Elementary school life math problems for help. Elementary school students asking for help with math

According to the regulations of the "Notice to Passengers" of the railway, the weight of passengers' carry-on items is free, 20 kg for adults and 10 kg for children. Carry-on items should be of sufficient length and volume to fit in the luggage compartment or under the seat.

Explosive, flammable dangerous goods, items that disturb public health and pollute vehicles are not allowed to be brought into the vehicle.

Carry-on baggage rules.

Baggage The free baggage allowance per person for full-ticket or half-ticket passengers on domestic routes is 20 kg for first-class and 15 kg for ordinary class if the ticket is purchased according to the first middle fare; If the ticket is purchased according to the second fare, the first class is 30 kg, and the ordinary class is 20 kg. Baggage exceeding the free allowance shall be paid according to the prescribed baggage tariff and can only be transported out of the same aircraft with passengers when there is free tonnage on the flight.

Carry-on items.

The weight is limited to 5 kg, and the volume of each piece shall not exceed 20*30*50 cm. Passengers on international flights are allowed to carry the following carry-on items free of charge per person: handbag, coat, umbrella, walking stick, small number of reading materials, baby food, baby bassinet and foldable wheeled seat for sick people.

By train, it is 20 kg for adults and 10 kg for children. The aircraft is 30 kg for first class and 20 kg for regular class.

20 kg for adults, 10 kg for children, 30 kg for class, 20 kg for ordinary class.

The two line segments A and B are 180 cm long.

If the line segment A is reduced by 20 cm, the total length of the two line segments A and B is reduced by 20 cm to 160 cm.

The B line segment is increased by 35 cm, and the total length of the two line segments A and B is increased by 35 cm, which is 160 + 35 = 195 cm.

Line A is 4 times that of line B.

That is, in 195 cm, B accounts for 1 part, A accounts for 4 parts, a total of 5 parts.

So after the change, A = 195 4 5 = 156 cm.

B = 195 1 5 = 39 cm.

The original length of the boot spike of A is 156+20=176 cm, and the original length of B is 39-35=4 cm.

Xiao Ning and Xiao Jie just now are the ones you put it, haha potato book judgment, you can really do it! In the same way as just now, change the 180 cm subtraction to 20 = 160 inside, plus 35 cm = 195 cm, which is the total length of the two of them in proportion to the present. Then divide it into 5 parts, divide 195 by 5 = 39, that is, the first line posture section is now 39 4 = 156 cm, and the B line is now 39 cm, and then make up just now, that is, A 156 + 20 = 176 cm.

B 39-35 cm = 4 cm finished! Said the next call, haha,!

Your child is 3 years old, and there should be no problem with the multiplication of Zen manuscripts, A + B = 180;

A-20=4 (B-burn tease +35);

A: 176, B: 4

Xiao Ning and Xiao Jie have a total of 160 + 122 = 282 stamps If all the stamps are divided into three parts, Xiao Jie's two copies, then Xiao Ning has only one copy, so Xiao Jie is twice as much as Xiao Ning.

1. Divided into three parts, the number of each copy = 282 3 = 942, Xiao Ning can only have one copy of 94 in the end, and he now has 160 pieces, so he should give Xiao Jie 160-94 = 66 sheets.

I also thought about it for a long time, and told the child that if Xiao Jie is twice as big as Xiao Ning, it means that after the two people add it together, it will be divided into three parts, Xiao Jie is 2 points, and Xiao Ning is 1 point.

So it's (160+122) 3=94 94*2=188

There are a total of 160 + 122 = 282 (sheets).

Divided into three parts, each is: 282 3 = 94 (sheets).

After Xiao Ning gave it to Xiao Jie, Xiao Ning should have 94 sheets.

160-94 = 66 (sheets).

160 + 122 equals 282 282 divided by 3 equals 94 160-94 equals 66

by"Xiao Jie's number of sheets is 2 times that of Xiao Ning"This sentence knows that after the delivery, the total number of philatelic collections by Xiao Ning and Xiao Jie should be three times that of Xiao Ning. Old.

160+122)/3=94...After Xiao Ning gave it to Xiao Jie, Xiao Ning had 94 cards, 160-94=66, so Xiao Ning gave Xiao Jie 66 cards.

Xiao Ning's later Zhang number.

94 (sheets) 160-94 = 66 (sheets).

Xiaojie is 2 times that of Xiao Ning, that is to say, the total number of stamps is 3 times that of Xiao Ning (160 + 122) 3 = 282 3 = 94 So Xiao Ning finally has 94 sheets.

Xiaojie has 94 2 = 188 sheets.

In this way, Xiao Ning gave Xiao Jie 160-94 = 66 sheets.

The reverse is 188-122 = 66 sheets, which is fine).

Add Xiao Ning and Xiao Jie's stamps together, divide them into three parts, Xiao Jie takes two points, Xiao Ning takes one point, so that the topic is satisfied, 160 + 122 = 282, 282 divided by 3 = 94, so Xiao Jie takes 94 2 = 188, Xiao Ning takes 94, so Xiao Ning gives Xiao Jie 160-94 = 66, it's over! Adopt it! Next time, bring a little bit of it!

I learned to solve equations in the third grade, and if there are no equations, I feel that this problem is difficult to understand, and it is very difficult to explain.

Equation: (160-x)*2=122+x

320-2x=122+x

3x=298

x=66 non-equation: (160*2-122) 3

Let the blank part be x meters long. then its area is x*(60 - 20) = 40x square meters.

The area of the entire rectangle is (30 + x) * 60 = (1800 + 60x) square meters.

So the shaded area is (1800 + 60x)- 40x = (1800 + 20x) square meters.

It can be seen that the area of the shaded part is related to the length of the blank part, that is, it is a function of long x, and the analytic formula: s = 1800 + 20x

In view of the fact that this is a primary school math problem, the aspect ratio of the rectangle of the blank part is consistent with that of the large rectangle, that is, 3:2 (obtained by the two conditions of 30 meters and 20 meters), so x = 60 meters, and the final answer is 3000 square meters.

Shaded part area = large rectangle - small rectangle = 90 60-60 40 = 3000 square meters.

If the figure is in the same proportion, then there is.

60×90-40×60=5400-2400=3000m²

Otherwise, there is no solution to the problem.

Since the length of the blank rectangle cannot be determined, A: I can't.

A: I can't do it. (Because the length of the white rectangle is not known).

Actually, your title is "(123 +121 * 372) (372 * 122-249)", right? ？

If so:

Denominator = 372x122-249

372x（121 +1）-249

372x121 +372-249

121x372 +123 = molecule.

Original = 1

One-third + 97/198 = 66/198 + 97/198 = 163/198 adoption.

The one upstairs is called the common denominator, which is to multiply the numerator and denominator by one at the same time, and you can make the denominator 198 and fill it up.

One hundred and sixty-three in one hundred and ninety-eight.

For the first encounter, the sum of the distance traveled by the two cars is equal to the distance between the two places AB;

In the second encounter, the sum of the distances traveled by the two cars is equal to three times the distance between the two places in AB.

Therefore, on the second encounter, the distance traveled by the two cars is also equal to three times the distance traveled at the time of the first encounter.

At the time of the first encounter, car A traveled 32 kilometers;

On the second encounter, car A traveled twice as long as 64 kilometers.

Therefore, the whole journey is twice reduced by 64 kilometers and is equal to 96 kilometers, and the whole journey is 80 kilometers.

The ratio of sugar to the number of sugar in A and B is 4:1, and after A and B eat 20 each, the ratio of the number of sugars in Jiayi and the number of sugars has become 6:1, that is, you can shoot your original cousin is x, so the fake cousin is four x, so there are four x minus 20 is equal to six brackets x minus twenty brackets, saying that x is equal to 50, and the nail art originally had 200 grams of sugar and 50 grams of sugar.

Let's say B now has x sugars and A has 6x.

6x+20)/(x+20)=4/1

x = 30 B now has 30 sugars and A has 180.

Let B have candy y, and A has sugar x;

According to the first condition: x=4y

According to the second and third conditions: (x-20) (y-20)=6, the two equations are solved synchronously: x=200, y=50.

So A now has 180 sugars, and B now has 30 sugars.