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Bunuel
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Sayan597
Hi Bunuel

As per the condition, if the car had been there 12 hours and 1 min, it would have been charged for a full 13 hours. Now as per the 1st statement the amount could have been 3.75 if the car had stayed for 30 more mins. The only way possible where the car paid less than 3.75 only if it leaves at max dot 12 hours. Even if we count that 30 mins from 12:00 hours still it is clear that the car was not there 'MORE THAN 12 hours'. Why can't we consider this statement as 'sufficient'?
Bunuel
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The parking fee at a certain lot is $1.00 for the first 2 hours, plus $0.25 for each additional hour or fraction of an hour beyond the first 2 hours. Did the car remain parked for more than 12 hours at the parking lot?

For any time exceeding 2 hours, the fee is calculated as \(1 + (⌈x⌉ - 2) * 0.25\), where \(⌈x⌉\) denotes the least integer greater than or equal to \(x\). The phrase "for each additional hour or fraction of an hour" means that if a car is parked for, say, 2.1 hours, it is rounded up and assumed to have been parked for 3 hours. Thus:

• If \(10 < x ≤ 11\), the fee would be \(1 + (11 - 2) * 0.25 = $3.25\).

• If \(11 < x ≤ 12\), the fee would be \(1 + (12 - 2) * 0.25 = $3.50\).

• If \(12 < x ≤ 13\), the fee would be \(1 + (13 - 2) * 0.25 = $3.75\).

Assuming the car was parked for \(t\) hours, the question asks whether \(t\) > 12.

(1) The total parking fee would have been $3.75 if the car had stayed 30 minutes longer.

This implies:

\(12 < t + 0.5 ≤ 13\)

\(11.5 < t ≤ 12.5\)

Not sufficient.

(2) The total parking fee would have been $3.25 if the car had stayed 90 minutes less.

This implies:

\(10 < t - 1.5 ≤ 11\)

\(11.5 < t ≤ 12.5\)

Not sufficient.

(1) + (2) Both statements provide the exact same information: \(11.5 < t ≤ 12.5\). Thus, even when combined, we still cannot determine if \(t > 12\). Not sufficient.

Answer: E

Not fully following what you're saying there, but the first statement implies 11.5 < t ≤ 12.5. You can easily test this yourself by adding 30 minutes to ANY time from that range and you'll get a $3.75 charge.
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