Help, solve these two Olympiad problems, 3Q!

Question 1. The construction period is sunny day x and rainy day y

1/29×x+1/29×4/5×y=1

1/32×x+1/32×9/10×y=1

5x+4y=145

10x+9y=320

y=320-145×2=30

x=5 Question 2.

Set B and C cooperation for x days, and A and C cooperation for y days.

x+y)/10+y/15=1

x+y)/12+x/15=6/5

3x+5y=30

9x+5y=72

6x=42x=7

1.For each rainy day, A has to add 1 5 extra time, and B has to add 1 10 more time

There are x rainy days, 29+x 5=32+x 10

x = 30 days.

After 30 rainy days, A completed 30 * (1 29) * (4 5) = 24 29 of the project

The remaining amount of work is 5 29, and A takes 5 days to complete, that is, 5 days on a sunny day.

2.A engineering quantity is 1, then B engineering quantity is 6 5

Set B and C to cooperate for x days, then.

A project A did x days alone, did x 10 of the project, and left: 1-x 10, i.e.: (10-x) 10;

By A and C, it takes days (10-x) 10 (1 10+1 15), i.e.: (30-3x) 5

B B B and C cooperated for x days, leaving 6 5-x*(1 12+1 15), i.e.: (24-3x) 20

To be done by B alone, it takes days: :(24-3x) 20 (1 12), i.e.: (72-9x) 5

The number of days is equal: x+(30-3x) 5=x+(72-9x) 5

x = 7 days.

If it's all sunny, A takes 32-29=3 days less than B to complete the work.

On rainy days, the ergonomics of A is 1-1 5=4 5

That is, 4 days of work on a sunny day, 5 days to complete on a rainy day, and 1 day more.

The ergonomics of B is the original 1-1 10 = 9 10

That is, 9 days of work on sunny days, 10 days to complete on rainy days, and 1 day more for rainy days.

For the same 10 rainy days, A can complete the work of 8 days on sunny days, and B can complete the work of 9 days on sunny days.

A completes 1 day of work on a sunny day less than B.

To complete the work at the same time, B has to complete 3 more sunny days than A in the same number of rainy days.

So rainy days are 3 1*10=30 days.

Sunny days are 29-30*4 5=5 days.

or 32-30*9 10=5 days.

The ergonomic ratio of A, B and C is 1 10:1 12:1 15 = 6:5:4, and the workload of project A is 1, then the workload of project B is 1 + 1 5 = 6 5, and the total workload is 1 + 6 5 = 11 5

No matter how to cooperate, if it is completed at the same time, the time of work of A, B and C is the same.

The ratio of the amount of work done is equal to the ergonomic ratio.

B has completed (11 5) * 5 (6 + 5 + 4) = 55 75b and 6 5-55 75 = 7 15, which is completed by C.

The time for C to complete these 7 15 is the time for B and C to cooperate, which is:

7 15) (1 15) = 7 days.

1, Xiao Ming has yuan on him, and Xiao Hong also has some money on him, after the two of them each bought the same story book, Xiao Ming's remaining money is more than Xiao Hong's remaining money, so Xiao Hong originally had ( ) money?

2. When Xiaohua calculates a decimal subtraction, he regards 5 on the decimated tenths as 3 and 1 on the decimated hundredth as 7, so is the wrong answer reduced or increased compared with the correct answer? How much do they differ?

If you look at 5 in the decimated decile as 3, here there is 20 less, and if you look at the 1 in the decimated hundredth place as 7, and there is 600 more, then 600-20 = 580, which is an increase of 580.

3. In the mid-term test, the average score of several homework subjects except mathematics was 94 points. If mathematics is included, with an average score of 95 in each subject, to know that he scored 100 points in mathematics, ask this student how many homework he has taken in total?

100-94 = 6 points, 6 points higher in mathematics, making the average score 1 point higher, it must have taken 6 homework.

4. Find the difference series 3, 8, 13, 18 ......The average of 803.

Number of items to be counted first: (803-3) 5=160.

Resum: (3+803)*160 2=6448

Average: 6448 160 = 403

5. Grade A sugar and grade B sugar are mixed together, and the average price is 7 yuan per kilogram. It is known that there are 4 kilograms of Grade A sugar, which are sold for 8 yuan per kilogram on average, and Grade B sugar is 2 kilograms, how many yuan is bought per kilogram on average?

There are 4 + 2 = 6 kg, then the total price is: 6 * 7 = 42 yuan, the total price of grade A is: 4 * 8 = 32 yuan, then the total price of grade B sugar is: 42-32 = 10 yuan, then the average price of grade B sugar per kilogram is 10 2 = 5 yuan.

6. The average score of the students in the third class of the fifth grade in the mathematics test is, and the subsequent review found that the 98 points of a student were mistakenly calculated as 89 points when calculating the score, and after recalculation, the average score of the whole class is a score, how many students are in the third class of the fifth grade?

98-89 = 9 points are undercounted, and the average score is increased by points, then 9 people.

2.It was reduced. They differ.

3.(100-95)/(95-94)+1=6.This student took a total of 6 homework.

5.(6 7-4 8) 2=5 yuan.

6.(98-89) (name.)

Money paid by someone: 20,000 + 12,000 * 5% = 20,600 yuan Income from handling goods: 12,000 * (30% + 10%) + 2000 * 75% * (1-30% - 10%) + 20000 * (1-75%) * 1-20%)) 70%

13,900 yuan.

Loss = 20600-13900 = 6700 yuan.

The consignment sale is divided into two parts.

Part 1: 40% of 75% of the consignment price, of which 30% is subject to 5% handling fee 12000 40%-12000 30% 5% = 4620 Part 2: 60% of 75% + remaining 25% (1-20%) = 80% 70% of the original price.

Fees received 20000 (75% 60% + 25% 80%) 70% = 9100

Loss 20000-(4620+9100)=6280

1.There is a. Second.

C weighed three people, known A. The average weight of B and B is 49 kg. The average body of C is 47 kg, A.

The average weight of C and C is 45Armor. Second.

What is the average weight of C3?

A + B = 49 * 2 = 98

B + C = 47 * 2 = 94

A + C = 45 * 2 = 90

Add the above three formulas: 2 * (A + B + C) = 98 + 94 + 90 A + B + C = 49 + 47 + 45

Average weight = (A + B + C) 3 = 141 3 = 472Xiaohua practiced calligraphy, an average of 85 words per day for the first four days, and he wanted to increase the average number of words written per day in the first five days to 87 words, so how many words did he have to write on the fifth day?

If the average of the first 5 days changes from 85 to 87, you need two more words per day, which means that you need more for the first five days.

87-85)*4=8 characters.

On the basis of 87 on the 5th day, 8 more words are needed, and the first 4 days should be given on average, so the sixth day needs to be written 87 + 8 = 95

1., A + B = 49 2 B + C = 47 2 C + A = 45 2 The above 3 formulas add up: (A + B) + (B + C) + (C + A) = 49 2 + 47 2 + 45 2 Simplify:

A + B + C) 2=(49+47+45) 2 It can be obtained: A + B + C = 49 + 47 + 45 = 141 The average weight is: 141 3=47

2.To make the first 5 days rise to 87, just write 87 + (87-85) 4 = 95 on the fifth day.

1.Let A's weight be x, B (y), C(z)(x+y) 2=49 (1); y+z)/2=47 (2); x+z)/2=45 (3)

1)+(2)+(3) = x+y+z=141, so (x+y+z) 3=47

2.Total number of words written in the first four days: 85*4=340

If he wants to increase the average number of words he writes per day for the first 5 days to 87 words, he should write: 87*5=435

So on day 5 to write: 435-340=95

1. (49 + 47 + 45) * 2 6 = 47 (kg).

5-85*4=95 (pcs).

Solution 2: Write 2 more than the original every day, a total of 2*5=10 more in 5 days, plus 85 that should have been written, so 85+10=95.

1.A + B = 49x2 = 98 B + C = 47x2 = 94 A + C = 45x2 = 90

Add the three A + B + B + C + A + C = 98 + 94 + 90 = 2822 (A + B + C) = 282 A + B + C = 141 (A + B + C) 3 = 141 3 = 47

1，1^2+2^2+3^2+..n^2=n*(n+1)*(2n+1)/6

So the original formula = 30 * (30 + 1) * (2 * 30 + 1) 6 = 9455

2. It is not difficult to find the pattern by writing a few more numbers.

Original = 1 + 2 + 3 + 4 + .100=5050

3, because it is the sum of 1 consecutive natural numbers, the sum is divisible by 11. The average of the 10 numbers is , so the difference between the sum and the division by 11 should be less than 11.

It was found that 298 plus 32 is a multiple of 11, and the average of 30 is close to it, which is verified to be accurate, and 32 is the missing number.

4. It is easy to get the answer by inverse extrapolation.

1) Finally, put the 1 7 of the current C pile into the A pile, indicating that the previous C pile left 6 7, 6 7 = 12, so there are 14 before the C pile, then there are 10 in the A pile.

2) Put the 1 5 of the current B pile into the C pile, indicating that 4 5 was left in the B pile before, and 4 5 = 12, so there were 15 before the B pile, then there were 11 in the C pile.

3) Put the 1 3 of pile A into pile B, indicating that 2 3 was left in pile A before, and 2 3 = 15, so there are 15 before pile A, then there are 10 in pile B.

In summary, there are 15 in A, 10 in B, and 11 in C.

Handmade...

It's OK to make a program and do the math.