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Bunuel
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Bunuel
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Bunuel
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Please explain:-
There are many ways to get a2+a3+a4=22 ,and it may or may not require one of the number to be 20, so 20 can be used 0 time or 1 time...so statement 1 is not sufficient...

The questions asks: how many integers of these four are greater than 20? We are given that the integers are positive and know from (1) that x + y + z = 22. None of the x, y, and z can be more than 20 here, because the least sum in this case would be 21 + 1 + 1 = 23. So, only one integer out of four is more than 20, the one which is given to be 98.

Hope it's clear.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I have the following doubt:

SUM (98, 19, 2, 1) = 120, but with none of the other 3 digits greater than 20. How can we say that the answer is A? Shouldn't it be E?
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I have the following doubt:

SUM (98, 19, 2, 1) = 120, but with none of the other 3 digits greater than 20. How can we say that the answer is A? Shouldn't it be E?

All possible sets, satisfying (1), including your set, give the same answer: there is only one integer greater than 20, which is 98.
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Oh ****, yes! Sorry, my bad. :D

Posted from my mobile device
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