Question 1006346
you are correct in that both bags will contain 96 batteries.
there will be 48 D and 48 C.


you are given that the first bag contains 60 batteries total and that the ratio of D to C is 3/7


let x = the common multiplier.


you get 3x + 7x = 60


combine like terms to get 10x = 60


solve for x to get x = 6


the common multiplier is 6.


3*6 + 7*6 = 18 + 42 = 60


the total number of batteries is 60.


the ratio of D to C is 18 / 42 = 3 / 7.


the batteries in the second bag are in the ratio of D / C = 5 / 1


let y be the common multiplier for the second bag.


you get 5y + 1y = z


z is the total for the second bag.


when you add the D batteries together and you add the C batteries together, you will get a ratio of 1/1.


you get:


(18 + 5y) / (42 + y) = 1/1 which becomes:


(18 + 5y) / (42 + y) = 1


multiply both sides of this equation by (42 + y) to get


18 + 5y = 42 + y


subtract y from both sides of this equation and subtract 18 from both sides of this equation to get:


5y - y = 42 - 18


simplify to get:


4y = 24


solve for y = get y = 6


when y = 6, the equation for your second bag of 5y + y = z becomes 30 + 6 = z which results in 36 = z


your second bag contains 36 batteries.
30 of them are D
6 of them are C.
the ratio of D/C = 30/6 = 5/1.


your combined bag contains 60 + 36 = 96 batteries.
your total D are 18 + 30 = 48
your total C are 42 + 6 = 48
the ratio of D to C is 1/1.