6th grade itinerary problems, 6th grade math itinerary problems

1.It was six o'clock in the afternoon, and for the first time the minute and hour hands were in a straight line.

The minute hand rotates 360 60 = 6° per minute

The hour hand, turning 6 12 = per minute

At 6 o'clock, the minute hand is 180° behind the hour hand (in a straight line).

The next time on the same line, I understand that the two needles coincide, it will take 180 (minutes.

2.If someone walks at a certain speed for a certain number of hours, if the speed increases by 2 5 and decreases the speed by 1 6, then the ratio of the original time to the present is ( ).

The meaning of the title is unclear, I can't understand it....

Check it out and ask again.

3.A car travels back and forth between A and B, it takes 4 hours to go, and when it comes back, the speed increases by 1 7, and how many hours does it take to come back.

The speed is the original 1 + 1 7 = 8 7, and the time is the original 7 8

It took 4 7 8 = hours to come back.

or 4 (1+1 7) = hours.

4.Liang Liang walked from home to school, 5 kilometers per hour, and when he went home, he rode a bicycle for 4 hours less than walking.

If the conditions are insufficient, check it and ask again.

5.A and B each walk a certain distance, their walking time ratio is 4:5, the speed ratio is 5:3, and the distance ratio is ( ).

The distance ratio is (4 5) :( 5 3) = 4:3

6.Two students A and B go home from school, A walks 1 5 miles longer than B, and B walks 1 11 less time than A, what is the ratio of the distance between A and B students returning from school.

This is a speed ratio, right?

The distance ratio of A and B (1+1 5): 1=6:5

The time ratio of A and B is 1:(1-1 11)=11:10

The velocity ratio of A and B is 6 11:5 10=12:11

7.From A to B, the bus takes 10 hours, the bus takes 15 hours, now the two cars drive out from the two places at the same time, when they meet, the bus is exactly 240 kilometers, and the two places are separated by how many kilometers.

The van takes 15 hours, right?

Passenger-to-freight speed ratio 15:10 = 3:2

The distance between the two places: 240 3 (3+2) = 400 km.

Children study hard.

1. Let the distance be S km, when we meet, the slow train drives 4s 9 km, and the fast train drives 5s 9 km, and the equation gets

5s/9-4s/9=15*5

Solution: s=675

So the distance between A and B is 675 kilometers.

2. The speed of A is 440 6 = 220 3 (kilometers per hour), and the speed of B is 440 5 = 88 (kilometers per hour).

The time from the departure of car B to the meeting t=(440-220 3*1) (220 3+88)=25 11 (hours).

Therefore, when the two cars meet, car A is still s=88*25 11=200 (kilometers) from place B

v A + v B) * 5 = 540, v A: v B = 5:4 solution gets: v A = 60 (km min), v B = 48 (km min) Note: This question seriously violates logic, can a person walk tens of kilometers per minute???

1. When two cars meet, the sum of A's distance and B's distance is the distance between A and B, that is.

S A + S B = S two places.

2. In addition, the speed of A and B, to find out what is the speed of A and what is the speed of B.

Solution: 440 (hours.)

A: After an hour the two cars met.

A540 5 (5+4) = 300

B540 4 (5+4)=240

3.A540 5 (5+4) = 300

B540 4 (5+4)=240

1 675 (km).

2 200 (km).

3 60 (km) 48 (km).

1 15 (1-4 9) 4 9-1 60 (km) 60-15 = 45 (km) (45 + 60) x 5 = 525 (km).

2 440 6=220 3 (km) 440 5=88 (km) (440-220 3x1) (220 3+88)=25 11 (h) 88*25 11=200 (km).

3 540 5 = 108 (km) 108 (4+5) = 12 (km) 12x4 = 48 (km) 12x5 = 60 (km).

Is this a 6th grade math problem?! I'm a 5th grader!

1. When the passenger car and the truck meet, the ratio of the journey is 5:3, then the travel time of the two vehicles is 6*5 8=15 4

The distance traveled by the truck is 36 * 15 4 = 135 A and B are separated by s = 135 * 8 3 = 360 kilometers.

2 Let the velocity of car A v1 and the velocity of car B v2 a, b and the distance between the two places s8(v1+v2)=s

8+6)(v1+v2)=2s-140

s = 560 km.

1.A: 360km

Let the speed of the passenger car be x, and the time when the passenger car and the truck meet is t; then xt:36t=5:3 gives x=60 ; 6x=360 （km）

2.A: 560km

Let the velocity of A be a, B be b, and the distance between s 8a + 6a = s ; 8b+6b+140=s According to the title, 8a=6b+140 ; 6a=8b.

1.The time taken at the time of the encounter is 6*5 8, and the truck travels 36*6*5 8 distance, and the distance can be obtained 36*6*5 8*8 3=360

2.Let the velocity of A and B be x,y a, b and the system of equations in the series of s s=8*(x+y).

s=14xs=14y+140

It can be solved with s=560 x=40 y=30

1, set the passenger car xkm h, the total distance is ykm, then y=6x, 3y (8x36)x=5y 8 y=360 2, let A xkm h, B ykm h, then 8(x+y)=6x+6y+140 (x+y)=70 are 560 apart

Let A do x every hour and B do y every hour

6x+12y=1

8x+6y=1

Solution: x=1 10; y=1/30

So A does three days: 3 10

7 10 B remaining: (7 10) (1 30) = 21 days.

A can be completed when A does 6 hours and B does 12 hours, A does 8 hours and B does 6 hours, so A does 8-6 = 2 hours of work, and B does 12-6 = 6 hours, that is, A does 1 hour and B takes 3 hours.

After A does 3 hours, the remaining workload is the workload of A doing 6-3 = 3 hours and B doing 12 hours and A doing 3 hours, B needs 3 * 3 = 9 hours.

So 9 + 12 = 21 hours remaining.

Let Xiao Wang walk at a speed of x meters per minute, according to the title:

400x5+2x=300x7

x=50 Mingming, Huahua, Lili before the three of them rode off Xiao Wang's walking distance is:

400x5-50x5=1750 (meters).

So Lili travels the following distance per minute:

1750+50x10) 10=225 (m).

Xiao Wang's speed: (300 7-400 50) (7-5) = 50 meters.

At first, the distance from Xiao Wang: 400 5-50 5 = 1750 meters.

Lili's speed: (1750+50 10) 10=225 meters.

Set: Xiao Wang's speed is x meters per minute.

400*5-5x=7*300-7x

2000-5x=2100-7x

2x=100

x=505*(400-50)=1750 (m)1750+50*10) 10=225 (m).

The speed of the express train passing through station A at a speed of 48 kilometers per hour, and the speed changes to 32 kilometers per hour due to snow cover after 30 minutes, and it passes through station B 30 minutes later than the scheduled time.