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# At a garage sale, the prices of all the items sold were different. The

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Math Expert
Joined: 02 Sep 2009
Posts: 60460
At a garage sale, the prices of all the items sold were different. The  [#permalink]

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27 Sep 2018, 05:32
1
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Difficulty:

65% (hard)

Question Stats:

48% (01:54) correct 52% (02:05) wrong based on 50 sessions

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At a garage sale, the prices of all the items sold were different. The items sold were radios and DVD players. If the price of a radio sold at the garage sale was the 15th highest price as well as the 20th lowest price among the prices of the radios sold, and the price of a DVD player sold was the 29th highest price as well as the 37th lowest price among all the prices of all the items sold, how many DVD players were sold at the garage sale?

(A) 30
(B) 31
(C) 32
(D) 64
(E) 65

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Manager
Joined: 14 Jun 2018
Posts: 211
Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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30 Sep 2018, 11:52
1
vipulshahi wrote:
Dear dave13,

Can you please explain the logic behind this question.

Why are you deducting lowest DVD from lowest Radio and Highest DVD from highest Radio.

Thank you

take a small example

A B C D E F

Here D is 4th highest and 3rd lowest. 4+3 - 1 = 6 items in total. We subtract 1 because D is being counted twice.
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Location: Kuwait
GPA: 3.2
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Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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27 Sep 2018, 07:05
Total number of radios sold = 15 + 20 - 1 = 34

Total items sold = 37 + 29 -1 = 65

65 - 34 = 31

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Manager
Joined: 14 Jun 2018
Posts: 211
Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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27 Sep 2018, 11:00
Radios = 20+15 -1 = 34 (That particular radio is being counter twice so -1)
Total items = 29 + 37 - 1 = 65

Total dvd's = 65-34 = 31
VP
Joined: 09 Mar 2016
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At a garage sale, the prices of all the items sold were different. The  [#permalink]

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27 Sep 2018, 12:04
Bunuel wrote:
At a garage sale, the prices of all the items sold were different. The items sold were radios and DVD players. If the price of a radio sold at the garage sale was the 15th highest price as well as the 20th lowest price among the prices of the radios sold, and the price of a DVD player sold was the 29th highest price as well as the 37th lowest price among all the prices of all the items sold, how many DVD players were sold at the garage sale?

(A) 30
(B) 31
(C) 32
(D) 64
(E) 65

At first glance I got confused but after some time i figured out how to solve it

so the logic here is to subtract lowest DVD from lowest Radio and Highest DVD from highest Radio

$$37 - 20 = 17$$
$$29 - 15 = 14$$

$$17+14 = 31$$
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Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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30 Sep 2018, 11:23
Dear dave13,

Can you please explain the logic behind this question.

Why are you deducting lowest DVD from lowest Radio and Highest DVD from highest Radio.

Thank you
VP
Joined: 09 Mar 2016
Posts: 1225
Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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30 Sep 2018, 12:22
vipulshahi wrote:
Dear dave13,

Can you please explain the logic behind this question.

Why are you deducting lowest DVD from lowest Radio and Highest DVD from highest Radio.

Thank you

vipulshahi thank you for your question sure I will gladly explain you my reasoning, though my logic is not standard it doesn't follow sometimes standard approaches and I don`t want to confuse you even more:)

HIGHEST PRICES 'DVDs $$x_1$$and RADIO $$x_2$$ ---> $$x_1 - x_2$$

LOWEST PRICES 'DVDs $$y_1$$ and RADIO $$y_2$$ ----> $$y_1 - y_2$$

thats it

i think approaches above are more logical than mine
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Joined: 25 Nov 2019
Posts: 50
Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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02 Dec 2019, 19:37
Turns out it doesn’t matter if you -1 as long as you are consistent

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CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 351
Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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03 Dec 2019, 00:15
This is an interesting question with a fair bit of logical reasoning built into the question.

An approach that always works in such questions is to use the number line to represent the prices and determine the exact positions of each item. The following diagrams depict how this can be done:

Attachment:

3rd Dec 2019 - Reply 1.jpg [ 47.9 KiB | Viewed 223 times ]

If a particular radio is the 15th highest, it means that there are 14 radios above it in terms of price; similarly, there are 19 radios that are below it since the same radio is the 20th lowest.
Therefore, total number of radios = 14 + 1 + 19 = 34.

Similarly, if a particular DVD player is the 29th highest article, it means that there are 28 articles above it in terms of price; similarly, there are 36 articles that are below it since the same DVD player is the 37th lowest article.
Therefore, total number of articles = 28 + 1 + 36 = 65.

Of 65 total articles, if 34 are radios, the remaining 31 have to be DVD players.
The correct answer option is B.

Hope that helps!
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Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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12 Jan 2020, 06:29
Bunuel wrote:
At a garage sale, the prices of all the items sold were different. The items sold were radios and DVD players. If the price of a radio sold at the garage sale was the 15th highest price as well as the 20th lowest price among the prices of the radios sold, and the price of a DVD player sold was the 29th highest price as well as the 37th lowest price among all the prices of all the items sold, how many DVD players were sold at the garage sale?

(A) 30
(B) 31
(C) 32
(D) 64
(E) 65

To understand the approach for solving this problem, consider the situation in which there are only 5 radios, listed in decreasing order of sales price: A B C D E. We see that radio D is the fourth highest in sales price, and it is also the second lowest in sales price. To calculate the total number of radios just from these two pieces of information, we see that we add the two values (4 and 2), but then we must subtract 1, since we double-counted radio D. Thus, we know that there are 4 + 2 - 1 = 5 total radios.

For the given problem, we see that there are 20 + 15 - 1 = 34 radios and 29 + 37 - 1 = 65 total items, so there are 65 - 34 = 31 DVD players.

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Re: At a garage sale, the prices of all the items sold were different. The   [#permalink] 12 Jan 2020, 06:29
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