唔該~我要清楚嘅式(+解釋) + ans~
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我要條式+ans!!!
2009-11-29 4:49 am
There are altogether twenty $5 coins and $2 coins in a purse. If half the value of the $5 coins ts $8 more than 4 times the value of the $2 coins, find the number of each kind of coins.
唔該~我要清楚嘅式(+解釋) + ans~
20點比曬你~
唔該曬!!!!!!!
唔該~我要清楚嘅式(+解釋) + ans~
20點比曬你~
唔該曬!!!!!!!
回答 (2)
2009-11-29 5:47 am
✔ 最佳答案
Let x be the number of $5 coins and y be the number of $2 coins,
{x+y=20..........(1)
5x/2-8=42y.....(2)
From (1),
x+y=20
x=20-y.........(3)
Substitute (3) into (2),
5(20-y)/2-8=42y
5(20-y)/2-8=8y
50-5y/2-8=8y
50-8=8y+5y/2
42=(8+5/2)y
42=10.5y
4=y
y=4
Substitute y=4 into (3),
x=20-4
x=16
∴The number of $5 coins is 16 and the number of $2 coins is 4.
2009-11-29 5:50 am
Let no. of $2 coins = x
no. of $5 coins = 20 - x
Value of $2 coins = 2x
Value of $5 coins = 5(20 - x)
so
5(20 - x)/2 - 4(2x) = 8
5(20 -x)/2 - 8x = 8
5(20 - x) - 16x = 16
100 - 5x - 16x = 16
21x = 84
x = 4 = $2 coins
no. of $5 coins = 20 - 4 = 16.
no. of $5 coins = 20 - x
Value of $2 coins = 2x
Value of $5 coins = 5(20 - x)
so
5(20 - x)/2 - 4(2x) = 8
5(20 -x)/2 - 8x = 8
5(20 - x) - 16x = 16
100 - 5x - 16x = 16
21x = 84
x = 4 = $2 coins
no. of $5 coins = 20 - 4 = 16.
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