6th Grade Math Olympiad Questions Note to write the process

1.Speed ratio = (1 + 1 5) 1:1 (1-1 11) = 12:112There are 8x large peaches and 5x small peaches

6x=510

x=85 large peaches = 8*85=680.

Small peach = 85 * 5 = 425.

3.Science & Technology Group: Composition Group: Mathematics Group = 9:10:14 Science & Technology Group = 72 (10+14) 9 = 27 students.

4.First bucket = 45 (1+2+3) = kg.

Second barrel = kilograms.

Third barrel = kilograms.

1. Let the distance of B be s, then the distance of A is (1+1 5)s=6s 5;

A walks to the time t, then B walks to the time of (1-1 10) t = 9t 10 So: A's velocity = 6s 5t, B's velocity = 10s 9t The velocity ratio of A and B = 54 50

2. Set 8 large peaches and 5 small peaches as one serving, and the total number of servings is:

The number of large peaches = 85 8 = 680 (pcs).

The number of peaches = 85 5 = 425 (pcs).

Number of students in the math group = 7 6 = 42 (people).

Number of people in the composition group = 5 6 = 30 (people).

Number of people in the science and technology group = 9 30 10 = 27 (people).

4. The weight of the first barrel of oil = 45 (1+2+3)=kg) and the weight of the second barrel of oil = 2 kg).

The weight of the third barrel of oil = 3 kg).

Question 1: Let B walk x meters, A walks six-fifths x meters, A's time is t, B's is 10/11 t, and then speed = distance divided by time, write the expression, compare, x and t can be approximated, the answer is 12:11 Answer...

Question 2: Set up 5x small peaches and 8x large peaches for sale. Column equation: multiply by 8x + multiply by 5x=510 to solve x=85

Big peach = 8x = 680, small peach = 5x = 425 A...

Question 3: The same is also done with x, modeled after question 2. (It's not that I don't want to write anymore, I hope you really master these questions, practice them yourself, it's not difficult.) (Six-grade equations should be taught, right?) ）

Question 4: Set the end of the set.

I. 2. The third barrel of oil is x, 2x, 3x kg. x+2x+3x=45, solve x=, and then the first barrel is kilograms, the second barrel: 2 times kilograms), the third barrel: 3 times kilograms) Answer: .

Note: I didn't write the unit, you have to add brackets to write, to answer!

1. Set the speed of A v1, B v2, v1 v2=(s1 t1) (s2 t2)=(s1 s2)(t2 t1)=[(1+1 5) 1]*[1-1 11) 1]=12 11

2. Set x large peaches and x small peaches (5 8).

x=680, 680 large peaches, 425 small peaches.

3. Set up the composition group x people, x (72-x) = 5 7, get x = 30, so the composition group is 30 people, and the science and technology group = 30 10 * 9 = 27 people.

4. The first barrel = 45 (1+2+3) = kilograms, let the second barrel x, the third barrel = because: (, get x =, so the third barrel = 20 kilograms.

Let the speed of 10 dogs pull y, and the two camps of A and B are x kilometers apart, then the time for 10 dogs to pull is :

x y=4+(x-4y) (7y 10)-2x y=4+21 y+(x-4y-21) (7y 10)-1 gives x=78

y=9, that is, the two camps of A and B are 78 kilometers apart.

Suppose that the two camps of A and B are x kilometers apart, and the speed of 10 dogs is y kilometers per hour, then, x y=(x-4y) (7y 10)+4-2x y=(x-4y-21) (7y 10)+4-1 solution: x=

As you can imagine, the route of the center of the circle is the circumference of a figure after the figure is reduced by 2 centimeters.

In this case, we can complete the graph. That is, add two small squares to the left and right to make the whole look rectangular.

In this case, the circumference is constant. The long side of the large rectangle is 30 cm and the short side is 20 cm.

The small rectangle (circle center path) is 26 (30-2*2) cm on the long side and 16 (20-2*2) cm on the short side.

So the path of the center of the circle is (26+16)*2=84 cm.

First calculate the length of the route excluding the corners: 8 + 8 + 6 + 26 + 6 + 8 + 8 + 6 = 76cm There are a total of 8 corners, each corner is a quarter of a circle, 8 is 2 circles, and the radius of the circle is 2cm: 76+

The length of the BC is required, and the area of the triangular ABC is actually required.

s(abc) The area of the semicircle minus the area of shadow 1 plus the area of shadow 2 The area of the semicircle - (shadow 1 minus shadow 2).

Substitute the numbers and calculate it.

628-108 520 (square centimeters).

Then use the triangle area formula, know the area and the base, and find the height.

520*2) 40=26 (cm).

There are two errors in the answers upstairs:

1. The area of the semicircle should be divided by 2 by the area of the circle.

2. The area of shadow one is larger than that of shadow two, not two to one, so the difference should be subtracted at the end, not added.

Let bc=x

The area of the blank part.

Solve x yourself.

Semicircular area:

Because: semicircle area - triangle area = shadow 1 - shadow 2 = 108 so: triangle area is.

bc：520×2÷40=26cm

As can be seen from the figure, the area of the semicircle - shadow 1 + shadow 2 = triangle ABC area, and because shadow 1 is 108 square centimeters larger than shadow 2, the area of the semicircle - shadow 1 + shadow 2 = triangle ABC area, that is, the area of the semicircle - 108 square centimeters = the area of the triangle ABC 40 2 = 20 cm.

20*20*cm².

628-108 = 520 square centimeters, which is the area of the triangle 520*2 40 = 26 centimeters is the length of BC.

The area of the semicircle = 20*20*

The area of the triangle abc = 628-108 = 520

bc=520*2/40=26

Let B's work efficiency be x, then A's work efficiency is 2x, then A and B respectively undertake the lead trapped in A and B's projects, and the time required is (A 2x) + B X

The time required for A and B to undertake the works of B and A respectively is a x + (b 2x), and the time is 50% higher than that of the original plan

Rule. A x+(B Wang Heng is good 2x)=[A 2x)+B x](1+50%), about the left and right sides of x.

a+ab=4, so the ratio of the engineering quantity of project a and project b is.

Olympiad math questions generally use some simple calculation methods to do the questions, depending on what the topic is.

There are many articles on Olympiad problem-solving techniques, which are very important. Skills are important, the cause of the mistake is actually very important, treat the mistake carefully to analyze it, find out why, such a mistake is the most meaningful.

In the process of solving the problem, there will be a mistake in the head of the forest, and the error is not necessarily expected, so the following is only an analysis of the more common types of errors.

1) Tampering with data.

There is a kind of mistake, it is the eyeball that tampered with the question, and people often make this mistake, either reading the numbers wrong, or reading the conditions of the question wrong, and some exam questions may be very similar to the questions that have been done, that is, this kind of mindset is the most terrible, it will lead you to the wrong up. This will cause carelessness, which is why many students can make the kind of difficult questions that they have never seen before, and they are very accurate, but they make mistakes for some common question types. That's a big loss.

Coping strategy: For this kind of mistake, we should pay attention to the subtleties from the usual, and develop a good habit when doing the questions, and I believe that you should not make it when you get the exam. The examination of the question must be carefully seen, what is the data, what are the conditions, what is the relationship between the conditions and the conditions, the students usually do the questions to develop the drawing, the conditions, the number of lines, the best to use the pen concisely to give the conditions in the question on the straw paper reflected.

If you accumulate a lot, you will be relaxed during the exam.

2) Answering questions that are not asked.

I believe that many people have made this kind of mistake, and they have made it many times. When people ask how much A is less than B, they answer how much A is, and so on. This is always called sloppy, and there is no sloppiness in the eyes of the teacher, only right and wrong, sloppiness is also wrong, sloppiness just won't.

1. Find the invariant as the unit "1", the invariant of this question is the sum of the inventory of A and B, so the total amount of the two warehouses of A and B is the unit "1".

Before A transferred 30 units, the inventory accounted for 7 10 of the total inventory

After A transferred 30 units, the inventory accounted for 3 5

Then 30 units account for 7 10-3 5 = 1 10 of the total inventory, so the total inventory is 30 1 10 = 300

Then the original inventory of A is 300 7 10 = 210 (sets).

B's inventory is 300 (1-7 10)=90 (sets).

You typed these problems into the computer one by one, why don't you make them yourself?