At a garage sale, all of the prices of the items sold were d : GMAT Problem Solving (PS)
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# At a garage sale, all of the prices of the items sold were d

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Math Expert
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At a garage sale, all of the prices of the items sold were d [#permalink]

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05 Feb 2014, 01:31
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Question Stats:

71% (01:53) correct 29% (01:23) wrong based on 380 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

At a garage sale, all of the prices of the items sold were different. If the price of a radio sold at the garage sale was both the 15th highest price and the 20th lowest price among the prices of the items sold, how many items were sold at the garage sale?

(A) 33
(B) 34
(C) 35
(D) 36
(E) 37

Problem Solving
Question: 80
Category: Arithmetic Operations with integers
Page: 72
Difficulty: 600

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Math Expert
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Kudos [?]: 92933 [0], given: 10528

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

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05 Feb 2014, 01:31
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SOLUTION

At a garage sale, all of the prices of the items sold were different. If the price of a radio sold at the garage sale was both the 15th highest price and the 20th lowest price among the prices of the items sold, how many items were sold at the garage sale?

(A) 33
(B) 34
(C) 35
(D) 36
(E) 37

As the price of an item was BOTH 15th highest price and the 20th lowest price then # of items is: 15 + 20 - 1 = 34.

Why do we have -1 there? In 15 + 20 the item is counted TWICE: in 15 (as it's 15th highest priced item) as well as in 20 (as it's 20th lowest priced item), hence we should subtract 1 to get rid of double counting.

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Kudos [?]: 81 [0], given: 29

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

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05 Feb 2014, 07:27
Ans B

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Kudos [?]: 56 [2] , given: 11

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

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07 Feb 2014, 12:36
2
KUDOS
I got (B) too. However, is it correct to use this approach?

20+15-1 (to avoid double-counting) = 34.
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Kudos [?]: 247 [0], given: 30

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

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07 Feb 2014, 20:42
I am not clear how to solve this problem.... Can any one detail out... thanks
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Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7067

Kudos [?]: 92933 [0], given: 10528

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

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08 Feb 2014, 03:14
SOLUTION

At a garage sale, all of the prices of the items sold were different. If the price of a radio sold at the garage sale was both the 15th highest price and the 20th lowest price among the prices of the items sold, how many items were sold at the garage sale?

(A) 33
(B) 34
(C) 35
(D) 36
(E) 37

As the price of an item was BOTH 15th highest price and the 20th lowest price then # of items is: 15 + 20 - 1 = 34.

Why do we have -1 there? In 15 + 20 the item is counted TWICE: in 15 (as it's 15th highest priced item) as well as in 20 (as it's 20th lowest priced item), hence we should subtract 1 to get rid of double counting.

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Joined: 07 Apr 2015
Posts: 188
Followers: 2

Kudos [?]: 56 [0], given: 185

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

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18 May 2015, 02:33
There are 14 objects that have a higher price as the radio, as well as 19 objects that have a lower price than the radio.

14 + 19 + 1 (the radio itself) = 34 different objects
Re: At a garage sale, all of the prices of the items sold were d   [#permalink] 18 May 2015, 02:33
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