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gmat2015p
Joined: 27 Dec 2014
Last visit: 29 Apr 2018
Posts: 43
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Concentration: Leadership, Technology
Schools: Insead July'17 MIT Sloan Fellows"18
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12 Dec 2015, 03:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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62% (01:34) correct
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The smallest of 7 positive integers is 1. Is the range of them greater than 3?
(1) The average (the arithmetic mean) of them is 4.
(2) The median of them is 4.
(1) The average (the arithmetic mean) of them is 4.
(2) The median of them is 4.
12 Dec 2015, 07:55
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gmat2015p
Solution:
STATEMENT 1) The average (the arithmetic mean) of them is 4.
\(Total/7=4---->Total=28\)
one of them is 1, then sum of remaining integers is 27.
if remaining integers are all 4, then 4*6=24 to get sum of 27 atleast one of them must be greater than 4.
So the range-----> \(integer(>4)-1=integer(>3)\)
SUFFICIENT.
STATEMENT 2) The median of them is 4.
1,2,3,4,4,4,4---> median=4 and range=4-1=3--> ANS: NO.
1,2,3,4,5,6,7--->median=4 and range= 7-1=6--> ANS: YES.
Hence INSUFFICIENT.
ANS. A
General Discussion
13 Dec 2015, 18:18
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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.
The smallest of 7 positive integers is 1. Is the range of them greater than 3?
(1) The average (the arithmetic mean) of them is 4.
(2) The median of them is 4.
In the original condition, there are 7 variables and 1 equation. In order to match with the number of equations, you need 6 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. In 1) & 2), when a question asks ‘greater than,’ you have to find a minimum value(because once find the minimum value, the rest is obviously going to be bigger than the minimum value). So, in order to find the minimum value of the range, the range can be minimum when 1,4,4,4,5,5,5. That is, range=5-1=4>3 -> yes, which is sufficient and therefore the answer is c. This is an integer question which is one of the key questions. When applying 4(A) of the mistake type, you have to find a minimum value in 1). Its range can also be minimum when 1,4,4,4,5,5,5. That is, range=5-1=4>3 -> yes, which is sufficient.
In 2), 1,4,4,4,5,5,5 range=5-1=4>3 -> yes, but 1,4,4,4,4,4,4 range=4-1=3>3 -> no, which is not sufficient.
Therefore, both C and A can be the answer. In this case, A is the answer and therefore A is the answer, You need to get this type of questions correct to reach the score range 50-51.
-> 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.
The smallest of 7 positive integers is 1. Is the range of them greater than 3?
(1) The average (the arithmetic mean) of them is 4.
(2) The median of them is 4.
In the original condition, there are 7 variables and 1 equation. In order to match with the number of equations, you need 6 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. In 1) & 2), when a question asks ‘greater than,’ you have to find a minimum value(because once find the minimum value, the rest is obviously going to be bigger than the minimum value). So, in order to find the minimum value of the range, the range can be minimum when 1,4,4,4,5,5,5. That is, range=5-1=4>3 -> yes, which is sufficient and therefore the answer is c. This is an integer question which is one of the key questions. When applying 4(A) of the mistake type, you have to find a minimum value in 1). Its range can also be minimum when 1,4,4,4,5,5,5. That is, range=5-1=4>3 -> yes, which is sufficient.
In 2), 1,4,4,4,5,5,5 range=5-1=4>3 -> yes, but 1,4,4,4,4,4,4 range=4-1=3>3 -> no, which is not sufficient.
Therefore, both C and A can be the answer. In this case, A is the answer and therefore A is the answer, You need to get this type of questions correct to reach the score range 50-51.
-> 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.
