Difference between revisions of "1993 AJHSME Problems/Problem 11"
Math Kirby (talk | contribs) (Created page with "== Problem 11 == Consider this histogram of the scores for <math>81</math> students taking a test: <asy> unitsize(12); draw((0,0)--(26,0)); draw((1,1)--(25,1)); draw((3,2)--(25...") |
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| − | == Problem | + | == Problem == |
Consider this histogram of the scores for <math>81</math> students taking a test: | Consider this histogram of the scores for <math>81</math> students taking a test: | ||
| Line 79: | Line 79: | ||
==Solution== | ==Solution== | ||
| − | Since <math> 81 </math> students took the test, the median is the score of the <math> 41^{st} </math> student, which is <math> \boxed{\text{(C)}\ 70} </math>. | + | Since <math> 81 </math> students took the test, the median is the score of the <math> 41^{st} </math> student. The five rightmost intervals include <math>2+3+6+12+16=39</math> students, so the <math>41^{st}</math> one must lie in the next interval, which is <math> \boxed{\text{(C)}\ 70} </math>. |
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1993|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:11, 5 July 2013
Problem
Consider this histogram of the scores for 
students taking a test:
![[asy] unitsize(12); draw((0,0)--(26,0)); draw((1,1)--(25,1)); draw((3,2)--(25,2)); draw((5,3)--(23,3)); draw((5,4)--(21,4)); draw((7,5)--(21,5)); draw((9,6)--(21,6)); draw((11,7)--(19,7)); draw((11,8)--(19,8)); draw((11,9)--(19,9)); draw((11,10)--(19,10)); draw((13,11)--(19,11)); draw((13,12)--(19,12)); draw((13,13)--(17,13)); draw((13,14)--(17,14)); draw((15,15)--(17,15)); draw((15,16)--(17,16)); draw((1,0)--(1,1)); draw((3,0)--(3,2)); draw((5,0)--(5,4)); draw((7,0)--(7,5)); draw((9,0)--(9,6)); draw((11,0)--(11,10)); draw((13,0)--(13,14)); draw((15,0)--(15,16)); draw((17,0)--(17,16)); draw((19,0)--(19,12)); draw((21,0)--(21,6)); draw((23,0)--(23,3)); draw((25,0)--(25,2)); for (int a = 1; a < 13; ++a) { draw((2*a,-.25)--(2*a,.25)); } label("$40$",(2,-.25),S); label("$45$",(4,-.25),S); label("$50$",(6,-.25),S); label("$55$",(8,-.25),S); label("$60$",(10,-.25),S); label("$65$",(12,-.25),S); label("$70$",(14,-.25),S); label("$75$",(16,-.25),S); label("$80$",(18,-.25),S); label("$85$",(20,-.25),S); label("$90$",(22,-.25),S); label("$95$",(24,-.25),S); label("$1$",(2,1),N); label("$2$",(4,2),N); label("$4$",(6,4),N); label("$5$",(8,5),N); label("$6$",(10,6),N); label("$10$",(12,10),N); label("$14$",(14,14),N); label("$16$",(16,16),N); label("$12$",(18,12),N); label("$6$",(20,6),N); label("$3$",(22,3),N); label("$2$",(24,2),N); label("Number",(4,8),N); label("of Students",(4,7),N); label("$\textbf{STUDENT TEST SCORES}$",(14,18),N); [/asy]](http://latex.artofproblemsolving.com/3/7/3/3738fa09e51884c2e5387d78a541bcc5b8f79acb.png)
The median is in the interval labeled
Solution
Since students took the test, the median is the score of the 
student. The five rightmost intervals include 
students, so the one must lie in the next interval, which is 
.
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
