fozzzy wrote:
Could you elaborate a bit more it isn't very clear. I approached this question differently. Let n = abc
abc
+ 8 in this scenarios b is 6 so c = 0,1,2,3 then we don't get a carryover b=6 but if c=4-9(range) then we get a carryover so b=5
Given statement 1 is sufficient since there will be carryover so c=5
Statement 2
abc
- 8 then in this case b=4 if c=9,8 then there is no carryover in that case c=4 but if c=0-7(range) then b=5
but my question in using this approach c can only be 4,5,6,7 ( I know its irrelevant for this question but I'm asking this for conceptual clarity?)
So can someone explain this part
Could you state exactly which part was not clear?
Anyways, here is another approach : From F.S 2, we know that \(40\leq{n-8}\leq{49} \to 48\leq{n}\leq{57}\)
Also, the question stem states that : \(60\leq{n+8}\leq{69} \to 52\leq{n}\leq{61}\)
The common intersection of both the in-equalities is \(\to 52\leq{n}\leq{57}\). So , yes, "c" as in your example can range only from 2 to 7.
Hope this is clear.
Note: When I say \(52\leq{n}\leq{57}\), it doesn't mean that n is a 2 digit number. It is just a scalable inequality for its last 2 digits.
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