When positive integer n is divided by 25, the remainder is 13. What is the value of n?
1) N < 100
2) When N is divided by 20, the remainder is 3
Yes, Sudhan, but it's not clear from your solution why (1&2) is sufficient. Here's a numerical approach.
The possible values of N are 13, 38, 63, 88, 113, 138, 163, 188, 213... what is n?
(1) Not enough information, as there are many values less than 100 in our list.
(2) So, N could be 3, 23, 43, 63, 83, 103, 123, 143, 163, 183, ... Again, we have at least two cases: 63 and 163. Insufficient.
(1&2) This narrows it down to exactly one value: 63. Sufficient.
Another way to deal with it is to express N as functions.
N = 25x + 13. What is N? (Here, the variable x can take on integer values. For each x, you'll get a corresponding value of N.)
(1) N < 100 is not sufficient. If x is 1, for example, N is 38 and if x is 3, N is 88.
(2) We are told that N can be expressed as 20y + 3 (where y is an integer). So we can attempt to solve the system:
N = 20y + 3
N = 25x + 13
Can't do it! Only two equations and three variables. Insufficient.
(1&2) Now let's go back to that system from (2). It yields:
20y + 3 = 25x + 13
20y = 25x + 10
So now it's a matter of finding values of x and y that make the equation true. If x = 1, the right side is 35, and that's not a multiple of 20 (as required by the left side). If x = 2, the right side is 60, and that *is* a multiple of 20. So x = 2 and y = 3 is a viable solution. That corresponds to a value of N = 63. Keep going, though. If x = 3, the right side is 85 -- no good, since 88 is not a multiple of 20. If you keep going with this exercise, the next value of x that works is x = 6, since y would equal 8. BUT, x = 6 corresponds to a value of N = 163, which is bigger than 100. Sufficient.
I think you'd agree that for these problems, a numerical solution is probably most efficient. But either way gets you there.