Elementary school arithmetic problems, elementary school arithmetic problems, how to do it?

Solution: From the meaning of the question, the number of A is more than the number of B, because the number of both of them is less than 10, then the number of A may be at most 9, after B gives A a copy, A may be up to 10, and the number of this into A is 4 times that of B, 10 4 = 2 and 2, indicating that the number of B is below 2 at this time, then the number of B before is at most 3, so the number of B is either 2 or 3, it cannot be one, because it is known by the title, if the number of B is 1, After B gives A a copy, B doesn't have any more, so B can only be 2 or 3 copies.

If B has 2 copies, then after B gives A a copy, B only has 1 copy left, and its 4 times is 4 copies, that is, A has 4-1=3 copies before, then when A gives B two copies, A has 1 book left, and B has 4 copies, which is not the same and does not meet the requirements. If B has 3 copies, when A is given a copy, there are 2 copies, and A is 4 times that of B, and there are 2x4=8 copies, then A should have 8-1=7 copies, and after giving B two copies, there will be 5 copies, and B will increase from 3 to 5 copies, just the same amount. Meets the requirements.

To sum up, A has 7 copies and B has 3 copies.

Note: First of all, I wish you a happy new year! I've been thinking about whether there is an easier way, but after thinking twice, I came up with the above method, I don't know if it's possible? Please criticize a lot!

From the multiplication table of 9 and 9, it can be seen that a number multiplied by 4 times is less than 10, only 1 and 2. There is also a question with the meaning of the title, it is B to A1 book, and it will be established after 4 times. All Bs may be 2 or 3.

Then A is 1*4-1=3 or 2*4-1=7, and then according to the title "A gives B two copies, A and B have the same amount", that is, B plus two copies will equal 4 or 5 and A minus two copies will equal 1 or 5. Therefore, according to all the conditions, the topic can only be satisfied under the condition of a = 3 and b = 7.

B gives A1 after the book.

Because A is 4 times that of B.

If B has at least 3 copies, A has at least 12 copies, and B gives A1 at least 11 copies.

All B gives A1 copies and B has a maximum of 2 copies.

If B gives A1 and then B has 1 and A has 4, then A and B each have 3 and 2.

If A gives B two copies, then A only has 1 copies, B has 4 copies, and A and B don't have the same amount, so no.

If B gives A1 and then B has 2 copies, and A has 8 copies, then A and B each have 3 and 7 copies, and if A gives B two copies, A and B are both 5 copies, which is eligible.

So A and B originally had 3 or 7 copies.

Due to"B gives A 1 book, and A is 4 times that of B"

In the first 10 numbers, only 4 is 4 times 1, and 8 is 4 times 2, which is divided into two cases.

When 4 is 4 times 1: A is 3 and B is 2

8 is 4 times 2: a is 7 and b is 3

Due to"A gives B two copies, and A and B have the same amount", then the initial a+b is an even number.

The result is a = 7, b = 3 to bring back to meet the requirements.

50 people voted for A, 60 people voted for B, and 10 spectators voted for both A and B.

The number of spectators is 50+60 10 100.

If I've been in City A for an hour, City B has passed.

1 17) days = 24 hours 17

Hour = 1 hour minute.

1 hour 24 minutes second.

Twenty-four hours a day, one hour and sixty minutes, so it is converted.