1. The area of the Olympic math problem in the fifth grade of primary school
1. A rectangle is divided into four different triangles. The area of the green triangle is 0. 15 times that of the rectangle, and the area of the yellow triangle is 2 1 cm2. Q: How many square centimeters is the area of a rectangle? Solution: For any point in a rectangle (vertex and edge in extreme cases), the sum of the areas of two triangles with opposite sides as the bottom is half of the area of the rectangle, which can be easily obtained according to the triangle formula. Then the area of the rectangle is 2 1÷(0.5-0. 15)=60 square centimeters.
The area of a rectangle is 60 square centimeters.
2. If the width of a rectangle remains the same and its length increases by 6 meters, its area will increase by 54 square meters. If the length remains the same and the width is reduced by 3 meters, its area will be reduced by 36 square meters. What is the original area of this rectangle?
The train of thought navigation shows that its width is 54÷6=9 (meters), which means: "The width is unchanged, the length is increased by 6 meters, and its area is increased by 54 square meters"; From "the length is constant, the width is reduced by 3m, so its area is reduced by 36m", we can know that its length is 36÷3= 12 (m), so the area of this rectangle is 12x9 = 108 (m). (36÷3)×(54÷9)= 108 (square meter)
The playground of Renmin Road Primary School is 90 meters long and 45 meters wide. After transformation, the length is increased by10m, and the width is increased by 5m. How many square meters has the playground area increased now?
Subtract the original area of the playground for thinking navigation to get the increased area. The current area of the playground is (90+ 10)×(45+5)=5000 (square meters), and the original area of the playground is: 90×45=4050 (square meters). So now it is 5000-4050=950 square meters more than before. (90+10) × (45+5)-(90× 45) = 950 (square meter)
2. The problem of the area of Olympic math problems in the fifth grade of primary school.
1, parallelogram parking lot, with a bottom of 63m and a height of 25m. Each car occupies an average area of15m2. How many cars can this parking lot park? 2. Trapezoidal vegetable fields with an upper bottom13m, a lower bottom15m and a height of 8m, average per square meter15kg of Chinese cabbage. How many Jin of Chinese cabbage can this land collect?
3. A triangular paddy field, with a bottom length of 32 meters and a height of 25 meters, has an average rice harvest of 1.2 kg per square meter. How many kilograms can this rice field harvest?
The triangle has an area of 22 square meters and a height of 4 meters. How long is its bottom?
5. A triangular steel plate, with a base length of 3.6dm and a height of 1.5dm, weighs 1.8kg per square decimeter. How much does this steel plate weigh?
3. The problem of the area of the Olympic math problem in the fifth grade of primary school
(1) After the parallelogram is cut and repaired into a rectangle, the area remains unchanged and the perimeter (). A. Expansion
B.it's shrunk
C. unchanged
(2) A parallelogram with an area of 56 square decimeters, a base of 14 decimeter and a height of ().
0.4 decimeter
B.2 decimeter
8 decimeters
(3) A parallelogram with the same base, three times higher and an area of ()
A. Zoom in 3 times
B. Zoom in 9 times
C. Reduce by 3 times
(4) Let the triangle have an area of 63 square decimeters, a height of 7 decimeters and a base of ().
A.4.5
18
C.9
(5) A triangle with five times the height of the same base has an area of ().
A. Zoom in 5 times
B. Zoom in 25 times
C. Reduce by 25 times
4. the divisibility of the number of Olympic math questions in the fifth grade of primary school
1, judge whether the nine digits of 123456789 are divisible by 1 1? Solution: The sum of odd numbers of this number is 9+7+5+3+ 1=25, and the sum of even numbers is 8+6+4+2=20. Because 25-20 = 5 and 1 15, it is 1638.
2. Is13574 a multiple of 1 1
Solution: The difference between the sum of odd digits and the sum of even digits of this number is: (4+5+ 1)-(7+3)=0. Because 0 is a multiple of any integer, 1 1|0. So 13574 is 164.
Characteristics of numbers divisible by 7 (1 1 or 13): The difference between the last three digits and the first three digits of an integer (which decreases greatly) can be divisible by 7 (1 1 3).
5. The divisibility of the number of Olympic math questions in the fifth grade of primary school.
1, in 1, 23, 4, 5, 15, 45, 65, 90 and 270, () is a multiple of 45, and () is a multiple of 15. The common factor of 15 and 45 is (), and that common multiple of 4 and 15 is (). 2. In 39, 47, 5 1, 63, 7 1, 2 1, 37, 53, 9 1, the prime number is (), and the composite number is ().
The factor of 3 and 42 is (). Among these factors, () is a prime number and () is a composite number. The prime factor of 42 is ().
4. The two digits that can be divisible by 3 and 5 at the same time are (); It is a multiple of 2 and a multiple of 3. The smallest three digits divisible by 5 is (), and its prime factor is ().
5. Among the ten numbers between 1 and 10, () and () are both composite numbers and prime numbers; () and () are both prime numbers and coprime numbers; One of () and () is a prime number and the other is a composite number, both of which are coprime relations. (No, just fill in one group. )
6. A two-digit number, which is divisible by 3 and is a multiple of 5. The number of digits is 0, and the minimum number of digits is ().
7. Spell a four-digit number with 5, 7, 8 and 0 to make it a multiple of 2. This number can be (), making it a multiple of 5. This number can be ().
6. the divisibility of the number of Olympic math questions in the fifth grade of primary school
From left to right, students numbered 1 to 199 1 line up, and from left to right, students numbered 1 to1report. Then the rest of the students count off from 1 to 1 1 from left to right, and the rest of the students get out of the queue; The remaining students count off for the third time from 1 to 1 1 from left to right. Students who reported to 1 1 stayed behind, and the rest of the students left the team. Then the first person from the left is (). Analysis: The original number of students left by the first report is a multiple of 1 1; These left students continue to count off, so the initial number of left students is11×11=121,and so on.
Solution: The initial number of students left after the first registration is a multiple of 1 1.
The initial number of students leaving after the second registration is a multiple of 12 1
The initial number of students left after the third report is a multiple of 133 1.
So there is only one student left in the end, and his initial number is 133 1.
A: The starting number of the first person from the left is 133 1.