F1 mathematics-要快(要步驟)
2007-10-17 3:41 am
1.Donna has three types ofcoins,$2coins and$5 coins.The numberof$1 coins is there times that of $ 2coins.The number of $2 coins is 5 more than the number of $5 coins. The total value of Donna's coins is $325. How many $2 coins does she have ?
回答 (5)
2007-10-17 3:57 am
✔ 最佳答案
Let the number of $2 coins is y, then the no. of $1 coins is 3y & $5 coins is y-5The equation :
3y+2y+5(y-5)=325
3y+2y+5y-25=325
10y-25=325
10y=325+25
10y=350
y=35
She has 35 $2 coins
2007-10-17 3:50 am
let the no. of $2 coins be X,no. of $1 coins be 3X,no. of $5 coins be (X-5)
2X+3X+5(X-5)=325
2X+3X+5X-25=325
10X=350
X=35
so she have 35 $2 coins
2X+3X+5(X-5)=325
2X+3X+5X-25=325
10X=350
X=35
so she have 35 $2 coins
2007-10-17 3:49 am
let the number of $2 coins be a
3a+5(a-5)+2a=325
5a+5a-5=325
10a-5+5=325+5
10a/10=330/10
a=33
應該岩
我都係f1 ga~!
希望幫到你
2007-10-16 19:50:56 補充：
o sor拆括號時miss 左x5添let the number of $2 coins be a3a 5(a-5) 2a=3255a 5a-25=32510a-25 25=325 2510a/10=350/10a=35
3a+5(a-5)+2a=325
5a+5a-5=325
10a-5+5=325+5
10a/10=330/10
a=33
應該岩
我都係f1 ga~!
希望幫到你
2007-10-16 19:50:56 補充：
o sor拆括號時miss 左x5添let the number of $2 coins be a3a 5(a-5) 2a=3255a 5a-25=32510a-25 25=325 2510a/10=350/10a=35
參考: 自己
2007-10-17 3:48 am
let x,y and z be the number of $1 conis, $2coins and $5coins respectively
x=3y
y-5=z
x+2y+5z=325
3y+2y+5(y-5)=325
5y+5y-25=325
10y=350
y=35
She has 35 $2coins
x=3y
y-5=z
x+2y+5z=325
3y+2y+5(y-5)=325
5y+5y-25=325
10y=350
y=35
She has 35 $2coins
2007-10-17 3:47 am
Let Donna has x $1 coins
So, she has x/3 $2 coins
she has (x/3 - 5) $5 coins
Total value = 325
i.e. x + 2(x/3) + 5(x/3 - 5) = 325
x + 2x/3 + 5x/3 - 25 = 325
10x/3 = 350
x = 105
So, she has 105/3 = 35 $2 coins
So, she has x/3 $2 coins
she has (x/3 - 5) $5 coins
Total value = 325
i.e. x + 2(x/3) + 5(x/3 - 5) = 325
x + 2x/3 + 5x/3 - 25 = 325
10x/3 = 350
x = 105
So, she has 105/3 = 35 $2 coins
參考: Myself~~~
收錄日期: 2021-04-13 13:57:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071016000051KK02960