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Comments
Let X and Y be the heights of the big and small cylinder respectively.
In figure B height of water in the small cylinder is inadvertently X−20.
So, the Volume occupied by water =π32X+π12(20−X)⇒V=20π+8πX
In figure C height of water in small cylinder is simply Y and that in the bigger one is28−Y
In figure C,
V=πY+9π(28−Y)⇒V=252π−8πY
Equating the two volume equations, gives us:
20π+8πX=252π−8πY⇒X+Y=29cm which is the height of the bottle.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Let X and Y be the heights of the big and small cylinder respectively. In figure B height of water in the small cylinder is inadvertently X−20. So, the Volume occupied by water =π32X+π12(20−X) ⇒V=20π+8πX In figure C height of water in small cylinder is simply Y and that in the bigger one is28−Y In figure C, V=πY+9π(28−Y) ⇒V=252π−8πY Equating the two volume equations, gives us: 20π+8πX=252π−8πY ⇒X+Y=29cm which is the height of the bottle.
(A) 29 cm of water. Using that volume of water is the same in figure b and figure c.