There are more than 3 Olympiad questions in the fifth grade, and 15 Olympiad questions in the fifth

1.Xiaohong walked 100 meters per minute after school, and at the same time, her mother also set off from home to pick up Xiaohong, walking 120 meters per minute. The two met 50 meters from the finish line. Ask for the distance from Xiaohong's house to school.

Solution: Because the two met 50 meters from the finish line.

So mom walks 50 2 = 100 meters more than Xiaohong.

Because Mom's speed is faster than Xiaohong's.

So mom walks 120-100=20 meters more per minute than Xiaohong.

So it takes a total of 100 20 = 5 minutes for the two to meet.

So the distance of Xiaohong's school is 5 (100+120)=1100 meters.

2.There is a two-digit number, and adding the number 1 in front of it gives a three-digit number, and adding it after it also gives a three-digit number, and the difference between these two three-digit numbers is 666. Find the original two-digit number.

Answer: From the place value principle, adding the number 1 in front of a two-digit number is equivalent to adding 100; Add the number 1 to a two-digit number, which is equal to multiplying the two-digit number by 10 and adding 1.

Let this two-digit number be x. Derived from the title.

10x+1)-(100+x)=666，10x+1-100-x=666，10x-x=666-1+100，9x=765，x=85。

A: The original two-digit number was 85.

3.Jiajia and Yaoyao love to ride bikes. One day, the two of them set off at the same time from the same place on the ring road, walking in opposite directions. It takes Jiajia 70 minutes to ride a lap, and the two meet 45 minutes after departure, so how many hours does it take for Yaoyao to ride a lap?

Answer: As shown in the figure below: Jiajia and Yaoyao depart at point A and meet at point B 45 minutes later. That is to say:

45 minutes, Jiajia + Yaoyao = one lap. Whereas.

70 minutes, Jiajia = one lap. So.

Jiajia walks (70-45=) 25 minutes = Yaoyao walks for 45 minutes. So Jiajia walks for 70 minutes = Yaoyao walks for 126 minutes (in the same proportion).

There are a lot of them on the Internet, and they are easy to find.

Primary School Olympiad Introduction] When solving Olympiad problems, we often have to remind ourselves whether the new problems encountered can be transformed into old problems to solve, turn the new into the old, through the surface, grasp the essence of the problem, and transform the problem into a familiar problem to answer. The types of conversions include conditional conversion, problem conversion, relationship conversion, and graph conversion. The following is the "15 Olympiad Questions for the Fifth Grade of Primary School" compiled by the test network, I hope it can help you.

[Part 1].1. A three-digit number can be divisible by 9, and after removing its last digit, the resulting two-digit number is a multiple of 13. 2. Xiao Ming has a bag of sugar, 4 grains are less than 3 grains, 5 grains are more than 2 grains, 3 grains are just right, how many grains are there in this bag of sugar?

3. A factory processes three batches of parts, the first batch of processing 123, the second batch of processing 162, the third batch of processing 260, each batch of parts is evenly distributed to the same group of workers for processing, respectively 3, 2 and 6, how many workers participate in the processing at most?

How many of the total approximations are there?

5. In the following equation, only add parentheses to make them all true.

[Part 2].

1. If the last five digits of 35018065 () are all 0, what is the minimum number of natural numbers filled in parentheses? 2. Divide the following 6 numbers into two groups equally, so that the two groups of numbers are equal to the product, and these 6 numbers are .

3. A six-digit number 546 9 is a multiple of 44, what is this number?

4. What is the first term in the series of equal differences, 444?

5. Calculate 1 2 3 4 5 6 7 8 9 58 59 60

[Part 3].

1. There are 20 red, blue and black pencils, of which the number of black pencils is 1 more than half of the red pencils, and the number of blue pencils is 1 more than half of the black pencils. *There is a blue pencil () stick. 2. In order to maintain the traffic safety of children and teenagers, the four classes of the first grade purchased a batch of small yellow hats.

Four classes give out as much money. When dividing hats, one shift is compared.

The second, third, and fourth classes take 8 less tops, thus.

The second, third, and fourth classes are each given one yuan. Then each little yellow hat () yuan.

3. A and B are walking opposite each other on the path next to the railway, and the speed is 1 meter per second. A train is coming towards A at a constant speed, and it takes 15 seconds for the train to pass by A's side, and then 17 seconds to pass by B's side. The length of this train is () meters.

4. The distance from home to school is 540 meters, and it takes 9 minutes for Xiao Ming to walk to school, and it takes 3 minutes less to go home than when he goes to school. Then Xiao Ming walks () meters per minute on average for a round trip.

5. The number of watermelons shipped from the fruit store is twice the number of white melons. If you sell 40 white melons and 50 watermelons every day, and after a few days, you will have 360 watermelons left. A total of watermelons and white melons from the fruit store.

One of A, B, and C is a pastor, one is **, and one is a gambler. Priests only tell the truth, ** only lies, and gamblers sometimes tell the truth and sometimes tell lies. A said:

C is a pastor. B said, "A is a gambler."

C said, "B is **." "So what are the occupations of A, B, and C?

Answer & Analysis:

A is a gambler, B is a priest, and C is **.

When a pastor tells the truth, it is impossible to say that someone else is a pastor, so A must not be a pastor. If B is a pastor, then A must be a gambler, then C is **, which is in line with the topic. If C is a pastor, then B refers to the bad luck as a gambler, and A is **, at this time A cannot say the truth of "C is a pastor", so it is contradictory.

Tips: This is a logical reasoning test, the key middle school exam is very willing to test such a type of question, when answering this kind of question, we must first sort out the relationship between the parts from the conditions given, and then analyze and reason, eliminate some impossible situations, gradually induct, find the correct answer.

The head is the same as the tail.

35x35=1225

46x44=2024

99x91=9009

The method is to multiply the number of ten digits by the number of the first year.

Multiply the single digits. Multiply the number by 11

12x11=132

123x11=1353

Method: Pull both sides and add them in the middle.

12345x11=135795 understood.

First course:

A = 75a, B = coprim.

a*b*75 = 450

a*b = 6

then |a - b|Min = |3-2|= 1These two numbers are 3*75 = 225 and 2*75 = 150 respectively.

100a + 10b + c

100a + 10c + b

100b + 10a + c

100b + 10c + a

100c + 10a + b

100c + 10b + a

It is easy to know that their sum = 222a + 222b + 222c = 3330 , a + b + c = 15

The three numbers are not equal to each other, and when you want to make the maximum three-digit number, the three numbers are , and the maximum three-digit number is 951.

To make the three-digit number smallest, the three-digit number is , and the maximum three-digit number is 159.

The third course is similar to two. There are no 0s, otherwise they make up less than 6 three-digit numbers.

Let this other three-digit number be ABC, then there is:

2234 + abc = 222(a+b+c)a + b + c = (2234 + abc) 222 11, i.e., from a multiple of more than 11 that is greater than 2234 and is more than 11 of 222, the solution is obtained abc = 652

The first question is 75 and the other is 150.

1) 150 and 225

450 = 75 * 2 * 3 or 450 = 75 * 1 * 6, the greatest common divisor is 75, so these two numbers may be 75 * 2 and 75 * 3, or 75 * 1 and 75 * 6, so that their difference is the smallest, then the two numbers are 75 * 2 = 150 and 75 * 3 = 225

2) The largest is 951, the smallest is 159

Let the three numbers be ABC, then the six numbers are ABC=100A+10B+C, ACB=100A+10C+B, BAC=100B+10A+C, BCA=100B+10C+A, CAB=100C+10A+B, CBA+100C+10B+A, and the sum of the six three-digit numbers is 2*100(A+B+C)+2*10(A+B+C+2(A+B+C)=3330, 2(A+B+C)(100+10+1)=3330, a+b+c=15, the highest hundredth of the three digits should be 9, followed by 5,1, so among the six numbers, the largest is 951 and the smallest is 159

The first two numbers are 75,150 or 150,225.

The second 8 cases are the largest and the smallest.

The reason 3330 is divisible by 3 and 5, so the sum of the 3 numbers taken out is divisible by 15) The third way is 652