數學問題,,唔識牙,,急

2008-12-31 9:58 pm
1.
David has 3 sons Albert, Ben and Curie. He wants to divided 34 gold conis among his sons so that the ratio of Albert gets to Ben gets is x:y and the ratio of Ben gets to Curie gets is x:z, where x, y and z are positive integers. Suggest how he can divide the gold coins into among 3 sons without melting of coins.

2.
Peter and Bob estimated the weight of a cat to be 1.3kg and 1283g. Bob claims that his estimate is more accurate as it has more significant figures. Do you agree with Bob's claim? Explain your answer.

唔該清楚寫步驟..謝謝

回答 (1)

2009-01-01 2:05 am
✔ 最佳答案
1) 以A,B,C代表三個仔所得的金幣量,現在求A:B:C,
由已知: A:B=x:y ; B:C=x:z ;
x : y
x : z
-----------------------------
x^2 :xy:zy
所以A : B : C = x^2 : xy : zy ,由於x,y,z都是positive integers,所以x^2,xy及zy都是positive integers.
x^2 : xy : zy這個比例式表示著34=17x2個金幣分法,所以:
x^2+xy+zy會等於34或17,因此可探求方程x^2+xy+zy=34及
x^2+xy+zy=17的解。兩方程的任何正整數解都是代表著一種分法,所以此題其實有很多分法。
例如:
x^2+y(x+z)=17
試x=2,則4+y(2+z)=17
y(3+z)=13,有y=1,z=11;所以 x^2 : xy : zy等於
2^2 : 2*1 :11*1 = 4 : 2 : 11, 這代表著34個金幣依 4 : 2 : 11分成
8個,4個及22個,即A : B = 4:2 = 2 : 1 , B : C = 2 : 11。
你可以試解x^2+xy+zy=34,代不同數入x,得出很多符合題目要求的解。

2)Not agree.假設隻cat重1.29999kg,則Peter比Bob準確。因為準確度高低不是看數位有多少,而是看估計的與真確值有多接近。

2009-01-01 14:44:13 補充:
y(3+z)=13改為

2009-01-01 14:45:19 補充:
y(2+z)=13


收錄日期: 2021-04-21 22:01:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081231000051KK00925

檢視 Wayback Machine 備份