Class 6 (1) packed 9 bags in one afternoon, and three fifths of the total number were left. Class 6 (2) continued to pack, and class 62 copper needed to pack 13 bags, and then two quilts were left. Question: 1. How much of the quilts were donated in a sack? 2. How many quilts were donated in our school

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1）
（1-3/5）÷9=2/45
2）
A total of 9 + 13 = 22 bags were made, accounting for 22 × 2 / 45 = 44 / 45 of the total number of quilts
2 ÷ (1-44 / 45) = 90
1. A sack of quilts is 2 / 45 of all donated quilts
2. Our school donated 90 quilts

There are 14 and 4 watermelons sold in the first time, and the remaining 12 and 2 watermelons sold in the second time. How many watermelons are there in this pile?

A: there are 16 watermelons in this pile

Two cases of oranges are sold, one case is 60kg, accounting for 6 / 13 of the total number sold, and oranges account for 1 / 5 of the total number sold. Q: how many kg of oranges are sold?

Oranges sold: 60 × 2 = 120 kg
Then the total number is: 120 △ 6 / 13 = 260 kg
Orange: 260 × 1 / 5 = 52kg

Two cars a and B set out from two places 360 kilometers apart at the same time, and after 4 hours, they met on the way. It is known that the speed of car a and car B is 5:4,
How many kilometers does car a travel per hour? How many kilometers does car B travel per hour?
The sum of subtraction and difference is 100, and the ratio of difference and subtraction is 1:4. What are the subtraction, subtraction and difference?

Speed sum: 360 △ 4 = 90 (km)
Speed of car a: 90 * 5 / (5 + 4) = 50 (km)
Speed of vehicle B: 90 * 4 / (5 + 4) = 40 (km) or 90-50 = 40 (km)
Subtracted = difference + subtracted = 100 △ 2 = 50
Poor: 50 * 1 (1 + 4) = 10
Subtraction: 50 * 4 / (1 + 4) = 40 or 50-10 = 40

If you move a decimal point one place to the right, you will get a new number 25.2 more than the original number. The original decimal point is ()

Let it be X. when the decimal point moves one place to the right, the value becomes 10 times of the original value
10*X-X=9*X=25.2
So x = 25.2 / 9 = 2.8

A passenger car and a freight car leave from a and B at the same time. The speed of the freight car is 45 times that of the passenger car. After 14 kilometers of the whole journey, the freight car runs 28 kilometers to meet the passenger car. How many kilometers are there between a and B?

A: the distance between a and B is 144 km

In triangle ABC, we prove that a = bcosc + ccosb, B = ccosa + acosc, C = acosb + bcosa

bcosC+ccosB=2RsinBcosC+2RsinCcosB
=2Rsin(B+C)
=2RsinA
=a
The rest is the same

The length of train a and train B are 144m and 180m, and train a runs 4m more per second than train B. the two trains are facing each other, and it takes 9s from meeting to staggering. What's the speed of the two trains?

Suppose car B travels XM per second, then car a Travels (x + 4) m per second. According to the meaning of the question, we get: 9 (x + X + 4) = 144 + 180, sort out: 2x = 32, solve: x = 16, then car a travels 20m per second, car B travels 16m per second

There are five numbers arranged from small to large. The average number is 39. The average number of the first three numbers is 35. The average number of the last three numbers is 40. What is the third number

The third number is 35 × 3 + 40 × 3-39 × 5 = 30

When they set out, their speed ratio was 3:2. After their first meeting, a's speed increased by 20%, and B's speed increased by 30%. In this way, when a arrives at B, B is still 14 kilometers away from a, so how many kilometers is the distance between a and B?

Suppose the distance between a and B is SKM, and the speed of a and B is 3x and 2x respectively. When a and B meet for the first time, the distance they take is 3s5 = 0.6skm and 2s5 = 0.4skm respectively. According to the time series equation of a to B after meeting: 0.4s3x (1 + 20%) = 0.6s − 142x (1 + 30%), s = 45km. A: the distance between a and B is 45km