There is a batch of rice in the rice shop. On the first day, 65,438+02 bags were sold, on the second day, the remaining third bag and 65,438+02 bags were sold, on the third day, the remaining quarter bag and 65,438+02 bags were sold, and on the fourth day, the remaining fifth bag and 65,438+02 bags were sold. How many bags of rice are there?

Solution: This problem is solved by reverse deduction.

On the fourth day, the 12 bag was sold, and one fifth remained on the third day, that is, the last 1- 1/5=4/5 on the third day, and 12 ÷ 4/5 on the third day.

The remaining 1- 1/4= 3/4 of the next day, that is, 12+ 15=27 (bag), and the remaining 27÷3/4= 36 (bag) of the next day.

The remaining 1- 1/3=2/3 on the first day is: 36+ 12=48 (bag), and the remaining on the first day is: 48 ÷2/3=72 (bag).

1/2 on the first day is: 72+ 12=84 (package),

This batch of rice is: 84÷ 1/2= 168 (bag).

Question 4: There is a column of figures. The first number is 105 and the second number is 85. Starting with the third number, each number is the average of the first two numbers. What should be the integer part of the 2002nd number?

It's 9 1

The fifth problem is that three people, A, B and C, each have some things to exchange with each other. Take a third from B first, and a tenth from C to A; Take one-eighth from A and one-ninth from C to B; Finally, take three tenths from B and two sevenths from A to C. At this time, there are 155 in A, 136 in B and 209 in C. How many such items do the first three people have?

Or use the backward method:

Finally, take three tenths from B and two sevenths from A to C. At this time, there are 155 in A, 136 in B and 209 in C.

B Remaining:1-3/11= 8/1. B before 3/ 1 1 is: 136 ÷ 8/65438+.

The remainder of A: 1-2/7=5/7, before A: 155÷5/7=2 17, and 2/7:2 17×2/7 = 62.

Before the last round of C: 209-5 1-62=96.

Penultimate round:

According to: "Party A takes one-eighth and Party C takes one-ninth and gives it to Party B"

The rest of A is: 1- 1/8=7/8. Before taking A, it was: 2 17÷7/8=248, and A's 1/8 was: 248×1.

The rest of C is: 1- 1/9=8/9. Before taking C, it was: 96÷8/9= 108, and C's 1/9 was:108×/kl.

Before the penultimate round B is:187-31-12 =144.

In the first round, according to: "Take a third from B first, and give a tenth from C to A;

The rest of b: 1- 1/3=2/3, the beginning of b: 144÷2/3=2 16, and b's1/3: 2/kloc-0.

The rest of c is: 1-110 = 9/10, and the beginning of c is:108 ÷ 9/10 =120.

A begins with: 248-72- 12= 164.