1. Test questions and answers of Olympic mathematics in the fourth grade of primary school
Grade 1, there are four classes in grade four. Except Class A, the total number of the other three classes is 13 1. The total number of the other three categories except category D is134; The total number of students in Class B and Class C is less than that in Class A and Class D 1. How many students are there in these four classes? Answer and analysis
The total number of other three classes is131; The total number of students in the other three classes except Class D is 134, that is,131+134 = 265. These 265 people include 1 A and 1D, and the sum of two classes B and two classes C. Therefore, 265- 1=264 is exactly the sum of three classes B and C, and 264÷3=88, which means that the sum of two classes B and C is 88 people. So the four classes add up to 88+89= 177 students.
Insert a number between two digits and it becomes three digits. For example, insert the number 6 in the middle of 72, and it becomes 762. Some two-digit numbers are inserted in the middle, and the three-digit number is nine times that of the original two-digit number. Find all these two-digit numbers.
Answer and analysis:
Analysis: Inserting a number in the middle of two digits results in three digits that are 9 times of the original two digits, that is, the digits of this number multiplied by 9 are still equal to the original digits, so the digits can only be 0 or 5. If it is 0, obviously not. Because 20×9= 180, 30×9=270, ... the unit can only be 5. Experiments show that 15, 25, 35 and 45 are numbers that meet the requirements.
2. Question and answer of the fourth grade Olympiad.
3 3 3 3 3 3= 100 3 3 3 3 3 3 3= 100
3 3 3 3 3 3 3 3= 100
3 3 3 3 3 3 3 3 3= 100
3 3 3 3 3 3 3 3 3 3= 100
Answer and analysis
( 1)(333-33)÷3= 100
(2)33÷3×3×3+3+3= 100
(3)33+33+33+3÷3= 100
(4)(33-3)×3+3+3+3+3÷3= 100
(5)3×3×3×3+3×3+(33-3)÷3= 100
3. Questions and answers about Olympic mathematics in the fourth grade of primary school.
1, students buy back four basketballs and five volleyballs at a cost of 185 yuan. A basketball is more expensive than a volleyball, and the unit price of basketball is _ _ _ _ _ _ _ _. Answer and analysis:
25 yuan.
Analysis: (185-4×8)÷(5+4)+8=25 (yuan). Because a basketball is more expensive than a volleyball, it is 8 yuan. If you subtract the total amount from the money spent on four basketballs, the remaining money is equivalent to the money spent on nine volleyballs. Find out the quantity of a volleyball, and the unit price of basketball can be easily calculated.
2. If 1988 is expressed as the sum of 28 consecutive even numbers, what is the even number?
Answer:
28 even numbers are grouped as 14, and two symmetrical numbers are grouped as a group, that is, the sum of the minimum numbers is:1988 ÷14 =142, and the difference between the minimum number and the number is 28-1=.
4. Question and answer of Olympic Mathematics in the fourth grade of primary school
The number 1. consists of five digits, namely 1, 2,3,4,5. At most, 1 digit can be repeated twice. For example, 1234, 1233 and 2454 meet the requirements, while 12 12 and 3454 are allowed.
Answer 1, do not repeat: 5*4*3*2= 120.
2. There are duplicates: c (5 5,3) * 3 * 3 * 2 = 360, ***480.
Second, counting.
Write down the natural numbers one by one from the beginning: 1 2345678910112 ... from left to right, after counting to what number, five consecutive/kloc will appear for the first time.
The answer to the five lines 1 appears in11,1 12.
One digit: 9 digits
Two digits: 90×2= 180.
Three digits: 100- 1 10, 1 1×3=33.
* * * There are 9+90×2+ 1 1×3=222 (pieces).
Third, counting.
Two thousand numbers are written in a line, and the sum of any three adjacent numbers is equal. The sum of these two thousand numbers is 53324. If 1, 1949, 1975 and the last number from the left are erased, the sum of the remaining numbers is 53236. Q: What is the 50th from the left among the remaining figures?
The answers are counted in groups of three from the left, and the sum of three adjacent numbers is equal.
The sum of the first two numbers in a group is (53324-53236)/2=44.
The sum of the first three numbers in a group is (53324-44)/666=80.
So the third number in a group is 80-44=36.
That is, the 50th number after erasing 1 from the left is 36.
5. Question and answer of Olympic Mathematics in the fourth grade of primary school
1, calculation 1 9999+1999+19 solution: all figures in this question are 9 except the number1,and rounding is still adopted. But this is 1. (For example, 199+ 1 = 200)
199999+ 19999+ 1999+ 199+ 19
=( 19999+ 1)+( 19999+ 1)+( 1999+ 1)+( 199+ 1)+( 19+ 1)-5
=200000+20000+2000+200+20-5
=222220-5
=22225
2. Use the number 0, 1, 2, 3, …, 9 10 to form five two-digit numbers, and each number is used only once.
It is required that their sum is odd, and the bigger the better, then what is the sum of these five two-digit numbers?
If the sum of five numbers is required to be odd, then there are odd numbers in the five numbers, and for the sum, first consider using 9, 8, 7, 7, 5 as the ten digits, and then the digits are 0, 1, 2, 3, 4, 5, so that only two of the five numbers are odd numbers. So the ten digits are adjusted to 9, 8, 7, 6, 4, so that one digit is 0, 1, 2, 3, 5, which meets the meaning of the question and is:
(9+8+7+6+4)× 10+(0+ 1+2+3+5)=35 1。