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wing yan

2007-12-26 10:47 pm

我唔識做啊~
識佢問D乜,但唔識做啊
1)
Alan has three types of coins:$1 coins, $2 coins and $5 coins. The number of $1 coins is three times that of $2 coins. The number of $2 coins is 5 more than the number of $5 coins. The total value of Alan's coins is $325. How many $2 coins does he have?

唔該你地~~

回答 (7)

Mandy

2007-12-26 11:16 pm

✔ 最佳答案

if $5 coins no.=y
$2 coins no.=y-5
$1 coins no.=3(y-5)
so 5y+2(y-5)+3(y-5)=325
5y+2y-10+3y-15=325
10y-25=325
10y=350
y=35
so $5 coins no.=35
$2 coins no.=y-5=35-5=30

參考: myself

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詳詳

2007-12-26 11:45 pm

let x be the no. of coins of $1
let y be the no. of coins of $2
let z be the no. of coins of $3

x = 3y --- i
y = z + 5
z = (x / 3) - 5 --- ii

(1) x + (2) y + (5) z = 325
(1) x + (2) x/3 + (5) [(x/3)-5] = 325
x + 2x/3 + 5x/3 - 25 = 325
3 (x + 2x/3 + 5x/3 - 25) = 325 x 3
3x + 2x + 5x - 75 = 975
10x = 975 + 75
x = 105

put x=105 int equation i :
105 / 3 = y
y = 35

put y=35 into equation ii :
y - 5 = z
35 - 5 = z
z = 30

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用戶26397

2007-12-26 11:38 pm

let no. of coins of $1 = x
let no. of coins of $2 = y
let no. of coins of $3 = z

x / 3 = y --- i
y - 5 = z -> z = (x / 3) - 5 --- ii

(1) x + (2) y + (5) z = 325
p.s. 括號內的數字是相應既coins既面值
　　因sum = total value，所以要倍大
(1) x + (2) x/3 + (5) [(x/3)-5] = 325
x + 2x/3 + 5x/3 - 25 = 325
3 (x + 2x/3 + 5x/3 - 25) = 325 x 3
3x + 2x + 5x - 75 = 975
10x = 975 + 75
x = 105

因 x 面值=1，不需除番開有幾多個coins
將 x 代入番equation i :
x / 3 = y
105 / 3 = y
y = 35

將 y 代入番equation ii :
y - 5 = z
35 - 5 = z
z = 30

你可以將個數 check 番，$1有105個．$2有35個．$5有30個
個面值係等於番$325.0

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用戶76421

2007-12-26 11:33 pm

let the number of $1 coins is X, then the number of $2 coins is 3X,the number of $5 coins is 3x+5,
so, X+(3*2)X+((3*2)X+5)*5=325,x=8
so, the number of $2 coins is 3*8=24

( * )這個括號里的星號表示乘

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山山

2007-12-26 11:03 pm

first you should make一條方程,將the numbers of $2 coins設為unknown,另外,你要知道$1,$2,$5之間嘅關係!

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