The output in March is 15% higher than that in February, that is to say, the part higher than that in February is 15%X. Then the output in March is the sum of X and the high part 15%X, and X+ 15%X is (1+. ,

The output in April is 65,438+00% higher than that in March, that is, (65,438+0+65,438+05%) x, so it is 65,438+00% higher than that in March, that is, (65,438+0+65,438+05%).

April is higher than February [(1+15%) x * (1+10%)-x] as the actual output, but the percentage is calculated based on February.

[( 1+ 15%)X *( 1+ 10%)-X]/X = 26.5%

2. Assume that the arrival time of Party A is x minutes, that is, Party A arrives 10 minutes earlier than Party B, and Party B arrives 10 minutes earlier than Party C, that is, Party B arrives slowly for (X+ 10) minutes.

C (X+ 10+ 10)=X+20 minutes.

Let the speed of C be y km/h, then B is 2+Y km/h and A is 2+3+y = 5+y km/h.

Distance = time * speed, the distance of the three capitals is the same,

There is a distance X*(5+Y)= B, a distance (X+ 10)*(2+Y)= C, a distance Y*(X+20).

In the process of solving equations, the three equations are equal. First, choose any two X*(5+Y)=Y*(X+20).

The calculated 5X=20Y, that is, X=4Y is substituted into (X+ 10)*(2+Y)=Y*(X+20) to calculate y =10; X=40

The distance is X*(5+Y)=600.

3. If the distance of each step of the hound is x meters, then the distance of each step of the rabbit is 5X=9* meters.

The distance of each step of the rabbit = 5x/9;

If the running time is y, the hound runs 2Y steps and the rabbit runs 3Y steps.

Set a distance of s meters and catch up with the rabbit. The hound found a running hare in front of it at 10 meters away.

Then the hound distance is 10+S, and the rabbit distance is S.

Distance = time * speed

X*2Y= 10+S

(5X/9)*3Y=S

Upper formula-lower formula = 1/3*XY= 10.

Then substitute XY=30 into the above formula S = 50m.

Is it detailed enough? If you don't understand, there is nothing I can do. I'm exhausted. I'm not sure whether this result is correct. Oh, do the math yourself.