(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
六年級數學急求
2006-10-28 1:40 am
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
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
(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
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
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#
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
=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
(2) 79
收錄日期: 2021-04-12 22:33:46
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