Primary 6 Olympiad Schedule Problems, Please Help!

From the meaning of the title, it can be seen that the distance required for butterflies to fly together, it is necessary to know the time and speed, the title has told that the speed is 15 miles per hour, in fact, the time of the butterfly line is the time of the two boys walking, then:

The time traveled by the two boys is: 20 (10+10)=1 hour, so the distance traveled by the butterfly is: 1 15=15 miles

The distance traveled by the butterfly is: 20 (10*2)*15=15 miles, I hope it can help you.

It's that simple ... 20 (10+10) calculates that the time it takes for the two cars to meet is 1 hour, which means that the saucer also flew for 1 hour, and then the time is multiplied by the speed to calculate the distance traveled by the butterfly 20 (10+10) 15=15 (miles) Answer: This saucer flew a total of 15 miles.

The speed at which two boys ride a bicycle: 2*10=20 (the speed at which two boys ride a bicycle per hour).

20 20 = 1 hour.

1*15=15 (the distance the butterfly flies).

--- = 1 (hour).

1*15=15 (miles).

A: This butterfly flew a total of 15 miles.

Do butterflies walk in a straight line?

Car B exceeds the midpoint by 30 kilometers, and car A travels 45 kilometers longer than car B, then car A exceeds the midpoint.

30 45 75 (km).

ab The two places are far apart.

75 (2 3 1 2) 450 (km).

？？？When car A travels to a distance of 2 3 from place b, it should be when car A travels to 3 molecules 2 of the full length.

If car B exceeds the midpoint by 30 km, and car A travels 45 km longer than car B, then car A passes the midpoint by 75 km.

Total length (30+45) (2 3-1 2)=450 km.

Let's do it algebraically.

Solution: Set the time of A to take the bus to x

A's distance is 40x km.

In that time, B walked 4x km.

The distance between A and B: 40x-4x=36x kilometers.

Then the car goes back to pick up B.

It takes to meet B: 36x (50+4)=2 3x hoursIn these 2 3 hours, A and B walked separately: 2 3x 4=8 3x kmA and B are still 36x km apart.

Then B gets in a car and catches up with A in the park.

Pursuit time: 36x (40-4) = x hours.

A walked again: 4x km.

A total walking: 8 3x+4x=20 3x kilometers, the whole journey is: 20 3x+40x=140 3x kilometers, the ratio of A's walking distance to the whole journey is 20 3x:140 3x=1:7

1. In the first stage, A takes a car and B walks. Let B travel 1 and A travel 10, a difference of 9.

2. In the second stage, A and B walk, and return empty. B and the car go in opposite directions, the speed ratio is 4:50, the total distance is 9, the distance of B is 2 3, A and B have the same speed, and A walks 2 3

3. In the third stage, A walks and B takes a car. The process is at the same stage. A walks 1

A walks 1+2 3, the total distance is 1+2 3+10, and the ratio is 1:7

Solution: Let the velocity of A be v1 and the velocity of B be v2, and the distance traveled by A in the first encounter is x, so that the first encounter (the same time taken): v1 v2=x (24-x) The second encounter (the same time):

v1/v2=[24+(24-x)+6]/[24+(24-(24-x)+6)]

The above equation is simplified as: v1 v2=(54-x) (6+x)) The above two equations are combined to obtain: x (24-x) = (54-x) (18+x) and simplified to: 96x=1296

Solve x = substitute x = into either of the above two equations: v1 v2 = Therefore, the velocity ratio of A and B is 9:7

6355535 SEF backbone militia returned to VM

Let the velocity of A be x, the velocity of B be y, the first encounter time t, and the second encounter time t to get (x+y)t=24

xt=2yt+6

yt=6 gives t=6 y

t=24/(x+y)

Simplification: x y=5 3 squares.

If A travels x kilometers in the first encounter, then B travels (24 x kilometers) at a speed of v1 and B at a speed of v2

v1/v2=x/(24-x)

v1 v2=[24-(x-60)] (x-6) gives x=15 km.

At the same time, A walked 15 kilometers and B walked 9 kilometers.

The speed ratio of A and B is 5:3

The first time the two met and walked a whole journey together, you can see the time used as a time; From the first encounter to the second encounter, it took 2 whole journeys, and it took 2 minutes; The distance A walks in two times is more than 6 kilometers that B travels twice the distance in one time, then the distance traveled in one time should be twice the distance B travels in one time and 3 kilometers more, that is to say, in the first encounter, A walks 3 kilometers more than B, which becomes a sum difference problem, (sum + difference) divided by 2 = large number, A's speed is (24+3) divided by 2= , then B's speed is, A B's speed ratio is::7

The time of the encounter is.

1 (1/10 1/15) 6 (hours) When we meet, the passenger car travels.

1/10 6 3/5

The van is gone. 1/15 6 2/5

The highway between the two cities is long.

96 (3/5 2/5) 480 (km).

Passenger cars travel 1 to 10 per hour of full length

Trucks travel 1 to 15 per hour of full length

Encounter time = 1 (1 10 + 1 15) = 6 hours.

Overall length = 96 [6 * (1 10-1 15)] = 480 km.

Solution: 1 (1 10 + 1 15) = 6 (6 hours in opposite directions).

1 10-1 15)*6)=1 5 (passenger cars go 1 5 more than trucks).

96 (1 5) = 480 km (distance between two cities).

The sixth-grade travel problem, on the one hand, introduces the score into the third- and fourth-grade travel problem, and on the other hand, emphasizes the loop problem, among which, the most representative is the clock problem, here is just one example:

There is a circular road, the circumference is 2 kilometers, A, B, C 3 people start from one point at the same time, each person circulars for 2 weeks, there are 2 bicycles, B and C set off by bicycle, A walks off, B and C get off and walk halfway, leave the bicycle for others to ride, it is known that the speed of A walking is 5 kilometers per hour, the speed of B and C walking is 4 kilometers per hour, the speed of 3 people riding bicycles is 20 kilometers per hour, please design a way of walking, so that 3 people and 2 cars reach the end at the same time, So how many minutes does it take to circumnavigate 2 weeks?

=Post a wrong solution first===

A: Ride x, walk 4-2x, ride x;

B: Walk x, ride 4-x;

C: Cycling 4-x, walking x;

4-x)/20+x/4=2x/20+(4-2x)/5

x= (km).

Hours = minutes.

=Paste another correct solution===

The above idea mistakenly sees the road as a straight road, and a circular runway should be considered, then:

First of all, B and C set off on a bicycle and chased A for a lap, at this time, a line X km, then B C line 2 + x km;

After that, A travels 4-x kilometers by bike, and B C takes turns traveling the remaining 4-(2+x) kilometers in one car.

1.encounter problems, total distance, speed, and encounter time; 2.Catch up with the problem Catch up with the distance Speed difference Catch up with the time.

Coaches and vans traveled in 3 hours.

Passenger cars and trucks are separated by encounters.

Journey 105 (27 20-1) = 300 (km) Here's an example Let's say the speed of a passenger car is x km h and the speed of a train is y km h.

Inspired by the title: 5x+

5x) y = (solution x = 75 km h

Let the encounter time be x, then x 5 = solution x = 6 (825 6) * [6 (5+6)] = 75

Let's assume that the speed of the passenger car is x km h and the speed of the train is y km h.

Inspired by the title: 5x+

5x)/y = (

The solution is x = 75 km h

It's a matter of encounter! It's a pity I don't know, I'm sorry! I can't help you!

Six years? Will there be fewer conditions?

Haven't you learned binary equations?