數學唔識做(5) 快........................... OTTL

2007-05-04 2:15 am
存款存入bank,年利率5%,2年後本利共9000$,求存款

ben 買了5$和每張2$的甲、乙兩種咭共50張,共142$,各有多少張?

x貨品八折後平了10$,原價是(50$)?要式

A set of number68,85 and 102 are divisible by a certain number. What is the greatest possible value of this number?

A piece of string is 30 cm longer than another piece. If the total length of the two piece string is 80 cm, find the lengty of the longer piece of string.

回答 (3)

2007-05-04 4:15 am
✔ 最佳答案
問題1) 存款存入bank,年利率5%,2年後本利共9000$,求存款。
回答:
設存款為n元,
n ( 1 + 5% )² = 9000
n ( 1 + 0.05 )² = 9000
n ( 1.1025 ) = 9000
n = 9000 / 1.1025
n = 8163.265306
n = 8163 (取至整數)
∴存款為8163元。
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問題2) Ben 買了$5和每張$2的甲、乙兩種咭共50張,共$142,各有多少張?
回答:
設Ben買了甲咭k張,乙咭m張:
5k + 2m = 142 --------(1) *[買了$5和每張$2的甲、乙兩種咭,共$142]
k + m = 50 *[甲、乙兩種咭共50張]
k = 50 - m -------------(2)

代 (2) 入 (1):
5 ( 50 - m ) + 2m = 142
250 - 5m + 2m = 142
250 - 3m = 142
250 - 142 = 3m
108 / 3 = m
m = 36

代 m = 36 入 (2),
k = 50 - 36
k = 14

∴ Ben買了甲咭 14 張 , 乙咭 36 張。
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問題3) 貨品八折後平了10$,原價是$50?
回答:
設貨品原價為n元,
n - n ( 80% ) = 10
n - n ( 0.8 ) = 10
n ( 1- 0.8 ) = 10
n = 10 / (1 - 0.8 )
n = 10 / 0.2
n = 50
∴貨品原價為$50。
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參考: 07年理科應屆會考生
2007-05-04 2:38 am
1. P = 本金(存款), R = 年利率 = 5%, T = 時期(年) = 2年, A = 本利和 = $9000
所以, PRT + P = A
P(RT + 1) = A,
P = A/(RT+1)
代D數入去搵P

2. 設有x張甲咭, y張乙咭
x + y = 50 --- (1)
5x + 2y = 142 ---(2)
從(1): y = 50 - x
代入(2):
5x + 2(50 -x) = 142
5x + 100 - 2x = 142
3x = 42
x = ( )
代x = ( )入(1):
( ) + y = 50
y = ( )

3. 設原價為y, 80%y = y - 10
y - 80%y = 10
20%y =10
y = ( )

4. 答案 = 3個數o既H.C.F.
68 = 2 x 2 x 17
85 = 5 x 17
102 = 2 x 3 x 17
所以, H.C.F. = ( )

5. 設較長的一條繩的長度為x cm, 則較短的一條繩的長度為x - 30 cm,
所以 x + x - 30 = 80
2x = 110
x = ( )
參考: Myself
2007-05-04 2:33 am
存款8181+9/11

甲咭14,乙咭36

x*(1-80%)=$10
x*20%=$10
x=$10/20%
x=50

The greatest value is 17

Longer piece of string is 55cm long.
參考: me


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