What is the “best” way to calculate the average of individual averages?

In preparing for GRE, I see questions looking for the mean of two or more individual means. Here is an example:

The average (arithmetic mean) of 100 measurements is 24, and the average of 50 additional measurements is 28

               Quantity A                          Quantity B
The average of the 150 measurements              26


A) Quantity A is greater.

B) Quantity B is greater.

C) The two quantities are equal.

D) The relationship cannot be determined from the information given.

My question regards the best way to determine the mean (average). I have found two methods that always basically solve it correctly, although the results differ by ~ 0.1.

One method is adding the weighted means: find sum of each sample, add the sums for a total sum, determine the proportion of each individual sum to the total sum to calculate the "weight", then add the products of each individual mean. In other words, Weighted mean = (proportion)(group A mean) + (proportion)(group B mean) + ....

So, to solve the example question: sum of Group A measurements = $100 \cdot 24 = 2400$ sum of Group B measurements = $50 \cdot 28 = 1400$ total sum = 3800 proportion of Group A to total = $2400 \div 3800 = 0.631579$ weighted mean of Group A = $24 \cdot 0.631579 = 15.157896$ proportion of Group B to total = $1400 \div 3800 = 0.368421$ weighted mean of Group B to total = $28 \cdot 0.368421 = 10.315788$ The average (mean) of the 150 measurements = 15.157896 + 10.315788 = 25.473684

The other method is to just divide the total sum by the 150 measurements, which is how a normal mean is calculated, to equal 25.333333.

Although both methods make B) the answer, they could be significantly different in other contexts. It seems to me that the first, weighted, method is better since it probably takes outliers, repeated measurement, etc. into account.

What is the "best" way to calculate the average of individual averages?

Your second method is correct and the mean is $25\frac 13$. The proportion in your first method should be the proportion of the measurements, not the proportion of the sum, so it would be $24\cdot \frac {100}{150}+28\cdot \frac {50}{150}=25\frac 13$, which agrees with the other.
The best way might be this: with $100$ sitting at $24$ and $50$ sitting at $28$, you need the balance point between the two. The balance point is closer to $24$ than to $28$, because there’s more mass sitting there. Therefore, it’s less than $26$.
If you want to be more precise, there’s half as much mass at $28$, so $28$ is twice as far away from the balance. You need the point “one third” of the way from $24$ to $28$, so that’s $24 + \frac43=25\frac13$.