Primary School Olympiad Questions, Urgent, Primary School Olympiad Answers. Hurry, hurry, hurry

The unit of water in the original pool B is 1

It turns out that A has: 1 (1-1 6) = 6 5

After the 1 5 of A is injected into B, A has: 6 5 (1-1 5) = 24 25 B has: 6 5 1 5 + 1 = 31 25

B has more than A: 31 25-24 25 = 7 25 B has water: 21 7 25 = 75 cubic meters.

Solution: If there is x cubic meters of water in the first pool, then there is (x-x 6) of water.

From the meaning of the title, 1-x 5=x 5+(x-x 6)-21

Just solve it.

Set up pool A and pool B with original water A and B cubic meters.

There is a title to know:

5/6 * a = b

1 5 * a + b -21 = 4 5 * a finally solve the equation to get: a = 80 ; b= 75 。

Solution: Let A be x, then B is 5x 6

5x/6+x/5-21=x-x/5

x=90, then 5x, 6=75

So B has 75 cubic meters of water.

21 (1-1 6+1 5-(1-1 5)) = 90 cubic meters.

90 (1-1 6) = 75 cubic meters.

Or: Suppose there is water x in pool A, then.

1-1 5)x=(1-1 6)x+1 5x-21 to obtain x=90 cubic meters.

75 cubic meters.

The original water storage capacity of pool B is less than 1/6 of that of pool A, "the original 6 parts of pool A can be set up, and the original 5 parts of pool B."

Now if 1/5 of the water stored in pool A is injected into pool B", then pool A has 24/5 and pool B has 31/5.

Then 21 cubic meters of water were pumped from pool B, and the amount of water in the two pools was exactly the same", indicating that seven-fifths of the water is 21 cubic meters.

then each serving is 15 cubic meters.

If there are 5 parts in Pool B, there will be 75 cubic meters of water.

When the two pools have the same amount of water, the amount of water is X cubic meters, and the idea is analyzed as follows:

A: x-(x-21)-x-21) (

B: x - (x+21) - x+21).

From this we get:

x-21）/(

Find x, and it's easy to find the problem.

If the original water x of pool A is set, then 5 6x + 1 5x-21 = 4 5x solution x = 90, so there is 75 cubic meters of water in pool B.

a=5/6b

4/5b=1/5a+b-21

Calculate a = 90 cubic meters.

b = 75 cubic meters.

Let A be pool A and B be pool B.

There were five children, and each two of them weighed themselves in turn, a total of 10 times. The average weight of each two people is , , 29, , 30, 31, 32, 33, 34 (in kilograms).

How many grams does the weight of the lightest child weigh in grams?

Let the weights of the five children be a, b, c, d, e, and a>b>c>d>ed+e=

c+e=a+b=34

a+c=33

c-d=1, b-c=1, then b-d=2

d+e=,c+e=,b+e=,a+b=34,a+c=33,a+d=32

There are also groups of a+e, b+c, b+d, c+d, c+d, because b-c=1, c-d=1, b-d=2, the number of lead is based on b+c>b+d>c+d, b+d=31, b+d=30, c+d=29

So a+e=

According to c + d = 29, c + e = to get d - e = because d + e = the weight of the lightest child is kg or 13,500 grams.

The second 1 2008 = 1 ?+1/?

The third one, since 7 is a natural number! That conjunction must be a natural number! And in the vicinity of 211 7 about = not right.

212 7 approx. = conformity.

So the answer is.

The first question is not thinking about the process for the time being.

Question 1:

By calculation, the sum of prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23 is 100, 281-100=181, and the sum of prime numbers 29, 31, 37, 41, 43 is 181This gives us an=43.

I don't understand what you mean by the second question.

The third question, I am willing to bow down.

Find the invariant as the unit 1

In this problem, B's amount of money has not changed, and B should be treated as unit 1Therefore, the amount of money of A was originally 1 + 1 4 = 5 4 of B, and later, B was 1 3 more than A, that is to say, B is equivalent to 1 + 1 3 = 4 3 of A, then A is equivalent to 3 4 of B, so the 360 yuan used by A is equivalent to 5 4-3 4 = 1 2, and 360 is divided by the corresponding fraction 1 2 is equal to 720 yuan, that is, the amount of unit 1, that is, the amount of money of B. A original 720 5 4 = 900 yuan.

Specific process: 1 (1+1 3)=3 4, (1+1 4)-3 4=1 2, 360 1 2=720 yuan, 720 5 4=900 yuan.

Before use, A:B = 5:4

After use, A:B = 3:4

Let's be clear, before using A is five-quarters of B, and after using it, A is three-quarters of BB's money: 360 (5 4-3 4) = 720 yuan, A's money: 720 * (1 + 1 4) = 900 yuan.

To put it bluntly.

Solution: Let B have 4x yuan, then A has 5x

5x-360+(5x-360)*1 3=4xThe result is 900 yuan for A and 720 yuan for B.

Give it a try and wish you: Merry Christmas and progress in your studies.

It's rare to write arithmetic.

Use B as x to slowly deduce the equation for A with known conditions.

It's not a difficult question, but it's not difficult in my elementary school.

Just think about it

Let B have x, then A has 5x 4, 5x 4-360=x-(1 3)*(5x 4)5x 4-360=x-5x 12

5x/4-360=7x/12

5x/4-7x/12=360

8x/12=360

x=5405x/4=540*5/4=675

According to the light to the heavy, the four people are called the number, weighed in pairs, only one pair is not weighed. There must be two weights and = the total weight of four people.

99 + 144 = 113 + 130 = 243 (total weight) can be known =99 [lightest = 113 [second light] Comparison shows that No. 3 is heavier than No. 2 113-99 = 14 [This is the difference between No. 3 and No. 2, the difference is even, indicating that their weight is the same as odd and even, and must be even].

Therefore, it can be seen that 125 is the weight of the number, [the sum of the weight of the number has been stated above is an even number, and 125 is not an even number].

The sum of number 3 and number 2 = 243-125 = 118

Knowing that the difference is 14 and the sum is 118, find the greater of it:

If the four children are A, B, C, and D, then A+B=99, A+C=113, A+D=125, B+C=130, B+D=144, and only C+D=? , subtract the first two equations to get c-b=14, combine b + c = 130 to get c = 72, b = 58, and then substitute the other equations to get a = 41, d = 86, c and d are large numbers compared to 86.

Black + White + Red + Blue = (60 + 80 + 90 + 70) 3 = 100 Black: 100-90 = 10 (kg).

White: 100-80 = 20 (kg).

Red: 100-70 = 30 (kg).

Blue: 100-60=40 (kg).

Black is 10, white is 20, red is 30, blue is 40