Three extraordinary math problems, three math problems are so difficult

Mr. Ma wants to use iron wire to make a cuboid with a length of 4cm, a height of 5cm, a width of 3cm and a cube with an edge length of 8cm.

4+3+5）×4=48cm

8×12=96cm

The sum of the edges and lengths of a cuboid is 56 cm, its length is 8 cm, its width is 4 cm, and how many centimeters is its height?

56÷4-8-4=2cm

How much does the total area increase when a cube with an edge length of 8 dm is cut into two identical boxes?

8 8 2 = 128 square decimeters.

1. Because the width and length of the rectangle can make the length of the two adjacent sides the same and the inner angle of the rectangle is 90 degrees, a square can be obtained.

2. y=-2x+b through a(1,1),b(2,1),c(2,2),d(1,2).

1=-2+b

b=32=-2+b

b=42=-4+b

b=61=-4+b

b=5, so the value range of b is 3<=b<=5

3. No picture.

The distance traveled by the passenger car is 5x, the distance traveled by the truck is 4x, the whole journey is 5x+4x=9x, and the speed of the truck = 9x 15

When they meet, time is equal, 5x 60 = 4x (9x 15).

x=80, so the distance between a and b is 9x=9*80=720 km.

Untie. 1) Set up to produce x products per month, and the profit obtained by both schemes is the same. Then according to the proposal:

Profit for option 1: (55-30)x-(

Profit of option 2: (55-30).

Because the profit is the same, so 24x-3000=18x, the solution: x=500Answer: When producing 500 products per month, the profit obtained by the two schemes is the same.

2) When 6,000 products are produced per month, the cost of plan 1 is: yuan.

The cost of option 2 is: RMB.

In the case of the same production quantity, option 1 costs less.

Therefore, on the premise of not polluting the environment and saving money, you should choose option 1.

Answer: When producing 6,000 products per month, if I am the factory manager, I should choose the first solution to treat sewage without polluting the environment and saving money. Complete.

Calculate yourself, there is a square karma, pay attention, "* is multiplied by the solution: let the average monthly growth rate of the plant's output in June and July be x."

500 * 1+x）²=648

Answer:.. Solution: Let's say the monthly growth rate is x

25 * 1+x）² =91

Answer:.. Solution: Suppose the average annual depreciation rate of the car in the second and third years is x12* (

Answer:..

(1) In May, it is 500 (1-10%)=450, and the growth rate is x

Then June is 450 (1+x).

July is 450 (1+x) (1+x) = 648

1+x)²=648/450=

1+x=x=20%

So the growth rate is 20%.

2) 250,000 in January, 25(1+x) in February and 25(1+x) in March

So 25+25(1+x) +25(1+x) 2=91(5x-1)(5x+16)=0 So x=1 5(3) solution: Let the car be the first.

2. The average depreciation rate for the third year is x.

12（1－20%）（1－x）^2＝

x1, x2 rounded).

A: The average depreciation rate is 10%.

1) 648 500 (1 10) [500 (1 10)]

2) Let the growth rate be x, 25 (1 1 x 1 2x x 2) 91 solution.

3) Let the depreciation rate x, 12 (1 20 ) (1 x) 2 solution.

In May it is 500 (1-10%)=450

Let the growth rate be x

Then June is 450 (1+x).

July is 450 (1+x) (1+x) = 648

1+x)²=648/450=

1+x=x=20%

So the growth rate is 20%.

1. Let the average growth rate be x, and the output in May can be seen to be 450 tons according to the question, so 450 (1 + x) 2 = 648, and the solution is x = 20% 2

25+25(1+x)+25(1+x) 2=91, (5x+16)(5x-1)=0, and x=20% is solved

3. From the question, it can be seen that the value of the car in the first year: 12 * (10,000, let the average annual depreciation rate be x, then:

Solution: x=

1.There are X people in the fifth grade and Y people in the sixth grade.

x+y=336, 5/7*x+3/7*y=188x=154, y=182

There were 154 students in the fifth grade and 182 in the sixth grade.

2.There are x first prizes, y second prizes, and z third prizes.

54z=2y, y=2x

543.Let this batch of *** machines have x units, the purchase price of *** machine is y, and the selling price of *** machine is x*(, x-10)*

x=150,y=40

So there are 150 *** machines in this batch.

1.Suppose the fifth and sixth grades are drawn 5 7, and the following will be drawn:

336 5 7 240 (person).

More than the actual number of people drawn:

240 188 52 (person).

Sixth-graders are:

52 (5 7 3 7) 182 (person).

Fifth-graders are:

336 182 154 (person).

pieces, 120x50%=60 pieces, the ratio of the first prize, the second prize, and the third prize is 1:2:(2x2)=1:2:4

1+2+4=7, so, get.

The number of first, second and third prize works should be in the range of 54 to 60, and at the same time it is a multiple of 7, then only 56 is compliant, and the first prize: 56 x 1 7 = 8 pieces.

2nd prize: 56x 2 7 = 16 pieces.

3rd prize: 56x 4 7 = 32 pieces.

3.There are x sets of *** machines in this batch.

Unit price: (6000 20%) x=30000 x6000*(1-96%)] 10=(30000 x)*(1+20%-90%)=9000 x

24=9000/x

x=375There are 375 *** machines in this batch.

1. There are x students in the fifth grade and y students in the sixth grade, according to the topic.

x+y=336

5/7x+3/7y=188

x=154 y=182.

2. The first prize is set to be x, according to the title, the second prize is 2x, and the third prize is 4x.

Because the total number of entries is 120 and the probability of winning the prize is between 45% and 50%, there are.

45%*120 x+2x+4x 50%*120 solution:

Because the prizes are integer, the first prize is 8, the second prize is 16, and the third prize is 32.

3. There are x units in this batch, and the purchase price is Y yuan.

Depending on the title, there is.

Divide by 2.

Substituting 2 formulas, you can get y = 1200 yuan, so as to get x = 25 units.

There are a total of 25 *** machines in this batch.

The answer to the third question seems to be wrong, in short, I calculated 200, neither 375 nor 150I'm also asking for verification, so let's ask the teacher around me. It's not the road to success yet.

120 times 45% equals 54, and 120 times 50% is 60 pieces.

Solution: Set the first prize x pieces.

x+2x+4x=y (y takes the value between 54 and 60), and 1+2+4=7,54 between 60 and the common multiple of 7 is 567x=56x=8

Question 2).

What grade questions are you in?

I'll give it a try. 1 The latter item is 12

2 5 to 63 6 to 5

There are four intervals of 8 seconds between five strikes, and 2 seconds for each interval, and 22 seconds for 11 intervals between 12 strikes.

At 5 o'clock, there are 5 knocks, there are 5-1 = 4 intervals, and each interval takes 8 4 = 2 seconds.

Hit 12 times at 12 o'clock, there are 12-1 = 11 intervals, and the time required is: 2 11 = 22 seconds.

Knock 5 times at 5 o'clock, knock it at 5 o'clock, and knock it last time at 5:08 seconds, with a time interval of 8 (5-1)=2 seconds.

Knock 12 times at 12 o'clock, and the time interval is 11 two seconds. The total time is 2 (12-1) = 22 (seconds).

This question is not difficult, it belongs to the Olympiad problem in the third grade of primary school. There are many more similar topics.

The big clock in the square struck 5 times at 5 o'clock and struck it in 8 seconds.

Then knock it out with 8 5 = seconds.

Knock 12 times at 12 o'clock, it takes seconds.

The interval between two rings is 2 seconds, and the interval between 12 strikes is 11 times for 22 seconds.