A sixth grade winter vacation math problem, urgent

Method 1:

A: The number of second prize winners accounts for 1 10

Method 2: 1-9 10 1 10 for the first prize and 1 5-1 10 for the second prize.

Method 3: Solution: Obtained.

1/5 of the total number of winners of the first and second prizes, and 1 1 5 = 4 5 of the total number of winners of the third prize.

Obtain. The second and third prizes account for 9/10 of the total number of winners, and the number of people who win the first prize accounts for 1 9 10 = 1 10.

The number of people who won the third prize accounted for 4 5 = 8 10.

The number of people who won the second prize accounted for 1 1 10 8 10 = 1 10.

A: The number of second prize winners accounts for 1 10 of the total number of people

Method 4: Take the total number of people 1 and subtract the gain.

The number of shares of the second and third prizes is equal to 1/10 of the number of shares of the first prize

Finally, take the total number of people 1 and subtract it.

The number of winners of the first and third prizes is determined by the number of winners of the second prize in the total number of winners.

Column: 1-9 10=1 10 1-1 5=4 5 1-1 10-4 5=1 10

A: The number of second prize winners accounts for 1 10 of the total number of people

Let's say that a total of 10 people win the award (just hypothetical

Then there are 4 people in the first and second prizes

There are 9 people in the second and third prizes

Then there is 1 person for the first prize

Then there are 3 people who won the second prize

Then the second prize is 1 10 la, 1 tenth).

There are still a lot of geniuses to start school, so what's the hurry. The third prize is three-fifths (two-fifths of the first and second prizes), and the first prize is one-tenth (nine-tenths of the second and third prizes), then the second prize is one-tenth minus one-tenth and then three-fifths to get three-tenths.

Set the first prize x number, the second prize y number, the third prize z number, and the overall 100 (good calculation) will be 30% according to the formula of the close series

x+y=40,y+z=90,x+y+z=100

The percentage of third prizes is: 1-

The ratio of first prizes is: 1

The proportion of the second prize is:

Turnover = m

m multiplied by 5% = 3000

m = 3000 divided by 5%.

m = 3000 divided by.

m=60,000 yuan.

Solution: Set 1000 kg of fruit.

Banana purchase price: yuan).

Pear price purchase price: yuan).

Total selling price: 2000 * yuan).

Purchase price: 1000 * 2 + 1000 * yuan).

A: Lost.

Honghong's father opened two fruit shops, and the workers paid 3,000 yuan in business tax in January this year, and the fruit store paid business tax at 5% of the turnover.

3000 50% = 60000 yuan.

The turnover of the two fruit shops in January was 60,000 yuan.

Let bananas be x yuan per kilogram and pear per kilogram y yuan, then x (1 1 4) yuan) y (1 1 4) yuan).

Solve the equation to get x=2 yuan, y=6 5 yuan.

Let's say the fruit weighs a kilogram.

The total amount sold at the original price is 2a 6 5a =

Counted as a new **.

Therefore, it is more to sell according to the original ** than to press the new **.

So it's a loss.

The number of people boarding at the starting point (the first stop) will be completed at the middle 9 stops and the last stop (10 stops in total). So there were 10 people at the beginning.

In the same way: at the second stop, 9 people get on the train and 1 person gets off.

At the third stop, 8 people got on the bus and 2 got off.

At the ninth stop, 1 person got on the train and 9 people got off.

By the tenth (and last stop), no one got on the train, and the number of 10 people getting off the train and getting on the bus decreased, and the number of people getting off the train increased.

Reach the maximum at the fifth stop.

10+9+8+7+6)-(0+1+2+3+4)=30 (pcs).

The number of people who get on the train at the first starting point (the first stop) will be completed at the middle 9 stops and the last stop (10 stops in total). So there were 10 people at the beginning.

It's the same thing: 9 people get on the train at the second stop, 1 person gets off the train, and by the third stop, 8 people get on the train and 2 people get off.

At the fifteenth (and last stop), no one got on the train, only 10 people got off the train and reached the maximum at the fifth station, so it should be calculated according to the number of people at the fifth station (10 + 9 + 8 + 7 + 6) - (0 + 1 + 2 + 3 + 4) = 30, I think it should be 30.

Solution: "Juvenile Daily" has 2 + (94-4) 8 (8 + 7) = 50 copies, and "Primary School Students' World" has 94-50 = 44 copies.

Idea: 2 copies of each of the two magazines, a total of 4 copies of unsubscribe, that is, there are 94-4 = 90 copies left, and the proportion of "Juvenile Daily" is 8: (8 + 7) = 8 15, so "Juvenile Daily" has 90 8 15 = 48 copies.

So there are 90-48=42 copies of "Elementary School Students' World".

It turned out that there were 48 + 2 = 50 copies of "Youth Daily" and 42 + 2 = 44 copies of "Primary School Students' World".

If you cancel two copies of each magazine, you will have a total of 4 copies unsubscribed, leaving 90 copies. At this time, the ratio of the number of copies ordered by the two magazines was 8:7, so it was calculated that there were 48 and 42 copies of the two magazines at this time, and then the original two copies were added respectively, so that the original order of "Juvenile Daily" was 50 copies, and the original order of "Primary School Students' World" was 44 copies.

94 returned 4 copies, so 94 2*2 90.

Juvenile Daily: Elementary School Students World 8:7, so Juvenile Daily 90 * 8 (8+7) 48 copies.

Elementary School Students 90 48 42 The question is how many copies of each of the two magazines were originally ordered, and the question was the original, so two copies were added respectively, and the articles were 50 and 44 copies.

After unsubscribing, there are a total of 94-2*2=90 copies, of which the ratio is 8:7, so that "Juvenile Daily" and "Primary School Students' World" each have *(7 15)=42 copies, and the original one only needs to add 2 copies of unsubscribed, that is

Number of unsubscribed copies: 2x2=4 (copies) 94-4=90 (copies) 8+7=15 90 divided by 15=6 (copies) "Juvenile Daily": 8x6=48 (copies) "Elementary School Students":

Set up a juvenile newspaper 8x+2 and a primary school student world 7x+2

then 8x+2+7x+2=94

x=6, so the juvenile newspaper 8*6+2=50 and the elementary school student's world 7*6+2=44

Set x parts for teenagers, y parts for elementary school, and column equations.

7*（x+2）=8*(y+2)

x+y=92

The solution is x=50 and y=44

50 juvenile newspapers. There are 44 books for elementary school students.

50 copies of the Juvenile Newspaper and 44 copies of the Elementary School Student World.

The ratio of number A to number B is 1:5, number B is two-thirds of number C, and the ratio of number A, B, and C is 2:10:15

The reasons are as follows: A is 1:5 than B, B is 2:3 than C, and the comprehensive result is: A is better than B than C = 2:10:15

A: B = 1:5;

B: C = 2:3;

Then: A: B: C = 2:10:15

Let A be 1 and B be 5

1 to 5 to 15 2 = 2 to 10 to 15