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Difference between revisions of "2021 JMPSC Sprint Problems/Problem 13"

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==Solution==
 
==Solution==
 
By the [[Pythagorean Theorem]], we have that the diameter of the cylinder's base is 15 units long. Thus, the cylinder's base has radius <math>\frac{15}{2}</math> units. Thus, the volume of the cylinder is <math>\left(\frac{15}{2}\right)^2\cdot8\pi=\boxed{450}\pi.</math>
 
By the [[Pythagorean Theorem]], we have that the diameter of the cylinder's base is 15 units long. Thus, the cylinder's base has radius <math>\frac{15}{2}</math> units. Thus, the volume of the cylinder is <math>\left(\frac{15}{2}\right)^2\cdot8\pi=\boxed{450}\pi.</math>
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~Lamboreghini
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== Solution 2==
 +
The diameter is <math>15</math>. Therefore,
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<cmath>\pi \left(\frac{15}{2}\right)^2 \cdot 8=450 \pi</cmath> So, <math>k=\boxed{450}</math>
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- kante314 -
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==See also==
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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 09:47, 12 July 2021

Problem

Grace places a pencil in a cylindrical cup and is surprised to see that it fits diagonally. The pencil is $17$ units long and of negligible thickness. The cup is $8$ units tall. The volume of the cup can be written as $k \pi$ cubic units. Find $k$.

Sprint14.jpg

Solution

By the Pythagorean Theorem, we have that the diameter of the cylinder's base is 15 units long. Thus, the cylinder's base has radius $\frac{15}{2}$ units. Thus, the volume of the cylinder is $\left(\frac{15}{2}\right)^2\cdot8\pi=\boxed{450}\pi.$

~Lamboreghini

Solution 2

The diameter is $15$. Therefore, \[\pi \left(\frac{15}{2}\right)^2 \cdot 8=450 \pi\] So, $k=\boxed{450}$

- kante314 -

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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