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macjas
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additionally as be see the mixture's overall % is closer to 16% ethanol mixture, this should show you that the 16% ethanol mixture's quantity is much higher than 12% mixture.
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Another way to do it (weighted averages):

30 of 200
12% = 15% 16%
A-------------Mixture--------------B
3 1

So 1 part of B and 3 parts of A make 200
So, A is 25% of the total 200 i.e 50
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To solve this problem using mixtures is also possible;
qty of A/qty of b = % of B (16%) - final %(15) / final %(15) - % of A(12%)

i.e. a/b = 1/4 and we know a+B = 200 hence a= 200/4 = 50
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@macjas thats a quick method as well! nice.
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sorry... was having trouble formatting my post correctly
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Hi,

The Allegation method is very useful in this case.

The percentage of ethanol in two mixtures = 12% & 16%
Final percentage in mixture = 30/200 * 100 = 15%

Following is the diagrammatic representation of the method:
Attachments

Allegations.jpg
Allegations.jpg [ 11.3 KiB | Viewed 13710 times ]

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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We can solve this problem using weighted averages method.

First we need to find 30 is what % of 200 => 15%;

12%(A) ------------ 15%Final Mix ------------------ 16%(B)
3 1

w1/w2 = 1/3 ;

gallons of fuel A added = 1/4(200) =50;
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Let's take Fuel of type A = A litre

Fuel of type B = B litre

Ethanol in type A = 12% => 0.12A
Ethanol in type B = 16% => 0.16B

Now,
A + B = 200
0.12A + 0.16B = 30

Multiplying first with 0.16 and subtracting second

0.16A - 0.12A = 32-30
0.04A = 2
A = 2/0.04=> A = 50

Answer is E
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