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If m is an even integer, v is an odd integer, and m > v> 0, which of

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If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?

A. \(\frac{m-v}{2} -1\)

B. \(\frac{m-v-1}{2}\)

C. \(\frac{m-v}{2}\)

D. \(m-v-1\)

E. \(m-v\)
[Reveal] Spoiler: OA

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If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 24 Jul 2017, 15:27
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carcass wrote:
If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?

A. \(\frac{m-v}{2} -1\)

B. \(\frac{m-v-1}{2}\)

C. \(\frac{m-v}{2}\)

D. \(m-v-1\)

E. \(m-v\)


Let m = 10 & v = 1...........There are 4 Even number less than 10 and greater than 1

A. \(\frac{10-1}{2} -1\)= Fraction number....Eliminate

B. \(\frac{10-1-1}{2}\) = 4.........................Keep

C. \(\frac{10-1}{2}\) = Fraction...................Eliminate

D. \(10-1-1\) = 8........................................Eliminate

E. \(10-1\)=9.............................................Eliminate

Answer: B

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 26 Jul 2017, 02:10
The fastest way is plugging in some numbers.

Can someone show how to solve the problem algebraically?

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 26 Jul 2017, 16:15
carcass wrote:
If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?

A. \(\frac{m-v}{2} -1\)

B. \(\frac{m-v-1}{2}\)

C. \(\frac{m-v}{2}\)

D. \(m-v-1\)

E. \(m-v\)


We can let m = 6 and v = 5. Since there are no (or zero) even integers less than m but greater than v, we see that the answer can be either B, (m - v - 1)/2, or D, m - v - 1, since either choice will produce 0 when we substitute m = 6 and v = 5.

Now, let’s let m = 8 and v = 5. Since there is 1 even integer (namely, 6) less than m but greater than v, we see that the answer must be choice B, since (8 - 5 - 1)/2 = 1 whereas choice D will yield 8 - 5 - 1 = 2. Thus, the correct answer choice must be B.

Answer: B
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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 27 Jul 2017, 01:28
Devbek wrote:
The fastest way is plugging in some numbers.

Can someone show how to solve the problem algebraically?


Hi Devbek,
I'll try to explain.

Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((a-b)/2)+1 (1 is added to include the first even number in the series)
In the current question we need to find number of even numbers "greater than v" and "less than m".
The even number after v(which is odd) is v+1 and the even number before m(which is even) is m-2. (This step is done to exclude both v and m from the total number of even integers)
Hence the number of even numbers between v and m (excluding v and m) should be ((m-2-v-1)/2) + 1) (1 added to include the number v+1)
Upon solving you will get (m-v-1)/2

Hope you got it.

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 27 Jul 2017, 03:06
Dkingdom wrote:
Devbek wrote:
The fastest way is plugging in some numbers.

Can someone show how to solve the problem algebraically?


Hi Devbek,
I'll try to explain.

Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((a-b)/2)+1 (1 is added to include the first even number in the series)
In the current question we need to find number of even numbers "greater than v" and "less than m".
The even number after v(which is odd) is v+1 and the even number before m(which is even) is m-2. (This step is done to exclude both v and m from the total number of even integers)
Hence the number of even numbers between v and m (excluding v and m) should be ((m-2-v-1)/2) + 1) (1 added to include the number v+1)
Upon solving you will get (m-v-1)/2

Hope you got it.


Still confusing, but thanks anyway :)

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 27 Jul 2017, 03:18
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Devbek wrote:
Dkingdom wrote:
Devbek wrote:
The fastest way is plugging in some numbers.

Can someone show how to solve the problem algebraically?


Hi Devbek,
I'll try to explain.

Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((a-b)/2)+1 (1 is added to include the first even number in the series)
In the current question we need to find number of even numbers "greater than v" and "less than m".
The even number after v(which is odd) is v+1 and the even number before m(which is even) is m-2. (This step is done to exclude both v and m from the total number of even integers)
Hence the number of even numbers between v and m (excluding v and m) should be ((m-2-v-1)/2) + 1) (1 added to include the number v+1)
Upon solving you will get (m-v-1)/2

Hope you got it.


Still confusing, but thanks anyway :)


Formula to find count of even numbers in a consecutive series between two numbers is ((a-b)/2) + 1. In this formula both a and b are included in the count.
let m = 10 (even), v = 3(odd)
What the question is asking : find the number of even numbers greater than v (3) and less than m (10).
Breaking the question down :1) We need to exclude m and v from the total count of even numbers. v obviously is out because it is odd. To exclude m we need to use an even number less than m.
Hence in order to apply the above stated formula we will use v+1 (4) and m-2 (8). (Since we need to find only the count of even numbers that's why it is easier to keep the starting and ending numbers as even digits.
Applying the formula : ((8-4)/2)+1) = 3 (i.e. 4,6,8) Remember :numbers are more than 3 and less than 10 and even.
The above equation can be written as ((m-2-v-1)/2)+1 = (m-v-1)/2

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 10 Sep 2017, 04:52
Devbek wrote:
The fastest way is plugging in some numbers.

Can someone show how to solve the problem algebraically?



You are better off just plugging in numbers, come the real test.

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 18 Sep 2017, 10:04
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m is even and v is odd -> (m-v) is odd -> (m-v)/2 is not an interger => elimate A, C imidiately

SOLVE: Between m and v, we have [(m-1) - (v+1) + 1] = (m-v-1) numbers, include both odd and even -> the number of even is half of (m-v-1)
=> The answer is B

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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 15 Oct 2017, 18:34
Hi,
when we use number picking for method for this problem, it does not give a consistent result. For ex. if m=2; v=1 then non of the answers stand true.

please let me know what am i missing here?

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If m is an even integer, v is an odd integer, and m > v> 0, which of [#permalink]

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New post 15 Oct 2017, 20:54
If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v?
We are looking for even integers x that m< x <v -> these three (m, v, x) are integers that means (m - v) cannot less than or equal to 1 (x cannot be an integer if it happens). So, m = 2 and v=1 are not the right numbers to pick.
Hope it helps!

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If m is an even integer, v is an odd integer, and m > v> 0, which of   [#permalink] 15 Oct 2017, 20:54
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