swerve wrote:If # is an operation which results in adding the digits of integer until a single digit is left, what is the probability that a number picked up in first 90 positive integers will have the result of # as an odd digit? (example, 99=9+9=18=1+8=9, so #=9)
(A) 4/10
(B) 4/9
(C) 1/2
(D) 6/10
(E) 5/9
The OA is E.
Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
From 1 to 9 number of odds are 5
Odd + Even = Odd ... (1)
Odd + Odd = Even ... (2)
Even + Even = Even ... (3)
11 to 19: We are adding the 1 of the tens place to the 5 odd and 4 even numbers from (1 to 9)
Using (1), (2) and (3), we will get 5 evens and 4 odds
1 + 1 = even, 1 + 2 = odd and likewise.
But note that the last number 1 + 9 = 10 (even) but when 1 + 0 of 10 is added, we get 1 which is odd.
Hence from 11 to 19: We have 1 even being converted to odd. Hence odds = 4 + 1 = 5.
21 to 29: We are adding 2(even) of tens place to 5 odds and 4 evens of (1 to 9)
Using (1), (2) and (3), we will get 5 odds and 4 evens
But note the last 2 numbers i.e 28 and 29.
28: 2 + 8 = 10 (even) and 1 + 0 = 1 (odd)
29: 2 + 9 = 11 (odd) and 1 + 1 = 2 (even)
Hence from 21 to 29: We have 1 even being converted to odd and 1 odd being converted to even. Hence odds = 5 + 1 - 1 = 5.
31 to 39: Extrapolating the logic, we will get 5 evens and 4 odds.
37: 3 + 7 = 10 (even), 1 + 0 = 1 (odd)
38: 3 + 8 = 11 (odd), 1 + 1 = 2 (even)
39: 3 + 9 = 12 (even), 1 + 2 = 3 (even)
Hence from 31 to 39: We have 2 evens being converted to odds and 1 odd being converted to even. Hence odds = 4 - 1 + 2 = 5.
We can extrapolate this logic and say that each series will have 5 odds.
Starting from 1 - 9 and ending at 81 - 89, we have 9 series having 5 odds each. Hence the total number of odds is 45.
Let us have 1 series of 10, 20, 30 .. 90. This series will also have 5 odds. So the total number of odds is 45 + 5 = 50
Total numbers = 90.
Hence, probability is 50/90 = 5/9
Ans is
E.
