# General Maths Problem 01

solve this problem

Note by Ajitesh Mishra
5 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 $$=\pi3^2X+\pi 1^2(20-X)$$ $$\Rightarrow V=20\pi+8\pi X$$ In figure $$C$$ height of water in small cylinder is simply $$Y$$ and that in the bigger one is$$28-Y$$ In figure $$C$$, $$V=\pi Y +9\pi (28-Y)$$ $$\Rightarrow V=252\pi-8\pi Y$$ Equating the two volume equations, gives us: $$20\pi+8\pi X=252\pi-8\pi Y$$ $$\Rightarrow X+Y=29 cm$$ which is the height of the bottle.

- 5 years, 2 months ago

(A) 29 cm of water. Using that volume of water is the same in figure b and figure c.

- 5 years, 2 months ago