f.1 manths
2007-03-08 7:10 am
1,.A bag contains some $5 and $10 coins. There are altogether 120 coins in the bag and the total amount is$810.Find the number of $5 coins and that of $10 coins in the bag.
回答 (6)
2007-03-08 7:17 am
✔ 最佳答案
let the no. of 5 coins is X5X + (120-X)*10=810
5X + 1200 - 10X = 810
5X = 390
X = 78
therefore, there are 78 $5 coins and (120-78) 42 $10 coins in the bag.
2007-03-08 8:30 pm
Let A be the number of $10 coins.
so,the bag have(120-A) $5 coins .
10A+5(120-A)=810
10A+600-5A=810
5A=810-600
A=210/5
A=42
Therefore,there are42 $10 coins and
(120-42) $5 coins
=78 $5 coins
so,the bag have(120-A) $5 coins .
10A+5(120-A)=810
10A+600-5A=810
5A=810-600
A=210/5
A=42
Therefore,there are42 $10 coins and
(120-42) $5 coins
=78 $5 coins
2007-03-08 7:25 am
Let x be the no. of $5 coins, then y b the no. of $10 coins.
Two equations
5x+10y=810......1
x+y=120......2
From 2, x=120-y......3
sub 3 into 1, 5(120-y)+10y=810
600-5y+10y=810
5y=810-600
5y=210
y = 42
Sub y=42 into 3
x=120-42
x=78
Therefore the no. of $5 coins and that of $10 coins are 42 and 78 respictively.
Two equations
5x+10y=810......1
x+y=120......2
From 2, x=120-y......3
sub 3 into 1, 5(120-y)+10y=810
600-5y+10y=810
5y=810-600
5y=210
y = 42
Sub y=42 into 3
x=120-42
x=78
Therefore the no. of $5 coins and that of $10 coins are 42 and 78 respictively.
參考: 自己
2007-03-08 7:18 am
If $5 coins have x,$10 coins have y,
x+y=120---------(1)
5x+10y=810-------(2)
(1),
x=120-y
(2),
5(120-y)+10y=810
600-5y+10y=810
600+5y=810
5y=210
y=42
(1),
x=120-y
x=120-42
x=78
so number of $5 coins are 78 and that of $10 coins have 42 in the bag.
x+y=120---------(1)
5x+10y=810-------(2)
(1),
x=120-y
(2),
5(120-y)+10y=810
600-5y+10y=810
600+5y=810
5y=210
y=42
(1),
x=120-y
x=120-42
x=78
so number of $5 coins are 78 and that of $10 coins have 42 in the bag.
2007-03-08 7:18 am
Llet the no of $5 and $10 coins be x and y respectively
x+y=120--------(1)
5x+10y=810----(2)]
(2)/5: x+2y=162----(3)
(3)-(1): y=42
x+42=120
x=78
Thetre are 78 $5 and 42$10 coins in the bag.
x+y=120--------(1)
5x+10y=810----(2)]
(2)/5: x+2y=162----(3)
(3)-(1): y=42
x+42=120
x=78
Thetre are 78 $5 and 42$10 coins in the bag.
2007-03-08 7:18 am
Let n be number of $5 coins
so, there is (120-n) $10 coins
5n + 10 x (120 - n) = 810
5n + 1200 - 10n = 810
5n = 390
n = 78
Therefore, the number of $5 coins is 78 and the number of $10 coins is 120-78 = 42
so, there is (120-n) $10 coins
5n + 10 x (120 - n) = 810
5n + 1200 - 10n = 810
5n = 390
n = 78
Therefore, the number of $5 coins is 78 and the number of $10 coins is 120-78 = 42
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