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Teach me math?
1. Solution: Let Party A be a person and Party B be a person. After transferring 2 1 person from Party A, the numbers of Party A and Party B are: A-21respectively; b+2 1

From this, the equations can be listed: a/b = 5/4; A-2 1/b+2 1=2/3, a = 75 and b = 60 can be obtained by solving the equation.

Reasoning solution: if the total number of people is X, then a+b=x, X can be divided into nine groups, A has five groups, B has four groups, and A has one more group than B, that is, one ninth of the total number. After transferring the number of Party A, the total number of Party A and Party B remains unchanged. At this time, it is divided into five groups, two for Party A and three for Party B. At this time, Party B is one fifth more than Party A. At this time, there is1/9x+1/5x = 2× 21(the number of Party A and Party B changes before and after the exchange), and the unary equation x = 6540 is solved.

2. Solution: Let the denominator be B and the numerator be A. After subtracting 1 from the numerator, the number is a- 1/b=2/7, and after adding 1, it is a+ 1/b=4/2 1.

By solving the equation, we can get: a=5, b=2 1.

Another solution, denominator unchanged, numerator changed. As long as the numerator is used for addition and subtraction, the denominator is 2 1.

3. Solution: Assuming that the sugar to be added is X grams, and the sugar content of 300 grams of sugar water with a sugar content of 10% is 300× 10% =30 grams, then the water after adding sugar accounts for 1-25% of the total weight.

4. Solution: Let the distance between Party A and Party B be s, the speed be V0, and the required time t0=S/V0. After the speed increase, it is V1= (1+0.15) v0, and the required time is t 1=S/V 1. The reduction time percentage is (t0-t1)/t0 =1-t1/t0, so the reduction percentage is11.15 = 0.

5. Solution: Let red ink be Box A, blue ink be Box B, a+b=65 ..., red ink be sold as 1 1 box, leaving A-1/and blue ink as (1-20%).

6. Solution: Assuming that the TV set originally planned to be produced every day is A, then the original planned demand is d= 1600/a, and the efficiency is improved by 25% after improving the process. At this time, the TV set produced every day is (1+25%)×a, and the remaining TV sets that are not produced are (1600-5a). At this time, it takes d1= (1600-5a)/(1+25%) days to complete the plan, and the number of days to complete the whole plan is reduced by 25%, that is, d-d1= (1-25%) d.

7. Solution: Suppose there are X people in Class Six 1, and the absenteeism of Class Six 1 is one sixth of the attendance. Divide the class into seven groups, with absenteeism accounting for 1 and attendance accounting for six groups. A person can be divided into six groups after taking leave, with absenteeism accounting for 1 group and attendance accounting for 5 groups. Since people are a whole, the number of people in this class is at least a multiple of 6×7=42, that is, n×42, where n is a non-zero positive integer, that is, 1, 2,3 ... The number of people in each group is 6n, so after one person takes leave, the number of people absent is 6n+ 1, and the number of people present is 36n-/kloc-.

8. Solution: Assuming that an exercise book is A Yuan, A takes 15 less than C, and B takes as many as C, then A also takes 15 less than B, C and B take 30 more, and C and B give a ***3 yuan, that is, 30a=3 yuan. The solution is A.

9. Solution: A cuboid * * * has six faces, including three pairs of identical faces. Their combinations are: 10 and 8, and the height at this time is h1= 6; 10 and 6, when the height is H2 = 8; 8 and 6, when the height is H3 = 10; The largest circular surface cut on a rectangular surface depends on the shortest side length, that is, the shortest side length of the three possible surfaces is the diameter of the circle, and there are three combinations of the diameter and height of the cylinder bottom surface, namely: d 1=8, h1= 6; d2=6,H2 = 8; D3=6, h3= 10. The volume of the cylinder V=πr2h, r= 1/2×d, π=3. 14. Substitute the three groups into the above formulas respectively, and when r2h is the largest, the volume of the intercepted cylinder is the largest. It can be calculated that the maximum value of the first group (d 1=8, h 1=6) is V=30 1.44 cubic decimeter.

10, solution: let the cylinder volume be V 1, the bottom area be S 1, and the height be h1; The volume of the cone is V2, the bottom area is S2 and the height is h2. That is V 1 =S 1×h 1, V2= 1/3×S2×h2, V 1: V2 = 2: 1, h1:.

Logical reasoning can't completely solve the above problems, only 2 yuan equation can be solved, but 2 yuan equation doesn't seem to be solved in primary school. I hope it works.