Show SOLUTION TO PROBLEM lri ko no 11 21 it if the first cup and cover weigh twice as much as the second cup cups the weight of the second cup added to that of the first cup and cover will equal three times the weight of the second cup consequently the of the second cup Is equal to one third of the weight of 0 the two cups and cover and if the second cup cop and cover weigh three times as much as the first cup the weight of 0 the first cup added to the weight of 0 the second cup and cover will equal tour four times the weight of the first cup consequently the weight of the first cup Is equal to one fourth of the weight of the two cups and cover then reducing these fractions to a common denominator tory torl dorthey they stand thus thua first twelfths second cup tour twelfths twelve twelfths equal the weight of the two cups and cover and after subtracting three twelfths plus tour foar twelfths the remain dlug five twelfths must equal the weight of the cover then if nive five twelfths I 1 equal to ten ounces one twelfth must he be one ona fifth of ten ounces ounce which Is two eunce ounce and abree twelfths must be three times two or six ounces weight of first cup and four foun twelfths must be four times two or eight ounces the v ft eight of second cup w fmc ar c A solution of the same problem hag has also been received from thos C llad Irad haddon don doD of san pete |