which the proportions of the different metals used are indicated by per cents. For example, brass usually contains about 65% copper and 35% zinc. Then, in 100 lb. of brass, there would be 65 lb. of copper and 35 lb. of zinc. Suppose, however, that instead of 100 lb. we wanted to mix a smaller amount, say 8 lb. The amount of copper needed would be 65% or .65 of 8 lb. .65x8 5.20 lb., or 5 lb., the copper needed. lb., the zinc needed. Sometimes, in dealing with very small per cents, we see a decimal per cent such as found in the specifications for boiler steel, where it is stated that the sulphur in the steel shall not exceed .04%. Now this is not 4%; neither is it .04; but it is .04%, meaning four one-hundredths per cent, or four onehundredths of one one-hundredth. This is 100 of 100=10000, 80 if we write this .04% as a decimal, it will be .0004. It is a very common mistake to misunderstand these decimal per cents, and the student should be very careful in reading them. Likewise, be careful in changing a decimal into per cent that the decimal point is shifted two places to the right. 92 100 46. Efficiencies.-Another common use of percentage is in stating the efficiencies of engines or machinery. The efficiency of a machine is that part of the power supplied to it, that the machine delivers up. This is generally stated in per cent, meaning so many out of each hundred units. If it requires 100 horse-power to drive a dynamo and the dynamo only generates 92 horse-power of electricity, then the efficiency of the dynamo is 1% or 92%. If the engine driving a machine shop delivers 250 horse-power to the lineshaft, but the lineshaft only delivers 200 horse-power to the machines, then the efficiency of the lineshaft is 280.80 =80%. The other 50 horse-power, or 20%, is lost in the friction of the shaft in its bearings and in the slipping of the belts. The efficiencies of all machinery should be kept as high as possible because the difference between 100% and the efficiency means money lost. The large amount of power that 'is often lost in line shafting can be readily appreciated when we try to turn a shaft by hand and try to imagine the power that would be required to turn it two or three hundred times a minute. 47. Discount.-In selling bolts, screws, rivets, and a great many other similar articles, the manufacturers have a standard list of prices for the different sizes and lengths and they give their customers discounts from these list prices. These discounts or reductions in price are always given in per cent. Sometimes they are very complicated, containing several per cents to be deducted one after another. Each discount, in such a case, is figured on the basis of what is left after the preceding per cents have been deducted. Example: The list price of in. by 1 in. stove bolts is $1 per hundred. If a firm gets a quotation of 75, 10 and 10% discount from list price, what would they pay for the bolts per hundred? Explanation: 75, 10 and 10% discount means 75% deducted from the list price, then 10% deducted from that remainder, then 10% taken from the second remainder. Starting with 100 cents, the list price, we deduct the first discount of. 75%. This leaves 25 cents. The next discount of 10% means 10% off from this balance. Deducting this leaves 22 cents. Next, we take 10% from this, leaving 20 cents per hundred as the actual cost of these stove bolts. 48. Classes of Problems.-Nearly all problems in percentage can be divided into three classes on the same basis as explained in Article 26. There are three items in almost any percentage problem: namely, the whole, the part, and the per cent. For example, suppose we have a question like this: "If 35% of the belts in a shop are worn out and need replacing, and there are 220 belts altogether, how many belts are worn out?" In this case, the whole is the number of belts in the shop, 220. The part is the number of belts to be replaced, which is the number to be calculated. The per cent is given as 35%. Any two of these items may be given and we can calculate the missing one. We thus have the three cases: 1. Given the whole and the per cent, to find the part. 2. Given the part and the per cent that it is of the whole, to find the whole. 3. Given the whole and the part, to find what per cent the part is of the whole. The principles taught under common fractions will apply equally well in working problems under these cases, the only difference being that here a per cent is used instead of a common fraction. In working problems, the per cent should always be changed to a decimal. One difficulty in working percentage problems is in deciding just what number is the whole (or the base, as it is often called). The following illustration shows the importance of this. If I offer a man $2000 for his house, but he holds out for $3000, then his price is 50% greater than my offer, while my offer is 33% less than his price. The difference is $1000 either way but, if we take my offer as the base, it would be necessary for me to raise it, or 50%, to meet his price. On the other hand, for him to meet my bid, he would only have to cut his price, or 33%. Examples of the three types of problems, before mentioned, may help somewhat in getting an understanding of the processes to be used. Example of Case 1: How many pounds of nickel are there in 1 ton of nickel-steel containing 2.85% nickel? 1 ton 2000 lb. 2.85%= .0285 .0285X2000=57 lb., Answer. Example of Case 2: Explanation: Here we have the whole (2000 lb.) and the per cent (2.85%) to find the part. After changing the 2.85% to a decimal fraction, the problem becomes a simple problem in multiplication of decimals. The machines in a small pattern shop require altogether 12 horse-power and are to be driven from a lineshaft by a single electric motor. If we assume that 20% of the power of the motor will be lost in the line shaft and belting, what size motor must we install? 100%-20%-80% Example of Case 3: Explanation: Here the per cent given (20%) is based on the horse-power of the motor, which is, as yet, unknown. The horse-power of the motor is 100% of itself and, if 20% is lost, then the machines will receive 80%, or .8 of the power of the motor. This is 12 horsepower. The whole will be 12÷.8=15 horsepower. This is the size of motor to install. If the force in a shop is increased from 160 to 200 men, what per cent is the capacity of the shop increased? 200÷160-1.25 1.25 125% Explanation: The present force is 1 or 1.25 of the old force, because 200 is 288 of 125%-100%=25%, Answer. 160, and 288=14. This is the same as 125%. The increase is, therefore, 25% of the former force. Note. To reduce the present force back to the old number would require a reduction of only 20%, because now the base is different on which to figure the per cent. 40÷200 .20, or 20%. = PROBLEMS 91. Write 4% as a decimal and as a common fraction. 92. Write 25% as a common fraction and reduce the fraction to its lowest terms. 94. If there are 240 men working in a shop and 30% of them are laid off. how many men will be laid off and how many will remain at work? 95. Out of one lot of 342 brass castings, 21 were spoiled and out of another lot of 547, 32 were spoiled. Which lot had the larger per cent of spoiled castings? 96. 500 lb. of bronze bearings are to be made; the mixture is 77% copper, 8% tin, and 15% lead. How many pounds of copper, tin, and lead are required? Note. This is a standard bearing mixture used by the Pa. R. R. and by some steam turbine manufacturers. 97. The boss pattern maker is given a raise of 25% on Christmas, after which he finds that he is receiving $130 a month. How much did he get per month before Christmas? 98. In testing a shop drive it was found that the machines driven by one motor required horse-power as follows: 60 in. mill... 3.31 horse-power .75 horse-power 2.42 horse-power 48 in. lathe. 42 in. by 42 in. by 12 ft. planer. 4.82 horse-power The total power delivered by the motor was 13.65 horse-power. What per cent of the total power was used in belting and lineshaft? What per cent by the machines? 99. A man who has been drawing $2.50 a day gets his pay cut 10% on May 1, and the following September he is given an increase of 10% of his rate at this time. How much will he get per day after September? 100. The following weights of metals are melted to make up a solder: 18 lb. of tin, 75 lb. of bismuth, 37.5 lb. of lead, and 19.5 lb. of cadmium. What per cent of the total weight is there of each metal? CHAPTER VII CIRCUMFERENCES OF CIRCLES; CUTTING AND GRINDING SPEEDS 49. Shop Uses.-In the running of almost any machine, judgment must be used in order to determine the speed which will give the best results. Lathes, milling machines, boring mills, etc., are provided with means for changing the speed, according to the judgment of the operator. Emery wheels and grindstones, however, are often set up and run at any speed which the pulleys happen to give, regardless of the diameter. If an emery wheel of large size is put on a spindle that has been belted to drive a smaller wheel, the speed may be too great for the larger wheel and, if the difference is considerable, the large wheel may fly to pieces. Every mechanic should know how to calculate the proper sizes of pulleys to use for emery wheels or grindstones, the correct speed at which to run the work in his lathe, or the most economical speeds to use for belts and pulleys. A little data on this subject may be useful and will afford applications for arithmetical principles. 50. Circles. To understand what has just been mentioned, it is necessary to get a knowledge of circles and their properties. The distance across a circle, measured straight through the center, is called the Diameter. Circles are generally designated by their diameters. Thus a 6 in. circle means a circle 6 in. in diameter. Sometimes the radius is used. The Radius is the distance from the center to the edge or circumference and is, therefore, just half the diameter. If a circle is designated by the radius, we should be careful to say so. Thus, there would be no misunderstanding if we said "a circle of 5-in. radius"; but unless the word radius is used, we always understand that the measurement given is the diameter. The Circumference is the name given to the distance around the circle, as indicated in Fig. 8. The circumference of any circle is always 3.1416 times the diameter. In other words, if we measure the diameter with a string and lay this off around the circle, it will go a little over three times. |