The nutritional content per ounce of three foods is presente
Solution
Let the quantity of the foods A, B, and C be x, y, and z oz respectively.
Then 100x+200y+400z = 1600, or, on dividing both the sides by 100, we have x+2y +4z = 16...(1)
8x+12y+13z = 78 ...(2) and
350x +750y + 250z = 3450 or, on dividing both the sides by 50, we have 7x+15y+ 5z = 69...(3)
On multiplying both the sides of the 1st equation by 7 and 8 respectively, we get 7x+14y +28z = 112...(4) and
8x+16y+32z = 128...(5)
Now, on subtracting the 3rd equation from the 4th equation, we get -y+23z = 43...(6)
Similarly, on subtracting the 2nd equation from the 5th equation, we get 4y+19z = 50...(7)
On multiplying both the sides of the 6th equation by 4, we get -4y+92z = 172...(8)
Now, on adding the 7th and the 8th equations, we get 111z = 222 so that z = 2.
Then on substituting z = 2 in the 6th equation, we get -y+46 = 43 so that y = 46-43 or, y = 3.
Finally, on substituting y = 3 and z = 2 in the 1st equation, we get x+6+8 = 16 or, x +14 = 16 so that x = 16-14 or, x = 2.
Thus the answer is x = 2,y = 3 and z = 2.

