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# The table above shows the discount structure for advanced purchase of

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The table above shows the discount structure for advanced purchase of [#permalink]

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05 Oct 2009, 23:05
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Difficulty:

75% (hard)

Question Stats:

61% (02:33) correct 39% (03:44) wrong based on 102 sessions

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ADVANCED PURCHASE DISCOUNTS FOR AIRLINE TRAVEL
days prior to departure-------------------percentage discount
0~6 days----------------------------------------0%
7~13 days--------------------------------------10%
14~29 days--------------------------------------25%
30 days or more---------------------------------40%

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 \$1,050. 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 days
B. 7 days
C. 13 days
D. 14 days
E. 29 days
[Reveal] Spoiler: OA

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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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06 Oct 2009, 00:13
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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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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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07 Mar 2010, 00:16
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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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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07 Mar 2010, 07:14
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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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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05 Sep 2011, 06:33
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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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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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06 Sep 2011, 12:57
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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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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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07 Sep 2011, 08:21
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.

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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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18 Jun 2017, 10:25
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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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18 Jun 2017, 13:55
D
You start from 40% and do the following: 1050/(1-0.4)*(1-0.25) - 1050=262.5
1050/(1-0.25)*(1-0.1)-1050= 210 no further trials are needed, this means that the passenger bought the ticket 1 day before the discount of 10% and that is the 14th day
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Re: The table above shows the discount structure for advanced purchase of [#permalink]

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11 Jul 2017, 11:01
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To be honest, this is an easy question.
Look at the options, Option A, C and E cannot be the answer. Reason is, if you decrease one day from each of these options, you still remain in same discount slab. For e.g., take option E which is 29 days. 14~29 have discount 25%. Subtract one day from 29, you get 28 days. Discount will still be 25%. Question tells that one day late and pay \$210 extra.
So we are left with option B and D which are 7 days and 14 days respectively.
Let 'x' be the original cost of ticket.
7 days discount is 10 % and changes to 0%. So change is 10%, then 10% of x = 210. So we get x = 2100. Subtract 10% of 2100 from 2100 and you won't get \$1050.
14 days discount is 25% and changes to 10 %. So change is 15%, then 15% of x = 210. We get x = 1400. Subtract 25% of 1400 from 1400 and you get \$1050. So this is the right answer.

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Re: The table above shows the discount structure for advanced purchase of   [#permalink] 11 Jul 2017, 11:01
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