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8 schools sent a total of 96 students to a math club competition, with [#permalink]
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17 Oct 2014, 15:33
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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.
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Re: 8 schools sent a total of 96 students to a math club competition, with [#permalink]
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17 Oct 2014, 23:57
Since no two schools sent same number of students, minimum number of students sent by last 5 school is 15 (1+2+3+4+5).
Statement 1 : The most number of students that were sent by any one school was 34. so max number of students that can be sent by 2nd and 3rd school combined = 961534=47 The school positioned 2nd can send max 33 students. The school positioned 3rd can send min 14 students (4733). so the 3rd positioned school might or might not send atleast 15 students. statement is insufficient.
Statement 2 : The second most number of students that were sent by any school was 33. so max number of students that can be sent by 1st and 3rd school combined = 961533=48 The school positioned 1st must send min 34 students, so school positioned 3rd can send max 14 students (4834). Hence there is no possibility of school positioned 3rd sending 15 or more students. Statement is sufficient.
Ans  B




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Re: 8 schools sent a total of 96 students to a math club competition, with [#permalink]
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18 Oct 2014, 00:02
earnit wrote: 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. B. 1) most = 34 possible arrangement = 1,2,3,4,5,15,32,34 or 1,2,3,4,5,14,33,34 inconclusive. 2) second most = 33 possible arrangement = 1,2,3,4,5,14,33,34 we can increase the last value but that would again mean the third highest value going down. in all cases it would be less than 15. hence this is sufficient.
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Re: 8 schools sent a total of 96 students to a math club competition, with [#permalink]
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22 Nov 2015, 02:00
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.
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Re: 8 schools sent a total of 96 students to a math club competition, with [#permalink]
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10 Oct 2017, 05:09
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.



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Re: 8 schools sent a total of 96 students to a math club competition, with [#permalink]
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12 Oct 2017, 18:15
earnit wrote: 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. 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
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Re: 8 schools sent a total of 96 students to a math club competition, with [#permalink]
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19 Apr 2018, 09:43
+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 !
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