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

KanishkM , with those choices, we also get a zero from option D.
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generis,

My reasoning backfired on me

I truly agree on that , said that,i have done the amendments.
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KanishkM - mmm, not exactly.
Your reasoning is really solid —including the part
in which you tell us, correctly, that we should check each option.

No matter. Mistakes are how we learn.
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generis

I fully support your statement.
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simply plugin m=6 and v= 3

4 ie. 1 digit even possible >3 <6 so
IMO B

Hi here are my two cents for this questions.

If this question were to be marked based on options here is how i would go about solving it .

Well,
Option A, C can be ruled out for following reason.

\(\frac{odd}{Even}\) is never Integer and we are looking for how many even integers would there be between v and m excluding m ,

and Option E gives number of integers between v and m, excluding either m or v

Which leaves us with choice B and D

Now choice B and D both gives us integers value but look at choice D . It actually gives us number of integers between v and m excluding both .

So our answer is Choice B

Lets do this algebraically.

if v is odd, then v+1 will be even . and if m is even

So we have number of integers between m and v+1 excluding one of the ends is = \(\frac{m-v-1}{2}\)


So we do have a formula for such type of questions


If we are counting in steps of x , from \(n_1\) to \(n_z\) including both end points we get \(\frac{n_z - n_1}{x}\) +1

If we are counting in steps of x , from \(n_1\) to \(n_z\) including only one end points we get \(\frac{n_z - n_1}{x}\)


If we are counting in steps of x , from \(n_1\) to \(n_z\) excluding both end points we get \(\frac{n_z - n_1}{x}\) -1
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Hi can you please help me understand the question? I am not clear with what is asked? Thanks in advance

Solution:

When it is not clear what is being asked in a question like this, you can always assign values to the variables and think in terms of those numbers, which will help you understand what is being asked.

For instance, we can let m = 4 and v = 1. Notice that 4 is an even number, 1 is an odd number and 4 > 1 > 0 is satisfied. For these values, the question now becomes “What is the number of even integers less than 4 and greater than 1?” We see that 2 is the only even integer less than 4 and greater than 1, so the answer to the question for m = 4 and v = 1 is “there is only one even integer between 4 and 1.”

As another example, let’s let m = 10 and v = 5. Again, 10 is even, 5 is odd and we have 10 > 5 > 0. In this case, the question is “What is the number of even integers less than 10 and greater than 5?” Well, there are two such integers: 8 and 6. The answer when m = 10 and v = 5 is “there are two such integers.”

Using the above examples (in addition to the ones I provided in my solution), we understand that we are being asked for the number of even integers between m and v, not inclusive.

To determine the answer, we can also use the new numbers I provided. For instance, if we choose to use m = 4 and v = 1, we need to determine which answer choice gives us 1 when we substitute m = 4 and v = 1. Substituting these values for m and v in the expressions given in the answer choices, we obtain 1/2, 1, 3/2, 2, and 3. As you can see, only the expression in answer choice B results in 1 when we substitute m = 4 and v = 1. That’s why B is the answer.

Answer: B

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substitution method works fine for this.

0<v<m

0<3<10

Rule: To Count INCLUSIVE, we use the following formula:

(Last Term of the Set - First Term of the Set) / Increment + 1


0 < V < M

where V = ODD
and
where M = EVEN

the questions asks us to find all the Even Numbers that will fall in between V and M, NOT INCLUSIVE of V and M

Since V is ODD, the very next Consecutive Integer AFTER V will be EVEN

thus (V + 1) is the First Even Term that should be included.

Since we can NOT Include M the Even Integer in our counting, the very next Even Integer Less than M will be exactly 2 Consecutive Integers away:

M = Even

M - 1 = ODD

M - 2 = Even


Count of all the Even Integers in between (and NOT Inclusive) of V and M:

First Term in Set = (V + 1)
LAST Term in Set = (M - 2)

Count = [ (M - 2) - (V + 1) ] / 2 + 1

Count = [ M - V - 3] / 2 + 2 / 2

Count = [ M - V - 3 + 2] / 2

Count = [M - V - 1] / 2

Answer -B-

Much Simpler to test answers though (and quicker)

It’s actually pretty easy ...

Since m>v and m is even and v is odd

m = v+1+2n (why?)

basically since v+1 = 1st even number just after v

Now we want to keep counting even numbers, so need to keep adding 2 so let’s assume now that nth number from v+1 is m so just solve and get value of n

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'm' is even and 'v' is odd and m > v > 0

Even - Odd = Odd and Odd divided by '2' is not an integer. Therefore, Eliminate A, C, and E.

Let us assume m = 8 and v = 1

Option D: m - v - 1 = 8 - 1 - 1 = 6 [From '1' to '8', we have only 3 even integers: 2, 4, and 6]

Option B: \(\frac{(m - v - 1)}{2 } = \frac{(8 - 1 - 1) }{ 2}\) = 3 [From '1' to '8', we have 3 even integers: 2, 4, and 6]

Answer B
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This formula
(Last even - first even)/1 +1 can solve the question easily.

Here,
Last even = m-2
First even = V+1

M is even ; V is odd.

If we have to find the number of integers between M and V, then the number of integers will be (M-V) - 1

We know that M is even and V is odd. So Even - Odd will always be odd. So we have (odd) - 1, which technically is an odd number - 1, which again is an odd number. So odd minus odd will always be even. So [(M-V)-1]/ 2 will give us the even numbers/ integers.

As the experts have explained this above, using different values, we can do the same.

Take the simplest case, how many even integers do we have between 3 and 4? 0 integers, 0 obviously is an even number.

Now take 3 and 6, we have 4 and 5 in between. So here, [(M-V)-1] = 2 ie 2 integers ie 4 and 5. 2 divided by 2 is 1, as we can see 4 is even, 5 is odd.

Similarly between 3 and 8, we have 4, 5, 6, 7 ie 4 integers out of which 4/2 ie 2 are even, and 2 are odd.

Hope this helps..

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\(Let \ m=6, \ and \ v=3\)

A. is out as the result will be a fraction

B. \(\frac{6-3-1}{2}=1, which \ is \ less \ than \ 3 \ or \ V\)

The answer is B
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Can someone explain why the answer is not A?

m is even and v is odd, so m-v = even-odd=odd.
Thus (m-v)/2=odd/2=fraction

But we cannot have number of even integers as a fraction
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There are 2 ways to approach this question.

Approach 1: Plugging values
Let’s say m= 8, n = 3
There are only 2 even numbers between 3 and 8 i.e., 4 and 6
So, when we substitute 8 and 3 as m and n in the answer options, you should get 2 as the answer.

A. (m−v)/2−1 ==> (8-3)/2 – 1 is a decimal number. So, its eliminated
B. (m−v−1)/2 ==> (8-3-1)/2 = 2
C. (m−v)/2 ==> (8-3)/2 is a decimal number. Eliminated
D. m−v−1 ==> 8-3-1 = 4. Eliminated
E. m−v ==> 8-3 = 5. Eliminated

Option B is the answer.

Approach 2:

If a and b are even numbers and a > b.
No of even numbers between a and b (including a and b) = (a-b)/2 + 1

Here m is an even integer, v is an odd integer, and m > v > 0.
V+1 is an even integer.
No of even number between m and v+1 (including m and v+1) =(m-(v+1))/2 +1

No of even numbers between m and v = (m-(v+1))/2 +1 -1 = (m – v – 1)/2
We subtract 1 from the total because m is not included the range

Option B is the answer.

Thanks,
Clifin J Francis,
GMAT SME
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Asked: 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 ?

Let m = 8; v = 5; m=8>v=5>0

The number of even integers less than m and greater than v = 1 ; Number 6
A. \(\frac{m-v}{2} -1 = \frac{8-5}{2} = 1.5 \) Incorrect

B. \(\frac{m-v-1}{2} = \frac{8-5-1}{2} = 1 < v=5\) Correct

C. \(\frac{m-v}{2} = \frac{8-5}{2} = 1.5 \)Incorrect

D. \(m-v-1 = 8-5-1 = 2 \)Incorrect

E. \(m-v = 8-5 = 3 \)Incorrect

IMO B
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