F.1 maths ~~ 急需請幫助!!

2007-12-25 7:46 pm
請問有人可以幫到手嗎?題目如下:

Three prizes worth $960 totally in a party. The value of the second prize is 4/5 of the value of the first prize and the value of the third prize is 3/4 of the value of the second prize. Find the value of each prize.
三份禮物共值 $960元。第二份是第一份的五分之四元,而第三份是第二份的四分之三元。找出每份禮份之價格。

回答 (5)

2007-12-25 8:00 pm
✔ 最佳答案
let y be the value of the first prize
y+ (4/5) y + (3/4)(4/5) y = 960
y(1+4/5+3/5) = 960
y = 960 x 5 / 12
y = 400

the first prize=$400
the second prize = 400x4/5=$320
the third prize = 320x3/4 =$240
2007-12-28 1:30 am
設第一份禮物是y元。
y+4/5y+(4/5 x3/4)y=960
y+4/5y+3/5y=960
12/5y=960
12/5y÷12/5=960÷12/5
y=400

第一份:$400
第二份:$400x0.8(五分之四)=$320
第三份:$320x0.75(四分之三)=$240
2007-12-26 4:38 pm
Let the value of the frist prize $ y
y+4/5y+(4/5 x3/4)y=960
y+4/5y+3/5y=960
12/5y=960
12/5y÷12/5=960÷12/5
y=400

The value of the frist prize $ 400
The value of the second prize $ 400x 4/5=$320
The value of the third prize $320x3/4=$240
2007-12-26 1:14 am
' 設第一份禮物是x元。
' x + x(4/5) + x(4/5) (3/4) = 960
' x + x(4/5) + x(3/5) = 960
' x + x(1.4) = 960
' 2.4x = 960
' x = 400
' ===

∴第一份禮物=$400
' 第二份禮物=$400x(4/5)=$320
' 第三份禮物=$400x(3/5)=$240
2007-12-25 10:20 pm
let $x be the value of the first prize
x+ (4/5) x + (3/4)(4/5) x = 960
x(1+4/5+3/5) = 960
x = 960 x 5 / 12
x = 400

the first prize=$400
the second prize = 400x4/5=$320
the third prize = 320x3/4 =$240


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