SOLUTION: I tried looking at some examples but i'm still confused. Can someone help please? thank you.
A chemical company makes two brands of antifreeze. The first brand is 20% pure anti
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A chemical company makes two brands of antifreeze. The first brand is 20% pure anti
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Question 1119385: I tried looking at some examples but i'm still confused. Can someone help please? thank you.
A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze and the second brand is 70% pure antifreeze. In order to obtain 180 gallons of a mixture that contains 40% pure antifreeze, how many gallons of each brand of antifreeze must be used?
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I tried looking at some examples but i'm still confused. Can someone help please? thank you.
A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze and the second brand is 70% pure antifreeze. In order to obtain 180 gallons of a mixture that contains 40% pure antifreeze, how many gallons of each brand of antifreeze must be used?
Let amount of 20% antifreeze be T
Then amount of 70% antifreeze to mix = 180 - T (amount of 20%, plus amount of 70% MUST equal 180 gallons)
We then get: .2T + .7(180 - T) = .4(180)
.2T + 126 - .7T = 72
.2T - .7T = 72 - 126
- .5T = - 54
T, or amount of 20% to mix:
Amount of 70% antifreeze to mix:
First, a solution by the traditional algebraic method: write and solve an equation that says the amount of antifreeze in one ingredient, plus the amount in the other ingredient, equals the amount in the mixture.
You want to make 180 gallons of the mixture; so you can use x for the number of gallons of the 20% antifreeze and (180-x) for the number of gallons of the 70% antifreeze. Then
(1) amount of antifreeze in x gallons of 20% antifreeze:
(2) amount of antifreeze in (180-x) gallons of 70% antifreeze:
(3) amount in the mixture; 180 gallons of 40% antifreeze:
The equation is then
The mixture should use x = 108 gallons of the 20% antifreeze and (180-x) = 72 gallons of the 70% antifreeze.
However, for problems like this, where just two ingredients are being mixed, there is another very different method that gets you to the answer with far less effort.
You should know the preceding algebraic method, because you will need it if the mixture problems get more complicated. But for mixture problems with just two ingredients, try using the method below.
(1) Find how far the percentage of the mixture is from the percentage of each ingredient:
70-40 = 30; 40-20 = 20.
(2) Find the ratio of those two differences:
30:20 = 3:2
That ratio is the ratio in which the two ingredients need to be mixed.
(3) Convert the ratio into fractions:
ratio 3:2 --> 3/5 of the mixture is one ingredient; 2/5 is the other
(4) Use logic to determine which ingredient is 3/5 of the mixture and which is 2/5; the larger portion has to be the ingredient with a percentage that is closer to the percentage of the mixture:
40% is closer to 20% than to 70%; so 3/5 of the mixture is the 20% ingredient and 2/5 is the 70% ingredient.
(5) Convert the fractions to numbers of gallons:
20% ingredient: 3/5 of 180 gallons = 108 gallons
70% ingredient: 2/5 of 180 gallons = 72 gallons
All the words of explanation make it seem like a long process. But without all those words, here are the actual calculations that are all that is required:
Let x = "how many gallons of the 70% brand to use", in gallons.
Then the volume of the 20% brand must be (180-x) gallons.
The balance equation for the PURE antifreeze is THIS
0.2*(180-x) + 0.7*x = 0.4*180. <<<---=== It is the key equation
Simplify and solve for x:
36 - 0.2x + 0.7x = 72
0.5x = 72 - 36 = 36.
x = = 72.
Answer. 72 gallons of 70% brand and (180-72) = 108 gallons of 20% brand.
Check. 0.7*72 + 0.2*108 = 72 gallons of pure antifreeze.
0.4*180 = 72 gallons of pure antifreeze. ! Correct !
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