六年級數學急求

2006-10-28 1:40 am
(1)某數去除140,60均可整除,求某數最大是多少,

(2)己知甲,乙箱共有橙182個,從甲箱內拿出12個放進乙箱,則與乙箱內現有的橙相等,問乙箱原有橙多少個.用方程式計

回答 (8)

2006-10-28 1:50 am
✔ 最佳答案
(1) 找 140與60的最大公因數= 20

(2)
設甲箱的橙數目為 x
設乙箱的橙數目為 y

x + y = 182----------------------1
x-12 = y +12 -------------------2

from 1
x= 182 -y -----------------------3

Sub 3 in 2

182 - y -12 = y +12
2y = 158
y=79 -------------------------------4

sub 4 in 1

x + 79 = 182
x = 103

因此,甲箱有橙103個,乙箱有橙79個
2006-10-28 4:02 am
1.420
2.79
2006-10-28 1:57 am
(1) 420

(2) x - 甲,y - 乙
x-12 = y+12
x = y+24

x+y = 182
y+24+y = 182
2y = 182-24
2y = 158
y = 79
x = 79+24 = 103
2006-10-28 1:52 am
1) 20
2) 乙箱原有橙 = (182 / 2) - 12
= 91 - 12
= 79
2006-10-28 1:52 am
1.8400(唔知岩唔岩...但最少一定係8400)

2.設乙箱原有橙A個
A+12x2  =182

A+12x2-12 =182-12

Ax2    =170

Ax2除2  =170除2

A     =85
乙箱原有橙85個
參考: 自己
2006-10-28 1:51 am
1.
140 = 10 * 2 * 7
60 = 10 * 2 * 3

So, HCF of 140 and 60 = 10 * 2 = 20#

2.
甲 + 乙 = 182 ( 1 )
甲 - 12 = 乙 + 12 ( 2 )

For ( 2 )
甲 = 乙 + 24

Sub 甲 = 乙 + 24 into ( 1 )
乙 + 24 + 乙 = 182
2 乙 = 158
乙 = 79#

And 甲 + 乙 = 182
甲 + 79 = 182
甲 = 182 - 79
甲 = 103#
2006-10-28 1:49 am
1)某數最大=140 x 60

=8400

2)設x=甲箱,y=乙箱

﹛x+y=182------------1

﹛x=y+12--------------2

2代入1

y+12+y=182

y=85

乙箱原有橙85個

2006-10-27 17:54:03 補充:
補充1) 140/20=7-------------160/20=3----------------------2因1同2再除就變分數so最大值係20
2006-10-28 1:48 am
(1) 420
(2) 79


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