Question

A solid metal cylinder of 10cm height and 14cm diameter is melted and re-cast into 2 cones in the proportion of 3:4 keeping the height 10cm what would be the percentage change in the flat surface area before and after

Answer 1

Heyy mate ❤✌✌❤

Here's your Answer..

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Let radius of cylinder, cone 1, cone 2 = R, r1, r2 respectively.

Let their volumes be v, v1, v2respectively.

Cylinder is melted and recast into cone 1 & cone 2

So V = v1 + v2.

Ratio of volumes of the two cones v1 : v2 is 3 : 4.
If v1 is 3x, v2 will be 4x

Hence, volume of cylinder V= 7x.

Therefore, The percentage change in the flat surface area is 50%.
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Answer 2

Let radius of cylinder, cone 1, cone 2 = R, r1, r2 respectively.

Let their volumes be V, v1, v2respectively.

Cylinder is melted and recast into cone 1 & cone 2

So V = v1 + v2.

Ratio of volumes of the two cones v1 : v2 is 3 : 4.
If v1 is 3x, v2 will be 4x

Hence, volume of cylinder V= 7x