Here's the algebra:

d = # of hot dogs at the beginning of the day
x = # of hot dogs sold during the day
h = # of hamburgers

The # of hot dogs at the beginning is double the # of hamburgers, so

d = 2h

and after we've sold x hot dogs, (d - x) is half the # of hamburgers, so

(d - x) = h/2

Now let's clean this up:

(d - x) = h/2
2d - 2x = h
2d = h + 2x

Since d = 2h, we substitute and get

2(2h) = h + 2x
3h = 2x

So x must be a multiple of 3. Thus we CANNOT have sold 2 hot dogs, as 2 isn't a multiple of 3.

red217 wrote:Q. A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. which of the following cannot represent the number of hot dogs he sold?

(a) 2
(b) 3
(c) 6
(d) 9
(e) 24

Let x = the multiplier for the original ratio and s = the number of hot dogs sold.

Original ratio of hot dogs to hamburgers = 2x/x.
After s hot dogs are sold, the remaining number of hot dogs = 2x - s.
Since the resulting ratio of hot dogs to hamburgers is 1 to 2, we get:
(2x-s)/x = 1/2
4x - 2s = x
3x = 2s
s = (3/2)x.

If x = 2, then s = (3/2) * 2 = 3.
If x = 4, then s = (3/2) * 4 = 6.
If x = 6, then s = (3/2) * 6 = 9.
The resulting values of s indicate that the number of hot dogs sold can be any positive multiple of 3.
Eliminate B, C, D and E.

The correct answer is A.