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an amusement park charges 5$ for admission and 2$ for each ride write an expression for the total cost of admission and r rides How Many Rides Can U Go On If U Have 16$??

User AmrataB
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2 Answers

5 votes

Final answer:

You can go on a maximum of 5 rides with $16 after paying for the $5 admission at the amusement park, where each ride costs $2.

Step-by-step explanation:

The amusement park charges $5 for admission and $2 for each ride. We can write an expression for the total cost as C = 5 + 2r, where C is the total cost and r is the number of rides. To find out how many rides you can go on with $16, we would set up the equation 16 = 5 + 2r.

Step-by-step solution:

  1. Start with the formula for the total cost: C = 5 + 2r.
  2. Replace C with $16, the total money you have, to get the equation 16 = 5 + 2r.
  3. Subtract $5 (the cost of admission) from both sides to find out how much money is left for rides: 16 - 5 = 2r, which simplifies to 11 = 2r.
  4. Divide both sides by $2 to find the number of rides r: 11 / 2 = r, which simplifies to 5.5.

Since you cannot go on half a ride, you could go on a maximum of 5 rides with your $16.

User Simon Streicher
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c = 5 + 2r

You can go on 5 rides.



First, create the expression for the cost.

c = ?



Since the price of admission is $5, that becomes a term in the expression, so the new expression is:

c = 5 + ?



And since you need to pay for each ride (r) that also becomes a factor in the expression. Giving you:

c = 5 + 2r + ?



There isn't anything else that costs you money, so you can get rid of the placeholder of + ? giving you a final expression of

c = 5 + 2r



Now you need to figure out how many rides you can go on for $16. Lets figure it out.



You first have to subtract the price of admission. So you get

16 - 5 = 11



So after being admitted, you now have $11 in which to pay for rides. Since each ride costs $2, simply divide 11 by 2.

11 / 2 = 5.5



Since you really can't go on half a ride, that means you can go on 5 rides after being admitted to the park.
User Spanky Quigman
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