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General Maths Problem 01

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Note by Ajitesh Mishra
3 years, 11 months ago

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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 · 3 years, 10 months ago

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(A) 29 cm of water. Using that volume of water is the same in figure b and figure c. Aditya Parson · 3 years, 11 months ago

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