A factory makes an oil mixture by mixing oils of grades A and B as follows. For every 2.3 liters of grade A oil, 1.2 liters of g
rade B oil are mixed in. How many liters of grade A oil and grade B oil will the factory need to make 1000 liters of their oil mixture? Explain your solution.
The ratio is 2.3:1.2 for A:B. Ratios will remain the same but can be scaled up or down by some factor/multiple. Like if you had a food recipe that you needed to double, the scaling factor would be 2 and all ingredients in the recipe would be doubled.
Since we don't know the scaling factor, we assign that multiple "x". Then the amounts of each ingredient oil would be:
amount of A = 2.3x amount of B = 1.2x
add them together to get the required mixture amount.
1000 = 2.3x + 1.2x 1000 = 3.5x 1000/3.5 = x 285.7 = x this is the scaling factor.
<span>To produce 1000L of mixture the factory will need 657 liters of grade A and 343 liters of grade B. To determine this you have to figure out the percentage of each grade in the mixture. The ratio is 2.3 liters to 1.2 liters. Therefor in this scenario the unit equaling 100% is 3.5 liters. To find the percentage of grade A you divide the amount used by the total amount of the unit 100%:
2.3 divided by 3.5 = .657 (multiply the answer by 100 to get 65.7%).
To find the percentage of grade B you divide the amount used by the total amount of the unit 100%:
1.2 divided by 3.5 = .343 (multiply the answer by 100 to get 34.3%)
To test your answer make sure both percentages add up to 100%:
65.75 plus 34.35 = 100%
To determine how much of grade A and grade B is needed for a set amount of liters you multiply the percentage by the liters needed.
For this situation you multiply 65.7% (when multiplying percentages you need to multiply in decimal form).
For grade A you multiply .657 (65.7%) by 1000 liters = 657 liters
For grade B you multiply .343 (34.3%) by 1000 liters = 343 liters
To test your answer you can use the same addition as you did to test the percentages:
657 liters plus 343 liters = 1000 liters</span>
