Join 60,000+ competitive exam aspirants
or with email
Your Cart
SubtotalтВ╣0
TotalтВ╣0
Examoogle тАв User тАв info@examoogle.com тАв EE-2024-8821
Chapter 1 of 12 тАв Page 1 of 248ЁЯФТ Protected PDF тАв Watermarked
Join 60,000+ competitive exam aspirants
If 1 is added to each odd digit and 2 is subtracted from each even digit in the number 485293, how many digits appear more than once in the new number?
1
2
3
0
2
Transform each digit individually according to the given parity rule, then count the frequency of each resulting digit to find duplicates.
Transform each digit individually according to the given parity rule, then count the frequency of each resulting digit to find duplicates.
Apply Rules to Each Digit
Take the number 485293 and apply the operations: subtract 2 from even digits (4, 8, 2) and add 1 to odd digits (5, 9, 3).
Determine Transformed Digits
The transformations are: 4 gives 2, 8 gives 6, 5 gives 6, 2 gives 0, 9 gives 10 (taking unit digit 0), and 3 gives 4. The new digits are 2, 6, 6, 0, 0, and 4.
Count Frequencies of New Digits
Frequencies: 0 appears 2 times, 2 appears 1 time, 4 appears 1 time, and 6 appears 2 times. Thus, digits 0 and 6 appear more than once, giving a total of 2 digits.
A: 1 is incorrect because both 0 and 6 appear twice, making the count 2. C: 3 is incorrect as there are only two repeating digits, not three. D: 0 is incorrect because digits 0 and 6 both repeat.
B is correct because transforming the digits of 485293 yields 2, 6, 6, 0, 0, 4, where digits 0 and 6 both appear twice (more than once), resulting in a count of 2.
Always pay close attention to whether operations on 9 result in single digits via modulo arithmetic and double-check digit frequencies carefully.
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download тАв 100K+ Aspirants