Difference between revisions of "1990 AJHSME Problems/Problem 19"

Revision as of 08:50, 10 December 2010

There are $120$ seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?

$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$

Every third seat is occupied resulting in 40 seats in addition to seating the first chair thus $\boxed{41}$

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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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 ==Solution==
-If there are <math>120/2=60</math> seats occupied, then it is possible to get an occupation of the seats where no person sits next to someone else.  However, the only way is through alternating seats.
+Every third seat is occupied resulting in 40 seats in addition to seating the first chair thus <math>\boxed{41}</math>
-Thus, if another person joins, two people must sit next to each other, so <math>\boxed{\text{D}}</math>.
 ==See Also==