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幾條中一數學題一問 +20
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求下條題目的中文解釋、列'色'+答案、教我計算方法1)Raymon spent 13/75 of his money on a book and spent 21/31of remainder on a watch. He then had $120 left.How much money did he have at first?PS:13/75=75分之13、21/31=31分之212)Give 2 numbers x and 198.If the H.C.F. is18 and L.C.M. is1386,find x.3)A rectangular... 顯示更多 求下條題目的中文解釋、列'色'+答案、教我計算方法 1) Raymon spent 13/75 of his money on a book and spent 21/31of remainder on a watch. He then had $120 left.How much money did he have at first? PS:13/75=75分之13、21/31=31分之21 2) Give 2 numbers x and 198.If the H.C.F. is18 and L.C.M. is1386,find x. 3) A rectangular sheet of paper 225 cm by 300 cm is cut into spuares of wpual size,each as large as possible. 3a) Wath will be length of the side of each square? 3b) Wath will be area of each square? 4)Three bells ring at intervals of 6,18 and 30 minutes respectively. 4a) If they ring at a certain moment,how many minutes will pass before they do so again? 4b)If they all ring at 1:00 pm,when they ring tothger again in the afternoon?(From 1:00 pm to 6:00 pm) 唔該大家識幾多答幾多, thx! 更新: 唔該比埋條題目ge中文意思我, thx
1) money he had at first = 120 / (1- 13/73)/(1-21/31) =120/ (31/10)/(73/60) =120x31x73/(60x10) =$452.6 2) x = 1386/198 x 18 = 126 3a) length = HCF of 225 & 300 = 75cm 3b) area = 75 x 75 = 5625 sq cm. 4a)minutes will pass ＝ LCM of 6, 18 & 30 = 90 minutes 4b)ring together at 2:30, 4:00 and 5:30 pm, 3 times
其他解答:
1. Let the first amount is x money spent on a book = 13x/75 money spent on a watch=(1-13x/75)*21/31 = 62x/75*21/31 = 14x/25 therefore total money spent = 13x/75 + 14x/25 = 11x/15 total money left = (1-11x/15) = 4x/15 = $120 ----> x = $450 answer = the first amount is $450 2. 1386/198 = 7 ----> 7*18 = 126 therefore x = 126 3a. the length of each square is 225 cm 3b. the area = 225cm * 225cm = 50625sq. cm. 4a. the L.C.M. of 6, 18, 30 is 90 therefore 90 minutes will pass before they do again. 4b. as they will ring together after each 90 minutes, therefore they will ring together at 2:30 p.m. , 4:00p.m. and 5:30 p.m. from 1:00p.m. to 6:00p.m.4E350C6F8B48ECA2
幾條中一數學題一問 +20
發問:
求下條題目的中文解釋、列'色'+答案、教我計算方法1)Raymon spent 13/75 of his money on a book and spent 21/31of remainder on a watch. He then had $120 left.How much money did he have at first?PS:13/75=75分之13、21/31=31分之212)Give 2 numbers x and 198.If the H.C.F. is18 and L.C.M. is1386,find x.3)A rectangular... 顯示更多 求下條題目的中文解釋、列'色'+答案、教我計算方法 1) Raymon spent 13/75 of his money on a book and spent 21/31of remainder on a watch. He then had $120 left.How much money did he have at first? PS:13/75=75分之13、21/31=31分之21 2) Give 2 numbers x and 198.If the H.C.F. is18 and L.C.M. is1386,find x. 3) A rectangular sheet of paper 225 cm by 300 cm is cut into spuares of wpual size,each as large as possible. 3a) Wath will be length of the side of each square? 3b) Wath will be area of each square? 4)Three bells ring at intervals of 6,18 and 30 minutes respectively. 4a) If they ring at a certain moment,how many minutes will pass before they do so again? 4b)If they all ring at 1:00 pm,when they ring tothger again in the afternoon?(From 1:00 pm to 6:00 pm) 唔該大家識幾多答幾多, thx! 更新: 唔該比埋條題目ge中文意思我, thx
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最佳解答:1) money he had at first = 120 / (1- 13/73)/(1-21/31) =120/ (31/10)/(73/60) =120x31x73/(60x10) =$452.6 2) x = 1386/198 x 18 = 126 3a) length = HCF of 225 & 300 = 75cm 3b) area = 75 x 75 = 5625 sq cm. 4a)minutes will pass ＝ LCM of 6, 18 & 30 = 90 minutes 4b)ring together at 2:30, 4:00 and 5:30 pm, 3 times
其他解答:
1. Let the first amount is x money spent on a book = 13x/75 money spent on a watch=(1-13x/75)*21/31 = 62x/75*21/31 = 14x/25 therefore total money spent = 13x/75 + 14x/25 = 11x/15 total money left = (1-11x/15) = 4x/15 = $120 ----> x = $450 answer = the first amount is $450 2. 1386/198 = 7 ----> 7*18 = 126 therefore x = 126 3a. the length of each square is 225 cm 3b. the area = 225cm * 225cm = 50625sq. cm. 4a. the L.C.M. of 6, 18, 30 is 90 therefore 90 minutes will pass before they do again. 4b. as they will ring together after each 90 minutes, therefore they will ring together at 2:30 p.m. , 4:00p.m. and 5:30 p.m. from 1:00p.m. to 6:00p.m.4E350C6F8B48ECA2
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