Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 12B Problems/Problem 19"

(→‎Solution)
(→‎Solution: removed redundancy: question already states that geometric sequence is increasing)
Line 10: Line 10:
 
<cmath>a(1+r+r^{2}+r^{3})=4a+6d+1</cmath>
 
<cmath>a(1+r+r^{2}+r^{3})=4a+6d+1</cmath>
  
−
Now we can start trying out some values of <math>r</math>. We start at <math>r=1</math> which gives that <math>6d=-1</math> which is a contradiction. Next we try <math>r=2</math>, which gives
+
Now we can start trying out some values of <math>r</math>. We try <math>r=2</math>, which gives
  
 
<cmath>15a=4a+6d+1</cmath>
 
<cmath>15a=4a+6d+1</cmath>

Revision as of 10:53, 30 December 2015

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 $100$ points. What was the total number of points scored by the two teams in the first half?

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution

Let $a,ar,ar^{2},ar^{3}$ 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 $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores is $4a+6d$. Now we can narrow our search for the values of $a,d$, and $r$. Because points are always measured in positive integers, we can conclude that $a$ and $d$ are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:

\[a(1+r+r^{2}+r^{3})=4a+6d+1\]

Now we can start trying out some values of $r$. We try $r=2$, which gives

\[15a=4a+6d+1\] \[11a=6d+1\]

We need the smallest multiple of $11$ (to satisfy the <100 condition) that is $\equiv 1 \pmod{6}$. We see that this is $55$, and therefore $a=5$ and $d=9$.

So the Raiders first two scores were $5$ and $10$ and the Wildcats first two scores were $5$ and $14$.

\[5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}\]

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_19&oldid=74102"