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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

This is a modified question of the the following OG question: https://gmatclub.com/forum/at-a-garage- ... 67090.html
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Imo C
From the given information we have DVD player sold was the 29th highest price as well as the 37th lowest price among all the prices of all the items sold Total number of items sold =29+37-1 =65
No radios sold =15+20-1=34
So no DVD players sold =65-34=31
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Bunuel
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.

Answer: B
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