# Lesson: Weighted Averages

## Comment on Weighted Averages

### Hello Brent,

Hello Brent,
Regarding the question https://gre.myprepclub.com/forum/list-x-and-list-y-each-contain-60-numbers-frequency-distrib-1642.html , how can we calculate total average of z as average of (X+Y)/2. Though when I solve the sum with the weighted average formula taught by you I get the same solution. Was this a mere coincidence.

Not a coincidence. The weighted average formula works perfectly for this question. Nice work!

I went about this a different way. Since we know there cannot be equal number of adult and children (group of 11), I selected to calculate the cost of 6 kids plus the 5 adults. And it worked out exactly. But if I chose to look at 6 adults the total cost was greater than \$5.90. Therefore, there has to be more children.

### NOTE: In the future, please

NOTE: In the future, please post the link of the question. Otherwise, I don't know which question you're referring to.

Great approach!!

Cheers,
Brent

### Hey Brent, you really changed

Hey Brent, you really changed my perception about the GRE exams with the way you teach and answer questions. You are indeed a master of your craft. Thank you.

### That's very nice of you to

That's very nice of you to say. Thanks!