Waste less time on Facebook — follow Brilliant.
×

General Maths Problem 01

solve this problem

Note by Ajitesh Mishra
4 years ago

No vote yet
2 votes

Comments

Sort by:

Top Newest

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. Aditya Parson · 4 years ago

Log in to reply

(A) 29 cm of water. Using that volume of water is the same in figure b and figure c. Aditya Parson · 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...