sandy wrote:
If j and k are even integers and j < k, which of the following equals the number of even integers that are greater than j and less than k ?
A. \(\frac{(k -j -2)}{2}\)
B. \(\frac{(k -j -1)}{2}\)
C. \(\frac{(k -j )}{2}\)
D. \(k -j\)
E. \(k -j -1\)
Practice Questions
Question: 15
Page: 206
sandy wrote:
Explanation
Since j and k are even integers, it follows that k = j + 2n for some integer n. Consider the sequence of even integers from j to k.
\(j, j(2)(1), j(2)(2), j(2)(3), ..............j(2)(n-1), j + 2n\)
Note that there are n – 1 integers in the sequence between j and k, and these are the even integers greater than j and less than k. Therefore the answer is n – 1, but the answer must be given in terms of j and k. Since k = j + 2n, you have
\(n= \frac{k-j}{2}\) and so
\(n-1= \frac{k-j}{2}-1= \frac{(k-j-2)}{2}\).
Thus the correct answer is Choice A.
Sandy, there will be (j+2n-j)/2 + 1 number of even integers will be there right?,
that will be n+1, how come you have taken n-1 integers, please explain?
even if we take j=2 and k=8,
there will be 2, 4, 6, 8
since j = k+2n
at n = 3 j=2+6=8, here there are 4 numbers are there n+1 = 3+1= 4.
on that case, if we write n+1 interms of j and k. n+1 = (j-k+2)/2, which is not given in answer!,
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43% (01:04) correct
