中一數學題~~~~急啊~~~~~!!

let (a) be the number of total oranges,
then, a/19 is the number of rotten oranges.

3 more rotten oranges are put into the box, the total number of rotten oranges is 1/10 of the total number of oranges--->
Total num of oranges = 3+a
number of rotten oranges = a/19+3 = (57+a)/19
the ratio is now 1/10, so (57+a)/19 / (3+a) = 1/10
(57+a)/(9+3a) =1/10
570+190a=9+3a
187a=561
a=3

Let x be the no. of rotten oranges.

3 + x ÷ 1/19 = (x + 3) ÷ 1/10
3 + 19x = 10x + 30
9x = 27
x = 3

因為 x ÷ 1/19 是原本橙的總數，(x + 3) ÷ 1/10 是加入 3 個橙後的總數

2007-12-28 12:42:09 補充：
上面兩位的計法都沒有錯！

Let n be the original number of total oranges, then
(n/19) = original number of rotten oranges

As 3 more rotten oranges are put into the box, so
new number of rotten oranges = (n/19) + 3
new number of total oranges = n + 3

(n/19) + 3 = (n + 3)/10
....
n = 57

So, the origianl number of rotten oranges = 57/19 = 3