smc

Let $a,ar,a{r}^2,a{r}^3$ be the quarterly scores for the Raiders. We know $r>1$ because the sequence is said to be increasing. We also know that each of $a,ar,a{r}^2,a{r}^3$ is an integer. We start by showing that $r$ must also be an integer. Suppose not, and say $r=m/n$ where $m>n>1$, and $\gcd (m,n)=1$. Then $n,n^2,n^3$ must all divide $a$ so $a=n^{3}k$ for some integer $k$. Then $S_R=n^{3}k+n^{2}mk+n{m}^{2}k+m^{3}k<100$ and we see that even if $k=1$ and $n=2$, we get $m<4$, which means that the only option for $r$ is $r=3/2$. A quick check shows that even this doesn't work. Thus $r$ must be an integer. Let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. Let $S_W=a+(a+d)+(a+2d)+(a+3d)=4a+6d$. Let $S_R=a+ar+a{r}^2+a{r}^3=a(1+r)(1+r^2)$. Then $S_R<100$ implies that $r<5$, so $r\in \{2,3,4\}$. The Raiders win by one point, so$$a(1+r)(1+r^2)=4a+6d+1.$$ If $r=4$ we get $85a=4a+6d+1$ which means $3(27a-2d)=1$, which is not possible with the given conditions. If $r=3$ we get $40a=4a+6d+1$ which means $6(6a-d)=1$, which is also not possible with the given conditions. * If $r=2$ we get $15a=4a+6d+1$ which means $11a-6d=1$. Reducing modulo 6 we get $a\equiv 5(\mathrm{mod}6)$. Since $15a<100$ we get $a<7$. Thus $a=5$. It then follows that $d=9$. Then the quarterly scores for the Raiders are $5,10,20,40$, and those for the Wildcats are $5,14,23,32$. Also $S_R=75=S_W+1$. The total number of points scored by the two teams in the first half is $5+10+5+14=\boxed{34}$. Note if you don't realize while taking the test that $r$ might not be an integer: since an answer is achieved through casework on the integer value of $r$ and since there is only one right answer, the proof of $r$ being an integer can be skipped on the test (it takes up time).