since the numbers sought in the solution of a problem are arithmetical rather than algebraic, when positive results are obtained, it is not likely that they will conflict with the conditions of the problem. Sometimes, however, even a positive result violates one or more of the conditions of a problem. In such cases the problem is impossible. 432. 1. A club consisting of 25 members raised the sum of $13 by assessing the men 80 cents each and the women 40 cents each. How many men were there, and how many women? Though the numbers found will satisfy the equation, yet since the number of men and the number of women cannot be fractional, the problem is impossible. 2. The second digit of a number expressed by two digits is twice the first, and 4 times the first digit is 9 greater than the second. What is the number? While these numbers satisfy the equation, they fail to satisfy the implied condition that the digits must be integers. Hence, the problem is impossible. NEGATIVE RESULTS 433. A few examples will suggest the methods to be employed in the interpretation of negative results. 1. If A is 40 years old and B is 30, in how many years will A be twice as old as B? - Though the result is algebraically correct, inasmuch as 20 substituted for a satisfies the equation, nevertheless it is arithmetically absurd. Hence, the conditions of the problem are inconsistent with each other. Had the result been +20, the statement that A would be twice as old as B in 20 years would have been arithmetically reasonable. However, since +20, the equation will give a result arithmetically reasonable, if is substituted for x; that is, if x is taken to represent the number of years since A was twice as old as B. x = -x The conditions of the problem should, therefore, be modified as follows: If A is 40 years old and B is 30, how many years ago was A twice as old as B? 2. How much money has A, if of his money is 5 dollars more than of it? position, made in the problem, that A has some money. The problem when modified to express conditions arithmetically reasonable will be: How much money has A, if of it is 5 dollars less than of it; or, if 60 dollars is interpreted 60 dollars in debt: How much money does A owe, if of what he owes is 5 dollars more than of it? 434. From the above discussions we may infer: 1. A negative result indicates that some quantity in the problem has been applied in the wrong direction. 2. A possible problem analogous to the given problem may be found by changing the absurd conditions to their opposites. PROBLEMS 435. Interpret arithmetically the negative results obtained by solving the following: 1. If A is 40 years old and B is 25, in how many years will B be half as old as A? 2. Find the numbers whose sum is 6 and difference 10. 3. What fraction is equal to 3, if 1 is added to its numerator or to if 1 is added to its denominator? 4. A boy bought some apples for 24 cents. Had he received 4 more for that sum, the cost of each would have been 1 cent less. How many did he buy? 5. A man worked 7 days, during which he had his son with him 3 days, and received 22 shillings. He afterwards worked 5 days, during which he had his son with him 1 day, and received 18 shillings. What were the daily wages of each? ZERO RESULTS 436. When the result obtained by solving a problem is zero, it may sometimes indicate that the problem is impossible, and sometimes it may be the proper answer to the question. 1. A dealer had two kinds of tea worth 75 and 60 cents per pound, respectively. How many pounds of each must he take to make a mixture of 45 pounds worth $ 27? This result means that no such mixture can be made. In fact, 45 pounds of the poorer tea is worth $27. 2. A is 48 years old, and B is 16 years old. years will A be 3 times as old as B? After how many This result indicates that A is now 3 times as old as B. 3. What number is equal to the square of itself? These results indicate that no number except 1 is equal to the square of itself. INDETERMINATE RESULTS 437. 1. A lady being asked her age replied, "If from 3 times my age you take 4 years and divide the difference by 2, you will have twice my age less half of my age 4 years hence." What was her age? Since (2) may be reduced to the identity 3x-4=3x-4, it may be satisfied by any value of x whatever. This relation, § 427, is expressed by (4). Hence, the problem is indeterminate. PROBLEMS 438. 1. If twice a certain number is subtracted from the square of the number, the result will be 1 less than the square of a number 1 less. What is the number? 2. A father is 30 years older than his son, and the sum of their ages is 30 years less than twice the father's age. What is the son's age? 3. The sum of the first and third of three consecutive integers is equal to twice the second. What are the integers? 4. A bought 400 sheep in two flocks, paying $1.50 per head for the first flock and $2 per head for the second. He lost 30 of the first flock and 56 of the second, but by selling the rest of the first flock at $2 per head and the rest of the second at $2.50 per head, he neither lost nor gained. How many sheep were there in each flock originally? 5. A and B receive the same monthly salary. A is employed 10 months in the year and his annual expenses are $600. B is employed 8 months in the year and his annual expenses are $ 480. If A saves as much money in 4 years as B saves in 5 years, what is the monthly salary of each? |