What is the parking cost from 12:45 p.m. to 6:15 p.m. if the cost is $1.50 for the first hour and 55 cents for each additional half hour?

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Multiple Choice

What is the parking cost from 12:45 p.m. to 6:15 p.m. if the cost is $1.50 for the first hour and 55 cents for each additional half hour?

Explanation:
To determine the total parking cost from 12:45 p.m. to 6:15 p.m., it's important to first calculate the total duration of the parking. The time frame is from 12:45 p.m. to 6:15 p.m., which totals 5 hours and 30 minutes. This can be broken down as follows: 1. The first hour (12:45 p.m. to 1:45 p.m.) costs $1.50. 2. After the first hour, there are 4 additional hours and 30 minutes (or 9 half-hour intervals) to consider for cost calculation. For the subsequent time: - Each half hour after the first hour costs $0.55. - Since there are 9 half-hour intervals, the cost for this duration would be calculated as: \[ 9 \times 0.55 = 4.95 \] Now, adding the cost of the first hour to the cost of the additional half hours gives us: \[ 1.50 + 4.95 = 6.45 \] However, rounding to the nearest quarter (if applicable in some pricing scenarios) could lead to an effective charge, but in this case

To determine the total parking cost from 12:45 p.m. to 6:15 p.m., it's important to first calculate the total duration of the parking.

The time frame is from 12:45 p.m. to 6:15 p.m., which totals 5 hours and 30 minutes. This can be broken down as follows:

  1. The first hour (12:45 p.m. to 1:45 p.m.) costs $1.50.

  2. After the first hour, there are 4 additional hours and 30 minutes (or 9 half-hour intervals) to consider for cost calculation.

For the subsequent time:

  • Each half hour after the first hour costs $0.55.

  • Since there are 9 half-hour intervals, the cost for this duration would be calculated as:

[

9 \times 0.55 = 4.95

]

Now, adding the cost of the first hour to the cost of the additional half hours gives us:

[

1.50 + 4.95 = 6.45

]

However, rounding to the nearest quarter (if applicable in some pricing scenarios) could lead to an effective charge, but in this case

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