F.1數學題 XDDDD

____________ RNY。

2008-11-23 7:19 am

以下的題目均須用方程來作答。
各位路過人士，如果您懂其中1條問題的解法也請告訴我。唔需要話全部識答先要答ge......
1. A write with a lengh of 72 cm is cut into pieces, and then bent to form two equilateral triangles with sides of x cm and 2x cm each.

(a) Express the perimeter of each triangle in terms of x.
(b)Find the perimeter of the large triangle.
2. A fast food shop provides two types of set meals, set A and set B. If William buys 5 sets A , he will owe $15; if he buys 6 sets B, he will have $10 left. Let $x be the price of each set B.
(a) If the price of each set B is $10 less than set A, express the price of each set A in terms of x.
(b) How much is set A and set B each?
(c) How much does William have?
3. There is a pile of candies which can be evenly distributed to 28 children. If the number of candies increases by 8, and the number of children decreases by 5, then each child will get 1 more candy. How many candies are there in the pile?
4. Mary has $66 and Jason has $10.5. If Mary gives some of her money to Jason, the amount Mary has will be 2 times that of Jason's. How much does Mary give to Jason?
5. Wilson took part in a competition where each participant had to answer x questions. 30 points were given for each correct answer, 20 points were deducted for each wrong answer, and no points were deducted for any question given up. Wilson answered 3/8 of all the questions correctly, and gave up 8 questions.
If Wilson got 130 points, find the value of x.

6. For a concert, a total income of $916000 was reveived from the sale of 5500 tickets. It is known that two types of tickets in $120 and $200 were availble for sale. Suppose x tickets in $120 were sold.

Find the value of x.

回答 (1)

Uncle Michael

2008-11-23 8:05 am

✔ 最佳答案

1.
(a)
The perimeter of the triangle with sides of x cm
= 3x cm

The perimeter of the triangle with sides of 2x cm
= 3(2x) cm
= 6x cm

(b)
Total length:
3x + 6x = 72
9x = 72
x = 8

Perimeter of the large triangle
= 6x cm
= 6(8) cm
= 48 cm

=====
2.
(a)
Price of each set A
= $(x + 10)

(b)
Amount that William has:
5(x + 10) - 15 = 6x + 10
5x + 50 - 15 = 6x + 10
-x = -25
x = 25
x + 10 = 35

Ans: Set A is $35 and set B is $25.

(c)
Amount that William has
= $(6x + 10)
= $[6(25) + 10]
= $160

=====
3.
Let n be the number of candies in the pile.

(n + 8)/(28 - 5) - n/28 = 1
(n + 8)/23 - n/28 = 1

Multiply all terms on the both sides by 644.
28(n + 8) - 23n = 644
28n + 224 - 23n = 644
5n = 420
n = 84

Ans: There are 84 candies in the pile.

=====
4.
Let $y be the amount that Mary gives to Jason.

66 - y = 2(10.5 + y)
66 - y = 21 + 2y
-3y = -45
y = 15

Ans: Mary gives $15 to Jason.

=====
5.
(3/8)x(30) - [(1 - 3/8)x - 8](20) = 130
(90/8)x - [(5/8)x - 8](20) = 130
(90/8)x - (100/8)x + 160 = 130
(90/8)x - (100/8)x = -30

Multiply all terms on the both sides by 8:
90x - 100x = -240
-10x = -240
x = 24

=====
6.
The number of tickets in $200 = 5500 - x

Total income:
120x + 200(5500 - x) = 916000
120x + 1100000 - 200x = 916000
-80x = -184000
x = 2300

=====
7.
(a)
Let n be the number of VIP tickets.
Then, number of standard tickets = 15n + 650

800(15n + 650) + 1500n = 7945000
8(15n + 650) + 15n = 79450
120n + 5200 + 15n = 79450
135n = 74250
n =550

Ans: Number of VIP tickets is 550.

(b)
Number of standard tickets
= 15n + 650
= 15(550) + 650
= 8900

(c)
Let y be the number of free VIP tickets.

1200(800) + y(1500) = 1140000
960000 + 1500y = 1140000
1500y = 180000
y = 120

Ans: The number of free VIP tickets is 120.
=

👍 0 👎 0