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28 Oct 2019, 13:50
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (02:26) correct
60%
(03:21)
wrong
based on 15
sessions
History
Date
Time
Result
Not Attempted Yet
Raju has forgotten his six-digit ID number. He remembers the following: the first two digits are either 1, 5 or 2, 6, the number is even and 6 appears twice. If Raju uses a trial and error process to find his ID number at the most, how many trials does he need to succeed?
(A) 972
(B) 2052
(C) 729
(D) 2051
(A) 972
(B) 2052
(C) 729
(D) 2051
23 May 2023, 02:33
Kudos
Bookmarks
The answer I got is 2052. Detailed solution below :
1 5 _ _ _ _
There are two cases possible. If the unit digit is 6 and if the unit digit is not 6.
(i) If the unit digit is 6
1 5 _ _ _ 6
6 can be placed in any of the three places, and the remaining can be arranged in 9*9 ways. Therefore, the total number of ways =3*9*9 = 243
(ii) If the unit digit is not 6
1 5 _ _ _ _
The last digit will have 4 possibilities. (0,2,4,8)
For the remaining digits, 2 digits will be 6, and the other can be any number other than 6.
Therefore, the total number of ways = 3*9*4 = 108
2 6 _ _ _ _
(i) If the last digit is 6
2 6 _ _ _ 6
Total number of ways = 9*9*9 = 729
(ii) If the last digit is not 6
2 6 _ _ _ _
The last digit has 4 possibilities, and the remaining numbers can be arranged in 3*9*9 ways.
Therefore, the total number of ways = 4*3*9*9 = 972
Maximum number of trials required = 108 + 243 + 729 + 972 = 2052
1 5 _ _ _ _
There are two cases possible. If the unit digit is 6 and if the unit digit is not 6.
(i) If the unit digit is 6
1 5 _ _ _ 6
6 can be placed in any of the three places, and the remaining can be arranged in 9*9 ways. Therefore, the total number of ways =3*9*9 = 243
(ii) If the unit digit is not 6
1 5 _ _ _ _
The last digit will have 4 possibilities. (0,2,4,8)
For the remaining digits, 2 digits will be 6, and the other can be any number other than 6.
Therefore, the total number of ways = 3*9*4 = 108
2 6 _ _ _ _
(i) If the last digit is 6
2 6 _ _ _ 6
Total number of ways = 9*9*9 = 729
(ii) If the last digit is not 6
2 6 _ _ _ _
The last digit has 4 possibilities, and the remaining numbers can be arranged in 3*9*9 ways.
Therefore, the total number of ways = 4*3*9*9 = 972
Maximum number of trials required = 108 + 243 + 729 + 972 = 2052
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