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

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

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New post 27 Sep 2018, 05:32
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Question Stats:

41% (01:26) correct 59% (01:48) wrong based on 22 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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Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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

Total items sold = 37 + 29 -1 = 65

65 - 34 = 31

Answer choice B

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

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New post 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
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At a garage sale, the prices of all the items sold were different. The  [#permalink]

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New post 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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New post 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
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Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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New post 30 Sep 2018, 11:52
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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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Re: At a garage sale, the prices of all the items sold were different. The  [#permalink]

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New post 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 :lol: 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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Re: At a garage sale, the prices of all the items sold were different. The &nbs [#permalink] 30 Sep 2018, 12:22
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