Please help me think about this IQ question

3 times. 1'Divide the balls into 4 groups of 3.

2'1 group and 2 groups ratio. If it is not balanced, the irregular balls are in groups 1 or 2. (3 and 4 sets of balls are standard) can choose any group and the 3rd group to compare, (assuming that the lighter group --- 1 group, the same is true for the heavier group).

There are two scenarios: a'Balanced, can push anomalous balls in 2 sets, and is heavier than standard balls. Then take any two of the two balls in the two groups, and the balance is the third.

Imbalance is heavy. b'If it is unbalanced, the anomalous ball is in group 1 and is the same as the weight of group 1, press A'The method can be found out.

3'If the ratio of 1 group to 2 groups, if balanced.

If it is known whether the difference between the balls is light or heavy, it is tripled; If you don't know whether the ball is light or heavy, you need 4 times.

4 times, because when the final weighing is determined to be among the three, it must be weighed twice again to determine which one is which, because it is not stated whether it is lighter than all or heavier than all.

a, at least 1 time, the probability of drawing a ball with abnormal weight is 1 12, but this does not mean that you will not be able to draw it the first time.

2 times, the first time to weigh just one of the two is special, the second time to weigh any of the first and any other one, the result is out.

Six at a time. Two times are called twice.

Each of the three times is called one.

A total of four times. So d: 4

If it's the least, it's once.

1> Divide the balls into 4 groups, each group of 3, numbered as: x1, x2, x3, x42> the first selection (assuming the abnormal ball is in the x1 group): x1 and x2 ratio, if the weight of x1 ≠ the weight of x2, you can number the 3 balls in x1 as:

x11, x12, x13, the three balls in x3 are numbered as: x31, x32, x33, take out x11 and x12 respectively to form the z1 group, take out x31 and x32 to form the z2 group, and make the second selection:

If z1=z2, then x13 is an anomaly. At this point, you only need to make 2 picks to find out.

If z1 ≠ z2, you need to make a third selection: take x11 to x31, if x11 = x31, then x12 is an anomaly, otherwise, x11 is an anomaly.

With this method, the anomaly ball can be pulled out at least 2 times, so the answer should be B.

3 times. First, all the balls are evenly divided into two groups (6 in each group) and put on the scale, and there is a ball at the light and heavy end; Then divide them evenly (3 in each group), and the light one has the ball; Finally, take one random from the group with balls, put it aside, and weigh the other two on the scale, if the two are not the same weight, you can find out; Like the kingdom, the rest is that. ★

A black cat and a white cat are walking on the road. Suddenly the white cat fell into the sewer, and it took half a day.

The length of the carriage is x meters, and the locomotive is long y meters.

Then y=x= i.e. y=

x= y=total length meters.

Let the carriage be x, then the locomotive length is = the length of the guard (m) +, since the length of the carriage is equal to the length of the guard plus the length of the locomotive, then x=, it is concluded that x is the length of the locomotive is equal to the meter.

Locomotive + Cabon + Carriage = Meters.

This is a primary school math problem, you can do without an equation, and that x can be used directly"Carriages"Denote.

Solution: According to the meaning of the topic, it can be obtained:

l locomotive = l cabooze + 1 2l carriage.

2L locomotive = 2L guard car + L car.

l carriage = l cabooze + l locomotive.

2L locomotive = 2L caboox + L cabooze + L locomotive.

l locomotive = 3l cabooze =

Carriage lCarriage=

l train =