Two positive numbers differ by 2001. When the larger number is divided by the smaller number, the quotient is 9 and the remainder is 41. The sum of the digits of larger number is?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 14
x - y = 2001 -(1) [where x=larger ,y=smaller]
x = 9y + 41 -(2)
from (1)
x = 2001 + y -(3)
putting (3) in (2)
2001 + y = 9y + 41
8y = 1960
y = 245
x = 2246
sum of the digits = 2 + 2 + 4 + 6 = 14