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06 May 2016, 08:57
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76% (01:20) correct
24%
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Each of 12 people chose an integer from 1 to 5, inclusive. Did at least one person choose the number 1?
1) No number was chosen by more than 3 people.
2) More people chose the number 5 than the number 4.
1) No number was chosen by more than 3 people.
2) More people chose the number 5 than the number 4.
24 Apr 2017, 09:35
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Quote:
Each of 12 people chose an integer from 1 to 5, inclusive. Did at least one person choose the number 1?
1) No number was chosen by more than 3 people.
2) More people chose the number 5 than the number 4.
1) No number was chosen by more than 3 people.
2) More people chose the number 5 than the number 4.
The language in (1) threw me off and I was not able to understand what was going on in the question.
For someone who is also confused by it, all it says that "a number can not be selected more than 3 times".
With that in mind, here is the solution:
(1) The constraint is that no number can be selected more than 3 times.
Case 1: Is it possible that 1 is not selected at all with this constraint?
Yes, that is possible. 12 people can pick numbers { 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5 } and still meet the constraint in the condition.
Case 2: Is it possible that 1 is selected at least once with this constraint?
Yes, this is also possible. 12 people can pick numbers { 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5 } and still meet the constraint in the condition.
Since both the cases are possible, this condition is INSUFFICIENT.
(2) This condition only tells you that 5 is selected more times than 4
Case 1: Is it possible that 1 is not selected at all with this constraint?
Yes, that is possible. Consider 12 people picking numbers { 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 } and still meeting the constraint in this condition.
Note that this is only 1 possible case out of many such cases in which 1 will not appear and we can have more 5s than 4s.
Case 2: Is it possible that 1 is selected at least once with this constraint?
Yes, this is also possible. 12 people can pick numbers { 1, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 } and still meet the constraint in the condition.
Note that this is only 1 possible case out of many such cases in which 1 will appear and we can have more 5s than 4s.
Since both the cases are possible, this condition is also INSUFFICIENT.
(1) + (2):
Since a number can only be selected at most 3 times and there are more 5s than 4s, consider this:
If 1 is not selected, the only possible selection that satisfies condition (1) is { 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5 } BUT this selection does not satisfy (2) condition. With similar reasoning, you can see that if you try to satisfy (2) condition you will break (1) without having 1 selected by at least 1 person.
Hence, for both the conditions to be true simultaneously, 1 should be selected by at least 1 person. SUFFICIENT
Answer: C
06 May 2016, 09:37
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Sallyzodiac
Each of 12 people chose an integer from 1 to 5, inclusive. Did at least one person choose the number 1?
1) No number was chosen by more than 3 people.
2) More people chose the number 5 than the number 4.
1) No number was chosen by more than 3 people.
2) More people chose the number 5 than the number 4.
1) No number was chosen by more than 3 people.
there are 12 people and they have to choose any of the 5 numbers..
worst case -- all 2, 3, 4 and 5 were choosen by 3 each.. so total = 4*3 =12..
hence no one chose 1.. NO
In all other scenarios, some one chose 1..YES
Insuff
2) More people chose the number 5 than the number 4
Nothing much
Insuff
combined..
it tells us that 4 was not choosen by 3 but less than 3..
so worst case 2, 3, and 5 were chosen by 3 each and 4 by2.. total 3*3+2 = 11..
so atleast ONE chose 1
Suff
C
