Advance purchase discounts for airline travel : GMAT Problem Solving (PS)
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Advance purchase discounts for airline travel

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05 Oct 2009, 22:05
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First colum specifies the No of Days prior to departure information. Second column specifies the Percentage discount offered.

0 - 6 days ==> 0%
7 - 13 days ==> 10%
14 - 29 days ==> 25%
30 days or more ==> 40%

Sorry could present the data in a tabular way.

The table above shows the discount structure for advanced purchase of tickets at a particular airline. A passenger bought a ticket at this airline for $1050. The ticket agent informed her that, had she purchased the ticket one day later, she would have paid$210 more. How many days before her departure did she purchase her ticket.

A) 6
B) 7
C) 13
D) 14
E) 29

This took me nearly 5 min and then finally followed POE to come with an ans.
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05 Oct 2009, 23:13
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You are given that the passenger paid 1050$for her ticket, and if she had waited one day, she would have paid$210 MORE. This constitutes a 20% (210/1050 = 1/5) raise in ticket price.

Now you simply look at the discount ranges and determine which range-to-range change constitutes a 20% increase in price:

0 - 6 days -> 0% -> 100% ticket price
7 - 13 days -> 10% -> 90% ticket price
14 - 29 days -> 25% -> 75% ticket price
30+ days -> 40% -> 60% ticket price

SO:

60% to 75% -> 75/60 = 1.25 -> 25% ticket increase (NOPE)
75% to 90% -> 90/75 = 1.2 -> 20% ticket increase (YES)

So she bought the ticket 14 days in advance. Therefore the correct answer is D.
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06 Mar 2010, 23:16
mrsmarthi wrote:
First colum specifies the No of Days prior to departure information. Second column specifies the Percentage discount offered.

0 - 6 days ==> 0%
7 - 13 days ==> 10%
14 - 29 days ==> 25%
30 days or more ==> 40%

Sorry could present the data in a tabular way.

The table above shows the discount structure for advanced purchase of tickets at a particular airline. A passenger bought a ticket at this airline for $1050. The ticket agent informed her that, had she purchased the ticket one day later, she would have paid$210 more. How many days before her departure did she purchase her ticket.

A) 6
B) 7
C) 13
D) 14
E) 29

This took me nearly 5 min and then finally followed POE to come with an ans.

Lets do it in less then one minute...

average discount increment is 15%.

Actual discount is 210 so the total amount in that case would be 1400.

He paid 1050 that means got a discount of 350 on 1400 or say 25%

further delay would have costed him 15% or 210 that means he purchased on 14th day getting 25% discount.

Hope this helps !!!
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07 Mar 2010, 06:14
mrsmarthi wrote:
First colum specifies the No of Days prior to departure information. Second column specifies the Percentage discount offered.

0 - 6 days ==> 0%
7 - 13 days ==> 10%
14 - 29 days ==> 25%
30 days or more ==> 40%

Sorry could present the data in a tabular way.

The table above shows the discount structure for advanced purchase of tickets at a particular airline. A passenger bought a ticket at this airline for $1050. The ticket agent informed her that, had she purchased the ticket one day later, she would have paid$210 more. How many days before her departure did she purchase her ticket.

A) 6
B) 7
C) 13
D) 14
E) 29

This took me nearly 5 min and then finally followed POE to come with an ans.

With POE it takes less than 2 minutes.
First thing we can make out is 40% discount is out.
Also A,C,E are not possible as if we purchase even one day later same discount is possible.
So now try first with 25%
75% of x = 1050
x = 1400
15% of 1400 = 210
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05 Sep 2011, 05:33
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210/1050 = 1/5 OR 20%
passenger would've to pay 20% more in case of delay.

let's find out what two categories have a difference of 20% between their price. let's asssume the original price is 100.

0% -> 100
10% -> 90
25% -> 75
40% -> 60

between 25%-40% -> 15/60 = 1/4 or 25%
between 25%-10% -> 15/75 = 1/5 or 20% .... this is it

so the passenger purchased 14 days prior to departure.
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06 Sep 2011, 11:57
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another method.. although many have provided shortcuts...

each slab decreases by 15 % hence if the guy would have been a day late.. the price wud have increased by 15% and he would have paid 210 more hence .15x = 210 = > x = 1400

but he paid only 1050... thats about 2/3 or close to 65%
40% discount (0.6 *1400) = 840
25 % discount = (0.75*1400) = 1050

so the answer is 14 days.
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07 Sep 2011, 07:21
mrsmarthi wrote:
First colum specifies the No of Days prior to departure information. Second column specifies the Percentage discount offered.

0 - 6 days ==> 0%
7 - 13 days ==> 10%
14 - 29 days ==> 25%
30 days or more ==> 40%

Sorry could present the data in a tabular way.

The table above shows the discount structure for advanced purchase of tickets at a particular airline. A passenger bought a ticket at this airline for $1050. The ticket agent informed her that, had she purchased the ticket one day later, she would have paid$210 more. How many days before her departure did she purchase her ticket.

A) 6
B) 7
C) 13
D) 14
E) 29

This took me nearly 5 min and then finally followed POE to come with an ans.

I took a slightly different approach.

First I figured out the percentage change from the immediate previous amount for each of the period.Thus it comes-

0 - 6 days ==> 0%
7 - 13 days ==> 10% (90% of the price)
14 - 29 days ==> 17% (approx) (went from 90% to 75%)
30 days or more ==> 20% (went from 75% to 60%)

If the passenger had to pay the extra amount, she had to pay=(210+1050)=1260

now 210 is somehow between 10% and 20% of 1260. So 17% is the choice here. And correct ans: 14 days(d).

Though we do not need to do so many things when we have the option of eliminating some of the choices.

And Kudos for the post.
Re: Advance purchase discounts for airline travel   [#permalink] 07 Sep 2011, 07:21
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