(Part A) Machinerys Handbook 31st Edition Pages 1-1484

Machinery's Handbook, 31st Edition

10 RATIO AND PROPORTION Inverse Proportion: Quantities with an inversely proportional relationship behave in such a way that as one increases the other decreases. For example, a factory employing 270 workers completes a given number of automotive components weekly, the number of working hours being 44 per week. If the hours are reduced, then more workers will be required to do the same amount of work. How many employees would be required for the same production if the working hours were reduced to 40 per week? The hours per week is inversely proportional to the number of workers; fewer hours per worker means more workers are required. Letting x be the number of workers needed when time is reduced, the inverse proportion is written: 270 : x :: 40 : 44 Thus 270 x ----- 40 44 = --- and x 270 44 × 40 = ----------- 297 workers = Problems Involving Both Direct and Inverse Proportions: If two groups of data are related by both direct (simple) and inverse proportions among the various quantities, a simple mathematical relation may be used to solve the problem as follows: Example: If a worker capable of turning 65 studs in a 10-hour day is paid $13.50 per hour, how much per hour should a worker be paid who turns 72 studs in a 9-hour day if compensated in the same proportion as the first worker? Solution: The first group of data in this problem consists of the number of hours worked, the hourly wage of the first worker, and the number of studs produced per day; the second group contains similar data for the second worker, except the hourly wage is unknown, so it is indicated by x . The labor cost per stud, as may be seen, is directly proportional to the number of hours worked and the hourly wage. These quantities, therefore, are used in the numerators of the fractions in the formula. The labor cost per stud is inversely proportional to the number of studs produced per day. (The greater the number of studs produced in a given time the less the cost per stud.) The numbers of studs per day, therefore, are placed in the denominators of the fractions in the formula. Thus, 10 ( ) 13.50 ( ) 65 --------------- 9 x 72 = --- x 10 ( ) 13.50 ( ) 72 ( ) 65 ( ) 9 ( ) = --------------------- $16.62 per hour = Percentage.— A percentage is a ratio expressed as a part of 100. For example, if out of 100 manufactured parts, 12 do not pass inspection, then 12 percent (12 of the 100) are rejected. The symbol % indicates percentage. The percent of gain (or loss) with respect to a base (original) quantity is found by divid­ ing the amount of gain (or loss) by the base quantity and multiplying the quotient by 100. For example, if a quantity of steel is bought for $2000 and sold for $2500, the profit is $500/2000 × 100, or 25 percent of the invested amount. Example: Out of a total output of 280 castings a day, 30 castings are, on average, rejected. What is the percentage of bad castings? 30 280 ----- 100 × = 10.71 percent Percent Change: Any increase or decrease in some measured quantity can be expressed as a percent change using the formula: final original amount original percent change − × = 100 . If in the Product of all directly proportional items in first group Product of all inversely proportional items in first group -------------------------------------------------------------------- Product of all directly proportional items in second group Product of all inversely proportional items in second group = -----------------------------------------------------------------------

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