標題:
數學~help
發問:
我唔識做啊~ 識佢問D乜,但唔識做啊 1) Alan has three types of coins:$1 coins, $2 coins and $5 coins. The number of $1 coins is three times that of $2 coins. The number of $2 coins is 5 more than the number of $5 coins. The total value of Alan's coins is $325. How many $2 coins does he have? 唔該你地~~ 更新: happyabbie1996 niki970612 你地[運吉]阿 要運吉行外D!!!!!!!!!
if $5 coins no.=y $2 coins no.=y-5 $1 coins no.=3(y-5) so 5y+2(y-5)+3(y-5)=325 5y+2y-10+3y-15=325 10y-25=325 10y=350 y=35 so $5 coins no.=35 $2 coins no.=y-5=35-5=30
其他解答:
let x be the no. of coins of $1 let y be the no. of coins of $2 let z be the no. of coins of $3 x = 3y --- i y = z + 5 z = (x / 3) - 5 --- ii (1) x + (2) y + (5) z = 325 (1) x + (2) x/3 + (5) [(x/3)-5] = 325 x + 2x/3 + 5x/3 - 25 = 325 3 (x + 2x/3 + 5x/3 - 25) = 325 x 3 3x + 2x + 5x - 75 = 975 10x = 975 + 75 x = 105 put x=105 int equation i : 105 / 3 = y y = 35 put y=35 into equation ii : y - 5 = z 35 - 5 = z z = 30|||||let no. of coins of $1 = x let no. of coins of $2 = y let no. of coins of $3 = z x / 3 = y --- i y - 5 = z -> z = (x / 3) - 5 --- ii (1) x + (2) y + (5) z = 325 p.s. 括號內的數字是相應既coins既面值 因sum = total value，所以要倍大 (1) x + (2) x/3 + (5) [(x/3)-5] = 325 x + 2x/3 + 5x/3 - 25 = 325 3 (x + 2x/3 + 5x/3 - 25) = 325 x 3 3x + 2x + 5x - 75 = 975 10x = 975 + 75 x = 105 因 x 面值=1，不需除番開有幾多個coins 將 x 代入番equation i : x / 3 = y 105 / 3 = y y = 35 將 y 代入番equation ii : y - 5 = z 35 - 5 = z z = 30 你可以將個數 check 番，$1有105個．$2有35個．$5有30個 個面值係等於番$325.0|||||let the number of $1 coins is X, then the number of $2 coins is 3X,the number of $5 coins is 3x+5, so, X+(3*2)X+((3*2)X+5)*5=325,x=8 so, the number of $2 coins is 3*8=24 ( * )這個括號里的星號表示乘|||||first you should make一條方程,將the numbers of $2 coins設為unknown,另外,你要知道$1,$2,$5之間嘅關係!|||||....sOrRy...我唔係好明白.. 你可以問下你呀媽架...=]|||||I Don't knowFAD2A23AB937987B
數學~help
發問:
我唔識做啊~ 識佢問D乜,但唔識做啊 1) Alan has three types of coins:$1 coins, $2 coins and $5 coins. The number of $1 coins is three times that of $2 coins. The number of $2 coins is 5 more than the number of $5 coins. The total value of Alan's coins is $325. How many $2 coins does he have? 唔該你地~~ 更新: happyabbie1996 niki970612 你地[運吉]阿 要運吉行外D!!!!!!!!!
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最佳解答:if $5 coins no.=y $2 coins no.=y-5 $1 coins no.=3(y-5) so 5y+2(y-5)+3(y-5)=325 5y+2y-10+3y-15=325 10y-25=325 10y=350 y=35 so $5 coins no.=35 $2 coins no.=y-5=35-5=30
其他解答:
let x be the no. of coins of $1 let y be the no. of coins of $2 let z be the no. of coins of $3 x = 3y --- i y = z + 5 z = (x / 3) - 5 --- ii (1) x + (2) y + (5) z = 325 (1) x + (2) x/3 + (5) [(x/3)-5] = 325 x + 2x/3 + 5x/3 - 25 = 325 3 (x + 2x/3 + 5x/3 - 25) = 325 x 3 3x + 2x + 5x - 75 = 975 10x = 975 + 75 x = 105 put x=105 int equation i : 105 / 3 = y y = 35 put y=35 into equation ii : y - 5 = z 35 - 5 = z z = 30|||||let no. of coins of $1 = x let no. of coins of $2 = y let no. of coins of $3 = z x / 3 = y --- i y - 5 = z -> z = (x / 3) - 5 --- ii (1) x + (2) y + (5) z = 325 p.s. 括號內的數字是相應既coins既面值 因sum = total value，所以要倍大 (1) x + (2) x/3 + (5) [(x/3)-5] = 325 x + 2x/3 + 5x/3 - 25 = 325 3 (x + 2x/3 + 5x/3 - 25) = 325 x 3 3x + 2x + 5x - 75 = 975 10x = 975 + 75 x = 105 因 x 面值=1，不需除番開有幾多個coins 將 x 代入番equation i : x / 3 = y 105 / 3 = y y = 35 將 y 代入番equation ii : y - 5 = z 35 - 5 = z z = 30 你可以將個數 check 番，$1有105個．$2有35個．$5有30個 個面值係等於番$325.0|||||let the number of $1 coins is X, then the number of $2 coins is 3X,the number of $5 coins is 3x+5, so, X+(3*2)X+((3*2)X+5)*5=325,x=8 so, the number of $2 coins is 3*8=24 ( * )這個括號里的星號表示乘|||||first you should make一條方程,將the numbers of $2 coins設為unknown,另外,你要知道$1,$2,$5之間嘅關係!|||||....sOrRy...我唔係好明白.. 你可以問下你呀媽架...=]|||||I Don't knowFAD2A23AB937987B
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