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Problem on Cylinder | AMC-10A, 2004 | Problem 11

Try this beautiful problem from AMC 10A, 2004 based on Mensuration: Cylinder. You may use sequential hints to solve the problem.

Try this beautiful problem from AMC 10A, 2004 based on Mensuration: Cylinder

Problem on Cylinder – AMC-10A, 2004- Problem 11


A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by \(25\%\) without altering the volume, by what percent must the height be decreased?

  • \(16\)
  • \(18\)
  • \(20\)
  • \(36\)
  • \(25\)

Key Concepts


Mensuration

Cylinder

Percentage

Check the Answer


Answer: \(36\)

AMC-10A (2004) Problem 11

Pre College Mathematics

Try with Hints


Let the radius of the jar be \(x\) and height be \(h\).then the volume (V) of the jar be\(V\)= \(\pi (x)^2 h\). Diameter of the jar increase \(25 \)% Therefore new radius will be \(x +\frac{x}{4}=\frac{5x}{4}\) .Now the given condition is “after increase the volume remain unchange”.Let new height will be \(h_1\).Can you find out the new height….?

can you finish the problem……..

Let new height will be \(H\).Therefore the volume will be \(\pi (\frac{5x}{4})^2 H\).Since Volume remain unchange……

\(\pi (x)^2 h\)=\(\pi (\frac{5x}{4})^2 H\) \(\Rightarrow H=\frac{16h}{25}\).

height decrease =\(h-\frac{16h}{25}=\frac{9h}{25}\).can you find out the decrease percentage?

can you finish the problem……..

Decrease Percentage=\( \frac {\frac {9h}{25}}{h} \times 100=36\)%

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