Primary school math problems, good plus points, primary school math urgent, there are extra points

80 pcs. Here's the answer:

1) Suppose there are x glass balls in the original bag.

After taking out the 3 4, there are still x 4 glass balls in the bag.

Put back 10 more and there are x 4 + 10 glass balls in that bag.

According to the inscription: in the bag is taken out 1 2.

So the equation is:

x/4 + 10 = 1/2 * 3x/4x/4 + 10 = 3x/8

It is possible to solve x = 80

2) The solution of the equation without the equation:

Let's take x and put back 10, and there are only (x-10) left, because the total should be x (3 4).

That is, the total number is (4 3)x, so there is a relational formula (4 3)x-(x-10)=1 2x

In this way, it can be calculated that x is 60, and the balls taken out are 60, so there are 80 balls in the bag, hehe.

Let's say there are x balls.

1/4x+10=3/4x*1/2

Solution: x=80

Column equation, there were x glass balls in the bag.

3/4x*1/2=（1-3/4）+10

x=80

Amateur type Local tariff Foreign tariff.

Within the first 100 grams, each weighing 20 grams.

Less than 20 grams will be counted as 20 grams].

The renewal weighs 101 to 2000 grams and weighs 100 grams each.

Less than 100 grams will be counted as 100 grams].

The first is that the waifu tariff is or.

The letter weight is 120 grams, and there are 20 grams beyond 100, which meets the second criterion.

More than 20 grams, although less than 100 grams, but also charged according to 100 grams, so the 20 grams need yuan.

And there's another 100 grams in 100 grams that meet the first criterion.

Every 20 grams of dollars so 100 grams need dollars.

A total letter itself needs meta.

It costs 3+ yuan in total.

The cost of less than 100 grams is yuan

The weight over 100 grams = 120-100 = 20 grams, and the cost of the part exceeding 100 grams is yuan

A total of dollar postage is required.

1>.300 km.

It can be seen from the figure that if the whole process is divided into 5 parts, A has completed the whole process, that is, 5 parts. And at the same time, A has 5 copies, then B has 4 copies, and there is still one part left unattended, this part of the question says: 60 kilometers, so 60 kilometers is 1/5 of the whole journey, 60x5 = 300, 300 is the whole journey.

I've also thought about the second question, if you choose me, I'll tell you.

1) Set the time to x

5x=4x+60

x=60 is 5x=300 km apart.

2) Set Xiao Ming to buy back x cards, the previous game board can be found, 56 times 5 8 equals 35 cards. The equation can be listed by the total number of discs. 35+x = (56+x) multiplied by 2 3x Ask for it yourself

The speed ratio of A and B is 5:4, and the ratio of distance traveled in the same time should also be 5:4. Divide the whole journey into 5 parts, when A has completed the journey and B has walked 4 parts, there is still 60 kilometers left, that is, 1 part, and the whole journey should be 60 5 = 300 kilometers.

In this problem, the number of game discs has increased, but the number of other discs has not changed, i.e. 56 (1-5, 8) = 21. Later, after buying back the game disk, other discs accounted for 1-2 3 = 1 3 of the total, then the later total was 21 1 3 = 63, and 63-56 = 7 was the number of game disks bought.

Game Disc Calculation Process.

35+x=（56+x）*2/3

35+x=56*2/3+(2/3)x

x=56*2/3-35+(2/3)x

x-(2/3)x=56*2/3-35

3/3)x-(2/3)x=56*2/3-351/3)x=56*2/3-35

x=（56*2/3-35）*3x=7

x=4x+60

x = 60 (the speed of A).

s=5x=300km

2. The original game disc.

35+x=2\3（56+x）x=7

675km7 sheets.

Just let people deceive, plus points to the process.

The least common multiple of a, b, and c is 60, the greatest common divisor of a and b is 4, and the greatest common divisor of a short b and c is 3

There must be a multiple of the values of a, b, and c, and in order to make the minimum value of a+b+c, we can obtain:

Not a multiple of 5, b = 12

2.Let c be a multiple of 5, then c=15, a=4, a+b+c=31 is a multiple of 5, then the dust is blind a=20, c=5 at this time, a>b+c, so discord is a.

Therefore, the minimum limb reserve of A+B+C is 31

Because abc is an integer, the greatest common divisor of A,B is 4,b,c The greatest common divisor is 3 So we can know that the factor of b must contain 3 and 4 But to make a+b+c the smallest, then we must take Wang Chun 12

A and b have the greatest common divisor of 4, so a must be a multiple of 4, so you can take 4, 8, 12, 16 ......But to make a+b+c the smallest, then you have to take 4

b and c have the greatest common divisor 3, which can be taken as 3,9, but 3+4 less than 12 cannot form a triangle, so c takes 9

4+9 12 So you can form a triangle, so a+b+c has a minimum value of 25

It's good, don't rob your points.

There are already people, but there are 2 answers, a simple analysis should be 31, 25 that says c is 9, but the common multiple of the three is 60