Miguels online banking password is nine characters long and includes

Question

Miguel’s online banking password is nine characters long and includes at least one character of each of five types: digits, punctuation marks, uppercase (capital) letters, lowercase (noncapital) letters, and certain other characters. The password begins or ends with one of these other characters, of which it contains exactly one. No consecutive characters in the password are of the same one of these types. (For example, no capital letter is preceded or followed by another capital letter.) The password contains more lowercase letters than characters of any one of the other four types. It also contains fewer punctuation marks than digits, lowercase letters, or uppercase letters. The first three characters of the password are M, ?, and G.

In the table, select a type of character that the fifth character in the password must be and a type of character that the seventh character must be. Make only two selections, one in each column.

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ID: 100324

In the table, select a type of character that the fifth character in the password must be and a type of character that the seventh character must be. Make only two selections, one in each column.

ID: 100324

bM22 · Accepted Answer

The password can contain: Lowercase(LC), uppercase (UC), digit(D), punctuation(P) & other characters(OC).
Now, the criteria are given as following: 
1. LC>any of the other types.
2. P< any of the other types.
3. No consecutive characters in the password are of the same one of these types.
4. The password begins or ends with one of these other characters (OC), of which it contains exactly one

What we are given: M ? G ___ _5th__ ___ __7th_ ___ ____

Now, since LC needs to be greatest => LC can be only 3, since if we take 4 LC's our 3 criteria is violated.

Then, no more UC, since no. of UC types cannot be greater that LC types, as per criteria 1.

No. of Digits(D) can be either 2 since it cannot be greater than LC. It cannot be 1 since it cannot be less than the no. of Punctuation type.

We can only use 1 OC which has to be added as the last character of the password, as per criteria 4.

No. of punctuation has to be the least when compared to no. of LC, UC or D, and " ? " is already present, so at max the no. of Punctuation types can only be 1, otherwise it would violate the 2nd criteria, since no. of digits is already 2.

So, we would exactly be needing: LC =3, D=2, OC= 1.
 
Next comes the order in which we have to place them, keeping in mind the criteria 3.

So the best combination that comes when we place the characters is:
M ? G _a__ __1_ _b__ _9__ __c_ _$__

Now, we can try arranging in any way possible, it would always violate either criteria 3 or 4. Therefore, 5th and 7th characters of the password will always be digits.

Ans. Digit and Digit

Apt0810 · Answer

Given that the first three characters are M,?,and G ,now since password begins or ends with the "other character" and the first three letters are known ,means the last letter is that "other letter"

Now the sequence could be M,?,G or G,?,M,_,_,_,_,_,'other character'

And we also know that lowercase letters are the most to be present and least ones are punctuations

So the probable car with which we are left is 2 digits and 3 lowercase letters since if we increase these numbers or put any other type in this condition it will not match the above given prerequisites,

The remaining five letter sequence would be lowercase,digit,lowercase,digit, lowercase and one thing to keep in mind here is that we can't keep the 4th digit as digit otherwise consequent repetition problem would come ,so the above is the only possible solution.Hence the fifth and seventh digit are digits

Posted from my mobile device

Sajjad1994 · Answer

Official Explanation

The passage indicates that Miguel’s password is nine characters long, contains at least one character of each of five different types—digits, punctuation marks, uppercase letters, lowercase letters, and "other characters"—and either begins or ends with one of the "other characters." We are also given the first three characters the password: M, ?, and G. So, because the password begins with an uppercase letter, it must end with one of the "other characters." We know the password contains more lowercase letters than characters of any of the other types, so it must contain at least three lowercase letters. We also know that there is at least one punctuation mark and fewer punctuation marks than digits, lowercase letters, or uppercase letters, so there must be at least two digits.

Given that the password is nine characters long, one can now determine the distribution of all the characters in the password across all five of the types: one other character, one punctuation mark, two digits, two uppercase letters, and three lowercase letters. We know that the two uppercase letters and the one punctuation mark are among the first three characters and that the "other character" is the ninth character of the password. Therefore, characters 4 through 8 of the password consist of two digits and three lowercase letters. The passage indicates that no consecutive characters are of the same type, and this entails that characters 4, 6, and 8 are all lowercase letters, otherwise two or more lowercase letters would appear consecutively. If characters 4, 6, and 8 are all lowercase letters, then both character 5 and character 7 must be digits.

The correct answer is Digit for both.

nisen20 · Answer

The passage said: The first three characters of the password are M,     ?    , and G.

This question mark represents a punctuation character, not an unknown character.

DN10 · Answer

It would've helped if the question mentioned "?" is a punctuation character.

I followed the following approach to solve this question

M?G _ _ _ _ _ _ (M is in position 1)

Most important part - The password begins or ends with one of these other characters,  of which it contains exactly one .

Each type of character occurs at least once.
Its given lowercase characters have the highest frequency ( should be greater than 2, because we know M and G are part of the password)
Also, its given frequency of punctuation marks is the lowest ( It has to be 1, so that frequency of digits is greater than 1)

From this, we can infer frequencies of each character type, lowercase=3, uppercase=2, punctuation=1, digits=2, other=1 and that last character type is "other" character.

As a result, lowercase characters are in position 4, 6 and 8.

Digit characters are in position 5 and 7.

Gemmie · Answer

D: Digit
P: Punctuation marks
U: Uppercase (capital) letters
L: Lowercase (noncapital) letters
C: Certain characters

(i) The first three characters of the password are M, ?, and G. 
 (ii) The password begins or ends with one of these other characters, of which it contains exactly one. 
 (iii) No consecutive characters in the password are of the same one of these types. 
 (iv) D + P + U + L + C = 9 
 (v) L > D, P, U, C
(vi) P < D, L, U

(i) and (ii) => C = 1; U = 2

=> Combined with (iv): D + P + L = 6

=> Combined with (v) and (vi):  L = 3 , D = 2; P = 1

=> Combined with (iii):

1: M  (U)
2: ?  (P)
3: G  (U) 
 4: (L) 
5: (D)
 6: (L) 
7: (D)
 8: (L) 
 9: (C)

=> Both 5th and 7th character are Digits

KarishmaB · Answer

Video Solution to this Question is here: https://youtu.be/WassWEtmcGo

Venu01298 · Answer

Appreciate all the inputs here. I took almost 5 minutes to solve this question. My question is how can we aim to answer these types of questions in under 2 mins (as they feel totally new)? Just understanding the question and putting the constraints together itself takes the 2 minute slot. Of course, if i see another exact similar question, i may know how to solve it faster. But generally what is the strategy with these types of questions? Is there any fundamental thinking approach change needed or we just have to work out and be familiar with as many types of questions as possible before taking the test?

Thanks

Thanks

tthannhnha · Answer

I'm confused. Is there any rule about the second character? there're other possible ways I can apply on these characters to other positions.

D: Digit
P: Punctuation marks
U: Uppercase (capital) letters
L: Lowercase (noncapital) letters
C: Certain characters

1: M (U)
2: ? (L)
3: G (U)
4: (D)
 5: (L) 
6: (D)
 7: (L) 
8: (P)
9: (C)

1: M (U)
2: ? (D)
3: G (U)
4: (L)
 5: (D) 
6: (L)
 7: (P) 
8: (L)
9: (C)

HarshavardhanR · Answer

Hey  tthannhnha ,

The second character is given to us as "?".

This is a Punctuation mark (P). Punctuation mark examples -> "," ";" "." "!"  "?"  etc.

So, consider that the second character is a Punctuation Mark (P) and then try solving!

HarshavardhanR · Answer

D -> Digits  
 P -> Punctuation marks  
 U -> Uppercase 
 L -> Lowercase 
 O -> Other characters

Password:    M   ?   G  _ _ _ _ _ _

i.e.,

Password:    U   P   U  _ _ _ _ _ _

->  The password begins  or  ends with one of the "other" characters  .

We can see that the password does not begin with an "other" character. So, it must end with an "other" character.

Password:    U   P   U  _ _ _ _ _  O

->   The password contains fewer punctuation marks than digits, lowercase letters, or uppercase letters .

Think  ->

(1) If #P in the password is even as high as 2,

Then, because of the above condition,

#D, #L, and #U in the password must all be  at least  3. Then,  the total number of characters in the password will exceed 9. Not possible .

(2) #P is at least 1 (given)

Thus,

#P = 1

So, the only "P" in the password is character 2 (?).

Password:   U   P   U  _ _ _ _ _  O

->  The password contains more lowercase letters than characters of any one of the other four types .
 
 Think  ->

#L in the passcode has to be >1 (because it has to be greater than #P, as seen above).

But this new condition also means that #L in the password cannot be 2.

Because -> given that #D and #U also need to be at least 2, #L will have to be greater than 2 to ensure that the password has more lowercase letters than #characters of any of the other 4 types.

With this, and given the constraint that the total number of characters is 9, we can arrive at the following ->
 
   D 
 P 
 U 
 L 
 O 
 
  2 
 1 
 2 
 3 
 1

Now, observe the password again ->

Password:   U   P   U  _ _ _ _ _  O

-> P and O are accounted for (char 2 and char 9).
-> Both the U's are accounted for (char 1 and char 3)
-> So, chars 4,5,6,7,8 are to be filled by 3 Ls and 2 Ds.

The only way we can ensure that no 2 Ls are together (remember -   No consecutive characters in the password are of the same one of these types )

is if we alternate the Ls, starting from char 4.

i.e.,

Password:   U   P   U   L    D    L    D    L   O

So, the 5th and 7th characters of the password are  both digits .

Hope this helps!

Harsha

RishiSpi · Answer

Quicker way to solve the problem as per me.

Step 1. Decide frequency

Write 1 in front of all

UpperC 1
LowC 1
Punct. 1
Dig 1
Other 1

Now keep adding 
UpperC 1 + 1. --- Given
LowC 1 +1 ---- Greater than Punc + 1 Since more than all 
Punct. 1
Dig 1 + 1----Greater than punt
Other 1
Total 9

Placing

M ? G - - - - - Oth
Now there is only one way to place 3 LowC, that is separated by 2 Dig
Thus 
M?GLDLDLO

In any case it would take min three mins

Fifth character	Seventh character
		Digit
		Lowercase letter
		Punctuation mark
		Uppercase letter
		Other character

Miguels online banking password is nine characters long and includes

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