Let m = 4 and v = 3

Always remember to check all the values, if in some case you find 2 options satisfying the values take another value and we should be good.

4>3>0, here the number of even integers less than m and greater than v = 0

Put back the values in the answer options

We can get a zero from B and D, Now we will to test for some other value for m and v

m = 8 v = 1

8>1>0, number of even integers less than m and greater than v = 3

(8-1-1)/2 = 3

D says there are 6 such values

So B wins.
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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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generis

I fully support your statement.
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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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