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List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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28 Dec 2012, 19:12
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List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A? (1) The integer 3 is in list A. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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28 Dec 2012, 20:46
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Archit143 wrote: List A contains 5 positive integers and the Mean of integers in the list is 7. If the integers in the list are 6,7, and 8, What is the range of the list A?
1. the integer 3 is in the list A. 2. The Largest term in A is greater than 3 times and less than 4 times the size of the smallest term.
Hello, can anyone help me in decoding the Statement 2...... One approach is to remember one key thing. Since its given that 7 is the mean, and 6 & 8 are on either sides of mean, therefore the other two numbers, namely x and y have to be equally spaced on either side of the mean. The effect of +1 of 8 is equally compensated by the effect of 1 of 6. Hence for 7 to reman the mean of the set, the other numbers have to do so as well. Statement 1) integer 3 is in the list A 3 is at a distance of 4 units from the mean on left hand side of number line. Therefore you are supposed to look for another number, but at the same distance on right hand side of the mean. \(7+4=11\) Hence the other number is 11. Now we know the smallest and the largest of the set. So range can be easily found out. Sufficient Statement 2) The Largest term in A is greater than 3 times and less than 4 times the size of the smallest term. One of the two numbers, the two that we don't know yet, have to be either side of the mean. Let the smaller one be x and the larger one be y. Since the set consists of positive numbers only, hence we have to check from 1 to 6. \(3x<y<4x\) Since only one number has to be on either side of the mean, hence the value of x comes out as 3 and y as 11. The smallest and the largest are known. Hence sufficient. +1D
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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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28 Dec 2012, 22:42
Thanx Marcab I got it if you can edit and one more line that is necessary constraint it will be easy i.e x+ y must be equal to 14.. I know its understood +1 dude Good going



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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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16 Jun 2015, 03:57



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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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16 Jun 2015, 04:18
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Bunuel wrote: List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A?
(1) The integer 3 is in list A. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
Kudos for a correct solution. First put everything together: 5 Integers * Average of 7 = Sum of 35 for the whole List A. 6, 7 and 8 included equals 21. Therefore 2 integers left with a value of 14. Statement 1: Sufficient The list contains 3, 6, 7, 8 and 11 as the last integer (all adding up to 35). Range = 113 = 8 Statement 2: Sufficient How can we split 14 so that the biggest Number of the List is > 3x smallest and <4x smallest. Only 3 and 11 fit. Range 113. Answer D.
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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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16 Jun 2015, 04:28
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avg of 5 integers is 7 means total is 7x5=35 if integers 6,7,8 are in list means sum of remaining 2 nos is 3521=14
option A: 3 is also in the list means other no is 11. so the range is 8.
Sufficient
Option B: take smallest term as 3 gives other term as 10 & 11. other term could only be 14 because sum of these 2 terms should be 14.
sufficient
Ans is  D
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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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16 Jun 2015, 07:34
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Bunuel wrote: List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A?
(1) The integer 3 is in list A. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
Kudos for a correct solution. Explaination : Let the two missing integers are x and Y. So list is x,y,6,7,8 and average is 7 Henc X + Y + 6+7+8 = 35 X + Y = 14. If X = 1 and Y = 13  range = 12 If X = 10 and Y = 4 Range = 6 Basically to find the range we need to find X and Y. Statement 1 : The integer 3 is in list A.So either X or Y is 3, hence Either X or Y is 11. Consider X = 3 and Y = 11 Range = 8  definite answer Hence statement 1 is sufficient. Statement 2 : The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.Consider extremes, consider Largest Term (Y) is equal to 3 time smallest term (x) X + y = 14 X + 3X = 14 4X = 14 X = 3.5 So one of the term has to less than 3.5 consider Largest Term (Y) is equal to 4 time smallest term (x) X + y = 14 X + 4X = 14 5X = 14 X = 2.8 so one of the term has to be more than 2.8 Only integer 3 satisfies these extremities, hence one of the missing integer is 3 and another is 11. Range = 8 Hence statement 2 is sufficient. Hence Option D, each of the statement is sufficient.



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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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16 Jun 2015, 09:09
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Bunuel wrote: List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A?
(1) The integer 3 is in list A. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
Kudos for a correct solution. Solution  Average of the list is (6+7+8+x+y) = 7*5 = 35 > x+y=14. Stmt 1  (6+7+8+3+y) = 35  > Missing number from the list is y=11 > Range = 113 = 8. Sufficient. Stmt 2  The largest term y in list is greater than 3 times and less than 4 times the size of the smallest term x. 3x<y<4x > After substituting y=14x in the inequality 3x<14x<4x > 4x<14<5x. The only integer x satisfy the given inequality is x=3 (9<14<15). Then Y=11. Range is 113=8. Sufficient. ANS  D Thanks, Please give me kudos.
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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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16 Jun 2015, 10:02
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Bunuel wrote: List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A?
(1) The integer 3 is in list A. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
Kudos for a correct solution. Given : Five Positive integers have 7 as average. If the integers 6, 7, and 8 are in list A, let us assume that other two integers are x and y. So, \((x + y + 6 + 7 + 8) / 5 = 7\) => \(x + y + 21 = 35\) => \(x+y=14\) So (x,y) can be (1,13), (2,12), (3,11), (4,10), (5,9), (6,8), (7,7), (8,6), (9,5), (10,4), (11,3), (12,2) or (13,1). Now, 1.) The integer 3 is in list A. So lets say, x = 3, that means y will be 11 (143). So the list A is (3,6,7,8,11) And Range of A is 8 (113). SUFFICIENT. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term. \(3*smallest term < largest term < 4*smallest term.\) Now we will check this condition for (x,y) as (1,13), (2,12), (3,11), (4,10), (5,9), (6,8) or (7,7). As we can see, out of the values of (x,y) specified above only for (3,11) or (11,3) the list A satisfies this condition. So the list A is (3,6,7,8,11) And Range of A is 8 (113). Hence again SUFFICIENT. So Answer is D.



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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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17 Jun 2015, 01:56
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5 Pos int in list A Avg = 7 6,7,8 are in list A
5x7 is 35, total in list A must be 35 6+7+8 = 21
1) 3 is in the list
3 + 6 + 7 + 8 = 24, must get to 35, so last number is 11
Range 3  11
S.
2) Largest number is greater than 3 times, less than 4 times smallest numbers
Test it!
6+7+8 = 21
1*3 = 3 1*4 = 4 not possible
2*3 = 6 2*4 = 8 not possible
3*3 = 9 3*4 = 12 possible
4*3 = 12 4*4 = 16 not possible
S.
D.



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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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22 Jun 2015, 04:55
Bunuel wrote: List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A?
(1) The integer 3 is in list A. (2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:The average of all five integers in the set is 7. Three of the integers in the set are given (6, 7, and 8) and they all have an average of 7. Therefore, the floating terms in this problem must also have an average of 7. If we assign x and y to represent these terms, we have: (x + y)/2 = 7; x + y = 14. Our rephrased question is thus, “Given that x + y = 14, what is either x or y?” Once we know one of the values, we can solve for the other, and thereby determine the range of the set. (1) SUFFICIENT: If 3 is one of the unknown integers, the other must be 11. The range is thus 11 – 3 = 8. (2) SUFFICIENT: This statement might seem a little too vague to be sufficient, but by visually listing the possible pairs that add up to 14, we can rule out pairs that don't fit the constraint from this statement: Notice that the pairings represent the constraint x + y = 14. Visually, this means that x and y are balanced around 7. Among these pairs: 8 is 1.33 times the size of 6 (the ratio is too low). 9 is 1.8 times the size of 5 (the ratio is too low). 10 is 2.5 times the size of 4 (the ratio is too low). 11 is 3.66 times the size of 3 (an acceptable ratio). 12 is 6 times the size of 2 (the ratio is too high). Only one pair of integers results in a ratio strictly between 3 and 4. The unknown terms must therefore be 3 and 11, and the range is 11 – 3 = 8. The correct answer is D.In this problem we saw the constraint x + y = 14: a fixed sum. Another common constraint is a fixed difference, such as a – b = 2. A fixed difference can be represented visually as a fixed distance between a and b on the number line, with a to the right because it is larger. That distance could move left or right: Attachment:
20150622_1550.png [ 28 KiB  Viewed 1383 times ]
Attachment:
20150622_1553.png [ 13.42 KiB  Viewed 1380 times ]
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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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17 Sep 2015, 09:25
(1) \(x+y=14\) \(3+y=14\) \(y=11\)
Sufficient.
(2) \(x+y=14\) \(x+3x<14<x+4x\) \(4x<14<5x\) \(\frac{14}{5}<x<\frac{14}{4}\) \(2.8<x<3.5\)
Since we are restricted to integers,
\(x=3\)
Sufficient.

Can anyone help explain how to algebraically get from: \(4x<14<5x\)
to
\(\frac{14}{5}<x<\frac{14}{4}\)
Or is it strictly intuitive?



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Re: List A contains 5 positive integers, and the average (arithmetic mean) [#permalink]
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26 May 2016, 11:44
(6+7+8+x+y)/5 = 7
Since the average of 6+7+8 is 7, the average of x+y must be 7 as well.
(x+y)/2 = 7 > x+y = 14
1.) The integer 3 is in the list A 3+y = 14 > y =11 so range = 311 SUFF
2.) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term In order for x+y = 14, x and y must be the smallest and largest numbers in the set. 3x = y > 4x = 14 x = 3.5 4x = y > 5x = 14 x = 2.8 If x is an integer then x = 3 and y = 11 SUFF
> Answer is D



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