# 1993 AJHSME Problems/Problem 11

## 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 $\boxed{\text{(C)}\ 70}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 