中學既數學問題..急要..[15點]

Question 1.
n+0.1/0.3 - n-0.1/0.2 = n+1.2/0.8 n=?
n +1/3 –n – 1/2 = n + 3/2
1/3 - 1/2 -3/2 = n

n = - 1 2/3 or – 1.6667 (4 decimal place)

Question 2.
Let x be new number of members in the club.
(x – 4) = original number of member in the club
New Number of males 1/3 (x - 4) + 4 = 2/5 (x)
1/3 (x) -4/3 + 4 = 2/5 (x)
Multiply by 5 to get rid of fraction
5x -20 + 60 = 6x
-20 + 60 = 6x – 5x
40 = x
x = 40

The new number of members in the club = 40.

Question 3.
Let x be the number of $2 coins.
The number of $1 coins = x + 7
The number of $5 coins = ½ (x + 7)
The total value = $1(x +7) + $2(x) + $5[1/2(x + 7)]
$85 = $1(x +7) + $2(x) + $5[0.5x + 3.5]
85 = x +7 + 2x + 2.5x + 17.5
85 = 5.5x + 24.5
5.5x = 85 - 24.5
5.5x = 60.5
x = 11

the number of $2 coins = 11

Question 4.
Let the number of newspapers sold = x
Let the number of magazines sold = 136 – x
Profit by selling newspapers = $2.40 (x)
Profit by selling magazines = $4.60 (136 - x)
Total profit = Profit from newspapers + Profit from magazines
$432 = $2.40 (x) + $4.60 (136 - x)
432 = 2.4x + 4.6 (136) – 4.6x
432 = 2.4x + 4.6 (136) – 4.6x
432 = 2.4x + 625.6 – 4.6x
4.6 x – 2.4x = 625.6 - 432
2.2x = 193.6
x = 88
Number of newspapers sold = 88
Number of magazines sold = 136 – 88 = 48
Profit from newspapers = $2.40 (88) = $211.20
Profit from magazines = $4.60 (48) = $220.80

Magazines sold today contributed a greater profit ($220.80 > $211.20)

Question 5.
Let x be the original number of marbles in bag A
Let (162 – x) be the original number of marbles in bag B
The new number of marbles in bag A = ¾ x
The new number of marbles in bag b = (162 – x) + 1/4x

¾ x – [(162 – x) + 1/4x] = 30
¾ x – [162 – 3/4x] = 30
¾ x – 162 + 3/4x = 30
¾ x + 3/4x = 30 + 162
1.5 x = 192
x = 192/1.5 = 128

Original number of marbles in bag A is 128
Original number of marbles in bag B is (162 – 128) = 34