Bunuel
Ten slips of paper, numbered 1 through 10, are placed in a bag. If three slips are chosen at random from the bag without replacement and one of the slips chosen is numbered 7, what is the sum of numbers on the three chosen slips of paper?
(1) The sum of two of the numbers chosen is 16.
(2) The sum of two of the numbers chosen is 14.
I didn’t find any of the offered explanations sufficient (pun-intended), and thought I’d add my approach in case it helped anyone understand this DI question a tad bit more...
1) The first statement alludes to any of the three chosen slips adding up to 16 (which two is unspecified). Thus, if the slips chosen are:
x, 7, y,
x+y=16, x+7=16 or 7+y=16
Many pairs of numbers could satisfy each equation, making this statement insufficient.
2) The second insinuates any of the three chosen slips add up to 14 (which two is unspecified). Thus, if the slips chosen are:
a, 7, b
a+b=14, a+7=14 or 7+b=14
a=b=7
Hence, the second is indeed sufficient.
This leads us to B as the correct answer, as stated by the experts above 😇