Difference between revisions of "1960 AHSME Problems/Problem 17"
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==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1960|num-b=16|num-a=18}} | + | {{AHSME 40p box|year=1960|num-b=16|num-a=18}} |
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| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 18:02, 17 May 2018
Problem 17
The formula 
gives, for a certain group, the number of individuals whose income exceeds 
dollars.
The lowest income, in dollars, of the wealthiest 
individuals is at least:
Solution
Plug for 
because is the number of people who has at least dollars.
![\[800 = 8 \times 10^{8} \times x^{-3/2}\]](http://latex.artofproblemsolving.com/5/8/5/58578afcec1d2d7a59d2038e119b387c58a8b4a5.png)
![\[10^{-6} = x^{-3/2}\]](http://latex.artofproblemsolving.com/8/2/4/824210d688a2019dfb05c15e5b1de91701457b9e.png)
In order to undo raising to the 
power, raise both sides to the 
power.
![\[(10^{-6})^{-2/3} = (x^{-3/2})^{-2/3}\]](http://latex.artofproblemsolving.com/8/8/0/8806d8a03a50beeac6ebd3cc572f5382921b3a4e.png)
![\[x = 10^4 \text{ dollars}\]](http://latex.artofproblemsolving.com/0/7/3/073b7d95aa47a63a06dca539a11bb59b29225427.png)
The answer is 
.
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
