Difference between revisions of "1950 AHSME Problems/Problem 33"
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We can represent the amount of water carried per unit time by cross sectional area. | 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>, | Cross sectional of Pipe with diameter <math>6 in</math>, | ||
| − | <cmath>\ | + | <cmath>\pi r^2 = \pi \cdot 3^2 = 9\pi</cmath> |
Cross sectional area of pipe with diameter <math>1 in</math> | Cross sectional area of pipe with diameter <math>1 in</math> | ||
| − | <cmath>\ | + | <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 | So number of 1 in pipes required is the number obtained by dividing their cross sectional areas | ||
Latest revision as of 22:19, 24 December 2023
Problem
The number of circular pipes with an inside diameter of 
inch which will carry the same amount of water as a pipe with an inside diameter of 
inches is:
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 
,
![\[\pi r^2 = \pi \cdot 3^2 = 9\pi\]](http://latex.artofproblemsolving.com/c/f/6/cf646ed6582355e8382beb405a642261eec6447a.png)
Cross sectional area of pipe with diameter 
![\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]](http://latex.artofproblemsolving.com/a/5/9/a59baa23757c9d6bc86383a88be7e8b582354b91.png)
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
![\[\frac{9\pi}{\frac{\pi}{4}} = 36\]](http://latex.artofproblemsolving.com/8/7/8/878d411debae59be7ca8f7d8e6a625e7daa786d8.png)
So the answer is 
.
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. 
