Help me do the math problems in the Primary School Math Olympiad! Calculate every step of the equati

1.Because: three-quarters of the first rod is exposed to the water, four-sevenths of the second rod is exposed to the water, and two-fifths of the C is exposed to the water.

So: A is 4 times longer than water, B is 7 3 times longer than water, C is 5 3 times longer than water, so.

A:B 12:7

B: C 7 5

So, A:B:C 12:7:5

Therefore, the length of the nail is 360*12 (12+7+5) and the depth of the water is 4 360*12 (12+7+5) 4 45 cm.

2.Because B completes 1 15 per day and C completes 1 20 per day, B and C complete it in 6 days

Remaining 1-(7 10) 3 10

Because A completes 1 10 every day, A completes 3 10 in 3 days and because it takes 6 days, A rests

6-3 3 days.

1.Solution: Set the water depth x cm.

From the meaning of the question: 4x+7 3x+5 3x=360 solves x=45

A: The water depth is 45 cm.

2.Solution: The armor has been stopped for x days.

From the meaning of the question: (6-x)*1 10+6*1 15+6*1 20=1 solution x=3Answer: A stopped for 3 days.

1. The length of A, B and C in water should be equal, that is, 1 4 of A = 3 7 of B = 3 5 of CB = A 7 12, C = A 5 12, A + B + C = 360, solution A = 180, water depth = A 1 4 = 45 meters.

2. If it is known that A does 10 days alone, B does 15 days, and C does 20 days, then A does 1 10 every day, B does 1 15 every day, and C does 1 20 every day, so that A stops for x days, that is, only B and C are doing it in that x day, and the rest of the (6-x) days are done by three people together, and the equation can be listed.

1/10）+（1/15）+（1/20）】*6-x）+【1/15）+（1/20)）*x=1

Solution x=3

If there are x tigers, there are 10-x in the lion bench, and the amount of food per animal per day in high temperature weather is reduced by 1 kg of wax fiber, then ten animals in three days are reduced by 30 kg, that is, 10 6x+10 4 (10-x)-30=490, simplified to 60x-40x=490-400+30,20x=120, and solved to x=6,10-x=4. That is, there are 6 tigers in the zoo and 4 lions in the sales brigade.

The answer is 700 meters.

Process: Traveling at a normal speed, the hour express train should have traveled half of the whole journey, and now, the express train has traveled 15 * kilometers more than the original.

The slow train also traveled kilometers longer than the normal hour.

This multi-line distance and (105 km) should be a slow (5 hours, hours) hour of normal speed.

So the normal speed of a slow train = 105 km/h.

Distance = 70 * 10 = 700 km.

Because each car travels 15 kilometers more per hour, the two cars travel 105 kilometers more per hour. If they meet at a normal speed for 1 (1 10 + 1 7) 70 17 hours.

Hours 119 34 hours.

70 17 hours 140 34 hours.

So: the combined speed of the two cars.

140 km-h

140 km).

A: A and B are 490 kilometers apart.

Normal speed for two cars takes 1 (1 10 + 1 7) = 70 17 hours.

After speeding up, it is actually equivalent to two cars walking together for an extra kilometer, which is equivalent to two cars originally walking 70 17-7 2 hours away, so there are:

Full journey = 105 * (70 17) (70 17-7 2) = 700 km.

Solution: The second time, because Grandpa Li played for half an hour, he set off.

So the arrival time was also half an hour earlier, and he and Boss Wang still arrived at the same time, indicating that Boss Wang was also ahead of schedule for the second time.

It took half an hour to arrive, and it would have taken him x hours.

Then there is: 60x=90(

Solution: x = so the distance from the vegetable garden to the market = 200 60 meters = 18 kilometers.

Distance from the city to the market = 60 km.

So the distance from the vegetable garden to the urban area = 18 + 90 = 108 km.

Hope it helps

If you don't understand, ask again

Comparing the two trips, assuming that the total length is x meters, under normal circumstances, according to the speed ratio of two people (200:1000), the distance from Boss Wang to the market should be 5 6 * x meters; The second time Grandpa Li set off half an hour early, which is equivalent to when Boss Wang set off, Grandpa Li had walked 6000 meters, but the two still went to the market at the same time, indicating that the distance Boss Wang walked still did not change, but at this time, according to the speed ratio of the two people (200:1500), Boss Wang should have walked 15 17* (x-6000).

That is, 5 6 * x = 15 17 * (x-6000), we can get x = 108000 meters, that is, 108 kilometers.

Set the simultaneous departure takes time t minutes to reach the market.

60 kmh = 1000 mmin, 90 kmh = 1500 mmin.

The distance from Grandpa Li to the market is: 200t, and the distance from Boss Li to the market is: 1000t, and the total distance of Boss Wang remains the same, but the time is 30 minutes less.

1000t=1500(t-30)

t = 90 minutes.

The same can be said about t: velocity ratio = inverse time ratio: 1500 1000 = t (t-30).

The distance from Grandpa Li to the market is: 200t = 200 * 90 = 18,000 meters = 18 kilometers.

The distance from Boss Wang to the market is: 1000t = 1000 * 90 = 90000 meters = 90 kilometers.

The total distance from the vegetable garden to the city is: 18 + 90 = 108 kilometers.

Grandpa Li's speed = 200 meters minutes = 12 kilometers per hour.

If Grandpa Li's vegetable garden to the market is x kilometers, then the distance from the urban area to the market is x 12 60 = 5x kilometers.

x/x/x/36=

x = 18 Grandpa Li's vegetable garden to the urban area = 18 + 18 5 = 108 km.

Grandpa Li's speed = 200 meters minutes = 12 kilometers per hour.

Boss Wang's speed: 60 kilometers per hour.

Let the general time be t, then the unit of the time of departure in advance is hours) so (12+60)*t=(12+90)*(solve the time t, so the distance l=(12+60)*t can be obtained.

Grandpa Li arrived half an hour early, and at the same time, Boss Wang also arrived half an hour early.

Set up the city XKM away from the market

x 90 + x = 90 then Grandpa Li rides 90 60 = hours to the market.

then the urban area to the vegetable garden 18 + 90 = 108km

1.The side length of the lawn is (80-1 4) 4=19 meters, and the lawn area is 19 19=361 square meters.

2.The length + width after amputation of 3 cm is (93-3 3) 3=28 cm, and the length + width before truncation is 28+3+3=34 cm.

The original circumference was 34 2 = 68 cm.

3.The small rectangle has one side equal to the square, the side length is 10 2 = 5 cm, and the area of the small rectangle is 35-5 5 = 10 square centimeters.

4.There are a total of 600 5% + 400 20% = 110 old workers, accounting for 110 (600 + 400) = 11% of the total workers

The number of workers to be transferred is 400 (20%-11%) = 36.

1.(x+2) squared - x squared = 80, x = 192x*y-(x-3)*(y-3)=93, x+y=34, circumference 2*(x+y)=68

3.x*y+y*y=35,2*(y+x+y)-2*(x+y)=10,y=5,x=2, the area sought is x*y=10

4.Veteran workers of engineering team A = 600 * 5% = 30, veteran workers of engineering team B = 400 * 20% = 80

30+x)/600=(80-x)/400，x=36

1.The small route consists of 4 small squares with a side length of 1 meter and 4 rectangles with a width of 1 meter, and the side length of the rectangle: (80 4) 4=19 is also the side length of the lawn.

2.The truncated area consists of four small squares with a side length of 3 cm and four rectangles with a width of 3 cm, so the sum of the lengths of the four rectangles is: (93 3 3) 4 21 cm, which is the circumference of the current plank 21 cm.

3.The circumference of the large rectangle is the two sides of the square compared to the original small rectangle, so the side length of the square is 10 2 5 cm, then the length of the large rectangle is 35 5 7 cm, the width of the small rectangle is 7 5 2 cm, and the area of the original small rectangle is 5 2 10 square centimeters.

4.If there are 600 5% 30 old workers in engineering team A, and 400 20% 80 old workers in engineering team B, then there are 30 + 80 110 old workers and 600 + 400 1000 workers in total, if the percentage of old workers in the two engineering teams is the same, the percentage is 11%, then there are 600 11% 66 old workers in engineering team A, and 400 11% 44 old workers in engineering team B, and 36 people should be transferred from engineering team B to engineering team A.

There are 4 piles of identical balls, 4 in each pile. It is known that three of the piles are **, one pile is defective, **28A car travels 135 kilometers in 3 hours, so how many kilometers does it travel in 8 hours? (Guilin, Guangxi).

The total number of winners is x

Since 60% of the total number of students receive awards in elementary and junior high schools, 40% of the total number of students receive awards in high schools.

And because the number of junior high and high school winners is 20 fewer than 80% of the total number of winners, the number of junior high school winners is 80%x-20-40%x, so 80%x-20-40%x=180

x=500 The number of winners of primary schools is 500*.

Elementary and junior high school awards account for 60% of the total number of awards, i.e.: high school: 1-60% = 40%.

The number of junior high and high school winners is 20 less than 80% of the total number of students, that is, the number of junior high school students is divided into 20 to elementary schools, then:

The ratio of junior high school and high school winners is 80% of the total number of students, and elementary school: 1-80% = 20% (including 20 students from junior high school).

The ratio of junior high school and high school awards to 80% of the total number of students, minus 40% of high school, is equal to 40% of junior high school (20 students less).

Total number of people: (180-20) 40%=400 (people).

Elementary and junior high school winners account for 60 per cent of the total number of winners, minus 40 per cent for junior high school, which equals 20 per cent (for primary schools, and 20 more).

Primary school students: 400 20%-20=60 (people).

a1：a2，a3 a2：a1，a4，a5，a6 a3：a1，a7，a8，a9

A4 can't shake hands with A1, A5, A6, otherwise 3 people have shaken hands with each other. Except for A2, the remaining 4 people (A3, A7, A8, A9) must have shaken hands with A4. The situation of A5 and A6 is the same as that of A4, and the same is true of A7. Namely:

a4：a3，a7，a8，a9 a5：a3，a7，a8，a9 a6：a3，a7，a8，a9

a7：a2，a4，a5，a6

A8 and A9 each shook hands with six people. For A8, in addition to A3, A4, A5, A6 who have already shook hands with him, there are still 2 people left, and he shakes hands with any two of the remaining 4 people (A1, A2, A7, A9), and 3 people will shake hands with each other.

The situation is the same for A9 as for A8. So no matter what, you will definitely find three people who have shaken hands with each other.

The number of people shaking hands with A1 plus the number of people in A2 plus the number of people in A3 ......And so on = 42 handshake events.

Suppose A1 shakes hands with eight of his people and A2 shakes hands with seven other people ......And so on and so forth, which adds up to 36

And the assumption is that there is no repeated handshake event, so there must be four groups of people who repeat the handshake.

1. This question is answered using the drawer principle.

1) In the grid of 8 8, the glyph "Tian" of "2 2" has the following

7 7 = 49 (pcs) – equivalent to 49 apples.

2) Fill in 1-4 in the field grid, and fill in 4 1s, and the minimum is: 1 4=4.

When filling in 4 4s, the sum is maximum: 4 4 = 16

There are 13 different situations from 4 to 16——— which is equivalent to 16 drawers.

3) Then, according to the drawer principle 1 knows: 49 = 13 3 10, at least 3 1 = 4 identical sums.

2. Solution: 4 2007 2=4018 (pcs).

1) When there is 1 point in the rectangle (black dot in the figure), this point can only connect the 4 vertices of the rectangle, and divide the rectangle into 4 triangles.

2) Because any three points are not on the same line, then the second point (the red dot in the figure) can only appear in one of the triangles, so that the triangles do not overlap each other, then this point can only connect the 3 vertices of the triangle, and divide the original 1 large triangle into 3 small triangles.

3) In the same way, the third point can only appear in one triangle, and one triangle is divided into three smaller triangles; Increase the number of triangles by 2. Point 4 and point 5 ......All the way up to the 2008th point, every point is like this.

4) That is, except for the first point that divides the rectangle into 4 triangles, the next 2007 points, each point will appear in the triangle, dividing a triangle into three, adding 2 triangles each time.

Therefore, a total of non-overlapping triangles can be made: 4 2007 2 = 4018 (pcs).