Primary school math Olympiad topic, who will solve it

Analyze the circumstances of each encounter.

The first time: B arrives at B and returns, meets A, time: 1100*2 (60+160)=10 minutes, distance B: 160*10-1100=500 meters.

The second time: B returns after arriving at A, A returns after arriving at B, time: 1100*4 (60+160)=20 minutes, distance B: 60*20-1100=100 meters.

The third time: B arrives at B and returns, catches up with A, time: 20+100*2 (160-60)=22 minutes, distance B: 100+60*2=220 meters.

Fourth: B arrives at A and returns, meets A, time: 1100*6 (60+160)=30 minutes, distance B: 60*30-1100=700 meters.

The fifth time: B arrives at B and returns, A arrives at A and returns, time: 1100*8 (60+160)=40 minutes, distance B: 1100-(60*40-2200)=900 meters.

Therefore, the closest distance to B at the time of the second encounter is 100 meters.

The simple way is to draw a graph, the abscissa time, the distance of the ordinate from b, and then look at the point of intersection of the two curves with the smallest ordinate.

I don't understand either, so I suggest you give more bounty points.

The breakthrough point of this problem is to see that the area of the triangle ADE is 24 square centimeters more than the area of the triangle DEF.

If AM and FN are higher from point A and point F to BC respectively, then AM=3FN is known from CF:AC=1:3, and so on

Let the height of the BC side in ABC be H

The height on the de side of the def is h

s△adh=s△ade-s△deh

s△efh=s△def-s△deh

s△adh-s△hef=s△adf-s△def=24h/h=cf/ac=1/3

It can be obtained from s adf-s def=24.

1/3*bc*h*1/2-1/3bc*h*1/2=241/9bc*h=24

1/2bc*h=108

s abc = 108 square centimeters.

As shown in the figure: the area of the triangle ABC is 3 times the area of the ADE, so only the area of the ADE can be found by requiring the ADE, and F is the AC trisect point, as an auxiliary line, you can know that the area of the triangle ADE is 3 times of the DEF, and ADH-EFH=ADE-DEF=24=2DEF=2 3ADE, then there is ADE=24X3 2=36, ABC=3XADE=3X36=108,

108 hope it will be helpful to you, hope to click"Satisfactory answer"

1 The whole formula should be: 28 + 35 + 37 + 40 - 45 * 2 = 5 Explanation: There are a total of 45 people, assuming that each person is applying for 3 subjects, and there are 135 subjects that have been reported, and in fact there are 140 subjects who have been registered, then the remaining 5 people will be those who have signed up for 4 courses - this is a limit method.

So at least 5 people signed up for 4 doors.

There are 45 students in a class, of which 28 are learning piano, 35 are learning computers, 37 are studying art, and 40 are attending Olympiad.

28 35 = 63 (person).

63-45 = 18 (people) [learn piano and computer at the same time] 18 37 = 55 (people).

55-45=10 (people) [learn piano, computer, art at the same time] 10 40 = 50 (people).

50-45=5 (person) [learn piano, computer, art, and Olympiad at the same time].

28 35 = 63 (person).

63-45 = 18 (people) [learn piano and computer at the same time] 18 37 = 55 (people).

55-45=10 (people) [learn piano, computer, art at the same time] 10 40 = 50 (people).

Piano and computer can be learned at least:

Piano, computer science, and art can be learned at least:

18+37-45 10 people.

Learn at least all four.

10+40-45 5 people.

40 + 37 = 77 (people).

77-45 = 32 (people) (while studying art, Olympiad mathematics) 32 + 35 = 67 (people).

67-45=22 (people) (Learn computer at the same time, art Olympiad, 22 + 28 = 50 (people).)

50-45 = 5 (people) (all four contents) at least 5 people.

28+35+37+40=140

At least 5 people have learned all four.

It's actually quite simple!! Ha multiplied by Ha In 0-9 only 1 digits can meet the condition Ha multiplied by Ya yes Ha is 1 Then it may be 0-9 is feasible But when Ha + Ha is equivalent to Ha Then only 0 is 0 to meet this condition!! Therefore, Ha is 1 yes is 0

Ha is 1, ah is 0, I am the first year of junior high school, believe me.

"Ha" stands for "1" and "Ya" stands for '0'

Idea: From the equation, we can know that Ha + Ya = Ha so Ya = 0

Ha Ya Ha multiplied by Ha = Ha Ya Ha so Ha = 1

101 x 11 = 1111

Ha x Ha = Ha So it can only be 1 or 5, and if it is 5, the product must be more than 4 digits. So it must be 1, which can be known from the result of 1111 and the multiplier of 11.

If you want to solve the problem, you have to have a question first, and I don't understand the question, how do you let us help solve it.

Math Olympiad questions are actually not difficult. The Olympiad questions are mainly a combination of the original old questions and the new ones. Read more questions and do more.

Didn't you write the answer below?

2. A:v B = 5:3

From the beginning to the time A and B walk, the distance is proportional to the speed.

s A: s B = v A: v B = 5:

3。Think of the distance between AB as "1", in which A walks 5 8 and B walks 3 8. In this way, A walked 5 8 2 = 5 16 in 1 hour, and B walked 3 8 2 = 3 16 in 1 hour.

A walks 5 16-3 16 = 2 16 more than B in 1 hour, and if A and B go in the same direction from ab and two places, A wants to catch up with B, and he has to walk one more distance between ab than B.

So the time it takes is:

1 2 16 = 1 16 2 = 8 (hours).

4. Set: Male student is x person, female student is Y person, according to the total score is equal to:

x+y)x92=95ⅹ+87y

92x+92y=95x+87y

95x-92x=92y-87y

3x=5y, both sides of which are divided by 3y, obtain.

x/y=5/3

x：y=5：3

Hours [(5+3) (5-3)]=8 hours.

4.Let two unknowns, let the boy x, the girl y, 92(x+y)=95x+87y, 92x+92y=95x+87y, 5y=3x, x:y=5:3

It is said that this is an elementary school math problem, but I did the math all night, is it because of my IQ?

Let's take a look at this topic.

You take 20,000 and he takes 20,000, and you put the money in the box together, and no one can touch the money inside.

Then, there are 40,000 yuan in it, and he sold me this box for 30,000 yuan, there are 40,000 in it, I can earn 10,000, and he can also earn 10,000 at the same time, which is really the best of both worlds.

On the Internet, it seems to be ** wrong, right?

A and B respectively set off from AB and traveled in the same direction, their speed ratio was 3:2 when they started, after the encounter, A's speed increased by 1 5, B's speed increased by 2 5, and when A arrived at B, B was still 26km away from A. How many kilometers are the two places apart?

Let AB and the two places be separated by x kilometers.

2/(3+2)x]/[3×(1+1/5)]=[3/(3+2)x-26]/[2×(1+2/5)]

x/9=3x/14-130/14

13x/126=130/14x=90

A, B, and C went shopping together, and 1 2 of A's money was equal to 1 3 of B's money, and 3 4 of B's money was equal to 3 5 of C's spending, and C spent 98 yuan more than A, and asked them how much money they had spent.

490 yuan. A and B run a 100-meter run (assuming their speed remains the same), and when A runs 75 meters, B runs 60 meters. So, how many meters did B run when A reached the finish line?

80 meters. 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192

Won't parse:

I haven't seen the title, of course not!

Math alphabet: Because the teacher said that I only need the teacher, I am an elementary school student.