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What is the average (arithmetic mean) of eleven consecutive [#permalink]
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23 Apr 2010, 17:05
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What is the average (arithmetic mean) of eleven consecutive integers? (1) The average of the first nine integers is 7 (2) The average of the last nine integers is 9
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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23 Apr 2010, 17:11
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Some how i got E
1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff?? 2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???
and combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.



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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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zz0vlb wrote: What is the average ( arithmetic mean ) of eleven consecutive integers? 1.The avg of first nine integers is 7 2. The avg of the last nine integers is 9 friends, please explain this to me. Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case \(mean=median=x_6\). (1) \(x_1+x_2+...+x_9=63\) > there can be only one set of 9 consecutive integers to total 63. Sufficient. If you want to calculate: \((x_65)+(x_64)+(x_63)+(x_62)+(x_61)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63\) > \(x_6=8\). OR: Mean(=median of first 9 terms=5th term)*# of terms=63 > \(x_5*9=63\) > \(x_5=7\) > \(x_6=7+1=8\) (2) \(x_3+x_4+...+x_{11}=81\) > there can be only one set of 9 consecutive integers to total 81. Sufficient. If you want to calculate: \((x_63)+(x_62)+(x_61)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81\) > \(x_6=8\). OR: Mean(=median of last 9 terms=7th term)*# of terms=81 > \(x_7*9=81\) > \(x_7=9\) > \(x_6=91=8\) Answer: D.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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24 Apr 2010, 06:11
Thanks for clarifying this Bunuel. +1Kudos to you.



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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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zz0vlb wrote: What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7 (2) The avg of the last nine integers is 9 Here is a neat little trick for such kind of problems: Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly \(\frac{d}{2} units \to\) The new Average = \(4\frac{1}{2} = 3.5\) Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.50.5 = 3. Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth. Back to the problem: From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient. From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 90.50.5 = 8. Sufficient. D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 22:15
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zz0vlb wrote: What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7 (2) The avg of the last nine integers is 9 As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1 1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient 2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient And is D



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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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28 Aug 2014, 08:43
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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09 Mar 2015, 18:15
I considered following approach
if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55>A if largest number in set is x ,then sum of 11 consecutive numbers=11x(1+2+10)=11x55
Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63 sum of 11 numbers =63+x+9+x+10>B
Equating A& B 11X+55=63+X+9+10 ,which can be solved to get x=3
statement I is sufficient similar approach for Statement II 11X55=8+2X19 ,can be solved to get X=13
statement 2 is sufficient
OA=D



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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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09 Mar 2015, 18:26
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Hi All, When you look at this question, if you find yourself unsure of where to "start", it might help to break down everything that you know into small pieces: 1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place "in line", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?) 2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7. What would have to happen for a group of consecutive integers to have an average of 7? Here are some examples: 7 6, 7, 8 5, 6, 7, 8, 9 Notice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern. With 9 total terms, that means there has to be 4 above and 4 below: 3, 4, 5, 6,.......7.......8, 9, 10, 11 Now we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT With this same approach, we can deal with Fact 2. The key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to "play around" with a question if you're immediately certain about how to handle it. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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29 Jun 2016, 02:10
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wow such complex explanations for such a simple problem?
given :
11 consec integers
let them be x,x+1,x+2,...,x+10
Q: what is their mean?
mean is (11x+55)/11 = x+5. Q becomes what is x+5
1) mean first 9 is 7.
so (9x+36)/9 = x+4 = 7 , so x+5 =8 ,> sufficient A or D
2) mean of last 9 is 9.
so (9x+54)/9 = x+6= > x+5=8, sufficient . so D
D



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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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14 Jul 2016, 23:10
zz0vlb wrote: What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7 (2) The average of the last nine integers is 9 For odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question (1) The average of the first nine integers is 7 7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7 we will have {3,4,5,6,7,8,9,10,11} now we can add last two consecutive integer after 11, they will be 12,13 our new set will become = {3,4,5,6,7,8,9,10,11,12,13} again since the number of total elements in the set is odd, Mean will simply be the middle value = 8 SUFFICIENT (2) The average of the last nine integers is 9 Again number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right Middle value will be {5,6,7,8,9,10,11,12,13} Add 3,4 at the start of the set new set = {3,4,5,6,7,8,9,10,11,12,13} Mean will be 8 Sufficient ANSWER IS D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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20 Dec 2016, 16:37
Nice Official Question> Here is my solution to this one => Set of consecutive integers => N N+1 N+2 . . . N+10
AP series with D=2 Hence Mean = Median = Average of the first and the last terms= N+5
So we just ned the value of N
Statement 1 N N+1 . . N+8
Mean => N+4=7 Hence N+5=> 8 So the mean of the original set will be 8 Hence Sufficient
Statement 2> p p+2 p+3 . . p+8 Mean = p+4=9 p=5 Hene p+8=>13 So N+10=>13 Hence N+5=>8 Hence the mean of the original data set must be 8
Hence Sufficient
Hence D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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26 Jan 2017, 04:52
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The key is that the integers are consecutive. So, if we can determine any one of the 11 (and know where it falls), we can answer the question. (1) The average of the first 9 consecutive integers is 7. We know that avg = sum of terms / # of terms. So, 7 = sum of terms/9 sum of terms = 63. Well, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9, we can certainly determine the last 2: sufficient. (2) The average of the last 9 terms is 9. Exact same reasoning as (1): sufficient. Both (1) and (2) are sufficient: choose (D).
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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13 Mar 2017, 04:39
The trick here is to catch the phase "11 consecutive integers". We know that an odd set of consecutive integers have the same median and mean {i.e. set 1,2,3 has a median and mean of 2}.
Based on this we can say the same for the statements:
s1) Represent the first 9 integers as: A+B+C+D+E+F+G+H+I. If the mean of this set is 7 then the median is also 7 so we found that the 7th number in the total set. We know they are consecutive and therefore we could count forward and backward to get the unknown numbers. Therefore statement is sufficient.
s2) Same concept as above. Statement is sufficient.




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