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## Comment on

Weighted Averages## Hi Brent,

I will suggest to move below question to geometry section.

I tried to solve this question by = volume of rectangular area/surface area of cone

But I got wrong ans. please let me know whats wrong with my approach.

https://gmatclub.com/forum/a-rectangular-box-with-dimensions-of-12-inches-by-18-inches-by-10-inc-240401.html

## Done!

Done!

Thanks for the heads up!

Did you say cone? If so, then that's the first problem :-)

If you meant cylinder, then I can see what happened.

Some students will find the volume of the box and the volume of one soup can, and then calculate how many soup can volumes will divide into the volume of the box.

The problem with this approach is that the soup cans are rigid structures, so their shapes remain the same.

Here's what I mean.

Let's say a soup can has a height of 20 inches and a radius of 1 inch.

Also, we have a box with dimensions 10x10x10

If you calculate the volume of the box and the volume of a soup can, you'll find that the volume of the soup can is about 63 cubic inches and the volume of the box is 1000 cubic inches. HOWEVER, we cannot fit any cans into the box, because each can is 20 inches tall, so the box is too small.

Does that help?

Cheers,

Brent

## Hi Brent,

Can we say that the weighted average will always lean towards the average of the group that has the larger number irrespective of the individual averages of the either group? Please could you explain logically why this is happening? Why the combined weighted average leans towards the average of 20 for 3 people rather than the average of 100 for a group of 2 people?

Thanks & Regards,

Abhirup

## Yes, the weighted average

Yes, the weighted average will always lean towards the average of the group that has the larger number, irrespective of the individual averages of the either group.

So, for example, if 3 dogs have an average weight of 20 pounds, 2 dogs have an average weight of 100 pounds, then the average weight of all 5 dogs combined will be closer to 20 pounds than it is to 100 pounds.

To see why this is, first notice that, if the 3 dogs have weights 10, 20 and 30, the average weight is 20 pounds.

Likewise, if the 3 dogs have weights 20, 20 and 20, the average weight is also 20 pounds.

So, let's say the first 3 dogs have weights 20, 20 and 20 (for an average of 20 pounds), and the last 2 dogs have weights of 100 and 100 (for an average of 100 pounds).

When we combine ONE dog from each group, we have one 20-pound dog, and one 100-pound dog, for an average weight of 60 pounds.

When we combine TWO dogs from each group, we have two 20-pound dogs, and two 100-pound dogs, for an average weight of 60 pounds.

From here, we've accounted for 4 dogs (2 from each group) for an average weight of 60 pounds.

At this point, we were to add a 60-pound dog to the group, the resulting average will REMAIN at 60 pounds.

So, if that last dog to be added weighs LESS THAN 60 pounds, the resulting average will be LESS THAN 60 pounds.

Conversely, if that last dog to be added weighs MORE THAN 60 pounds, the resulting average will be MORE THAN 60 pounds.

Does that help?

Cheers,

Brent

## Thanks Brent for that

There is only one thing I would like to point out. In the statement "So, let's say the first 3 dogs have weights 20, 20 and 20 (for an average of 100 pounds)....." there seems to be a typo. The average should read 20 pounds for first 3 dogs having weights of 20 each.

## Good idea, and good catch. I

Good idea, and good catch. I've edited my response accordingly.

Cheers,

Brent

## Good idea, and good catch. I

Good idea, and good catch. I've edited my response accordingly.

Cheers,

Brent

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