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初中數學問題~~~~~~~~~
發問:
1) The total age of Mr. Chu and his son is 51. Three years ago, he was 8 times as old son, How old is his son now?2) The difference between the present ages of Mr. Chan and his son is 32. 6 years later, their total age will be 50. How old is Mr. Chan' s son now?3) Sam cycled at 2 different speeds: 10 km/h... 顯示更多 1) The total age of Mr. Chu and his son is 51. Three years ago, he was 8 times as old son, How old is his son now? 2) The difference between the present ages of Mr. Chan and his son is 32. 6 years later, their total age will be 50. How old is Mr. Chan' s son now? 3) Sam cycled at 2 different speeds: 10 km/h and 12 km/h. He rode 67 km in 6 hours. How long had he cycled at the slower speed? 更新: 4) There are 3 types of concert tickets: $400, $200 and $100. The number of $200 tickets sold is 4 times that of $400 tickets. The number of $100 tickets sold is 380 more than that of $200 tickets. If total income from the tickets is $86000, find the number of $400 tickets sold.
最佳解答:
1.Let the son be x years old. Then the father is (51-x) years old. 8(x-3)=51-x-3 8x-24=48-x 9x=72 x=8 Therefore, the son is 8 years old now. 2. Let Mr. Chan's son be x years old Then Mr. Chan is 32+x years old. x+6+32+x+6=50 44+2x=50 2x=50-44 2x=6 x=3 Therefore, Mr. Chan's son is 3 years old now. 3. Let he cycled x hrs at the slower speed. Then he cycled (6-x) hrs at the faster speed. 10x+12(6-x)=67 10x+72-12x=67 -2x=--5 x=2.5 Therefore, he cycled 2.5 hrs at the slower speed. 4. Let the number of $400 tickets sold be y. Then the number of $200 tickets sold be 4y and the number of $100 tickets sold is 4y+380. 400y+(200x4y)+100(4y+380)=86000 400y+800y+400y+38000=86000 1600y=86000-38000 16000y=48000 y=30 Therefore, the number of $400 tickets sold is 30. 2007-05-08 18:40:44 補充： I hope you can understand the rationale of the answers. Follow the steps in your other assignments instead of just copying the answers to your homework.Best wishes!!
1.) Let x is age of Mr. Chu and y is his son now. x + y = 51---(1) Three years ago, x - 3 = 8(y - 3)----(2) x = 8y -24 + 3 x = 8y -21-------(3) sub (3) into (1) 8y - 21 + y = 51 y = (51+21) / 9 therefore, 8years old of his son 2.) Let x is age of Mr. Chan and y is his son now. x - y =32-----(1) 6 years later, (6+ x)+ (6+ y) = 50 x + y = 38-----(2) (2) - (1) 2y = 6 y = 3 Therefore, 3 years old of Mr. Chan's son 3.) Let x be a hours on10 km/h and y be a hours on 12km/h 10x + 12y = 67----(1) x + y = 6-----(2) y= 6 - x-----(3) sub y into (1) 10x + 12(6 - x) = 67 -2x = -5 x = 5/2 Therefore, he cycled at the slower speed 2.5hours 4.) Let x be number of $400 ticket and y be number of $200 ticket and z be number of $100 ticket 4x = y y + 380 = z z = 380 + 4x 400x + 200y + 100z = 86000 400x + 200(4x) + 100(380 + 4x) = 86000 1600x + 38000 = 86000 x = 48000/1600 x = 30 Therefore, number of $400 tickets is 30|||||1. 8 2. 3 3. 2.5 hr 4. 30
初中數學問題~~~~~~~~~
發問:
1) The total age of Mr. Chu and his son is 51. Three years ago, he was 8 times as old son, How old is his son now?2) The difference between the present ages of Mr. Chan and his son is 32. 6 years later, their total age will be 50. How old is Mr. Chan' s son now?3) Sam cycled at 2 different speeds: 10 km/h... 顯示更多 1) The total age of Mr. Chu and his son is 51. Three years ago, he was 8 times as old son, How old is his son now? 2) The difference between the present ages of Mr. Chan and his son is 32. 6 years later, their total age will be 50. How old is Mr. Chan' s son now? 3) Sam cycled at 2 different speeds: 10 km/h and 12 km/h. He rode 67 km in 6 hours. How long had he cycled at the slower speed? 更新: 4) There are 3 types of concert tickets: $400, $200 and $100. The number of $200 tickets sold is 4 times that of $400 tickets. The number of $100 tickets sold is 380 more than that of $200 tickets. If total income from the tickets is $86000, find the number of $400 tickets sold.
最佳解答:
1.Let the son be x years old. Then the father is (51-x) years old. 8(x-3)=51-x-3 8x-24=48-x 9x=72 x=8 Therefore, the son is 8 years old now. 2. Let Mr. Chan's son be x years old Then Mr. Chan is 32+x years old. x+6+32+x+6=50 44+2x=50 2x=50-44 2x=6 x=3 Therefore, Mr. Chan's son is 3 years old now. 3. Let he cycled x hrs at the slower speed. Then he cycled (6-x) hrs at the faster speed. 10x+12(6-x)=67 10x+72-12x=67 -2x=--5 x=2.5 Therefore, he cycled 2.5 hrs at the slower speed. 4. Let the number of $400 tickets sold be y. Then the number of $200 tickets sold be 4y and the number of $100 tickets sold is 4y+380. 400y+(200x4y)+100(4y+380)=86000 400y+800y+400y+38000=86000 1600y=86000-38000 16000y=48000 y=30 Therefore, the number of $400 tickets sold is 30. 2007-05-08 18:40:44 補充： I hope you can understand the rationale of the answers. Follow the steps in your other assignments instead of just copying the answers to your homework.Best wishes!!
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其他解答:1.) Let x is age of Mr. Chu and y is his son now. x + y = 51---(1) Three years ago, x - 3 = 8(y - 3)----(2) x = 8y -24 + 3 x = 8y -21-------(3) sub (3) into (1) 8y - 21 + y = 51 y = (51+21) / 9 therefore, 8years old of his son 2.) Let x is age of Mr. Chan and y is his son now. x - y =32-----(1) 6 years later, (6+ x)+ (6+ y) = 50 x + y = 38-----(2) (2) - (1) 2y = 6 y = 3 Therefore, 3 years old of Mr. Chan's son 3.) Let x be a hours on10 km/h and y be a hours on 12km/h 10x + 12y = 67----(1) x + y = 6-----(2) y= 6 - x-----(3) sub y into (1) 10x + 12(6 - x) = 67 -2x = -5 x = 5/2 Therefore, he cycled at the slower speed 2.5hours 4.) Let x be number of $400 ticket and y be number of $200 ticket and z be number of $100 ticket 4x = y y + 380 = z z = 380 + 4x 400x + 200y + 100z = 86000 400x + 200(4x) + 100(380 + 4x) = 86000 1600x + 38000 = 86000 x = 48000/1600 x = 30 Therefore, number of $400 tickets is 30|||||1. 8 2. 3 3. 2.5 hr 4. 30
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