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In a box of oranges, the number of rotten oranges is 1/19 of the total number of oranges. If 3 more rotten oranges are put into the box, the total number of rotten oranges is 1/10 of the total number of oranges.
(a) Find the original number of rotten oranges in the box.
(b) Find the final number of oranges in the box.
form 1方程文字體1條 急.....! 5分
2006-11-24 4:15 am
回答 (2)
2006-11-24 4:31 am
a.let y be the total number of oranges in the box
y/19 + 3 = Y+3 / 10
10y+570=19y+57
y= 57
so the total number of oranges in the box is 57
the original number of rotten oranges in the box= 57乘 1/19
= 3
b the final number of oranges in the box = 57+3 / 10
= 6
2006-11-23 20:32:24 補充：
第2度計錯左-.-係57+3 = 60
y/19 + 3 = Y+3 / 10
10y+570=19y+57
y= 57
so the total number of oranges in the box is 57
the original number of rotten oranges in the box= 57乘 1/19
= 3
b the final number of oranges in the box = 57+3 / 10
= 6
2006-11-23 20:32:24 補充：
第2度計錯左-.-係57+3 = 60
參考: myself
2006-11-24 4:29 am
let x be the original number of rotten oranges
let the original number of orange = y
the original number of rotten oranges= y/19
the new number of rotten orange = (x+3) =( y+3)/10
x= y/19
y=19x
x+3 = (19x+3)/10
x+3 = (19x+3)/10
10x+30=19x+3
9x=27
x=3
the original number of rotten orange is 3
b the final number of orange
=y+3
=19x+3
=19(3)+3
=57+3
=60
the final number of orange is 60
let the original number of orange = y
the original number of rotten oranges= y/19
the new number of rotten orange = (x+3) =( y+3)/10
x= y/19
y=19x
x+3 = (19x+3)/10
x+3 = (19x+3)/10
10x+30=19x+3
9x=27
x=3
the original number of rotten orange is 3
b the final number of orange
=y+3
=19x+3
=19(3)+3
=57+3
=60
the final number of orange is 60
收錄日期: 2021-04-23 16:03:25
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