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Difficulty:
15%
(low)
Question Stats:
81% (01:09) correct
19%
(01:04)
wrong
based on 70
sessions
History
Date
Time
Result
Not Attempted Yet
On a certain day, a news vendor began the day with P papers. Between opening and noon, he sold 40 percent of the papers, and between noon and closing, he sold 60 percent of the papers that remained. What percent of the original P papers did he sell?
(A) 0%
(B) 20%
(C) 24%
(D) 76%
(E) 100%
_________________
(A) 0%
(B) 20%
(C) 24%
(D) 76%
(E) 100%
_________________
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Let total number of papers P= 100
Papers sold between opening and noon= 40
Remaining paper= 100-40= 60
Papers sold between noon and closing= 60% of 60= 36
Total numbers of paper sold= 40+36= 76
Percent of original P paper sold= \(\frac{76}{100}\)= 76%
Answer: D.
Papers sold between opening and noon= 40
Remaining paper= 100-40= 60
Papers sold between noon and closing= 60% of 60= 36
Total numbers of paper sold= 40+36= 76
Percent of original P paper sold= \(\frac{76}{100}\)= 76%
Answer: D.
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 27 Aug 2026
Posts: 1,228
Location: United States (MO)
GMAT 1: 710 Q49 V36
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GMAT 1: 710 Q49 V36
Posts: 1,228
Bunuel
Let P = 100
he sold 40% b/w opening and noon = 40
Remaining = 60
Now he sells 60% of rem b/w noon and closing = 60x60/100 = 36
papers sold 40+36 = 76
remaining = 24
sold 76%
(d) imo
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