What should I do with a 5g and a 7g weight, which can only be used 3 times, dividing 140 into 50g an

Yes: Weigh 12g first, at this time there are 12g, 128g 2 parts.

The second time the 128g will be divided equally, at this time there are 3 parts: 64, 64, 12 The last time (key): the left plate of salt + 12 grams of salt + 7 grams of weights, the right plate of 64 grams of salt and 1 weight of 5 grams.

At this time, the total mass of the right disc is 69

Go out to the left plate of 12 grams of salt and 7 grams of weights, the left plate weighed out a pile of salt is just 50g, so that 3 times can be solved.

I can only do it 4 times.

For the first time, it is easy to separate 2 70s.

The second time put the 12g weight aside and grab 6g from 70 to the other side to separate 64 and 76g

64g and then 2 7g is 50g, and I have to use it 4 times.

I hope you can figure it out from my mind!

Note that I didn't use the small readout bar of the balance, maybe there is a way here. Forget to forget sorry!

Step 1: Distribute 140 grams on both sides of the balance to balance it, so that you get 70 grams on one side.

Step 2: Take one side and distribute it on both sides of the scale to balance it, so that you get 35 grams on one side.

Step 3: Use a 5 weights mixed with a 35-gram object, distribute the balance on both sides, and balance it, so that you get 20 grams on one side, mix 20 grams without weights with 70 grams you weighed before, get 90 grams, and the rest is 50 grams.

Use a large scale, put 1 5g on one side, and 5 2g on one side, a total of 15 grams.

5+5x2=15, put 15 grams on the other side and you're good to go.

The working reference group and standard weights are usually divided into kilogram group (1-20kg), gram group (1-50g) and milligram group (1-500mg), and there can also be microgram group or other weight combinations (such as the weight increase group used on the bench scale) as needed. The combination of weights usually has and .

With a large balance, put 5 2G on one side, and 2 2G on the other 1 5G, because the balance is large, the 1G gap is completely negligible.

Or let the 5g weights rust and make them even.

It is also possible to cheat with a balance nut.

1) Weigh out 9g of salt with a weight of 7g+2g, 140g-9g=131g

2) Use 7g + 2g to divide 131g of salt into two parts of 61g and 70g (61 + 2 + 7 = 70).

3) Divide 61g of salt into two parts: 50g and 11g of dandan, with 2g weights and 9g of salt (61-9-2=11, 11+50=61).11g+9g+70g=90g

For this problem, the last remaining part is a loss of 100g, so as to ensure that the number of times to weigh is the least, so as to move forward, and the penultimate time must meet the condition of 100g, 100 = 30 + 30 + 30 + 5 + 5 = 30 + (30 + 5) * 2

That is, 30g of weight and 35g of weight are required, that is, 35g of weight is needed for the penultimate time, which is exactly the weight of the existing weight.

The first time you weigh 35g, you can get 35g of items.

The second time, the item +30g weight = 65g, and the weight of the item taken out this time plus the original one, is exactly 100g

The third time, use the items obtained the second time as a weight, you can get a batch of items, the weight of which is 100g

The rest of the items weigh 100g, so there is no need to divide them. It takes three times in total.

With a balance with 7 2 gram and 1 5 gram weight, can you weigh 15 grams of sugar at a time with all the weights?

Hello, the main test of the balance is that the weight of the left and right sides is equal, it will maintain a horizontal balance, so if there are 7 two grams and one 5 grams want to hold out 15 grams of sugar, of course, it is also possible, that is, the left and right sides are balanced, then we can use a 5 grams on the left. Add 5 two-gram weights to add up to 15 grams, then put 15 grams of sugar on the right, and the left and right sides will be balanced.

According to the title, it can be obtained:

5+30=35（g）；Hail closed.

100 35 = 2 (times)....30（g）；

The remaining 30g can be weighed again with a 30g weight;

Therefore, if you want to weigh 100g of sugar from hail potatoes, first weigh 35g twice, and then weigh 30g once from the source, that is, at least 3

Therefore, b

(1) Weigh 9g with a weight of 7g + 2g, 140g-9g = 131g (2) Use 7g + 2g weight to divide 131g into two parts: 61g, 70g (61 + 2 + 7 = 70).

3) Divide 61g into two parts, 50g and 11g, with 2g weight and 9g (61-9-2=11, 11+50=61).

11g+9g+70g=90g

Teacher Xu of Hebei Huatu will answer this question for you. This question is similar to the weighing problem in Hebei Province in 2011, and it is completely possible to do this problem with the previous problem-solving thinking. (For details, please log in to Hebei Huatu**: Try to get closer to the target every step you take.)

The first time: using the balance, divide 30g weight and 300g monosodium glutamate into two parts, that is, 330 2 = 165, you can get a 165g monosodium glutamate, a 135g monosodium glutamate + 30g weight.

Because the average is divided into three parts, each serving is 65 different from 100 and 100The next target is to weigh 35g and 65g.

The second time: with 5g+30g weights, put on one side of the scale, put MSG on the other side, you can weigh 35g from 135g of MSG, exactly 100g left.

The third time: with 30g of weights and the second time to weigh 35g of monosodium glutamate, put on one side of the scale, the other side can weigh 65g from 165g of monosodium glutamate, leaving 100g.

Finally, put the second 35g and the third weighing 65g together to get 100g of monosodium glutamate, and complete the task of this weighing, that is, three times to achieve the purpose.

Through the above explanation, I hope to help you understand how to solve the weighing problem in the overall planning problem. Hebei Huatu (in the future, we will continue to solve problems for everyone, I wish you all a "public"!)

For the first time, put 30 grams of weights on one side of the scale, add 300 grams of powder to balance both sides, and the weight of powder on the side with weights is (300+30) 2-30=135 grams; The second time, put the 3g and 5g weights on one side, and take 35g from the 135g on the other side to balance the balance on both sides; 135-35 = 100 grams; The third time you weigh 100g of the remaining 200g with 100g. Is the answer wrong 3 times? Aren't people asking the least?

Weigh with two weights at a time: 30 5 35 (grams).

Weigh 3 times in a row: 35 * 3 105 (g).

Weigh (withhold) 5 grams with another 5 gram weight to get one serving.

The second part is weighed in the same way, and the remaining third part does not need to be weighed again.

Therefore, a total of 3 1 3 1 8 (times) was weighed

Summary. 2.Put 200 weights on the right side, if the balance is balanced, then the big wood is 190g, four have a balance and two weights, one weight 200 g, one weight 10g, how.

Hello classmates, can you add a question to the teacher.

I'll send it to you, you help me write the answer.

May I? If you can, send it first.

What grade are we<>

Fifth grade. Question 4.

Place the rice and 10 weights on the left side of the scale.

Put 200 weights on the right side, and if the balance is balanced, the big wood is 190g.

You write it out and send it to me.

Answer Write directly on paper:1Place the rice and 10 weights on the left side of the scale.

2.Put 200 weights on the right side, if the balance is balanced, the big wood is 190g, and the above is the standard answer.

A balance is an equal-arm lever and also a weighing instrument, an instrument used to measure the mass of an object. It is made according to the principle of lever, at both ends of the lever there is a small plate, one end of the weight, the other end of the object to be weighed, the lever ** is equipped with a pointer, when the two ends are balanced, the mass of both ends is equal. <>

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Modern balances include ordinary balances, analytical balances, constant analytical balances, micro analytical balances, semi-micro analytical balances, <>