1.

Distance problem (encounter)

Formula: At the moment of meeting, all the distances have passed. Divide by the sum of the speeds and you get the time.

For example, A and B walk in opposite directions from two places 120km apart. Party A's speed is 40km/h and Party B's speed is 20 km/h. How long did they meet?

At the moment we met, the distance disappeared. That is, the distance traveled by Party A and Party B is exactly 120km. Divide by the sum of the speeds and you get the time. That is, the total speed of Party A and Party B is 40+20 = 60 (km/h), so the meeting time is 120 ÷ 60 = 2 (h).

2.

Distance problem (catching up)

Formula: slow birds fly first, fast birds chase after. Divide the distance you go first by the speed difference, and the time is right.

Brother and sister go to town from home. My sister walked at a speed of 3 km/h, and after walking for 2 hours, my brother rode at a speed of 6 km/h. When will he catch up?

The distance ahead is 3×2=6 (km) speed difference, that is, 6-3 = 3 (km/h). So the catch-up time is: 6 ÷ 3 = 2 (hours)

3.

The chicken and the rabbit are in the same cage.

Formula: Suppose all chickens, suppose all rabbits. How many feet are there? How many feet are missing? Divided by the foot difference, it is the number of chickens and rabbits.

For example: chickens come out of the same cage freely, with head 36 and feet 120. Find the number of chickens and rabbits. When seeking rabbits, it is assumed that they are all chickens, and the number of exemptions = (120-36× 2) ÷ (4-2) = 24. When seeking chickens, it is assumed that they are all rabbits, and the number of chickens = (4× 36-120) ÷ (4-2) =1

4.

Sum-difference problem

Given the sum and difference of two numbers, find these two numbers. Formula: sum plus difference, the bigger it becomes; Divided by 2, it is big; And subtract the difference, the smaller the reduction; Divided by 2, it is small.

Example: It is known that the sum of two numbers is 10, and the difference is 2. Find these two numbers. According to the formula, large number = (10+2) ÷ 2 = 6 and decimal number = (10-2) ÷ 2 = 4.

5.

Concentration problem (dilution with water)

Formula: sugar should be added before adding water, and sugar water should be added after adding sugar. Sugar water minus sugar water is the added water.

For example, there is 20 kg of sugar water with a concentration of 15%. How many kilograms of water are added, and the concentration becomes 10%. Get the sugar before adding water. The original sugar content is 20× 15% = 3 (kg).

When the sugar is used up, how much sugar water should there be when the concentration is 10%, 3 ÷ 10% = 30 (kg) sugar water MINUS sugar water, and the amount of sugar water after that minus the original amount of sugar water is 30-20 = 10 (kg).