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Difference between revisions of "1993 AJHSME Problems/Problem 11"

(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 11 ==
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== 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:
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==Solution==
 
==Solution==
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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>.
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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}}

Revision as of 21:36, 22 December 2012

Problem

Consider this histogram of the scores for $81$ 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]

The median is in the interval labeled

$\text{(A)}\ 60 \qquad \text{(B)}\ 65 \qquad \text{(C)}\ 70 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 80$

Solution

Since $81$ students took the test, the median is the score of the $41^{st}$ student. The five rightmost intervals include $2+3+6+12+16=39$ students, so the $41^{st}$ one must lie in the next interval, which is $\boxed{\text{(C)}\ 70}$.


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
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