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QOTD#18 If j and k are even integers and j < k, which of the

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QOTD#18 If j and k are even integers and j < k, which of the [#permalink] New post 12 Sep 2016, 12:07
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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
[Reveal] Spoiler: OA

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Re: QOTD#18 If j and k are even integers and j < k, which of the [#permalink] New post 12 Sep 2016, 12:13
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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.
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Re: QOTD#18 If j and k are even integers and j < k, which of the [#permalink] New post 20 Sep 2016, 05:45
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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. \(k -j -1\)



Another approach is to use the INPUT-OUTPUT approach.
Let's choose some nice input values of j and k and see what the answer (output) to the question becomes.
Let's try j = 0 and k = 2
So, the number of even integers that are greater than 0 and less than 2 = 0
So, in this case, when we input j = 0 and k = 2, the output (answer to our question) is 0

Now we'll check the answer choices to see which one yields an output of 0, when we plug in j = 0 and k = 2

A. \(\frac{(k -j -2)}{2}\) we get: (2 - 0 - 2)/2 = 0 KEEP

B. \(\frac{(k -j -1)}{2}\) we get: (2 - 0 - 1)/2 = 1/2 ELIMINATE

C. \(\frac{(k -j )}{2}\) we get: (2 - 0 )/2 = 1 ELIMINATE

D. \(k -j\) we get: 2 - 0 = 2 ELIMINATE

E. \(k -j -1\) we get: 2 - 0 - 1 = 1 ELIMINATE

Answer:
[Reveal] Spoiler:
A


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Re: QOTD#18 If j and k are even integers and j < k, which of the [#permalink] New post 30 Jun 2017, 20:16
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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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Re: QOTD#18 If j and k are even integers and j < k, which of the [#permalink] New post 03 Jul 2017, 03:19
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In the series 2, 4, 6, 8 here j=2 and k=8

n= 3 as \(k = j + 2*3\).

The numbers between 2 and 8 are 4 and 6 or 2 terms.


The reason you get n+1 terms is because you have added the first term (2) and final term (8). That is incorrect!
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Re: QOTD#18 If j and k are even integers and j < k, which of the   [#permalink] 03 Jul 2017, 03:19
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