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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

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