Alice buys three burgers and two sodas for $\$3.20$, and Bill buys two burgers and a soda for $\$2.00$. How many cents does a burger cost?
Let's work with this problem in cents, not dollars, because the answer calls for a number in cents. So, Alice's three burgers and two sodas cost 320 cents and Bill's food costs 200 cents. Let a burger cost $b$ cents and a soda cost $s$ cents. We are trying to find the value of $b$. We can set up a system of two equations to represent the given information. These equations are:

\begin{align*}
3b + 2s &= 320 \\
2b + s &= 200 \\
\end{align*}We are solving for $b$, so we want to eliminate $s$ from the equations above. Multiplying both sides of the second equation by 2, we get $4b+2s = 400$, or $2s = 400 - 4b$. Substituting this equation into the first equation above to eliminate $s$, we get that $3b + (400-4b) = 320$, or $b=80$. Thus, a burger costs $\boxed{80}$ cents.
80
Level 2
Algebra
Problem: Alice buys three burgers and two sodas for $\$3.20$, and Bill buys two burgers and a soda for $\$2.00$. How many cents does a burger cost?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the cost of a burger be denoted as b cents and the cost of a soda as s cents. 

From Alice's purchase, we can formulate the equation: 
3b + 2s = 320. 

From Bill's purchase, we can formulate the second equation: 
2b + s = 200. 

Now we have a system of two equations:

1) 3b + 2s = 320 
2) 2b + s = 200 

To solve for b and s, we can manipulate the second equation to express s in terms of b:
s = 200 - 2b. 

Next, we substitute this expression for s into the first equation:
3b + 2(200 - 2b) = 320. 

Expanding this gives:
3b + 400 - 4b = 320. 

Combining like terms results in:
-b + 400 = 320. 

Isolating b, we subtract 400 from both sides:
-b = 320 - 400, 
-b = -80. 

Multiplying both sides by -1 gives:
b = 80. 

Thus, the cost of a burger is 80 cents.<|END_OF_LATENT|> The final answer is $80$.