2. Because 210 = 2 * 3 * 5 * 7 =14 *15, the circumference =2( 14+ 15)=58.
3. The rest is 1/4, which is 2/4= 1/2 more than the rest. Therefore, sand ***5/8*2=5/4 tons.
4. Set x tons of sand. 5/8-(x-5/8)=3x/4 .X=5/7 tons.
5. Set x tons of sand. (x-5/8)-5/8=3x/4 .X=5 tons.
6.20kg is 20% less than (25) kg, and (6) m is 20% longer than 5m.
X( 1-20%)=20 gives x = 25,5 (1+20%) = 6.
7.4* A = B *20% A: B = 1: 20。
8.6 1 A work started, originally scheduled to be completed in one month. In actual construction, the task was completed on June 25th and exceeded by (20)% on June 30th.
Let the total workload be "1", the actual production efficiency be 1/25, and the overproduction from 26th to 30th =5* 1/25= 1/5.
9. It can be known that length: width: height = 2: 1: 1, and surface area =4000. It is easy to find length 40, width 20 and height 20. If two such long bodies are spliced, only the smallest one needs to be butted, and the combined surface area is the largest.
More than two surface areas and-two butting areas =8000-800=7200 square centimeters.
10. The motor rotates 3600 revolutions per minute, and the rotation distance is the same. Total circumference =0. 16π3600 m/min, thresher wheel rotation =0. 16π3600/π0.24=2400 rpm.
1 1. The number of trees of species A is the sum of the other three 1/2, the number of trees of species A is (60-A) ÷2, and the number of trees of species A is 20; The number of trees in species B is 1/3 of the sum of the other three people, B =(60- B) ÷3, and b is15; Bingzhi
The number of trees is the sum of the other three people 1/4, C =(60- C) ÷4, and c is12; D = 60-20-15-12 =13 trees
12. Assume that the number of people who originally planned to participate in voluntary labor is X, and the total number of students in the class is 5x. According to the meaning of the question, the equation x+2=[5x-(x+2)]* 1/3, and the result is x=8.
13.*** is set to x, the first workshop is 2x/7, and the third workshop is (2x/7+ 1600)/2=x/7+800.
According to the meaning of the question, 2x/7+ 1600+x/7+800=x, and x=4200.
14. The two digits with the greatest common divisor of 9 and the smallest common multiple of 360 are (45) and (72) respectively.
360/9=40= 1*40=5*8
Because it is a two-digit number, it is:
9*5=45
9*8=72
15. When a two-digit number is divided by 7, the quotient and the remainder are the same. The minimum value of this two-digit number is (8) and the maximum value is (42).
Let the remainder be x, then 7x+x=8x. When x= 1, the number is 8. When x=6, the number is 42. X is less than 7.
There are x students in class 16 (1). X (1+12.5%) = 45. x = 40 students, and Class 6 (1) is 45-40=5 students less than Class 6 (2).