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Shirley has three types of coins: $1-coins, $2-coins and $5-coins. The number of $5-coins is more than the number of $1-coins by 3 , while the number of $2 is 8 more than twice the number of $5-coin. If Shirley has 97 coins altogether, find the total value of the coins.
Please help me about Maths. F.1(20)
2006-11-27 4:46 am
回答 (2)
2006-11-27 5:27 am
Let D be the number of $1coins,
T be the number of $2-coins,
F be the number of $5-coins
From the question, we can formulate 2 equations as follow:
F=D+3
T=2F+8
F+T+D=97
(D+3) + (2F+8) +D =97
D+3+2(D+3)+8+D = 97
D+3+2D+6+8+D = 97
4D=97-17
4D=80
D=20
The total value of the coins:
=$1-coins+$5-coins+$2-coins
=20+5(20+3)+2 [ 2(20+3)+8 ]
=20+115+46+8
=189
T be the number of $2-coins,
F be the number of $5-coins
From the question, we can formulate 2 equations as follow:
F=D+3
T=2F+8
F+T+D=97
(D+3) + (2F+8) +D =97
D+3+2(D+3)+8+D = 97
D+3+2D+6+8+D = 97
4D=97-17
4D=80
D=20
The total value of the coins:
=$1-coins+$5-coins+$2-coins
=20+5(20+3)+2 [ 2(20+3)+8 ]
=20+115+46+8
=189
2006-11-27 5:12 am
Let x be the number of $1-coin, the number of $5-coin = x+3 and the number of $2-coin = 2(x+3)
Solution:
x+2(2(x+3))+5(x+3)=97
x+4x+12+5x+15=97
10x+27=97
10x=70
x=7
The total value of $1-coin = $1(7) = $7
The total value of $2-coin = $2(2(7+3)) = $40
The total value of $5-coin = $5(7+3) = $50
Solution:
x+2(2(x+3))+5(x+3)=97
x+4x+12+5x+15=97
10x+27=97
10x=70
x=7
The total value of $1-coin = $1(7) = $7
The total value of $2-coin = $2(2(7+3)) = $40
The total value of $5-coin = $5(7+3) = $50
參考: myself~ please vote me :)
收錄日期: 2021-04-12 18:11:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061126000051KK05536