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1.Tina has some $1 coins and $5 coins. The total value of the coins is $47. If there are19 coins altogether, how many coins of each kind does she have?2.There are some mike candies of 1.6g each and some coffee candies of 2.3g each in a pile of candies with the weight of 88.6g. If all milk candies and coffee... 顯示更多 1.Tina has some $1 coins and $5 coins. The total value of the coins is $47. If there are19 coins altogether, how many coins of each kind does she have? 2.There are some mike candies of 1.6g each and some coffee candies of 2.3g each in a pile of candies with the weight of 88.6g. If all milk candies and coffee candies are grouped into packs of 3 and 2 respectively, then there are 19 packs of candies in total. How many candies are there?
最佳解答:
1. Tinahas some $1 coins and $5 coins. The total value of the coins is $47. If thereare 19 coins altogether, how many coins of each kinddoes she have? Let a be the number of $1 coins, and b be that of $5 coins. a + 5b = 47 ...... (1) a + b = 19 ..... (2) (1) - (2): 4b = 28 b = 7 Put b= 7 into (2): a + 7 = 19 a = 12 Hence, there are 7 $1 coins and 12 $5coins. = = = = = 2. Thereare some milk candies of 1.6g each and some coffee candies of2.3g each in a pile of candies with the weight of 88.6g. If all milk candiesand coffee candies are grouped into packs of 3 and 2 respectively, then thereare 19 packs of candies in total. How many candies are there? Let m be the number of milk candies, and c be that of coffee candies. 1.6m + 2.3c = 88.6 ...... (1) (m/3) + (c/2) = 19 ...... (2) From (1): 16m + 23c = 886 ...... (3) From (2): 48 * [(m/3) + (c/2)] = 48 * 19 16m + 24c = 912 ...... (4) (4) - (3): c = 26 Put c = 26 into (2): (m/3) + (26/2) = 19 m/3 = 6 m = 18 Hence, there are 18 milk candies and26 coffee candies.
其他解答:
1. Let y and z be the number of $1 coins and $5 coins respectively. y+z=19 ---- (1) y+5z=47 ----(2) by elimination: (1) - (2) 4z=28 z=7 by substution: Put z=7 into (1) y+7=19 y=12 Thus, there are 7 $5 coins and 12 $1 coins. 2. Let y and z be the number od milk candies and coffe candies respectively. 1.6y+2.3z=88.6 --- (1) y/3+z/2=19 ---- (2) From (1) 16y+23z=886 ---- (3) From (2) 6*y/3+6*z/2=19*6 2y+3z=114 2y*8+3z*8=114*8 16y+24z=912 --- (4) (4) - (3) z=26 Put z=26 into (1) 1.6y+2.3*26=88.6 1.6y=28.8 y=18 Thus, there are 18 mile candies and 26 coffee candies.
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中二聯立方程 15點發問:
1.Tina has some $1 coins and $5 coins. The total value of the coins is $47. If there are19 coins altogether, how many coins of each kind does she have?2.There are some mike candies of 1.6g each and some coffee candies of 2.3g each in a pile of candies with the weight of 88.6g. If all milk candies and coffee... 顯示更多 1.Tina has some $1 coins and $5 coins. The total value of the coins is $47. If there are19 coins altogether, how many coins of each kind does she have? 2.There are some mike candies of 1.6g each and some coffee candies of 2.3g each in a pile of candies with the weight of 88.6g. If all milk candies and coffee candies are grouped into packs of 3 and 2 respectively, then there are 19 packs of candies in total. How many candies are there?
最佳解答:
1. Tinahas some $1 coins and $5 coins. The total value of the coins is $47. If thereare 19 coins altogether, how many coins of each kinddoes she have? Let a be the number of $1 coins, and b be that of $5 coins. a + 5b = 47 ...... (1) a + b = 19 ..... (2) (1) - (2): 4b = 28 b = 7 Put b= 7 into (2): a + 7 = 19 a = 12 Hence, there are 7 $1 coins and 12 $5coins. = = = = = 2. Thereare some milk candies of 1.6g each and some coffee candies of2.3g each in a pile of candies with the weight of 88.6g. If all milk candiesand coffee candies are grouped into packs of 3 and 2 respectively, then thereare 19 packs of candies in total. How many candies are there? Let m be the number of milk candies, and c be that of coffee candies. 1.6m + 2.3c = 88.6 ...... (1) (m/3) + (c/2) = 19 ...... (2) From (1): 16m + 23c = 886 ...... (3) From (2): 48 * [(m/3) + (c/2)] = 48 * 19 16m + 24c = 912 ...... (4) (4) - (3): c = 26 Put c = 26 into (2): (m/3) + (26/2) = 19 m/3 = 6 m = 18 Hence, there are 18 milk candies and26 coffee candies.
其他解答:
1. Let y and z be the number of $1 coins and $5 coins respectively. y+z=19 ---- (1) y+5z=47 ----(2) by elimination: (1) - (2) 4z=28 z=7 by substution: Put z=7 into (1) y+7=19 y=12 Thus, there are 7 $5 coins and 12 $1 coins. 2. Let y and z be the number od milk candies and coffe candies respectively. 1.6y+2.3z=88.6 --- (1) y/3+z/2=19 ---- (2) From (1) 16y+23z=886 ---- (3) From (2) 6*y/3+6*z/2=19*6 2y+3z=114 2y*8+3z*8=114*8 16y+24z=912 --- (4) (4) - (3) z=26 Put z=26 into (1) 1.6y+2.3*26=88.6 1.6y=28.8 y=18 Thus, there are 18 mile candies and 26 coffee candies.
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