Primary School Olympiad Different solutions, several primary school Olympiad problems,

If you use the ternary equation, it is easy to get the answer is the sum of 1/8 + 1/9 + 1/18, divide by two and then cut 1/8 to get 1/48Of course, you don't know how to do it just by looking at this answer, and you may not know how to use ternary equations, and using equations is not good for your Olympiad thinking.

Now I'll tell you what I think.

As you know, the sum of the speeds of A and B is 1/8, B and C are 1/9, and A and C are 1/18.

So I'm going to add up these 3 scores now.

That is, the speed of A, B, B, C, and C is 24/7 at the speed at which 2 A, 2 B, and 2 C do things together.

Now let's halve this rate, 48/7It's equivalent to 1 A, 1 B, and 1 C.

Then we cut out 1/8, which is the velocity of 1 A and 1 B.

Get 1/48, which is the velocity of one C.

So C does it for 48 days.

So the equation for this problem is very simple:

1/8 + 1/9 + 1/18) divided by 2 and subtracting 1/8 to get 1/48

A and B and 1 8, B and C and 1 9, A and C and 1 18, if you add up the 3 numbers, it is 2 A, 2 B, 2 C, 1 8 + 1 9 + 1 18 = 7 24Then 1 A, 1 B, and 1 C use (7 24) 2=7 A, B, C-A, B = C) C's working time is 48 days.

1/8 + 1/9 + 1/18 = 7/24 This is the efficiency of two A, B, C (six people).

7/24 divided by 2 = 7/48 This is the efficiency of A, B, and C.

7/48-1/8 = 1/48 This is propylene efficiency.

1 divided by 1/48 = 48 days.

1/8 + 1/9 + 1/18 = 72/72 (9+8+4) 21/72 = 7/24

7/24 2 = 7/48 (the sum of the efficiency of A, B and C) 1 8 = 1/8 (the sum of the efficiency of A and B).

7/48 - 1/8 = 1/48 (efficiency of C) 1 (1/48) = 48 (days).

1. Solution: Let the total time be x hours, and the distance between the two places is s.

The two of them walked a total of 3 s, so 5x+

Zhang Mingxing is less rice, so.

Find out that x=1 8000 hours, Zhang Ming walked kilometers.

2.17 for the first item, 17+3 for the second item, and 17+3+3 for the third .........Item 101 is 17 + 100 3s, that is, 17 + 300 = 317

3 x cypress trees and 4x pine trees.

There are 5 classes in total, 4x-7 (x 5)=65 to find x=25, so there are 100 pine trees and 25 cypress trees.

Since the time spent on the two voyages is not equal, the least common multiple of the two times is taken first, and the two voyages of the same time are equivalent

The least common multiple of 8 and 12 is 24, so that the first voyage with a current of 98 3 = 294 km and a countercurrent voyage of 42 3 = 126 km is equal to the time taken by the second voyage of 72 2 = 144 km and 108 2 = 216 km against the current, i.e. 24 hours

Thus, in the same time, the first voyage travels 150 km more downstream and 90 km less counter-current than the second, indicating that 150 km downstream and 90 km counter-current take the same time, so.

The downstream velocity is 150 90 5/3 times the countercurrent velocity.

The first 8-hour cruise was considered a sailing downstream.

98 42 5 168 km 3/3.

The downstream velocity is: 168 8 = 21 (kmh).

The speed of the boat is: (21+ kmh).

If the two piers are 315 km apart, the time it takes to make a round trip between the two piers is:

315 21+315 hours).

Directly set the downstream flow to x and the countercurrent to y

98/x+42/y=8

72/x+108/y=12

Calculate x=21,y=

The speed of the boat is (time 315 21+315

In the second question, the whole pool is 1, and the individual potencies are a b c d ea+b=1 2, c+d=1 6, e+a=1 10, d+e=1 3, b+c=1 15

Find: a=c+13 30; b=c+17/30,d=c+20/30;e=c+5/30

Find: c=-1 4 (the highest water discharge efficiency) e=-1 12 (water discharge) d=5 12 (the highest water injection efficiency) b = 19 60; a=11/60

Fill with water: 1 (5 12) = 12 5 hours.

Full water: 1 (1 4) = 4 hours.

1.Let the ship speed v0 and the water speed v, and the time taken is t1, t2, t3, and t4 in the order of (v0+v)*t1=98

v0-v)*t2=42

t1+t2=8

v0+v)*t3=72

v0-v)*t4=108

t3+t4=12

(v0+v)*t1=98

v0-v)*(8-t1)=42

v0+v)*t3=72

v0-v)*(12-t3)=108

then t1 t3 = 49 36

8-t1)/(12-t3)=7/18

Then t1 = 14 3 t2 = 10 3 t3 = 24 7 t4 = 60 7

v0= v=

Then the speed of the ship is.

2.Let the time t be for the water and the time t for the water t0

v0+v)*t=315

v0-v)*t0=315

then t=15 t0=25

t+t0=40

If the two piers are 315 km apart, it will take 40 hours for a round trip.

1.It turns out that the denomination of the check is yuan, and the exchanger mistakenly looks at it as yuan, and it should be returned to the yuan The method I used involves a binary equation, I don't know if it meets your requirements?

Let the digit of the yuan be x, the digit of the cent is y (the digit of the angle is represented by the ten digits of y), the original denomination of the check is x yuan y cent, the exchanger sees it as y yuan x cents, we may as well use cents as the unit of measurement.

Then the original denomination of the check was (100x+y) cents, and later the exchanger saw it as (100y+x) cents.

After Li Lin spent 350 points, that is, the remaining (100y+x-350) points, which was twice as much.

Therefore, the column equation is 100y+x-350=2*(100x+y) shift, y=(350+199x) 98=25 7+199x 98 because x, y are integers within 100, and 199 is a prime number, 98=7*14, so x must be a multiple of 14, and the value of x divided by 14, the sum of 25, is divisible by 7.

25 7 = 3 remaining 4

199 7 = 28 remainder 3

So the value of x divided by 14 must be 1 or 8

Substituting the verification, we get x=14, y=32

So it turned out that the cheque was denominated in yuan.

2.I don't understand the second question, please explain, why is it a and b in front and x in the back?

Solution: If the number of people queuing up before the start of ticket sales is y, the number of people coming per minute is n, and 1 window can sell tickets to person A in 1 minute, then according to the description of the title:

1), the original number of people + the number of people who came in 30 minutes = the number of people who sold tickets in 5 windows in 30 minutes, that is: y+30n=5 30 a....y+30n=150a

2), the original number of people + the number of people who came in 20 minutes = the number of people who sold tickets in 6 windows in 20 minutes, that is: y + 20n = 6 20 a....y+20n=120a

3), the original number of people + the number of people who came in 10 minutes = the number of people who sold tickets in 10 minutes in Y window, that is: Y + 10N = 10ya

1)-2) Get: 10n=30a, n=3a, y=60a substitute 3) get: 60a+30a=10ay, y=9 that is: 9 ticket windows need to be opened.

This title is a variation of the "Cows Eat Grass" problem. Analyzing the problem well is the key to solving it.