2005 CEMC Gauss (Grade 7) Problems/Problem 19

Problem

Chris and Pat are playing catch. Standing $1 m$ apart, Pat first throws the ball to Chris and then Chris throws the ball back to Pat. Next, standing $2 m$ apart, Pat throws to Chris and Chris throws back to Pat. After each pair of throws, Chris moves $1 m$ farther away from Pat. They stop playing when one of them misses the ball. If the game ends when the $29th$ throw is missed, how far apart are they standing and who misses catching the ball?

$\text{(A)}\ 15 m, Chris \qquad \text{(B)}\ 15 m, Pat \qquad \text{(C)}\ 14 m, Chris \qquad \text{(D)}\ 14 m, Pat \qquad \text{(E)}\ 16 m, Pat$

Solution

Solution

At each distance, two throws are made: the $1st$ and $2nd$ throws are made at $1 m$, the $3rd$ and $4th$ are made at $2 m$, and so on, with the $27th$ and $28th$ throws being made at $14 m$. Therefore, the $29th$ throw is the first throw made at $15 m.$ At each distance, the first throw is made by Pat to Chris, so Chris misses catching the $29th$ throw at a distance of $15 m$. The answer is therefore $A$.

See Also

2005 CEMC Gauss (Grade 7) (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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CEMC Gauss (Grade 7)
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