seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 31 % ryegrass , what percent of the weight of the mixture is from mixture x ?
"31 % is 6 % - points above 25 % and 9 % - points below 40 % . thus the ratio of mixture y to mixture x is 3 : 2 . the percent of mixture x is 2 / 5 = 40 % . the answer is c ."
divide(subtract(31, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100)))