x +37=56 x=56-37 858. While these exercises are so simple that they can be worked without a pencil, they should be used to show the steps generally taken in more complicated equations. In 1, for instance, the work should take the form here indicated, only a single step being taken at a time. In 19, the first step is to clear the equation of fractions by multiplying by 6; the second step is to transpose the unknown. quantities to the left side of the 2x-6=16+ X X= 19 2 3 equation, and the known quantities 12x3696+3x-2x to the right; the third step is to com 12x-3x+2x=96+36 bine the unknown quantities into 11x=132 X= 12 one, and to make a similar combination of the known quantities; the last step is to find the value of x. After a little more familiarity with exercises of this kind, the pupil can take short cuts with less danger of mistakes; for the present, however, it will be safer to proceed in the slower way. 859. 5. x+(x+75)+x+(x+75) = 250. NOTE. The parentheses used here are unnecessary. They are employed merely to show that x + 75 is one side of the field. 13. Let x= price of a horse; 4x= cost of four horses; 3x X 4x+3x-240 = 635, 7x=635+ 240=875, x=125, price, in dollars, of a horse; - 80 = 45, price, in dollars, of a cow. Other pupils may solve the problems in this way: pieces; 10x = value of dimes (in cents); 5x+55 = value of five-cent pieces. 10x+5x+55= 100. x+x+ 2400+ x + 2400 + 2400 = 18000. 18. Let x = less; x+33 = greater. x+33-3x=11. Bringing known quantities to the left side of the equation, and the unknown quantities to the right, This problem may also be worked in this way: xless; 3x+11= greater. 19. x = number of 5-cent stamps; x+15= number of 2-cent stamps; x+30= number of postal cards. 5x+2x+30 + x + 30 = 100. 20. x = number of horses; x+17= number of Cows; 2x+39 = number of sheep. x+x+17+2x+39 88. XIV NOTES ON CHAPTER ELEVEN With this chapter begins the regular work in percentage, and it is important that the pupils obtain, as soon as possible, a correct idea of what is meant by the term per cent. Many of the various subdivisions of this topic found in some books, are taken up only incidentally, while others are omitted altogether, the aim being to give the pupils a foundation upon which they can subsequently build, rather than to scatter their energies over too great a diversity of subjects. 860. The reduction of a common fraction to a per cent, consists in changing the former to a decimal of two places. In reducing to a decimal, the result is .5, or 5 tenths; in changing to an equivalent per cent, the result is 50 per cent, or 50 hundredths. In reducing to a decimal, the answer is given in three places, .125, or 125 thousandths; in changing it to a per cent, the division is stopped at the second place, and the remainder written as a fraction, 12 per cent, or 12 hundredths. The denominator of a per cent being always the same, 100, the comparative value of several per cents is known at sight. To compare and as common fractions, they must be changed to 25 and 24; if a further comparison is to be made between these and a new common denominator must be employed, and the fractions reduced to 5%, 12%, and 7%. Changing the fractions to decimals, 625 thousandths, 6 tenths, and 58 hundredths, simplifies the comparison; but it is still easier to determine their relative value when they are expressed as 62 per cent, 60 per cent, and 58 per cent. 75 70 The teacher must not be discouraged if the pupil fails to grasp at once the full meaning of percentage. Definitions will not help materially; much practice in working examples is necessary to give the knowledge desired. 863. Many children find it difficult to distinguish between 1% and 50%. If the former is read in the business way, of one per cent, it may make the distinction plainer. 864. Per cents being generally given in two figures, scholars hesitate to give the correct answers: 300%, 250%, 125%, 16331%, 420%, 910%. 865. While pupils will find 33% of 81 cows, by dividing 81 by 3, they should understand that they are really multiplying 81 by 33 hundredths, or 81 by . In 6, 6% of 150, or 18 of 150, may be obtained by multiplying 150 by 6 and cutting off two ciphers; or by dividing 150 by 100, obtaining 14, and multiplying this quotient by 6; or by reducing 6% to, and finding of 150. In 9, the pupil should find 1% of $ 640 and take onehalf of the result. The scholars should be permitted to use their own method of solving these problems, the different analyses given by the pupils furnishing their class-mates an opportunity to select a simpler method. 866. Although every pupil may not be able to determine at once the shortest way of calculating a given example, no one should be allowed to work 3, by multiplying by 33. When the multiplication by has been performed, the answer has been obtained, except as to the location of the decimal point, and the waste of time in multiplying by 3, repeating this product, and adding three columns should not be tolerated. No fault should be found with the average pupil for failing to recognize in 1, that 6% is; or that in 12, 31% is. The general method should be to multiply by the figures given to represent the per |