1950 AHSME Problems/Problem 33

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.

A circular pipe with diameter 1 inch and height h has a volume of $\pi \left(\frac{1}{2}\right)^2h=\frac{\pi h}{4}$ . A pipe with diameter 6 inches and height h has volume $\pi \left(\frac{6}{2}\right)^2h=9\pi h$ . To find how many 1-pipes fit in a 6-pipe, we divide: $\frac{9\pi h}{\frac{\pi h}{4}}=\frac{9*4\pi h}{\pi h}=\frac{36\pi h}{\pi h}=36 \textbf{(D)}$

If the ratio of similar length of similar shapes is x, then the ratio between area is $x^2$ . Therefore, since the ratio between diameters is $1/6$ , the ratio between area is $1/36 textbf{(D)}$ , so $36$ pipes of diameter $1$ are required.

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.

Page

Toolbox

Search

1950 AHSME Problems/Problem 33

Problem

Solution

See Also