A 3rd grade math problem 2 and a 3rd grade math problem

One day, a young man came to the Mi Bu children's shoe store and bought a pair of shoes. The cost of the shoes is 15 yuan, and the list price is 21 yuan. The result was that the young man shelled out 50 yuan to buy the shoes.

The Mi Bu children's shoe store had no change at that time, so it used the 50 yuan to exchange 50 yuan for change from the neighborhood and found 29 yuan for the young man. But the neighborhood later found out that the 50 yuan was a counterfeit banknote, and the Xiaomi Butong shoe store had no choice but to return the 50 yuan to the neighborhood. Now the question is:

How much money did Xiaomi Bu Children's Shoe Store lose in this transaction?

Money to be earned: 21-15 = 6 (yuan).

Counterfeit money lost: 50 (yuan).

Loss: 50-6 = 44 (yuan).

A: In the end, I lost 44 yuan.

$29 + 1 pair of shoes.

The price of the shoes is difficult to set because they have not been sold yet.

If you count what you should have earned, that's a loss of 50

It's dizzy, I don't know how to count it at first, it's 36. Later, I found out that it was 44, and then I looked at the updated building and thought it was 94. Now it doesn't change, it's 44. -

How much money did Xiaomi Bu Children's Shoe Store lose in this transaction?

This time the loss was 29 + 15 = 44 yuan (not counting profits), but the total loss was 44 + 50 = 94 yuan!

Let's look at A first, it turns out to be 10 times the addition of ten digits, 1 times of the single digit is actually 1 times of the addition of the ten digits, and the 10 times of the single digit number is less than the correct result: 10-1 = 9 times of the ten digit number minus 10-1 = 9 times the single digit.

The difference between the ten digits and the single digit is: 81 9 = 9

Among the numbers, there is only 9 and 0 with a difference of 9

So the first ten digits of the number is 9

B copied 9 into 4, and the result was less: (9-4) 10=50 The correct answer is: 9797+50=9847

Solution: Let the single digit of the first addition be x, the ten digit be y, the second number b, and the resulting addition number c.

Then: x+10y+b=c 110x+y+b=c-81 2

40+x+b = 9797 3

Eq. 1-2 yields: 9y-9x=81

i.e.: y-x=9

Since x and y are both single digits of 0 9, it is only possible for x to be 0 and y to be 9 to bring in x=0 and y=9: 90+b=c 140+b=9797 3

So c=9847