1. the cost of a ruler is four times that of a rubber.if the total cost of a ruler and a rubber is $15,find the cost of a rubber.
2. jim's father is now 10 times jim's age.in eight year's time,his age will be four times jim's age.how old is jim now?
3. jane has 15 coins which include $2-coins and $5-coins only.if the total value of $2-coins is less than that of $5-coins by $33,how many $5-coins does jane have?
maths ^^^^vvv^^^
2010-11-03 3:13 am
回答 (2)
2010-11-03 3:45 am
✔ 最佳答案
1. let x be the cost of a rubber
4x + x = 15
5x = 15
x = 15/5
x = 3
therefore, the cost of a rubber is $3
2.
let y be Jim's age
4*(y+8) = 10y+8
4y+32 = 10y+8
32-8 = 10y-4y
24 = 6y
y = 24/6
y = 4
therefore, Jim now is 4 years old
3.
let x be the number of $5 coins
2(15-x) + 33 = 5x
30-2x+33=5x
30+33 = 5x+2x
63 = 7x
x = 63/7
x = 9
therefore, Jane has 9 pieces of $5 coin
2010-11-02 19:47:30 補充：
回覆001
第二條唔知你問咩==in eight year's time????
in eight year's time = 8年後
2010-11-03 3:22 am
1.Let x be the cost of a rubber
4x+x=15
x(4+1)=15
x(5)=15
x=3
3.Let x and y be the No.of $2-coins and $5-coins respectively
x+y=15...(1)
5y-2x=33...(2)
(1)X2,
2x+2y=30...(3)
(2)+(3)
5y-2x+(2x+2y)=33+30
5y-2x+2x+2y=63
7y=63
y=9
Put y=9 into (2)
5(9)-2x=33
45-2x=33
-2x=-12
x=6
So:Jane has 9 $5-coins.
第二條唔知你問咩==in eight year's time????
4x+x=15
x(4+1)=15
x(5)=15
x=3
3.Let x and y be the No.of $2-coins and $5-coins respectively
x+y=15...(1)
5y-2x=33...(2)
(1)X2,
2x+2y=30...(3)
(2)+(3)
5y-2x+(2x+2y)=33+30
5y-2x+2x+2y=63
7y=63
y=9
Put y=9 into (2)
5(9)-2x=33
45-2x=33
-2x=-12
x=6
So:Jane has 9 $5-coins.
第二條唔知你問咩==in eight year's time????
收錄日期: 2021-04-19 23:29:38
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