Maths問題(10 points!!)

Let x be original number of rotten orange, then the number of orange is 19x
when 3 rotten oranges are added,
the number of rotten orange will be x+3, and the number of oranges will be 19x+3
(x+3)/(19x+3) = 1/10
10(x+3) = 19x+3
10x+30 = 19x+3
9x = 27
x=3
so the answer of Q.1 is 3
the final number of orange = 19x+3 = 19(3)+3 = 57+3 = 60
so the answer of Q.2 is 60

(不需要理會呢 d 箭咀 ----->, 因為要防止數目字移位, 我想條算式可以清楚 d 表達出黎)

Let X be the original number of oranges:

1/19 X + 3 = 1/10 (X + 3)

3 = (X + 3)/10 - X/19

---->19 (X + 3) - 10 X
3 = ------------------------
---------->190

---->19 X + 57 - 10 X
3 = ------------------------
---------->190

----> 9 X + 57
3 = ------------
------->190

3 (190) = 9 X + 57

570 = 9 X + 57

9 X = 570 - 57

9 X = 513

X = 513/9

X = 57

Since the original rotten oranges is 1/19 X, therefore,
1/19 X = 57/19 = 3

Final number of oranges in the box: X + 3 = 60

Assuming X = the original number of rotten oranges.
and Y = the original total number of oranges.
X/Y = 1/19
X = Y/19

Adding 3 more rotten oranges into the box:
X+3/Y+3 = 1/10
Because:
X = Y/19
Therefore:
(Y/19 + 3)/Y+3 = 1/10
10(Y/19 + 3) = Y+3
10(Y+57/19) = Y+3
10Y + 570 = 19Y + 57
570 - 57 = 19Y - 10Y
513 = 9Y
Y = 57

Therefore, the original number of rotten oranges in the box:
X = 57/19
X = 3
Answer of question #1 is 3.

Therefore, the final number of oranges in the box:
Y + 3
57 + 3
= 60
Answer of question #2 is 60.

let R be the no. of rotton oranges at the beginning
let T be the total no. of oranges at the beginning

before:
R = T x (1/19)
T = 19R

after:
R+3 = (T+3) x (1/10)

sub,
R+3 = (19R+3) x (1/10)
10 x (R+3) = 19R + 3
10R + 30 = 19R + 3
19R - 10R = 30 - 3
9R = 27
R = 3

therefore, original no. of rotton oranges = 3
final no. of oranges = (3+3) x 10 = 60

=]

let p be the original no. of rotten oranges

the total no. of orange= p / (1/19)
=19p

when 3 rotten oranges are added
the new no. of rotten oranges = p + 3
the new total no. of oranges = 19p + 3

so
( p+3 ) / ( 19p + 3 ) = 1 / 10
10 * ( p+3 ) = 19p + 3
10p + 30 = 19p + 3
27 = 9 p
p = 3
so there is 3 rotten oranges originally

the final no. of oranges = 19 * 3 + 3
=60
因為final number of oranges in box係要加埋果三個rotten apple

57..................................