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by rahul.s » Sun Jan 31, 2010 7:46 am
What is the value of the least integer in a set of k consecutive even integers, if the average (arithmetic mean) of the k consecutive even integers is equal to 17?

1) The range of the k integers is 18.
2) The greatest of the k integers is 26.

OA: D
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Source: — Data Sufficiency |
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by papgust » Sun Jan 31, 2010 8:31 am
The set is consecutive even integers. Average of the set is 17.

In evenly-spaced sets such as this set, average of the set is (lowest+highest)/2.

From the question stem, we know that (L + H)/2 is 17. which is nothing but L + H = 34 ....... (1)

But we don't know the number of integers. Let's go to each statement.

1) The range of the k integers is 18.

This means that H - L = 18 [Where H is highest value, L is lowest value]
Solving with help of eqn (1), we could find the lowest/least integer.
Sufficient.

2) The greatest of the k integers is 26.

H = 26

Sub. H in (1),
L can be found. Sufficient.

Hence D.
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by thephoenix » Sun Jan 31, 2010 8:38 am
let the series be n1,n2,n3.....nk upto kth term
now
nk=n1+(k-1)d[ where d is common diff =2]
nk-n1=(k-1)2
summation=(n1+nk)k/2=17k=[2n1+(k-1)2]k/2
task find n1
s1)nk-n1=range=18--->(k-1)2=18or k=10
summation=17k--->[2n1+18]*5=170
n1=8
suff
s2)nk=n1+(k-1)2---17k=[n1+26]k/2----17*2=n1+26--->n1=8
suff
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