Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1950 AHSME Problems/Problem 33"

(→‎Solution)
(→‎Solution 2 (Solution 1 but easier))
 
(8 intermediate revisions by 3 users not shown)
Line 12: Line 12:
 
It must be assumed that the pipes have an equal height.
 
It must be assumed that the pipes have an equal height.
  
−
A circular pipe with diameter 1 inch and height h has a volume of <math>\pi \left(\frac{1}{2}\right)^2h=\frac{\pi h}{4}</math>. A pipe with diameter 6 inches and height h has volume <math>\pi \left(\frac{6}{2}\right)^2h=9\pi h</math>. To find how many 1-pipes fit in a 6-pipe, we divide: <math>\frac{9\pi h}{\frac{\pi h}{4}}=\frac{9*4\pi h}{\pi h}=\frac{36\pi h}{\pi h}=36 \textbf{(D)}</math>
+
We can represent the amount of water carried per unit time by cross sectional area.
 +
Cross sectional of Pipe with diameter <math>6 in</math>,  
 +
<cmath>\pi r^2 = \pi \cdot 3^2 = 9\pi</cmath>
  
−
If the ratio of similar length of similar shapes is x, then the ratio between area is <math>x^2</math>. Therefore, since the ratio between diameters is 1/6, the ratio between area is 1/36, so 36 pipes of diameter 1 are required.
+
Cross sectional area of pipe with diameter <math>1 in</math>
 +
 
 +
<cmath>\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}</cmath>
 +
 
 +
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
 +
 
 +
<cmath>\frac{9\pi}{\frac{\pi}{4}} = 36</cmath>
 +
 
 +
So the answer is <math>\boxed{\textbf{(D)}\ 36}</math>.
 +
 
 +
==Solution 2 (Solution 1 but easier)==
 +
Knowing that <math>d^2 \alpha A</math>, where <math>d</math> is the pipe diameter and <math>A</math> is the cross-sectional area we simply get <math>6^2 = 36 = \boxed{\textbf{(D)}\ 36}</math>.
 +
 
 +
This works because the diameter of one of the other pipes is <math>1</math>, which is not affected by powers.
 +
 
 +
~PeterDoesPhysics
  
 
==See Also==
 
==See Also==

Latest revision as of 23:56, 25 August 2024

Problem

The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:

$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$

Solution

It must be assumed that the pipes have an equal height.

We can represent the amount of water carried per unit time by cross sectional area. Cross sectional of Pipe with diameter $6 in$, \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]

Cross sectional area of pipe with diameter $1 in$

\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]

So number of 1 in pipes required is the number obtained by dividing their cross sectional areas

\[\frac{9\pi}{\frac{\pi}{4}} = 36\]

So the answer is $\boxed{\textbf{(D)}\ 36}$.

Solution 2 (Solution 1 but easier)

Knowing that $d^2 \alpha A$, where $d$ is the pipe diameter and $A$ is the cross-sectional area we simply get $6^2 = 36 = \boxed{\textbf{(D)}\ 36}$.

This works because the diameter of one of the other pipes is $1$, which is not affected by powers.

~PeterDoesPhysics

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_33&oldid=225793"