If the average(arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?
1) The range of n integers is 14.
2) The greatest of the n integers is 17.
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Verbal Review 2nd ed DS question 66
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pharmxanthan
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Say, the least of the n odd integers is a and the greatest is b.pharmxanthan wrote:If the average(arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?
1) The range of n integers is 14.
2) The greatest of the n integers is 17.
Therefore, sum of n consecutive odd integers = (n/2)*(a + b)
Average of these n consecutive integers = [(n/2)*(a + b)]/n = (a + b)/2
Given the average is 10 => (a + b) = 20
Statement 1: Range of n integers is 14 => Greatest - Least = 14 => b - a = 14 => b = a + 14
As, (a + b) = 20 => (a + a +14) = 20 => (2a + 14) = 20 => 2a = 6 => a = 3
Thus, least of the integers is 3.
Sufficient.
Statement 2: Greatest of the n integers is 17.
As, (a + b) = 20 => (a + 17) = 20 => a = 3
Thus, least of the integers is 3.
Sufficient.
The correct answer is D.
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goyalsau
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I think answer should be Bpharmxanthan wrote:If the average(arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?
1) The range of n integers is 14.
2) The greatest of the n integers is 17.
1) Not sufficient.
But still i will not be surprised if somebody comes up with the solution
Because we know the difference in two terms that is 2,
We know the average of terms as 10.
But i am not able to figure out what how to use the information. That why for me its not sufficient
2) last term is 17 and it is the greatest
Sum of first 10 odd terms from [ 1 3 5 . ...19 ] = 100
average is 100/10 = 10
But the last term is 17 ( if we consider terms from ( 1 3 5 ..... 17 )
Sum will be equal to 81 { so the average will be 81 / 9 = 9 }
If we take 1 out then sum we will be 80 and average will be 10
80/ 8 = 10 First term is 3
Hence Sufficient.
What the OA?
Saurabh Goyal
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goyalsau
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Great Work,Rahul@gurome wrote: Say, the least of the n odd integers is a and the greatest is b.
Therefore, sum of n consecutive odd integers = (n/2)*(a + b)
Average of these n consecutive integers = [(n/2)*(a + b)]/n = (a + b)/2
I didn't knew that the sum of consecutive integers is = n/2 * { a + b }
I think we can use this formula then the different between two integers is constant { whether 2 or 3 or 4 or 5 ........ }
what is know is this is this formula n/2 * { n + 1 }
We use this formula when the difference between two integers is 1.
Please correct me Rahul Sir, if i am wrong.....
Saurabh Goyal
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Rahul@gurome
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We can use this formula whenever the common difference is constant, i.e. the numbers are in arithmetic progression. This formula is just an rearrangement of an well known formula.goyalsau wrote:Great Work,Rahul@gurome wrote: Say, the least of the n odd integers is a and the greatest is b.
Therefore, sum of n consecutive odd integers = (n/2)*(a + b)
Average of these n consecutive integers = [(n/2)*(a + b)]/n = (a + b)/2
I didn't knew that the sum of consecutive integers is = n/2 * { a + b }
I think we can use this formula then the different between two integers is constant { whether 2 or 3 or 4 or 5 ........ }
what is know is this is this formula n/2 * { n + 1 }
We use this formula when the difference between two integers is 1.
Please correct me Rahul Sir, if i am wrong.....
We know, the sum, S of first n numbers of any arithmetic progression with common difference d and first term A1 is given by,
S = (n/2)*[2*A1 + (n - 1)*d]
Now, n-th term of an arithmetic progression, An = A1 + (n - 1)*d
Therefore, S = (n/2)*[A1 + A1 + (n - 1)*d] = (n/2)*[A1 + An]
Rahul Lakhani
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+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
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pharmxanthan
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