2025 October(J08) Math Question 27
Perimeter, Area, Volume
- Other Shapes
Volume of the cone of sugar: V=13πr2h=13π(3)2(2)=6πV = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(3)^2(2) = 6\piV=31πr2h=31π(3)2(2)=6π
Weight of leaked sugar =6π×0.025≈0.471= 6\pi \times 0.025 \approx 0.471=6π×0.025≈0.471 lb
Sugar remaining =1.5−0.471≈1.029= 1.5 - 0.471 \approx 1.029=1.5−0.471≈1.029 lb
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