Stumped my 3rd grade math problem, 3rd grade math problem, help

I've done this topic, so let's analyze it first. If it is a 4-digit number, then the first digit of this number must not be 0, and the last digit cannot be 0Then multiply 9 by 4 digits, then the first digit must be 1, if it is 2, multiplied by 9, it will become 5 digits, then, the first digit after multiplying 9 must be 9, then the fourth digit of the first number is 9, now it becomes 1bc9, and then analyze b, if b multiplied by 9, it is a two-digit number, according to the algorithm of every 10 into 1, multiply 9 and then become a 5-digit number, because the first digit has reached 9, as long as it is carried forward, it will be greater than or equal to 10

So b can only be 0 or 1, then the number could be 11c9 and 10c9

If it's 11c9, the first two digits after multiplying 9 are 99, and the number is 1199, but the result of multiplying this number by 9 is not 9911, so this is not right, then there is only 10c9 left.

The result of multiplying this number by 9 should be 9c01, and the calculation process of multiplying 10c9 by 9 is, multiply the last digit first, 9*9 to get 81, and get 1 into 8, then the result of c*9+8 should be c*10, that is.

9c+8=10c

Solve the equation to get c = 8

So the number should be 1089

i.e. a=1 b=0 c=8 d=9

abcd*9=dcba(1089×9=9801)

First analysis, ABCD*9 DCBA is a four-digit number, ABCD is also a four-digit number, so ABCD must be 1 in the first place A (if it is 0, it is not a four-digit number, if it is greater than 1, then *9 gets a five-digit number), the same reason, b=0 (different letters are different numbers, b cannot be 1), and because a=1, the single digit of d*9 = a, so d=9. Three numbers have been determined, and C can be found with a few numbers.

There are 240 people in the lower grades of Guangming Elementary School, the number of middle grades is 7 9 of the lower grades, and the number of senior years is 2 3 of the middle grades.

280 (person).

There were 60 students in Class 61, and the number of boys was seven-twelves of the class, and after a few girls were transferred, the number of boys was five-ninths of the class. How many girls are transferred?

Original number of boys: 60 7 12 = 35.

Class size after transfer of girls: 35 5 9= 63.

So the number of girls transferred: 63-60=3.

A store sells both A and B commodities for 480 yuan for the sake of ** funds, and it is known that commodity A has earned 20% and commodity B has lost 20%, what is the profit and loss result of the store? The cost** of both goods is:

1000 yuan.

1000-480 2=40 yuan.

The loss is 40 yuan.

The two trains depart from A and B at the same time, the fast train travels 150 kilometers per hour, the slow train travels 90 kilometers per hour, they meet at a distance of 240 kilometers from the midpoint, how many kilometers are A and B apart?

Solution: The fast train travels 150-90 = 60 kilometers more per hour than the slow train, then the fast train travels a total of 240 2 = 480 kilometers more than the slow train, and the meeting time is 480 60 = 8 hours.

Distance between A and B (150+90) 8=1920 km.

There is a circular pool in the park, and Uncle Li runs around the pool 20 times a day, if the radius of the pool is 38 meters, how many meters does Uncle Li run every morning?

Solution: Run 2 meters a day.

Hello, 1 basket of apples - 1 basket of oranges = 15 kg (1) a basket of pears + a basket of apples = 75 kg (2) a basket of pears + a basket of oranges = 60 kg (3) (1) + (3).

A basket of pears + a basket of apples = 75 kg, is the condition (2) so there are actually only two equations, two equations to solve three unknowns can not be solved, less conditions are happy to answer for you, I wish you progress in learning! If you don't understand, you can ask! If you agree with my answer.

Suppose apple x, orange y, pear z

Then x -y=15

x+z=75 y+z =60x> 15

z<60

Just take an integer. The specific solution is not unique, because the conditions of the topic are not enough.

A three-way equation can be set up: a basket of apples x kilograms, a basket of oranges y kilograms, and a basket of pears z kilograms.

x-y=15

z+x=75

z+y=60

It seems that the solution cannot be solved, and the unknowns are eliminated, which means that there are many groups of solutions.

1 basket of apples — 1 basket of oranges = 15 kg.

One basket of pears + one basket of apples = 75 kg.

One basket of pears + one basket of oranges = 60 kg.

Pear; 75-60=15 kg

Tangerine; 60-15=45 kg

Apple; 75-15=60 kg

There are countless answers to this question.

For example, apples 40kg, oranges 25kg, pears 35kg or apples 30kg, oranges 15kg, pears 45kg.

Here are my personal insights. Only 60 is a whole so take the third one. There may be two cases of pears and oranges, 1 is a whole ten.

2 is at the end of 5, so you can follow 15 and 75 aboveConnected. There are more apples than oranges, and pears and apples are comparable.

According to the distribution of the whole ten in the first way, the orange is ten. It makes just the right amount to bring in.

30 25 750 km This is the total distance that can be traveled with a full tank of fuel, and the total distance is 728 km.

First of all, let's be clear, there is a two-fold quantitative relationship.

For example, if A has 6 flowers, then A will give B 6 flowers, and B will have 12 flowers, which means that there is a 2-fold relationship.

After figuring out this relationship, this question uses the backward method:

The total number of flowers is clear, 72 in total.

After the C split, the number of flowers of the three is: A 24 B 24 C 24 Total 72

According to the two-fold relationship, A has 12 flowers and B has 12 flowers before the C flower, and according to the total number of 72 flowers, C has 48 flowers.

That is, before the C flower (that is, after the B flower), the number of flowers of the three people is: A 12 B 12 C 48

In the same way, it can be determined that the number of flowers of the three people before the second division is: A6 B42 C24

Further, the number of flowers of the three people before the division of A is determined as: A39 B21 C12

That is, the original number of flowers of A, B and C was 39, 21 and 12 respectively.

The number of flowers in A, B and C was 39, 21 and 12 respectively.

Solution: Let A, B, and C originally have x, y, and z duos.

1) After A gives B C:

A: x-y-z

B: 2yC: 2z

2) After B gives A and C:

A: 2 (x-y-z).

B: 2y-(x-y-z)-2z=3y-x-zC: 4z

3) After C is given to A and B:

A: 4 (x-y-z).

B: 2 (3y-x-z).

C: 4z-2(x-y-z)-2(3y-x-z)=2z-y 4(x-y-z)=24 (a).

2（3y-x-z）=24 （b）

2z-y=24 (c) Solve the system of equations composed of abc to obtain: x=81;y=42；z=33, that is, A has 81 flowers, B has 42 flowers, and C has 33 flowers.

Solution: A gives B according to the existing number of paper flowers in B, and then gives C the existing number according to C, B gives the same number of flowers to A, at this time A has the remaining 2 times, and C gives the same number of flowers, at this time A is the remaining 2 * 2 = 4 times, at this time A has 24 flowers, then the remaining flowers after A gives B and C are 24 4 = 6, and it is just the same as the flowers in B C's hand.

There are a total of 24 3 = 72 flowers, so the flowers given out are (72-6) 2 = 33 then A, there are 33 flowers + 6 = 39 flowers, so there are 33 flowers in the original B and C.

Because there are 24 flowers in the end, before C gives flowers to A and B, A and B have 12 flowers each, and C's flowers are 72-12-12 = 48 flowers.

Then the original flowers of C are 48 2 2 = 12 flowers.

B's original flowers were 33-12=21.

Answer: It turns out that A, B and C each have 39, 21 and 12 paper flowers.

The ninth time I took notes.

Obviously, there were eight intervals, each with an interval of three hours, and then 24 hours passed.

The second is right, the first is right, it's all 9 o'clock. See for yourself, they're specific enough.

The first explanation is more reliable, because the first record was made at 9 o'clock, and the second record was made at 12 o'clock after three hours, and the interval was three hours, and two records had actually been made, and so on the third record was made with an interval of 2 three hours, and the fourth record was made with an interval of 3 three hours......The interval between the 9 records is 8 3 hours, and the total time is 24 hours, so the hour hand points to 9 o'clock, understand?

The first recording is 3 hours away from the start of the recording, the ninth recording is 3 9 = 27 hours after the start of the recording, and the first recording is 3 hours after the start, that is to say, the first recording time is exactly 24 hours before the ninth recording, so the first hour hand points to 9 o'clock.

There are 8 intervals for 9 recordings, each with an interval of 3 hours, 8 3 = 24 hours, the clock hand turns exactly 2 times, and the first time it is recorded, it also points exactly to 9 o'clock.

The title is really messy. The number of questions he did was 6 more than planned, and the number of days to do 6 more questions a day was 30-8-4=18, which is equivalent to the number of questions he did in these 18 days was equal to the amount of questions he originally planned for 4 days, 18*6=108, so the number of questions he planned to do every day, 27, and the total number of questions was not 27*30=810.

Assuming that the original plan is to do A questions every day, then there are a total of 30A questions in the exercise book, and there are 30A=8A+(30-8-4)(A+6), and A=27 is obtained, so the exercise book has 30*27=810 questions.