F.1 Maths Algebra (20 PoInT)

2009-04-19 8:19 pm

1) In a certain football Leaguem a team will score 3 points for a win, 1 point for a draw, and no points if the team loses the match. In this season, team Meanchester has 4 times as many draws as losses, and 8 times as many wins as losses. Write an algebraic expression to show the total points of Meanchester this season in each of the following situations

a) Let x be the number of losses
b) Let y be the number of draws

2) Solve the following equations (請寫埋個process)

a) 0.2(1-m)- 0.3(m+1)+0.1=0
b) 7- 1/2[ x-1/5(6-x)]= 3x+22

3) a) A bus leaves the bus terminal with y passengers. When the bus comes to the first stop, 3/5 of the passengers get off the bus and 4 passengers get on it. Write an algebraic expression in y for the number of passengers on the bus when it leaves the first shop.

3)b) At the second shop, 1/4 of the passengers get off the bus and 9passengers get on it , If the numbers of passengers on the bus now is only half that when the bus left the bus terminal find the value of y.

4) Playing games at Fun Land requires some special tokens. Kenny has bought fifty tokens. Having played 'Bowling Machine' twice, 'Monsters Shooting ' three times , 'Car Racing' twice and ' Space Journey' four times, he still has five tokens left.He won some coupons from the above games with which he can get special prizes.

a) The number of tokens required for each game is the same except for the ' Space Journey' which requires twice the usual number of tokens. How many tokens are required for playing the ' Bowling Machine' once?

b) Kenny needs 5 more coupons to get two special prizes. Or,together with his friend 's 40 coupons , he can get three special prizes. How many coupons is one special prize worth?

*所有文字題列式,同埋解釋*

回答 (1)

老爺子

2009-04-19 9:19 pm

✔ 最佳答案

1)
a)
No. of losses = x
No. of draws = 4x
No. of wins = 8x

Total points
= 8x(3) + 4x(1) + x(0)
= 28x

b)
No. of draws = y
No. of losses = y/4
No. of wins = 8(y/2) = 2y

Total points
= 2y(3) + y(1) + (y/4)(0)
= 7y

2)
a)
0.2(1 - m) - 0.3(m + 1) + 0.1 = 0

Multiply all terms on the both sides by 10:
2(1 - m) - 3(m + 1) + 1 = 0
2 - 2m -3m - 3 + 1 = 0
-5m = 0
m = 0

b)
7 - (1/2)[x - (1/5)(6-x)] = 3x + 22

Multiply all terms on the both sides by 10:
70 - 5[x - (1/5)(6-x)] = 30x + 220
70 - 5x + (6 - x) = 30x + 220
70 - 5x + 6 - x = 30x + 220
36x = -144
x = -4

3)
a)
The number of passengers on the bus when it leaves the first shop
= y - (3/5)y + 4
= (2/5)y + 4

b)
[(2/5)y + 4] - (1/4)[(2/5)y + 4] + 9 = (1/2)y

Multiply all terms on the both sides by 20:
20[(2/5)y + 4] - 5[(2/5)y + 4] + 180 = 10y
8y + 80 - 2y - 20 + 180 = 10y
-4y = -240
y = 60

4)
a)
Let n be the number of tokens needed for playing the 'Bowling Machine' once.
Then, no. of tokens needed for playing the 'Monsters Shooting' once = n
and no of tokens needed for playing the 'Car Racing' once = n
and no. of tokens needed for playing the 'Space Journey' once = 2n

2n + 3n + 2n + 4(2n) + 5 = 50
15n = 45
n = 3

Ans: 3 tokens are needed for playing the 'Bowling Machine' once.

b)
Let y be the number of coupons that one special prize is worth,
and let m be the number of coupons that Kenny has.

Kenny needs 5 more coupons to get two special prizes:
m + 5 = 2y ...... (1)

Together with his friend 's 40 coupons , he can get three special prizes
m + 40 = 3y ...... (2)

(2) - (1):
y = 35

Ans: One special prize is worth 35 coupons.
=

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