Question
Download Solution PDFSolve the given puzzle:
If 1111 = R,
2222 = T,
3333 = E,
4444 = N
5555 = ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe logic followed here is:
Logic: Sum of the number and Last letter of the number is Code.
For, 1111 = R,
⇒ 1 + 1 + 1 + 1 = 4(FOUR) = R.
For, 2222 = T,
⇒ 2 + 2 + 2 + 2 = 8(EIGHT) =T.
For, 3333 = E,
⇒ 3 + 3 + 3 + 3 = 12 (TWELVE) = E.
For, 4444 = N
⇒ 4 + 4 + 4 + 4 = 16(SIXTEEN) = N.
Similarly,
5555 = ?
⇒ 5 + 5 + 5 + 5 = 20(TWENTY) = Y.
Thus, 5555 = Y.
Last updated on Jul 4, 2024
-> The JK Police SI applications process has started on 3rd December 2024. The last date to apply is 2nd January 2025.
-> JKSSB Sub Inspector Notification 2024 has been released for 669 vacancies.
-> Graduates between 18-28 years of age who are domiciled residents of Jammu & Kashmir are eligible for this post.
-> Candidates who will get the final selection will receive a JKSSB Sub Inspector Salary range between Rs. 35,700 to Rs. 1,13,100.