harsh.champ wrote:One number, n, is selected at random from a set of 10 integers. What is the probability that ½ n + 13 = 0 ?

1. The largest integer in the set is 13.
2. The arithmetic mean of the set is zero.

(1) We know that the largest integer of the set is 13 but we know nothing about the lowest number, which could be up to - infinity. INSUFF.
(2) We know the mean of the set is 0 but nothing about the distribution of the numbers. INSUFF.
(1) AND (2) tell us that the largest integer is 13 and the mean is 0. So all the remaining 9 numbers have to be < 13. I have the feeling that we could answer this question now, even though I'm not able to provide a clear solution.

I would gamble C, but definitely I need a better explanation from someone out there!

harsh.champ wrote:One number, n, is selected at random from a set of 10 integers. What is the probability that ½ n + 13 = 0 ?

1. The largest integer in the set is 13.
2. The arithmetic mean of the set is zero.

1/2 n +13 =0 --> n = -26
St. 1 is insufficient. We don't know anything about what exactly other numbers are apart from numbers<13
St. 2 again insufficient. arithematic mean of set is zero ...a lot of combinations can do this .....by "a lot" i mean infinite.
combined: here also a lot of combination can make the arithematic mean 0 as we have a whole lot of no (-infinity,13)
so insufficient
I will go with E

cunazza wrote:

harsh.champ wrote:One number, n, is selected at random from a set of 10 integers. What is the probability that ½ n + 13 = 0 ?

1. The largest integer in the set is 13.
2. The arithmetic mean of the set is zero.

(1) We know that the largest integer of the set is 13 but we know nothing about the lowest number, which could be up to - infinity. INSUFF.
(2) We know the mean of the set is 0 but nothing about the distribution of the numbers. INSUFF.
(1) AND (2) tell us that the largest integer is 13 and the mean is 0. So all the remaining 9 numbers have to be < 13. I have the feeling that we could answer this question now, even though I'm not able to provide a clear solution.

I would gamble C, but definitely I need a better explanation from someone out there!

I think cunazza you are going right way but for "combined" how could u say that friend
I wanna show u some combinations...
1,2,3,4,13,-13,-4,-3,-2,-1
2,3,4,5,13,-13,-5,-4,-3,-2
1,2,3,4,5,6,7,8,9,10,13,-68
......................................
......................................and so on


so there could be a lot of such combination which could be formed..please correct me if I m wrong..

You are perfectly right, it's such a silly mistake I did. If I read it again now it's pretty obvious what you say.
I was timing myself and I was already over 1 minute when I came to analyze combined (1) and (2) so I did speed up my logic too much...
Your answer is correct.

cunazza wrote:You are perfectly right, it's such a silly mistake I did. If I read it again now it's pretty obvious what you say.
I was timing myself and I was already over 1 minute when I came to analyze combined (1) and (2) so I did speed up my logic too much...
Your answer is correct.

I appreciate you cunazza that you are using stop watch, a new tool by BTG moderators to solve problem in time...
Well I think you should give 2 min time for these problems.. And if ur taking a few seconds more than that, then also it s mangeable ..as there are lots of questions which you solve in few seconds only as your mind is very active that time....Hope I m correct..