Here greg mentions option 1 can’t be true. But I think I have proved it below. Can someone confirm it
You can have c = 39, b = 1 and a = 29. Remember that c does not include b but does include a, so you cannot determine a uniquely.
Oh I get it, a+b+c+n = 80
Thanks!
Just following up on this since I am still confused. I don’t understand why A is not valid answer here because isn’t value b also derived from the equation? We aren’t choosing numbers here right because if the logic that we are using to invalidate A is this because there is no unique solution then even in the answer choice C there isn’t a unique solution because we are saying that b= 7 and d=11 by that logic -
a+7+c+11 = 80
a+c= 62
Now, you can get combination of 62 in different ways without violating any assumption so I still don’t understand the logic behind rejecting A. Could you explain this a bit more @Leaderboard ?
Extremely sorry for the inconvenience.
That’s the problem - the problem implies that you must be able to get a unique answer.
Yes, this is what I am asking. We want to find unique solutions to this and based on that we are rejecting A but when we look at option C that also doesn’t provide unique solutions so why is C chosen as a correct answer and A not?
As I highlighted, even in option C : a+c = 62 that means c and c can be anything they just need to be sum upto 62 how is this unique but not A?
The thing with C is that a + c = 62 - and you would almost be correct, except that here the question is asking for a + c - and that is unique. If, for instance, you could find two values of a + c for option C, that would not work.
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ohhhh…that makes total sense. Thank you!