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25 Dec 2023, 02:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
50% (02:52) correct
50%
(03:01)
wrong
based on 111
sessions
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How many times does the digit 5 appear if you list the numbers from 189 to 5004?
A. 1460
B. 1462
C. 1466
D. 1480
E. 1500
A. 1460
B. 1462
C. 1466
D. 1480
E. 1500
Aryaa03
Joined: 12 Jun 2023
Last visit: 19 Feb 2026
Posts: 210
Given Kudos: 159
Location: India
Schools: ISB '27 (A) IIM Calcutta '26 (A) IIMA '26 (D)
GMAT Focus 1: 645 Q86 V81 DI78 
25 Dec 2023, 22:48
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Bunuel
Sol: IMO- Option-C
Number Range - 189 to 5004
or
189-199 + 200-4999 + 5000-5004
1) Total no. of 5s between 189-199 = 1 (195)
2) Total nos. of 5 between 5000 - 5004 = 5
3) Total nos. of 5 between 200 - 4999 :
Total no. between 200-4999 = 4999-200+1 = 4800
a) At unit's place : 10 nos. of 5s every 100 (05, 15, 25...) , therefore, total nos. of 5s = 10*4800/100 = 480
b) At ten's place : 10 nos. of 5s every 100 (51, 52, 53..) , therefore, total nos. of 5s = 10*4800/100 = 480
c) At hundred's place : 100 nos. of 5s every one thousand (501, 502,..), therefore, total nos. of 5s = 100*5 = 500
Therefore, Total nos. of 5s = 1+480+480+500+5 = 1466
General Discussion
25 Dec 2023, 11:18
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The problem becomes easier if we consider the following case:
* \(189 - 200 \to1 (195)\)
* \(5000-5004 \to 5 (5000, 5001, 5002, 5003, 5004)\)
Hence, check for a number which ends with \(6\), i.e. \(1466\)
\(0-9 \to 1\)
\(0-99 \to 20\)
\(100-199 \to 20\)
\(500 - 599 \to 120\)
\(1000 - 1999 \to 300\)
\(\cdots\)
\(4000 - 4999 \to 300\)
\(0 - 999 \to 180 + 120 = 300\)
\(1000-4999 = 300 \times 4 = 1200\)
For this problem we can do,
\(1 + 20\times 7 + 120 + 300 \times 4 + 5 = 1466\)
C
* \(189 - 200 \to1 (195)\)
* \(5000-5004 \to 5 (5000, 5001, 5002, 5003, 5004)\)
Hence, check for a number which ends with \(6\), i.e. \(1466\)
\(0-9 \to 1\)
\(0-99 \to 20\)
\(100-199 \to 20\)
\(500 - 599 \to 120\)
\(1000 - 1999 \to 300\)
\(\cdots\)
\(4000 - 4999 \to 300\)
\(0 - 999 \to 180 + 120 = 300\)
\(1000-4999 = 300 \times 4 = 1200\)
For this problem we can do,
\(1 + 20\times 7 + 120 + 300 \times 4 + 5 = 1466\)
C
