2 math problem solving with steps and results speed

It turns out that A accounts for 5 (5+3) = 5 8 of the total

After 90 barrels are shipped, A accounts for 2 (2 + 3) = 2 5 of the total, so the total is 90 (5 8-2 5) = 400 barrels.

It turns out that B has: 400-400*5 8=150 barrels.

A's 2 9 is equal to B's 1 4

According to the nature of the ratio, B is A's: (2 9) (1 4) = 8 9 So A gets: 850 (1 + 8 9) = 450 yuan.

B gets: 850-450 = 400 yuan.

1. If the original x barrel of warehouse B is set, the original 5x 3 barrel of warehouse A will be set up.

5x/3-90=2(x+90)/3

Find x=150

Therefore, the B warehouse originally had 150 barrels of oil.

2. If A gets x yuan, then B gets 850-x yuan.

2x/9=(850-x)/4

Find x=450, 850-x=400

So A gets $450 and B gets $400.

1.If there are x barrels in warehouse B, then the original (5 3)*x barrels in warehouse A, after 90 barrels are transported from warehouse A and put into warehouse B, warehouse A has 5x 3 - 90 barrels, and warehouse B has 90 + x barrels.

So (5x 3 - 90): 90 + x = 2 :32*(90+x) = (5x 3 -90)*3180 + 2x = 5x - 270

3x = 450

x = 150

That is, the original oil of the B warehouse is 150 barrels.

2.If A has $x, then B has $850-x.

Another 2 9 * x = 1 4 * 850-x) 8x = 850 * 9 - 9x

17x = 850 * 9

x = 50 * 9

x = 450

Therefore, A has 450 yuan, and B has 850-450 = 400 yuan.

Set up the original oil x barrel of B and A Y barrel. So, y:x=5:3··· 1）y-90):(x+90)=2:3··· 2) There is (1) (2) to get x=150, y=250

Let A get x, then B gets (850-x).

2 9x=1 4(850-x), the solution is x=450, so B gets (850-x)=400

2, 2 root No. 3, Bei Chuan Ling 6 root No. 3

4.In the known parallelogram ABCD.

OA=OC The circumference of the triangle AOB is 8cm longer than the circumference of the BOC of the 3 Nawu angles, and the circumference is set to BC=X, AB=X 8

2(x+8)+2x=60

x=11x+8=19

So ab=19, bc=11

2, 2 root No. 3, Bei Chuan Ling 6 root No. 3

4.In the knowledge of the parallelogram imitation of the Qi form ABCD.

OA=OC Because the circumference of the triangle AOB is 8cm longer than the circumference of the triangle BOC, find the length of AB and CD.

So let bc=x, ab=x=8

2(x+8)+2x=60

x=11x+8=19

So ab=19, bc=11

I didn't understand the first question, is it x 2 times the back? That is, (x 2-4) (a-1) = (x + 2) (x -2) (a-1).

In the second question, using the formula of square difference, we can get (1-1 2) (1+1 2) x (1-1 3) (1+1 3) ......

Then put the bad items and the sum items together to get (1 2x2 3x3 4x......x9/10)

x(3/2x4/3x5/4x……x11/10)=1/10x11/2=11/20

(1) the waiting time of these 15 passengers (0,25);

2) The number of passengers who have waited for less than 10 minutes among these 60 passengers: 2+6=8

3) 2 people from the 3rd and 4th groups of 6 people in the above table were selected for further questionnaire survey, a total of c(6,2)=15 different results, and the results of the two people drawn happened to be from different groups c(4,1)*c(2,1)=8 kinds, so the probability of the two people drawn from different groups = 8 15

Solution: (1) The waiting time of these 15 passengers is between (0 and 25);

2) Because the number of people who wait for less than 10 minutes in 15 passengers is 2+6=8 people, so the waiting time of less than 10 minutes in 60 passengers is: 60*(8 15)=32 people.

3) The probability that two people happen to be from different groups is 8 15 (refer to Hedushuma's answer).

A school carried out tree planting activities, and it was known that the ratio of the number of trees planted in the seventh and eighth grades was 5:4, the ratio of the eighth grade and the ninth grade was 2:3, and the sum of the number of trees planted in the eighth and ninth grades was 150 more than that of the seventh grade.

7th Grade: 8th Grade = 5:4, 8th Grade: 9th Grade = 2:3=4:6

So: 7th Grade: 8th Grade: 9th Grade = 5:4:6

Per serving: 150 (6+4-5) = 30 plants.

30 (5+4+6)=450 trees.

A certain farm originally had 180 hectares of grain land and 50 hectares of vegetable land, but now it is planned to convert part of the vegetable land into grain field, so that the vegetable land area accounts for 15 percent of the grain field area.

Solution: The vegetable field to be converted into a grain field should be X hectares.

50-x)÷(180+x)=15%

50-x)÷(180+x)=3/20

180+x) ×3=（50-x）×20

x = 20 The two factories A and B completed 120 percent and 110 percent of the planned tasks respectively last year, and produced a total of 4,000 tons of food, which was 400 tons more than the sum of the original two factories.

According to the topic: A exceeds 120%-1=20%, and B exceeds 110%-1=10%.

20% of A + 10% of B = 400, that is.

A 20% + B 10% = 400, multiply it by 10 on both left and right sides, and get: A 2 + B = 4000

And the question tells us: A + B = 4000-400 = 3600

Subtract "A + B =3600" from "A2 + B =4000" where B is offset to get:

A = 4000-3600 = 400 tons.

1 question planted trees, a total of 450 trees.

Question 2: 20 hectares.

Question 3: The original production task of Field A was 400 tons.

1.Solution: Let the 8th grade be x, the 7th grade be 5 4x, and the 9th grade be 3 2x, so there is:

x+3 2x=5 4x+150, get x=120, and plant trees in each grade: x+5 4x+3 2x=450

1. Seventh grade: Eighth grade = 5:4

8th Grade: 9th Grade = 2:3=4:6

So 9th grade: 8th grade: 7th grade = 6:4:5

Grades 8 and 9 planted 150 more trees than grades 7 combined.

The sum of the number of trees planted in grades 8 and 9 is more than that in grades 7 (6+4)-5=5, and in grades 7 it is exactly 5, so the number of trees planted in grades 7 is 150, and in grades 8 it is 150 4 5=120.

In the ninth grade, it is 120 3 2 = 180 trees.

2. There are a total of 180 + 50 = 230 hectares.

The area of vegetable fields accounts for 15% of the area of grain fields

Therefore, the total area of vegetable field and grain field is 115 of the area of grain field, so after transformation, the area of grain field is 230 115 = 200 hectares, so the vegetable field changed to grain field is 200-180 = 20 hectares 3, let the planned task of A last year be x, and the planned task of B last year is y120 x + 110% y = 4000

20%x+10%y=400

The solution is x=400 and y=3200

150 for 7th grade, 120 for 8th grade, and 180 for 9th grade

20 hectares. 3640 tons.

Use C or C++ programming software to program it and it will come out. 1+1/2+1/3...1/n=ln(n)

c where c =

Then it's as simple as that.

If you use the approximate formula 1 + 1 2 + 1 3 + 1 4 ...1／100=

LN100+, this formula. This is an approximation.

1.= 1/2 + 2/3 + 3/4... One hundred and ninety-nine (after all add a fraction to make one) = 1+1+1+1...

99) - (half + third + quarter...) One hundredth) = 99 - amount, it seems to be wrong.

I didn't understand the first question, is it x 2 times the back? That is, (x 2-4) (a-1) = (x + 2) (x -2) (a-1).

In the second question, using the formula of square difference, we can get (1-1 2) (1+1 2) x (1-1 3) (1+1 3) ......

Then put the bad items and the sum items together to get (1 2x2 3x3 4x......x9/10) x (3/2x4/3x5/4x……x11/10)=1/10x11/2=11/20

If the speed of the motorcycle is a kilometer per hour, the speed of the repair vehicle is kilometers per hour. The time required for the motorcycle to complete the journey is 30 hours, and the time required for the repair vehicle to complete the journey is (30 hours, if the two vehicles arrive at the destination at the same time, the time taken by the two vehicles is equal. That is, 30 a = 30 stool disturbances, and the solution is a = 40 kilometers per hour.

A: The speed of the motorcycle is 40 kilometers per hour, and the speed of the late repair is 60 kilometers per hour.

The speed of the motorcycle is x kilometers per hour, and the speed of the repair car is kilometers per hour, and the unit is unified and debated, 15 minutes = 15 60 hours = hours.

The time taken by the motorcycle.

30 x hours, the time spent repairing the car is (30 hours, the time spent repairing the car can also be expressed as (30

Hour. Column equations.

Solution: x=40

Answer: The speed of the motorcycle is 40 kilometers per hour, and the speed of the repair vehicle is 60 kilometers per hour.

This is a fractional equation problem with the following answers.

The speed of the unfastened motorcycle is x kilometers per hour, including the state car, and the repair vehicle kilometers per hour.

15 minutes = 1 4 hours.

The first is the speed of the motorcycle, the middle one is the speed of the repair car, the oldest such as the latter is the speed of the repair car faster than the motorcycle, 1 4 hours, and the final solution is x = 40, the speed of the motorcycle is 40 kilometers per hour, the speed of the repair car is 60 kilometers per hour, the answer is absolutely correct, do not believe in the calculation. . . Just look at how hard I've worked.

If the speed of the motorcycle is x kilometers per minute, then the repair car is known as the sail is kilometers per minute. The distance is 30 kilometers, the motorcycle has to walk 30 x minutes, and the repair car has to walk 30 x minutes, because the motorcycle walks for 15 minutes first, which is 30 x = 15 ten 20 x. Get x = 2 3 km per minute.

It is 40 kilometers of hail per hour, and the repair car is 60