(2) Yes.

(3) It is possible to weigh it out at one time. Because if you put four baskets at each end of the balance, the balance is just balanced, and the rest of the baskets are eaten by the little squirrel, then you can find out the basket of pine cones that the little squirrel has only eaten once; If it is unbalanced, it will take two or three times to weigh it out, so this method is the best.

2.

Divide the 15 box into three powders evenly, and weigh it at least three times to ensure that you can find the lighter box of biscuits.

3.

7+4/5=43/35 5/ 12+5/8=25/24 7/9+7/6= 103/63 1-2/7-3/7=2/7

1/4+ 1/6+3/4=7/6 1/2- 1/6= 1/3

4.

Analysis: The age difference between Dad and Xiaoming is the same, and it is the same now and three years later.

Answer: Suppose Xiaoming is X years old, then his father is (X+24) years old.

X+24+x=34

X=5

So Xiao Ming is 5 years old and his father is 5+24=29 (years old).

5.

Balance →→→→ 4 (1,1,2) 3 times.

Divide the 12 bag of sugar into 3 parts, with 4 bags for each part →→→→→→→ Put 4 bags on each side of the balance-

Unbalance →→→→ 4 (1,1,2) 3 times.

6.

Put a bag of sugar on both sides of the balance for the first time. If the balance is balanced, the remaining bags are defective. If the balance is unbalanced, one of the two bags must be defective. You can take off the light (or heavy) bag and put the rest on the balance. If the balance is balanced, the light (or heavy) package is defective. If the balance is unbalanced, the heavy (or light) package is defective.

7.

Analysis: There were 6 people who didn't participate in both groups. Subtract the six people who didn't participate in the total, and the rest is the number of people who participated in the extracurricular group.

The number of participants in the two extracurricular groups is

12+ 10=22 (person)

The number of people actually participating in extracurricular groups is 19, so there are two groups 19.

25-6= 19 (person)

12+ 10-9=3 (person)

Do you know that?/You know what? Do you know that?/You know what?

(1) In order to ensure that the defective products can be called out six times, the number of tested objects may be 244 ~ 729.

(2) From the table, it can be found that as long as the number of tested objects is between 3n- 1+ 1 ~ 3n, it only needs to be tested for n times at most, so that the defective products can be found. In the calculation process, we should know that n in 3n is the product of several 3s: for example, three 6 = 3× 3× 3× 3× 3× 3× 3.