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DS - Nan car park

by harsh.champ » Wed Feb 17, 2010 5:57 am
Yesterday Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan's total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged?

(1) Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2) If the charge for the first hour had been $1.00, Nan's total parking charge would have been $4.00.
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by shashank.ism » Wed Feb 17, 2010 7:15 am
harsh.champ wrote:Yesterday Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan's total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged?

(1) Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2) If the charge for the first hour had been $1.00, Nan's total parking charge would have been $4.00.
1. 0.75 + (x-1) 0.50 = 3.75 will give x ----sufficient
2. 4- 3.75 = 0.25 so rate for 1st hour = 1-0.25 =0.75
1+ (x-1)y = 4
0.75 +(x-1)y = 3.75 === not sufficeit
Ans A
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by Osirus@VeritasPrep » Wed Feb 17, 2010 7:18 am
shashank.ism wrote:
harsh.champ wrote:Yesterday Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan's total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged?

(1) Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2) If the charge for the first hour had been $1.00, Nan's total parking charge would have been $4.00.
1. 0.75 + (x-1) 0.50 = 3.75 will give x ----sufficient
2. 4- 3.75 = 0.25 so rate for 1st hour = 1-0.25 =0.75
1+ (x-1)y = 4
0.75 +(x-1)y = 3.75 === not sufficeit
Ans A
Since your math skills seem impeccable, could you explain in more detail why you used (x-1) in statement 1?
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by shashank.ism » Wed Feb 17, 2010 7:32 am
osirus0830 wrote:
shashank.ism wrote:
harsh.champ wrote:Yesterday Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan's total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged?

(1) Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2) If the charge for the first hour had been $1.00, Nan's total parking charge would have been $4.00.
1. 0.75 + (x-1) 0.50 = 3.75 will give x ----sufficient
2. 4- 3.75 = 0.25 so rate for 1st hour = 1-0.25 =0.75
1+ (x-1)y = 4
0.75 +(x-1)y = 3.75 === not sufficeit
Ans A
Since your math skills seem impeccable, could you explain in more detail why you used (x-1) in statement 1?
I have considered x as the total no. of hours of parking .. For 1st hour rate is 0.75 in St.1 and 0.50 for other hours
so the eq. 0.75 + (x-1) 0.50. I hope I am correct enough to be called "impeccable."
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by money9111 » Wed Feb 17, 2010 7:34 am
yeah I think you are... thanks for the post!
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by harsh.champ » Thu Feb 18, 2010 1:00 am
osirus0830 wrote:
shashank.ism wrote:
harsh.champ wrote:Yesterday Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan's total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged?

(1) Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2) If the charge for the first hour had been $1.00, Nan's total parking charge would have been $4.00.
1. 0.75 + (x-1) 0.50 = 3.75 will give x ----sufficient
2. 4- 3.75 = 0.25 so rate for 1st hour = 1-0.25 =0.75
1+ (x-1)y = 4
0.75 +(x-1)y = 3.75 === not sufficeit
Ans A
Since your math skills seem impeccable, could you explain in more detail why you used (x-1) in statement 1?
Well,shashank has already posted the reason to take (x-1) in the equation but i guess the soln. approach will seem easy if we just use n terms and form a series.
That way the calculation will also be quite simple.

Ex:- 0.75 + 0.50 + 0.50+ ........(n times 0.50)
Well,I would say "to each his own" but I specifically prefer the above method.What-say ppl??
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by missrochelle » Thu Aug 05, 2010 3:50 pm
For statement (2), I don't get why we are supposed to assume that the hypothetical first hour value of $1.00, gives us the REAL value of the first value by plugging in the HYPOTHETICAL value into the question?
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by Fiver » Fri Aug 06, 2010 10:22 am
I would have chosen E for an answer to this one.

We know that the total charges are 3.75

We need to know the total hours gievn that the 1st hour parking is a fixed cost > the parking cost of subsequent hrs.

St1] Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.

Here the total cost of subsequent hrs not essentially an integer is 3.
Given this we could have a total of 7 hrs or a value between 6 & 7 hrs.

Insuff

St2] If the charge for the first hour had been $1.00, Nan's total parking charge would have been $4.00.
This suggests only that the total parking charges for subsequent hrs = 3.
Insuff.

Both statements together is as good as St1].
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