My mother bought 6 boxes of milk and 5 packets of biscuits in the supermarket, which cost 27 yuan in total. The money of 3 boxes of milk and 2 packets of biscuits is equal. How much is each box of milk and 1 packet of biscuits? How to calculate,

My mother bought 6 boxes of milk and 5 packets of biscuits in the supermarket, which cost 27 yuan in total. The money of 3 boxes of milk and 2 packets of biscuits is equal. How much is each box of milk and 1 packet of biscuits? How to calculate,

Three cartons of milk is equal to two packs of biscuits
Six cartons of milk is equal to four packs of biscuits
27 yuan is equivalent to buying 4 + 5 = 9 packets of biscuits
Money for a packet of biscuits
27 △ 9 = 3 (yuan)
How much does it cost to buy milk
27-5 × 3 = 12 yuan
Money for a pack of milk
12 △ 6 = 2 (yuan)

There are biscuits,
How many people? How many biscuits?
All right, speak again~

With X biscuits, there are two equal ways to express the number of people (x + 25) / 9 and (x + 7) / 6
(x+25)/9=(x+7)/6
2x+50=3x+21
x=29
(x+7)/6=6
So there are 29 biscuits for six people

There are 54 students in a class, and at least 2 students have their birthdays in the same week?

This is called the drawer principle. There are 52 weeks in a year. 54 people put 52 squares, and 2 of them must be in the same square

There are 54 students in class 6 (2). How many of them have their birthdays in the same week?

365/7=52.1
That means at least two people have birthdays in the same week

There are 49 students in the class, and at least 5 students have their birthdays in the same month. Why

000000161,
Twelve months a year
49 △ 12 = 4 (Group), 1 (person)
Four groups indicate that at least four students have the same birthday, and there is one person left behind. This person must also have the birthday in one month of 12 months, so there should be at most 4 + 1 = 5 students who have the same birthday in one month

There are 13 students in group 1 and group 2 of class 6 (1). How many of them have their birthdays in the same month?
Xiao Ming put five pencils into four pencil boxes. What conclusions would you draw?
To calculate the formula Oh, thank you!

First question: at least two people's birthdays are in the same month
Second question: the conclusion is that at least two pencils should be put in the same pencil case
This is a kind of drawer problem. There is no formula

There are 13 students in group 1 and group 2 of class 61. At least several of them have their birthdays in the same month,

At least 0 students, birthday in the same month. (this problem is a trap problem, 12 months a year, and now 13 students, people mistakenly think that at least one month. The question is at least, if the 13 students in the same month, the rest of the month is 0, so at least 0 students). Is this answer satisfactory?

There are 37 students in a certain class, at least 30 students______ Two students have their birthdays in the same month

Establishing drawers: 12 months in a year are regarded as 12 drawers, 37 △ 12 = 3 1, 3 + 1 = 4, answer: at least 4 students have their birthdays in the same month

There are 37 boys in class 61. At least () students have their birthdays in the same month

4. (drawer principle)

There are 37 students in a certain class, at least 30 students______ Two students have their birthdays in the same month

Establishing drawers: 12 months in a year are regarded as 12 drawers, 37 △ 12 = 3 1, 3 + 1 = 4, answer: at least 4 students have their birthdays in the same month