Hi, this is my approach,
Since N is a three digits number, 0 < hundreds digit <= 9.
So, hundreds digit could be (1,2,3,4,5,6,7,8,9)
St#1: The hundreds digit of N+120 is 7
As we do not have information about tens digit of N, so the hundreds digit of N could be 5 or 6. --->Not sufficient.
St#2: The tens digit of N+15 is 9
As we do not have information about de units digits of N, so the tens digit of N could b 7 or 8. And this does not give information about the hundreds digit of N ---> Not sufficient.
St#1 & St#2
Combining the 2 statements, we have that, N could be 57U, 58U, 67U or 68U, where U= units digits of N.
If I take 9 as units digit of N to to complete the number 57U we have: 579. Adding 579+120 we have 699 so, that contradicts St#1: The hundreds digit of N+120 is 7 , ( 6)
If I take 9 as units digit of N to to complete the number 58U we have: 589. Adding 589+120 we have 709 so, that confirms St#1: The hundreds digit of N+120 is 7 (7). .
If I take 9 as units digit of N to to complete the number 67U we have: 679. Adding 679+120 we have 799 so, that confirms St#1: The hundreds digit of N+120 is 7 , ( 7)
If I take 9 as units digit of N to to complete the number 68U we have: 689. Adding 689+120 we have 809 so, that contradicts St#1: The hundreds digit of N+120 is 7 (8).
---> Not sufficient because we have 5 or 6 in the hundreds digit of N
Answer : E
Do you think that my approach is correct?
Thanks