.Mzcw.NzIxMg

(27) to each other as 2000x12, 660x7 & 1255x5, or as, 24000, 4620, & 6275.

This problem may be solved like the 2d question §82, or by several rules of three. For the sum of the numbers 24000, 4620, & 6275, or 34895 34895:1560::2400:x = the share of the first. Again, 34895:1560::4620:y = share of the second. And lastly, 34895:1560::6275:z = share of the third = 1560 - x - y.

87. Hence questions relating to the rules of underline fellowship /underline in arithmetic may be solved.

88. VI. Suppose two sorts of silver, the first at 60d. oz. & the second at 68d. oz: How many ounces of each sort might to be taken to make up a mixed mass of 50. oz. at 65d. an ounce? Let x = number of ounces of the first sort y = of the Second sort. Then x + y = 50. And y = 50 - x.