142 



THE POPULAR EDUCATOR. 



extent of each of the triangles represented by that of the plan 

 ABC, there'iore we know the length of the edges of the inclined 

 triangles, of which A B D is the plan of one, B D c of the second, 

 and A D C of the third. Of course the vertex of the pyramid 

 will be perpendicularly above its plan in the centre D, therefore 

 we must rabat the perpendicular, that is, turn it down upon the 

 paper, and thus form the right angle B D E. From B with the 

 distance B A or B C cut the perpendicular D E in E, join B E, 

 which will represent the rabatted and inclined edge of the py- 

 ramid, whilst D E will represent the height of the pyramid. We 

 may, perhaps, make it clearer in this way : that as the line 

 B D must be the plan of an inclined edge of the triangle A B D, of 

 which B D is the plan, and because B E, the rabatted edge, is 

 equal to B A, and D E perpendicular to D B, therefore D E must 

 be equal to the height of E, the vertex from the ground. To 

 represent the elevation draw B B', A A', and c c', at right angles 

 with x y (the axis of the plane of projection), produce D D' to 

 any length and make D' E' equal to D E ; draw from E' lines 

 to B', A', and c', which will represent the vertical projection or 

 elevation of the pyramid. To draw the plan, and ascertain the 

 height of the pyramid by the rabatment of the right-angled tri- 

 angle B r> E, will be all that is necessary to prepare the subject 

 for the perspective representation. We have added the ortho- 

 graphic elevation, trusting it may assist the pupil to understand 

 that the height is not equal to one of the edges. 



To proceed with the perspective elevation, draw the plan as in 

 Pig. 42, find its height by Fig. 43, and set off that height from 

 I to L (Fig. 42). For the rest proceed as in Fig. 42. We will 

 give another question similar in character to the last problem, 

 for the pupil to work out by himself, without any accompanying 

 explanation except the figure. 



PROBLEM XXIII. (Fig. 44). Give a perspective view of a regular 

 pyramid on an hexagonal base, the height of tlie pyramid being 

 equal to three times the length of one of the edges of its base. 

 Assume that it is seen from a point to the right of it, and at a 

 height above the horizontal plane equal to * tlie lieight of tlie 

 pyramid. 



We will merely add that as no definite scale is given with the 

 above problem, the pupil can please himself as to the size, only he 

 must take care to observe the proportions mentioned. The ex- 

 pression " the horizontal plane " means the ground upon which 

 it stands. The question is taken from one of the Military 

 Examination Papers. 



LESSONS IN ARITHMETIC. XL. 



AVERAGES, ETC. 



3. THE Average, or Mean, of any set of numbers is the number 

 obtained by dividing their sum by the number of different quan- 

 tities forming the set. It is that number which, if placed in 

 the position of each of the quantities forming the set, will, 

 when added together, give the same result as the original quan- 

 tities, when treated in the same manner. 



EXAMPLE 1. A man spends, in seven successive years, the 

 following sums : .200, .250, .300, .320, .180, .330, 210. 

 What is his average or mean annual expenditure during that 

 time ? 



. 200 + 250 + 300 + 320 + 180 + 330 + 210 

 The answer is = , 



which is 255-?-. 



If he spent <255| every year, ho would in seven years have 

 spent the same sum as he actually spent in that time. 



EXAMPLE 2. Out of 20 men, 6 die at 25 years of age, 3 at 

 30, 4 at 35, and 7 at 40. What is the average duration of 

 their lives ? 



The total number of years they live is 



(6 x 25) + (3 x 30) + (4 x 35) + (7 x 40), 



which is 660 years. 



Hence the average required is - n .-, or 33 years. Answer. 

 4. EXAMPLE. How may a mixture of tea at 3s. 3d. a pound 

 and tea at 4s. 9d. a pound be made so as to produce a mixture 

 worth 4s. 6d. a pound ? 



We shall call 4s. 6d. the average price. 

 The tea at 3s. 3d. a pound is Is. 3d. a pound cheaper than the 



mixture ; 

 The tea at 4s. 9d. a pound is 3d. a pouad dearer than the mixture. 



It is evident that the amount by which the cost of the number of 

 pounds of the first kind is less than the cost of the same num- 



ber at the average price, must be equal to the amount by which 

 the cost of the number of pounds of the second kind exceeds 

 the cost of the same number at the average price. Hence, 



Number of first kind x 15d. = number of second kind x 3d. 

 Hence the required number of pounds are in the ratio of 3 : 15, or 

 1 : 5, i.e., inversely, as the difference of the prices from the ave- 

 rage price. 



Hence, any pairs of numbers of pounds in this ratio will form a 

 mixture of the required kind. 



5. When there are more than two quantities at different prices, 

 which are to be mixed ? 



This kind of question can be solved by the same principles. 



Separating the quantities to be mixed into two sets, one 

 cheaper and the other dearer than the average price, we must 

 evidently have the sum of tho quantities which are cheaper 

 than the average, multiplied each by tho amount by which its 

 price is less than the average price, equal to the sum of the 

 quantities which are dearer, multiplied each by the amount by 

 which its price is greater than the average price. 



There is one difference, however, in this case and that of only 

 two ingredients. In the latter, the ingredients must be mixed 

 in one given ratio ; in the former, an infinite number of numbers 

 can be found, so that ingredients mixed proportionately to them 

 will satisfy the required condition. 



6. EXAMPLE. How may teas at 3s. 6d., 4s., and 4s. 6d. a 

 pound respectively, be mixed so as to form a mixture worth 

 4s. 2d. a pound ? 



Here the first two are less than the average price, and the last is 



greater ; and the differences are 8d., 2d., and 4d. respectively. 

 Therefore, 



No. of Ibs. of first x 8d. + no. of Ibs. of second x 2d. = no. of Ibs. 

 of third x 4d. 



Take, then, any number of pounds of the first and any num- 

 ber of pounds of the second, at pleasure. We can then deter- 

 mine what number of pounds of the third must be mixed with 

 these, so as to satisfy the required condition. As the simplest 

 case, suppose we take 1 pound of each of the first two kinds. 

 Then we have 



Number of pounds of the third x 4d. = 8d. + 2d. = lOd. ; or, 



number of pounds of the third = V = 2-J-. 

 Now 1 : 1 : f are in the proportion of 2:2:5. 



Heuce 2 pounds of the first, 2 pounds of the second, and 5 pounds of 

 the third will form a mixture worth 4s. 2d. a pound. 



7. EXAMPLE. How may wines at 15s., 18s. 6d., 20s., 26s., 

 and 30s. a gallon respectively, be mixed so as to form a mixture 

 worth 24s. a gallon ? 



The first three are in defect, the differences being 9s., 5s. 6d., and 4s, 

 respectively ; and the last two are in excess, the differences being 

 2s. and Cs. 



Taking 1 gal. of each of the first four kinds, we must have 

 9 + 54 + 4 = 2 + no. of gals, of last x C. 



jjence will give the number of gallons of the last kind, 



which, mixed with one gallon of each of the others, will produce 

 a mixture worth 24s. a gallon. 



This reduces to V : 



Honce 4 gallons of each of the first four kinds mixed with 11 gallons 

 of the last will fulfil the required condition. 



8. These examples will sufficiently explain tho following 

 Rule for determining a proportion of Ingredients at Different 



Prices, to form a Mixture at a Given Price. 



Divide the differences from the average price into two sets 

 those which are in excess, and those which are in defect. 



(1.) When there are only 3 ingredients. 



Add together the two differences which are of the same kind, 

 and divide this sum by the third difference. This will give tho 

 quantity of the last, which must be mixed with one unit of each 

 of the former. 



(2.) When there are more than 3 ingredients. 



Add one set together, and subtract from the result the sum of 

 the other set with one difference omitted. This result, divided 

 by the omitted difference, will give the quantity of this last, 

 which must be mixed with one unit of each of tho former. 



EXERCISE 61. 



1. Find the average of 3, 4, 5, 6, and 7. 



2. Find the average of 25, 34, 19, 0, 12, and 5. 



3. Find the average of 11|, 4J-, 93jj, 0, 3'625, and 41. 



4. A master pays his labourers as follows : 20 receive 7s. a weelt, 



