1) Mothers' age is 4 times of her daughters' age. 10 years ago, the sum of their age is 60, Find the daughter's age now?
2) 5 rings and 8 necklaces cost $640, The price of a ring is $50, The selling prices of neackes is first deducted by $10 and is halved afterwards Find the marked price of the necklace.
F1 MATHS
2008-11-11 12:33 am
回答 (2)
2008-11-15 6:26 am
1)
Let X be the daughter's age now.
(4X-10)+(X-10)=60
4X+X =60+10+10
5X=80
X=80/5
X=16
The daughter's age now is 16.
2)
Let $Y be the marked prices of the necklace.
8*((Y-10)/2)=640-50*5
4Y-40=640-250
4Y=390+40
4Y=430
Y=430/4
Y=107.5
The marked prices of the necklace is $107.5.
Let X be the daughter's age now.
(4X-10)+(X-10)=60
4X+X =60+10+10
5X=80
X=80/5
X=16
The daughter's age now is 16.
2)
Let $Y be the marked prices of the necklace.
8*((Y-10)/2)=640-50*5
4Y-40=640-250
4Y=390+40
4Y=430
Y=430/4
Y=107.5
The marked prices of the necklace is $107.5.
參考: myself
2008-11-11 12:38 am
1)
如果女兒今年歲數是A
咁(4A-10)+(A-10)=60
5A-20=60
5A=80
A=16
如果女兒今年歲數是A
咁(4A-10)+(A-10)=60
5A-20=60
5A=80
A=16
收錄日期: 2021-04-29 19:07:36
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