The Correct Answer and Explanation is:
Daniel had
4 and the rides cost $1.50 each, what is the greatest number of rides Daniel can go on? A. Write an inequality that represents Daniel’s situation. B. How many rides can Daniel go on?. Justify your answer.
The Correct Answer and Explanation is:
To solve this problem, let’s break it down step by step:
Part A: Writing an inequality
Daniel has $4, and each ride costs $1.50. We need to figure out how many rides he can afford. Let rrr represent the number of rides Daniel can go on.
The total cost for rrr rides is 1.50×r1.50 \times r1.50×r. Since Daniel has only $4, the total cost should be less than or equal to his available money. So, the inequality is:1.50×r≤41.50 \times r \leq 41.50×r≤4
This inequality represents the situation where the cost of the rides must be less than or equal to the $4 that Daniel has.
Part B: How many rides can Daniel go on?
To find how many rides Daniel can afford, we need to solve the inequality for rrr:1.50×r≤41.50 \times r \leq 41.50×r≤4
First, divide both sides of the inequality by 1.50:r≤41.50r \leq \frac{4}{1.50}r≤1.504
Now, calculate 41.50\frac{4}{1.50}1.504:r≤41.50=400150=2.67r \leq \frac{4}{1.50} = \frac{400}{150} = 2.67r≤1.504=150400=2.67
Since Daniel can’t go on a fraction of a ride, we round down to the nearest whole number. Therefore, r≤2r \leq 2r≤2.
Conclusion:
Daniel can go on 2 rides. Here’s why: Although the calculation shows that he can afford about 2.67 rides, he can only purchase whole rides, so the greatest number of rides he can go on is 2. The total cost for 2 rides is:1.50×2=3.001.50 \times 2 = 3.001.50×2=3.00
which is within his $4 budget. If he tried to go on 3 rides, the cost would be:1.50×3=4.501.50 \times 3 = 4.501.50×3=4.50
which exceeds the $4 he has. Therefore, 2 rides is the maximum number Daniel can afford.
