Two Olympiad questions to process thank you 10

Starting at the same time, the ratio of the distance traveled by A, B, and C is 8:6:9, and the middle of A and B is 7

Add the 9 that C does and the whole process is 16

When C is in the middle of A and B, A's distance is 8 16 = 1 21600 2 80 = 10 minutes.

It took a minute to set up the encounter.

It took A minutes from the beginning to the meeting, and B took 9 minutes to walk the distance, and the ratio of the speed of A and B was 9:a

It took a minute for B to go from the beginning to the meeting, A took 4 minutes to walk this distance, and the ratio of the speed of A and B was a:4

9：a=a：4

a=6 for row A for a week with 6+4=10

Line B uses 6+9=15 for a week

1. In the middle, which is half of the sum of the speed of the two people, 1600 ((80+60) 2+90) = 10 minutes.

1600-90x=(80x+60x)/2

1600=160xx=10

v B * 9) v A = v A * 4) v B v A * v A (v B * v B) = 9 4 v A v B = 3 2

A4+9*2 3=10 minutes.

B 9 + 4 * 3 2 = 15 minutes.

In the first question, the distance between A and B is 1600At time T, C is between A and B. Then it means that the distance between C and B at this moment is exactly half of the distance between A and B, and (80t-60t) 2+90t=

In the second question, A is 10 minutes, B is 15 minutes, and the process is as follows

1. a+b=6, (a+b)^2=36=a^2+2ab+b^2=26+2ab, 2ab=10

a-b)^2=a^2-2ab+b^2=26-10=16

a-b|=4

2.9 2=81>78, so a, b, and c are all less than 9 and are prime numbers. Then the possible values are 2, 3, 5, 7

According to the carry-in method, a, b, and c are 2, 5, and 7

If a=7, b=5, c=2, 49-25=24=2d 2, d 2=12 are rounded.

If a=7, b=2, c=5, 49-4=45=5d 2, d 2=9, d = 3

If a=5, b=2, c=7, 25-4=21=7d 2, d 2=3, d=3 are rounded.

In summary, a=7, b=2, c=5, d=3

3. 2x-3a=0, x=3a/2

3x+a-7=0, x=(7-a)/3

3a/2=-(7-a)/3, 9a=2a-14, a=-2

4. x^2-4y^2=(x+2y)(x-2y)=2011

2011 is a prime number and can only be decomposed into 2011*1

x+2y=2011, x-2y=1, x=1006, y=1005 2, not an integer, rounded.

x+2y=1, x-2y=2011, x=1006, y=-1005 2, not an integer, rounded.

Select A, 0 group solution.

5. (x+y)^2=5^2=25=x^2+2xy+y^2=x^2+y^2-22, x^2+y^2=47

x-y)^2=x^2+y^2-2xy=47+22=69

x+y)^3=x^3+y^3+3x^2y+3xy^2=x^3+y^3+3xy(x+y)=5^3=125

x^3+y^3=125-3xy(x+y)=125-3*(-11)*5)=125+165=290

6.Let the length be s, the uphill time t1=s v1, and the downhill time t2=s v2

Total time elapsed t=t1+t2=s v1 + s v2=s(v1+v2) v1v2

The average speed is v=2s t=2v1v2 (v1+v2).

So much. Add QQ to talk to you slowly

Let the speed of car A when it departs is a, the speed of car B when it departs is b, and the distance of ab is s

The analysis shows that A is faster than B, A has walked S+4S 5, and B has only finished S, so there is.

s/a + 4s/5)/( = s/b

Hence b= 3a 5

A and B met an hour later, and there was.

Therefore, there is s a=27 22 (this is the time of the departure journey).

The time for the return trip is 5 to 6 of the outbound trip

Therefore, it takes (1 + 5 6) * 27 22 = 27 12 hours for car A to go round trip between AB, that is, 2 hours and 15 minutes.

There are two cases in total.

where the speed of car A is x and the speed of car B is y, then when returning, the speed of car A is.

The speed of car B is 5s from A to B

Scenario 1: If A is faster than B, then there is.

Equation 1: (5s y) = (5s x) + (4s

Equation 2: Solution (s x) =

The time for a round trip between AB is (5s x) + (5s hours.

Situation 2: If B is faster than A, then there is.

Equation 1: (5s y) = (s x).

Equation 2: Solution (s x) =

The time for a round trip between AB is (5s x) + (5s hours.

Note: The equal signs in this solution are all approximate equal signs.

The answer upstairs is really speechless, I don't know who is faster in the analysis, and I don't know how the upstairs resolutely concludes A.

Faster than B. The watchtower owner should consider the problem comprehensively.

It is known that when B arrives at place B, A is 1 5 (return) from place A at a distance from AB

It can be seen that A's speed is faster than B's.

The two cars of A and B meet before B arrives at place B, and A returns to meet B after AB.

Let A take time x to go from A to B at the original speed

A is 1 5 of the distance AB from place A, which can be decomposed into the original velocity from a to b for time x plus the return walk.

AB distance (1-1 5) = 4 5 After the speed is increased by 20%, then the time needs to be 4 5x divided by 120% to equal 2 3x

The time for A and B to walk is x+2 3x=5 3x hours.

In other words, when B reaches place B, the time taken is 5 3x, and the speed is naturally 3 5 times that of A.

Immediately after arriving at place B, the speed is increased by 20%, and the two cars meet an hour after departure.

Let the velocity of A be 1, then the velocity of B is 3 5

A takes time from A to B at the original speed, and the distance is x ab

Return to meet B A time distance (

B hour walking distance and also ab

Then x=( x=27 22 hours.

A return speed is increased by 20%, then the time is only 5 6

How many hours does it take for a car A to make a round trip between ABs? 27 22 * (1 + 5 6) = 2 and 1 4 = hours.

Two hours and 15 minutes.

It is clear that A's speed is faster than B's.

Let A go at the original speed AB distance is 1 time required for 1 A speed, after the speed of A increases by 20% the speed is the original 6 5, the whole time of walking after speeding up (1-1 5) = 4 5 5 is not speeding up 4 5 (6 5) = 2 3 A speed, the total time = 5 3 A speed, equal to the time of B to complete the journey 1 B speed hour, that is, the ratio of A and B speed is 5 3, the two cars meet, for the hour after departure, at this time B line distance = B speed = 9 10 * A speed, let A travel at the original speed The time required for the whole journey is t, and the time required for round trip =t(1+5 6)=11t 6, (A velocity + 9 10 * A velocity = t A velocity, solution t = 27 22

Round-trip time = t(1+5 6)=t*11 6=2 hours and 15 minutes.

There are two cases in total.

If the speed of car A is x and the speed of car B is y, then when returning, the speed of car A is car B, the speed of car A is 5s, and the distance from A to B is 5s

Scenario 1: If A is faster than B, then there is.

Equation 1: (5s y) = (5s x) + (4s Equation 2:

Solution (s x) =

The time for car A to make a round trip between AB is (5s x) + (5s hours, situation 2:

If B is faster than A, then there is.

Equation 1: (5s y) = (s x).

Equation 2: Solution (s x) =

The time for car A to make a round trip between AB is (5s x) + (5s hoursNote: The equal signs in this problem solving process are all about equal signs.

The answer upstairs is really speechless, and I don't know who is faster than A and B, and I don't know how Upstairs resolutely concludes that A is faster than B. The watchtower owner should consider the problem comprehensively.

1. Calculate how many combinations they have, 5 * 4 2 = 10 kinds, each 90 10 = 9 minutes. 4 plays per person, 9*4=36 minutes per person.

2. The minimum score is 0 points, so the total score of the remaining 8 cards is the highest 9*68, so the average score is 9*68 8

1. At least 11 people will board the train at the first station and 10 people at the second station,。。 All the way up to the 11th station, 1 piece. Before arriving at the terminal, there may be a maximum of 66 people on the bus, a minimum of 11 people, and a minimum of 11 seats on the bus.

2. The number of male students is more than that of female students is (65*100 - 63*100) 5 = 40

The average number of numbers is (28 + 36 + 42 + 46) * 3 4 = 114

4. The water consumption is (23 - 5*2* =

5. The time required for reverse water is 180 [180 15-(180 9-180 10)] = 18

6. If the velocity of A and B is x,y, then 4x y = (6+1 4)y x, and the solution is x y=5 4, because x-y=20, so x=100, y=80

ab = 4*100+(6+1/4)*80 = 900

-1 = 11 (pcs).

(70-63)*100 (70-60)-100=40 (person)3, (28+36+42+46) 4=38

4. ( T ) 2 * 5 + 6 = 16 (T ) 5 (180 10-180 15) 2 = 3 (km h) 180 10-3 = 15 (km h) 180 9-15 = 5 (km h) 180 (15-5) = 18 (h).

6. (4*20+kmh) 100*4+(100-20)*km) Give it to me, I'm only in the fifth grade of primary school. That's a good thing to do!

(1): 36 seats.

2): 70 boys, 30 girls, 40 more.

4): 16 tons.

If part of the water fee is not exceeded, 5*2*yuan.

Excess part of the water fee, 23-11 = 12 yuan.

Excess part of water, 12 2 = 6 tons.

Therefore, the total water consumption is 6 + 5 * 2 = 16 tons.

5): 18 hours.

It takes 10 hours to go downstream, so boat speed + water speed = 18

It takes 15 hours to go downstream, so boat speed - water speed = 12

So get, the boat speed is 15 and the water speed is 3

The water speed increases, and it only takes 9 hours to follow the water, that is, the boat speed + water speed = 20, so the current water speed is 5, and the time required against the water is 180 (15-5) = 18 hours (6): let the speed of A be x, the speed of B is y, the meeting place is C, then the ac distance is (25 4) y, and the BC distance is 4x When arriving at C, A and B take the same time, so.

25/4)y]/x=4x/y

Get x y = 5 4

And because x-y=20

So, x=100, y=80

ab The distance between the two places is (25 4)y+4x=900 km.

4, yuan).

12 2 = 6 (ton) 6 + 2 * 5 = 16 (ton).

A: 16 tons

5. Downstream speed: 180 10=18 (kmh) Reverse water speed: 180 15=12 (kmh).

Water velocity = (18-12) 2 = 3 (kmh) hydrostatic velocity = 18-3 = 15 (kmh).

Now the speed of the water: 180 9 = 20 (kmh) and the speed of the water = 20-15 = 5 (kmh).

Backwater velocity = 15-5 = 10 (km/h) Backwater time = 180 10=18 (h).

6. You should know the equation.

Let A's velocity be x km/h, then B's velocity is (x-20) km/h.

4x=x=9/500

ab distance = 4 * x = 4 * = 2000 9 (km).

Okay, I'm sure the answer must be right, so let's give it points! I'm tired ...