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Another way to quickly eliminate options instead of substituting values in all options:
m is odd, so m/2 is not an integer, so eliminate A, B
We are asked number of even integers between m and 2m, we know the number of even integers cannot be 2m+1, since 2m + 1 is larger than 2m, eliminate E

Then check with m = 1, 2m = 2 for C and D, turns out C is the right answer.

Hence Option C!
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Bunuel
If m is an odd integer, which of the following expresses the number of even integers between m and 2m inclusive?


A. m/2 + 1

B. m/2 - 1

C. (m + 1)/2

D. (m - 1)/2

E. 2m + 1

i picked number 3 also tested 5 in both cases C was correct
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If m is an odd integer, which of the following expresses the number of even integers between m and 2m inclusive?


A. m/2 + 1

B. m/2 - 1

C. (m + 1)/2

D. (m - 1)/2

E. 2m + 1

Assume value of m to be 3 , so m+2m = 3-6 total 2 even integers , now using value of m check in options where we are getting value as 2 , only option C suffices
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is this really a 600 level question ?
just asking
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The best way to do this question is from the math book notes:

# of terms: (last term - first term)/2 + 1; we use 2 as a divisor since we are looking for even numbers.

last term: 2m
first term: m+1 (since m is odd)

plug and chug and you get (m+1)/2
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Bunuel
If m is an odd integer, which of the following expresses the number of even integers between m and 2m inclusive?


A. m/2 + 1

B. m/2 - 1

C. (m + 1)/2

D. (m - 1)/2

E. 2m + 1

looking at ratios of number of even integers, e, to m:
1:1
2:3
3:5
note that e=(m+1)/2
C
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Bunuel
If m is an odd integer, which of the following expresses the number of even integers between m and 2m inclusive?


A. m/2 + 1

B. m/2 - 1

C. (m + 1)/2

D. (m - 1)/2

E. 2m + 1
Solution:

We can let m = 3 and thus 2m = 6. We see that there are 2 even integers between 3 and 6, inclusive, namely 4 and 6. Now let’s see which answer choice yields the number 2 when m = 3.

A. 3/2 + 1 = 5/2 → No
B. 3/2 - 1 = 1/2 → No

C. (3 + 1)/2 = 2 → Yes

D. (3 - 1)/2 = 1 → No

E. 2(3) + 1 = 7 → No

Alternate Solution:

Let’s apply the formula [(greatest - smallest)/common difference] + 1. Notice that 2m is even, thus greatest = 2m. Since m is odd, m + 1 is even and that’s the smallest even integer between m and 2m; thus smallest = m + 1. Finally, common difference = 2 since consecutive even integers differ by 2. We have:

[(greatest - smallest)/common difference] + 1

[(2m - (m + 1))/2] + 1

[(2m - m - 1)/2] + 1

(m - 1)/2 + 2/2

(m - 1 + 2)/2

(m + 1)/2

Answer: C
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