a feed store sells two varieties of birdseed : brand a , which is 40 % millet and 60 % sunflower , and brand b , which is 65 % millet and 35 % sunflower . if a customer purchases a mix of the two types of birdseed that is 50 % sunflower , what percent of the mix is brand a ?
"yes there is a simple method : consider the following method brand a : 40 % millet and 60 % sunflower brand b : 65 % millet and 35 % sunflower mix : 50 % sunflower here the weighted average is 50 % , now brand a has 60 % sunflower , which is 10 % more than the weighted average of mix = + 0.10 a - - - - - - - - - - - - - - - i similarly , brand b has 35 % sunflower , which is 15 % less than the weighted average of mix = - 0.15 b - - - - - - - - - - - - ii now , both brand a and brand b are combined to give a 50 % mix containing millet , so equate i and ii implies , 0.10 a = 0.15 b therefore a / b = 0.15 / 0.10 = 3 / 2 a : b : ( a + b ) = 3 : 2 : ( 3 + 2 ) = 3 : 2 : 5 we have to find , percent of the mix is brand a i . e . a : ( a + b ) = 3 : 5 = ( 3 / 5 ) * 100 = 60 % here is a pictorial representation : brand a = 60 % - - - - - - - - - - - - - - - - - - - - - - - - 10 % or 0.10 above average , a times - - - - - - - - - - - - - - - - - total below = + 0.10 a - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - average = 50 % or 0.50 brand b = 35 % - - - - - - - - - - - - - - - - - - - - - - - - - - 15 % or 0.15 below average , b times - - - - - - - - - - - - - - - - - total above = - 0.15 b since the amount below the average has to equal the average above the average ; therefore , 0.10 a = 0.15 b a / b = 3 / 2 a : b : total = 3 : 2 : 5 therefore a / total = 3 : 5 = 60 % answer : e"
multiply(divide(const_3, add(const_3, const_2)), const_100)