solution: We take the inequalities one at a time. Adding $144$ to both sides of the first inequality, we get $$144\ge 54p,$$implying $$\frac{144}{54}\ge p.$$Reducing the fraction and switching the sides (along with the direction of the inequality), we get $p\le\frac{8}{3}$.
  
  
  To solve the second inequality, we add $20p$ to both sides: $$20p > 12$$Dividing both sides by $20$, we get $$p>\frac{12}{20}.$$Reducing the fraction gives $p>\frac{3}{5}$.
  
  
  We are looking for $p$ which satisfy both inequalities. The intersection of the solutions above is $\boxed{\left(\frac{3}{5},\frac{8}{3}\right]}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Find all $p$ which satisfy both the inequalities $0\ge 54p-144$ and $0>12-20p$. Express your answer in interval notation, reducing any fractions in your answer.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: \left(\frac{3}{5},\frac{8}{3}\right]
extra_info.question: Find all $p$ which satisfy both the inequalities $0\ge 54p-144$ and $0>12-20p$. Express your answer in interval notation, reducing any fractions in your answer.
extra_info.split: test