Five math problems that you can t do, please help the masters, and rescore additional rewards!!

1.Solution: If the bicycle travels x meters when the tires are replaced, if it is not exchanged, the front wheels will be scrapped after another 5000-x meters, and the rear wheels will be scrapped after another 3000-x.

At this time, the tires are exchanged, and the front wheel can only be driven again (5000-x) * (3000 5000) = 3 (5000-x) 5, and the rear wheel can be driven again (3000-x) * (5000 3000) = 5 (3000-x) * 3 when the front wheel is replaced to the rear wheel. So there is: 3 (5000-x) 5 = 5 (3000-x) * 3, x = 1875 meters, that is, the bicycle traveled 1875 meters when changing tires.

After changing the tires, the bike went again: 3 (5000-1875) 5 = 5 (3000-1875) * 3 = 1875 meters.

In this way, the bike traveled a total of 1875 + 1875 = 3750 meters.

2.Solution: Prime numbers are all odd numbers except 2, 3 times of odd numbers are odd numbers, and 2 times are even numbers. The sum cannot be 2000, so the one with 3 times is 2, then the one with 2 times is 997.

So the sum is 997 + 2 999

3.No, it won't.

4.Solution: Let the copper content of A be and the copper content of B B, and cut off x kilograms.

Then because the final copper content is the same, the equation can be obtained:

xa+b(4-x)]/(x+4-x)=[(6-x)a+bx]/(6-x+x)

The solution is x=5Solution: Assuming that each person is allowed to carry x kg of luggage free of charge, the excess amount will be charged at 26 (150-x) per kilogram.

Then 26 (150-x) * (150-3x) = 18

Solution x=20

A: Each person can carry 20 kg free of charge.

Hey, my brain is degenerate.

Never mind. I.

Defeated. I.

Defeated to you.

I think: let me help you look at the third question, which is a question of modulo 7 congruence; You divide 1-2008 into 7 categories, which are divisible by 7, divided by 7 and left 1, divided by 7 and left 2...In addition to 7 more than 6

mod 0: 7, 14,21,28...2002;--286

mod 1:1,8, 15,22,29...2003;--287

mod 2:2,9, 16,23,30...2004;--287

mod 3:3,10,17,24,31...2005;--287

mod 4:4,11,18,25,32...2006;--287

mod 6:6,13,20,27,34...2008;--287

Therefore, the combinations that cannot be present at the same time are mod 0, mod 1, mod 6;

mod 0,mod 2,mod 5;

mod 0,mod 3,mod 4;

mod 1,mod 1,mod 5;

mod 1,mod 2,mod 4;

mod 1,mod 3,mod 3;

mod 2,mod 2,mod 3;

mod 2,mod 6,mod 6;

mod 3,mod 5,mod 6;

mod 4,mod 5,mod 5;

mod 4,mod 4,mod 6;

mod 0:3c;mod 1/2/3/4/5/6:5c;

286(mod 0)+287(mod 1)+287(mod 2)=861

286(mod 0)+287(mod 2)+287(mod 3)=861

286(mod 0)+287(mod 3)+287(mod 5)=861

Hey, it seems that there is still a bit of a problem, my mind is not clear, I can't do it, you just use it for reference!!

1.The turnover of the food city in June is 3 million yuan, and if the business tax is paid according to 5% of the turnover, then the business tax should be paid in June [ ] 10,000 yuan. 300x5%=15 (10,000 yuan).

2.Xiao Ming's family deposits 5,000 yuan in the bank for a fixed period of one year. According to the annual interest rate, how much after-tax interest can his family get after maturity? The interest rate is 5

3.Xiao Gang deposited 2,000 yuan of New Year's money in the bank, with a regular annual interest rate of three years, and had to pay an interest tax of 5. How much money can Xiaogang get back at maturity?

4.Xiaolian's mother deposited 30,000 yuan in education deposits, the deposit period is three years, the annual interest rate is, and it can be withdrawn in a lump sum after expiration.

How much interest can I get on this deposit when it matures?

If it is not deposited as an education deposit, how much should I pay interest tax? The interest rate is 5

2916x5%=

13 million times 5% = 150,000.

2,5000 times (interest) multiplied by (100%-5%) = 3,2000 times (100% + 5%) cubic = yuan interest is unpaid tax) multiplied by 95% = yuan there are yuan.

4, "1" 30,000 times (100% + cubic = yuan.) The interest-bearing yuan is about 3012 yuan, and "2" 3012 times 95% = yuan.

, what's not going to do that.

1.The average number of households per floor is solved.

8x=120

x=15A. 2.Untie Xiao Ming and give Xiao Fang x sheets.

260+x=300-x

2x=560

x=280A. 3。The Yangtze River is x meters long.

x+371=6670

x=6299

Answer.. x=8A.

5.The diameter of the Earth is about x kilometers.

x=A. 6.The original price is $x.

x÷2=9x=18

Answer.. 7.The length of the solution is x meters.

12x=180

Who's going to do the math for you!! It's so lazy (can't you do all of this?) Think for yourself! )

1 solution: Let A and B be separated by xkm v passenger = 1 6x v cargo = 1 8x ( x -240) ( 1 6x) = 240 (1 8x) solution : x = 560 A:

A and B gather 560km2** Yes Chen Gang's thinking: everyone tells a truth and a lie, Yin Hua, no matter what he says, the conclusion is that "the prisoner can't be him", so the remaining two people are excluded, Chen Gang and Wang Chun have two situations: The prisoner is Wang Chun, then her first sentence is false, then the second sentence is true, "Chen Gang did not have Sun Fuhong to do this", so the three people are not**, so this situation is excluded The prisoner is Chen Gang, then Chen Gang's book is true, "Yin Hua did not do this either." "In line with the reasoning, Wang Chun's truth can only be:

I didn't do it"This situation was established, so the prisoner was Chen Gang.

1) For the same distance, 6 hours for passenger cars and 8 hours for trucks, so their speed ratio is 1 6:1 8, which is 4:3. When they meet, they take the same amount of time and have a distance ratio of 4:3.

Because the truck traveled 240 kilometers for 3 parts, the passenger car traveled 320,000 strips of chaotic rice, and the result came out, it was 560 kilometers. (2) Assuming that Chen Gang's previous stupid sentence is true, then Yin Hua did it, obviously, what Yin Hua said was not what he did is a lie, I don't know (I don't know what I did) is also a lie, which does not match the meaning of the accompanying topic, and is discarded. Then there is only one other situation, and what Chen Gang said is the truth, that is, he has done it.

So that's what Chen Gang did.