8 schools sent a total of 96 students to a math club competition, with

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

8 schools sent a total of 96 students to a math club competition, with each school sending at least one student. If one school sent the third most number of students to the competition (and no two schools sent the same number of students), did that school send at least 15 students?

(1) The most number of students that were sent by any one school was 34.

(2) The second most number of students that were sent by any school was 33.

There are 8 variables (8 schools) and 2 equations (total=96, at least=min=1) in the original condition, but only 2 equations are given by the conditions, so there is high chance (E) will be our answer.
Looking at the conditions together,
if we let T be the school that is the third from top sending students,
1+2+3+4+5+15+33+34=97>96, so T cannot be 15, so it is always 'no' and the conditions is sufficient, making the answer seem like (C); this is an integer type of question, question with commonly made mistakes, so if we apply mistake type 4(A),
For condition 1, 1+2+3+4+5+14+33+34(no), 1+2+3+4+5+16+31+34 (yes), so this is insufficient.
For condition 2, the answer is always 'no' in 1+2+3+4+5+15+33+34, so this is also sufficient, and the answer becomes (B).

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

This question can be approached in the following way: By calculating the maximum number of students sent by the school that sent the third highest number of students which can be done by minimizing the number of students sent by all other schools and by calculating the minimum number of students sent by the school which sent the third highest number of students which can be done by calculating the maximum number of students sent by all other schools.

For this question we have to assume that the minimum number of students sent by a school is 1.

Statement 2 is sufficient because it provides enough information to get a definite no to the question- is the minimum number of students sent by the school which sent the third highest number of students >=15.

We are given that 8 schools sent a total of 96 students to a math club competition, with each school sending at least one student. We also know that no two schools sent the same number of students. We need to determine whether the school that sent the third most number of students sent at least 15.

Statement One Alone:

The most number of students that were sent by any one school was 34.

The information in statement one is not sufficient to answer the question. For example, we can have the following scenarios of students sent, from greatest to least:

34 + 30 + 15 + 9 + 4 + 3 + 1 = 96

or

34 + 30 + 14 + 10 + 4 + 3 + 1 = 96

We see that the school that sent the third most students could have sent at least 15 students or fewer than 15 students.

Statement Two Alone:

The second most number of students that were sent by any school was 33.

The information in statement two is sufficient to answer the question. Let’s say the greatest number of students sent by a school was 34, the smallest number of students sent by a school was 1, the second smallest was 2, the third smallest was 3, the fourth smallest was 4, and the fifth smallest was 5, while the third most (the one in the equation) is x. We can create the following equation:

34 + 33 + x + 5 + 4 + 3 + 2 + 1 = 96

x + 82 = 96

x = 14

We see that, in this case, the third greatest is 14 students, which is less than 15. In the equation above, we can modify the numbers (except 33). However, any numbers we modify have to be larger; for example, if the greatest number of students is not 34, then it has to be 35 or larger. This will cause the value of x to be even smaller (i.e., less than 14). Thus, we will never have at least 15 students sent by the school with the third greatest number of students.

Answer: B

+1 for option B.

St 1 - We only know the maximum nos of students. The third highest can't be determined. NS
St 2 - The second highest is 33. Take the highest as 34. We have the max value of third highest as 14. An increase in highest will only reduce the third highest qty. Hence it will never be 15. Sufficient.

Option B it is !

These “scenario-driven” min/max problems require that you consider extremes in order to find the asked about limit. The question is asking whether the school that sent the 3rd highest number of students must have sent at least 15 students.

In statement 1, you learn that the school with the most students sent 34 to the math club competition. Remove those 34 from 96 to see that the remaining 7 schools must account for 62 students. This can be done with the 3rd highest school having at least 15 (giving a yes answer to the question) or it can be done with the 3rd highest having less than 15 (giving a no answer to the question), so this statement is not sufficient. To show this, prove that it can work with 15 and with 14. If the third highest has 15 students, then the second highest could have 16 and now you have accounted for 34 + 16 +15 or 65 for the top 3 schools, leaving 31 for the remaining 5 schools to be distributed any number of ways. Likewise, the top three could be 34, 16, and 14 accounting for 64 students, leaving 32 students to be distributed any number of ways among the remaining 5 schools.
The second statement is trickier. If the second most school sent 33 students, then the smallest number that the most could have sent is 34 students. Statement 2 means that at a minimum, the top two schools have accounted for 67 students, leaving 29 to be accounted for by the remaining 6 schools. If the third highest school accounted for 15 of those 29 students, then the remaining 5 would have to account for 14 students, and that is not possible! The smallest numbers that the remaining 5 schools could have sent would be 1,2,3,4,5 or 15 total students. This guarantees that the third highest school must have sent 14 or fewer and this statement is sufficient as it proves definitely that the third highest school did NOT sent at least 15 students.

8 schools - have difft points
Min points for each school will 8,7,6,5,4,3,2,1 that is in total 36
Stmt 1
the highest point is 34
So from the balance we can distribute a minimum of 26 (if added to 8) and max of 33(if added to 1)
We are left with 34 or 27 as the case may be
Adding 26 to 7 =33; we are still left with 7 points which when added to 6 gives us 13
This distribution can go either ways resulting in the third highest school greater than or less than 13

Stmt II gives the number of second highest school which means the first highest bare minimum would be 34. And the above scenario can be followed
We can then definitely say that 15 is not possible
Sufficient

Official Explanation:
These “scenario-driven” min/max problems require that you consider extremes in order to find the asked about limit. The question is asking whether the school that sent the 3rd highest number of students must have sent at least 15 students.

In statement 1, you learn that the school with the most students sent 34 to the math club competition. Remove those 34 from 96 to see that the remaining 7 schools must account for 62 students. This can be done with the 3rd highest school having at least 15 (giving a yes answer to the question) or it can be done with the 3rd highest having less than 15 (giving a no answer to the question), so this statement is not sufficient. To show this, prove that it can work with 15 and with 14. If the third highest has 15 students, then the second highest could have 16 and now you have accounted for 34 + 16 +15 or 65 for the top 3 schools, leaving 31 for the remaining 5 schools to be distributed any number of ways. Likewise, the top three could be 34, 16, and 14 accounting for 64 students, leaving 32 students to be distributed any number of ways among the remaining 5 schools.

The second statement is trickier. If the second most school sent 33 students, then the smallest number that the most could have sent is 34 students. Statement 2 means that at a minimum, the top two schools have accounted for 67 students, leaving 29 to be accounted for by the remaining 6 schools. If the third highest school accounted for 15 of those 29 students, then the remaining 5 would have to account for 14 students, and that is not possible! The smallest numbers that the remaining 5 schools could have sent would be 1,2,3,4,5 or 15 total students. This guarantees that the third highest school must have sent 14 or fewer and this statement is sufficient as it proves definitely that the third highest school did NOT sent at least 15 students. Answer is (B).

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