1.some sweets are shared by a gorup of children. if each child gets 5 sweets,8sweets will be left.if each child wants to get 7 sweets,10 more sweets will be needed.how many sweets are there?
2.find two consecutive integers such yhat 3/5 of the larger number exceeds 1/4 of the smaller one by 5又2分1
2題英文數學~幫幫手!
2007-09-03 2:54 am
回答 (2)
2007-09-03 3:10 am
✔ 最佳答案
1)let x be the no. of children
5x+8=7x-10
5x=7x-10-8
7x-5x=10+8
2x=18
x=9
because 5(9)+8
=45+8
=53
therefore , there are 53 sweets.//
2)
唔肯定...
驚教錯...
參考: myself
2007-09-03 3:12 am
1. Let x be the no. of sweets,
x/ 5 + 8 = x/7 - 10
x/ 5 - x/ 7= 10-8
7x - 5x = 70
x = 35
Therefore, there are 34 sweets.
2. Let x be the smaller no. and x + 1 be the larger no.
( 3 / 5 )( x + 1 ) + ( x / 4 ) = 11/2
(3x + 3)/ 5 + x/4 = 5.5
(12x+12+5x)/20 = 5.5
7x =98
x = 14
Therefore, the smaller integer is 14 and the larger one is 15.
2007-09-02 19:13:14 補充：
* Therefore, there are [35] sweets.
2007-09-02 19:21:02 補充：
Sorry 我計錯左第一題-____-重做::Let x be the no of sweets,y be the no. of children{5y 8 = x ---(1){7y - 10 = x ---(2) 5y 8 = 7y - 10 2y = 18y = 9Putting y = 9 into (1)x = 5(9) 8x =53Therefore, there are 53 sweets. =]
x/ 5 + 8 = x/7 - 10
x/ 5 - x/ 7= 10-8
7x - 5x = 70
x = 35
Therefore, there are 34 sweets.
2. Let x be the smaller no. and x + 1 be the larger no.
( 3 / 5 )( x + 1 ) + ( x / 4 ) = 11/2
(3x + 3)/ 5 + x/4 = 5.5
(12x+12+5x)/20 = 5.5
7x =98
x = 14
Therefore, the smaller integer is 14 and the larger one is 15.
2007-09-02 19:13:14 補充：
* Therefore, there are [35] sweets.
2007-09-02 19:21:02 補充：
Sorry 我計錯左第一題-____-重做::Let x be the no of sweets,y be the no. of children{5y 8 = x ---(1){7y - 10 = x ---(2) 5y 8 = 7y - 10 2y = 18y = 9Putting y = 9 into (1)x = 5(9) 8x =53Therefore, there are 53 sweets. =]
參考: =]
收錄日期: 2021-04-13 18:52:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070902000051KK03906