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A2A2
Vasya wrote down several consecutive integers. Which of the following could not be the percentage of odd numbers among them?

(A) 40
(B) 45
(C) 48
(D) 50
(E) 60

Quick approach..

Well I started solving with the approach that everybody has followed here..but midway..I realized something..
Let's start..

There are in total Three possibilities

1. Odd and Even numbers are equal --> So (D) out.

2. Even numbers one more than Odd numbers

So Odd numbers by Total numbers..

\(\frac{x}{2x+1}\)

3. Odd Numbers one more than Even numbers..

\(\frac{x+1}{2x+1}\)

So the percentage values in both the cases will respectively be..

\(\frac{x}{2x+1}*100 = N\)


\(\frac{x+1}{2x+1}*100 = N\)

In both the expressions, the numerator and denominator of x terms will never have any common factor(as you can easily see in the first that the denominator will leave a remainder of 1 when divided by the numerator, so no canceling out will be possible; and in the second the numerator and denominator are consecutive integers, with denominator being an odd number).
Thus, 2x+1 has to be Odd and a factor of 100 for N to be an integer, in other words it will be an odd multiple of 5 that also divides 100. What's important to notice is that the expression

\(\frac{100}{2x+1}\) will always give us an Even Integer.

Thus, N can never be an odd number, and your answer is (B) :)
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Tried a different approach

If n = odd, it could be n+1/2 or n-1/2
If n = even, then it will be n/2

(A) 40 = 40/100 = 2/5. Here n=5 and 5-1/2 = 2 odds are possible out of 5
(B) 45 = 45/20 = 9/20. For n= 20, it would be 20/2 = 10 odds not 9, This is the answer
(C) 48
(D) 50
(E) 60
A2A2
Vasya wrote down several consecutive integers. Which of the following could not be the percentage of odd numbers among them?

(A) 40
(B) 45
(C) 48
(D) 50
(E) 60
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