In a candy dish, the ratio of red to blue candies is $3: 5$. When 3 blue candies are removed, the ratio of red to blue candies becomes $2: 3$. How many more blue candies than red candies were in the dish before any candies were removed?
(A) 10
(B) 12
(C) 6
(D) 8
(E) 14
The original ratio of red candies to blue candies is $3: 5$, and so the number of red candies was a positive integer multiple of 3 , and the number of blue candies was the same positive integer multiple of 5 .

For example, there could have been $3 \times 1=3$ red candies and $5 \times 1=5$ blue candies, or $3 \times 2=6$ red candies and $5 \times 2=10$ blue candies, or $3 \times 3=9$ red candies and $5 \times 3=15$ blue candies, and so on.

We continue these possibilities in the table below, and consider the number of red and blue candies and the resulting ratio after three blue candies are removed from the dish.

| Possible num  and blue candi | 3 red,  5 blue | 6 red,  10 blue | 9 red,  15 blue | 12 red,  20 blue | 15 red,  25 blue | 18 red,  30 blue |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| umbers of rec  after 3 blu | 3 red,  2 blue | red,  7 blue | red,  12 blue |  |  |  |
| New  of $\mathrm{r}$ | $\overline{3: 2}$ | $6: 7$ | $9: 12$  $=3: 4$ | $12: 17$ | $15: 22$ |  |

If the dish originally contained 18 red candies and 30 blue (note that $18: 30=3: 5$ ), then when three blue candies are removed, the ratio of the number of red candies to blue candies
becomes $18: 27$ which is equal to $2: 3$, as required.

Therefore, there were $30-18=12$ more blue candies than red candies in the dish before any candies were removed.

ANswer: (B)