1. **State the problem:** We have 3 gallons of milk, 6 bottles of water, and 5 snack-size bags of chips. The total cost is 25.50 before tax. 2. **Define variables:** Let $c$ be the cost of one bag of chips. 3. **Express other costs in terms of $c$:** - Cost of one bottle of water is twice the cost of a bag of chips: $2c$. - Cost of one gallon of milk is $1.60$ more than a bottle of water: $2c + 1.60$. 4. **Write the total cost equation:** $$3(2c + 1.60) + 6(2c) + 5(c) = 25.50$$ 5. **Simplify the equation:** $$3(2c + 1.60) = 6c + 4.80$$ $$6(2c) = 12c$$ So, $$6c + 4.80 + 12c + 5c = 25.50$$ Combine like terms: $$6c + 12c + 5c + 4.80 = 25.50$$ $$23c + 4.80 = 25.50$$ 6. **Solve for $c$:** $$23c = 25.50 - 4.80$$ $$23c = 20.70$$ $$c = \frac{20.70}{23} = 0.90$$ 7. **Answer:** The cost of one snack-size bag of chips is $0.90$. **Final answer:** $\boxed{0.90}$