A primary school math Olympiad problem, please point it with high fingers

When the father is 9 times the age of the younger brother.

The sum of the ages of the three is 9 + 1 = 10 times the younger brother plus 9 years old.

By this year, the age of three years has changed by the same number of years.

10 times the age of the younger brother, plus 9, plus 3 times the positive integer, is equal to 10 times the age of the younger brother of 55, which is divisible by 10.

Then add 3 times the positive integer to equal 6, 16, 26, 36 only 6 and 36 satisfy.

2 years ago, younger brother: (46-6) 10=4 years old.

Dad: 4 9 = 36 years old.

This year, Dad: 36 + 2 = 38 years old.

Older brother: 9 + 2 = 11 years old.

Younger brother: 4 + 2 = 6 years old.

12 years ago, younger brother: (46-36) 10=1 years old.

Dad: 1 9 = 9 years old (not in question).

So this year my father is 38 years old, my brother is 11 years old, and my younger brother is 6 years old.

Calculation of the exclusion method according to the actual situation:

The answer is: my father is 38 years old, my brother is 11 years old, and my brother is 6 years old. Two years ago, my father was just 9 times older than my younger brother, and my brother was just 9 years old.

I don't know if you're satisfied.

When the father's age is 9 times that of the younger brother, the elder brother is just 9 years old, then the younger brother's age is x, the father is 9x, and the elder brother is 9, and the sum of the ages of the three after n years is (x+n)+(9+n)+(9x+n)=55, that is, 10x+3n=46 Because x, n are integers, the solution is: x=4 n=2, or x=1, n=16 (this situation is obviously not true, because the father is only 9 years old).

In summary: x=6 n=2

The substitution results in the younger brother x + n = 6 years old, the elder brother 9 + n = 11 years old, and the father 9x + n = 38 years old.

When the father is 9 times the age of the younger brother and the elder brother is just 9 years old, then the sum of the father and the younger brother is 10 times the age of the younger brother. Since age is a whole number, 10 of the younger brother's age is a whole ten. We use the sum of the three people this year 55-9 = 46, obviously because it is not in the same year, 46 is not a whole ten, because it is not a year, all three of them have to be one year older, so 46 should be subtracted by a multiple of 3 and become a whole ten, then it is their respective ages at that time, so 46-2*3=40, then the younger brother is 4 years old, the father is 36 years old, and the elder brother is 9 years old.

The age of the three at this time is 2 years away from this year (2 years younger), and if you go back, each plus two years old is the age of this year: father: 38, younger brother 6, elder brother 11 years old.

There are other mathematical possibilities in multiples of 46-3, but they do not correspond to the reality of life, so there is only one solution.

This will be the same ternary equation, 6, 11, 38

The least common multiple of 100 and 80 is 400

Every 400 cm, the footprints overlap.

Daddy Run: 400 100 = 4 steps.

Xiaohua Run: 400 80 = 5 steps.

For every 400 cm, the footprints left are: 4 + 5-1 = 8 square circumference: 1256 8 400 = 62800 cm = 628 meters Square diameter: 628 meters.

Solution: The least common multiple of 100 and 80 is 400, i.e., every 400 meters there is a footprint that coincides. For every 400 meters, Dad ran four steps, Xiaohua ran five steps, and one footprint coincided, so there were a total of 8 footprints on the snow.

So 8 refers to the number of footprints between the distance between each footprint overlap and the last footprint. 4 is the distance between each footprint overlap and the last footprint coincidence.

That is, 8 footprints means that you have run 4 meters.

That is, the diameter of this square is 100 meters.

Good luck with your studies.

Solution: The least common multiple of 100 and 80 is 400, that is, for every 400 cm run, Dad has 400 100 4 (footprints), Xiaohua has 400 80 5 footprints, one of which is coincidental, and there are actually 4 5 1 8 (() footprints, so the circumference of the circle is 1256 8 400 62800 (cm).

So the diameter is 62800 cm) 200 (meters).

8 bicycles and 11 tricycles.

Suppose a full bicycle 19 * 2 = 38 49-38 = 11 or a full tricycle 19 * 3 = 57 57-49 = 8 elementary school students do this and can do equations later.

8 bicycles and 11 tricycles.

8 taels for bicycles and 11 taels for tricycles.

X bikes, Y tricycles.

x+y=19

2x+3y=49

The solution is x=8y=11

That is, 8 bicycles and 11 tricycles.

If it's all bicycles, there are 38 wheels, and now there's 11 less, and one tricycle has one more wheel than a bicycle, so there's 11 tricycles, and there's 8 bikes left. I hope you're satisfied.

Typical P&L issues.

First remove the 3-minute walk, a total of 3 * 80 = 240 meters at a speed of 80 meters, 3 minutes late, that is, a loss of 3 * 80 = 240 meters at a speed of 110 meters, 3 minutes in advance, that is, a profit of 3 * 110 = 330 meters (profit + loss) Two distribution difference = number of units (that is, time) (240 + 330) (110 + 80) = 19 minutes (19 + 3) * 80 + 240 = 2000 meters.

Solution: Let the total distance be x meters.

Get: x 80 (x 3 80) 110 = 6: x 2000

A: ......

Let the total distance be x meters, and after the first 3 minutes of walking, there will be y minutes left before the class time

Yes: x = 80 * 3 + 80 * y+3) x = 80 * 3 + 110 * y-3) solution: x=2000

Set the time to t

According to the question, the late arrival time is (t+3); The early arrival time is (t-3) because the distance is certain.

s=80*3+80*(t+3)=80*3+110*(t-3) The solution is t=19

Substituting it, the distance is s=2000

After 3 minutes, there are x minutes to be late.

80*(x+3)=110(x-3)

x=19s=(19+3)*80+3*80=2000

Solution: Let the length of the journey be s and the time is t

80*（t+3）=s

80*3+110*（t-3）=s

The two formulas are combined, and the solution is t=22 minutes.

s=2000

So the distance is 2000 meters.

If you haven't learned the equation algorithm, then remember not to calculate the distance first, let both sides of the equation equal the distance, solve the time, and then you can do it. This is the general method of this kind of problem.

Solution: Set x minutes away from class.

80(x+3)-80*3=110(x-3-3)80x+240-240=110x-660

80x-110x=-660

30x=-660

x=22, so there are 80 (x+3) - 80 * 3 - 110 (x - 3-3) = 80 * (22 + 3) = 80 * 25 = 2000 (m) from the student's home to school

A: The distance from the student's home to the school is 2000 meters.

The distance from the student's home to the school is S meters, which can be seen from the title.

s/80－（s－240）/110=9

solution, s=2000

The time it takes for naughty to get to grandma's house = 800 50 = 16 minutes.

At the same time, the dog ran for 16 minutes, so the distance the dog ran = 16 * 200 = 3200 meters.

The distance the dog runs is the speed of the dog The time that the dog runs, and the time that the dog runs is the time it takes for the naughty to walk from the naughty house to the grandmother's house, so the dog runs a total of 200 (800 50) = 200 16 = 3200 meters.