Difference between revisions of "2010 AMC 12B Problems/Problem 19"
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We need the smallest multiple of <math>11</math> (to satisfy the <100 condition) that is <math>\equiv 1 \pmod{6}</math>. We see that this is <math>55</math>, and therefore <math>a=5</math> and <math>d=9</math>. | We need the smallest multiple of <math>11</math> (to satisfy the <100 condition) that is <math>\equiv 1 \pmod{6}</math>. We see that this is <math>55</math>, and therefore <math>a=5</math> and <math>d=9</math>. | ||
− | So the Raiders first two scores were <math>5</math> and <math>10</math> and the Wildcats first two scores were <math>5</math> and <math>14</math>. | + | So the Raiders' first two scores were <math>5</math> and <math>10</math> and the Wildcats' first two scores were <math>5</math> and <math>14</math>. |
<cmath>5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}</cmath> | <cmath>5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}</cmath> |
Revision as of 14:51, 3 April 2016
Problem 19
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?
Solution
Let be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let be the quarterly scores for the Wildcats. The sum of the Raiders scores is and the sum of the Wildcats scores is . Now we can narrow our search for the values of , and . Because points are always measured in positive integers, we can conclude that and are positive integers. We can also conclude that is a positive integer by writing down the equation:
Now we can start trying out some values of . We try , which gives
We need the smallest multiple of (to satisfy the <100 condition) that is . We see that this is , and therefore and .
So the Raiders' first two scores were and and the Wildcats' first two scores were and .
See also
2010 AMC 12B (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 AMC 12 Problems and Solutions |
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