Past Simple(simple past tense)- one of the most common tenses in the English language. By using Past Simple we can convey events that happened in the past, and in some cases, events in the present. Therefore, at any level of proficiency English you will learn something new about Past Simple.
How is the past simple formed?
Statement
In order to form Past Simple, we add the ending to the verb - ed, if the verb is correct ( work - worked, learn - learned, ask - asked). If the verb is irregular, then we look for the form for Past Simple in the second column tables of irregular verbs. Most likely, at first it will be difficult for you to immediately determine which verb is in front of you. You will find the irregular verb in the table. If it's not there, it's correct. Negation
In a negative sentence in Past Simple auxiliary verb appears did(second form of irregular verb do) and particle not. In an English sentence, only one verb can be in the past tense, so as soon as did, the main verb takes the form of the infinitive without a particle to (go, look, feel). In colloquial speech did And not combine to form a shortened form didn't:

• He didn't tell.
• We didn't discuss.

Question
To ask a question in Past Simple, we put it first did, followed by the subject, then the main verb. Verbto beVPast Simple
As you already know, to be- this is an irregular verb, so you need to remember the past tense form. But in the past tense it does not have one form, like all other verbs, but two: was(for singular nouns and pronouns) and were(for plural nouns and pronouns). In denial was (were) combines with not and forms the shortened form:

• I wasn't at work.
• They weren't happy.

Using the Past Simple
Let us immediately note that the main functions Past Simple And Present Simple match. We only need to transfer the action from the present to the past.

1. Past Simple shows a fact in the past or a single action that ended in the past.

Words that indicate the duration of action are often used here: yesterday(yesterday), two weeks ago(two weeks ago), the other day(the other day), a long time ago(for a long time), last month(last month) in 2010 (in 2010), on Monday(on Monday) during my holidays(during the holidays), etc. These words must necessarily denote a completed period of time.
I saw him yesterday. - I saw his yesterday. (single action in the past, yesterday already ended)
The Titanic sank in 1912. - Titanic sank 1912. (fact)
He went to Italy last month. - He traveled to Italy last month. (single action in the past)
Words indicating the duration of the action may not be used in this feature.
She spoke to him in a low voice. - She's quiet spoke with him. (single action in the past)
Aivazovsky painted"The Ninth Wave". - Aivazovsky wrote painting "The Ninth Wave". (fact)

1. Also Past Simple used to describe a condition in the past.

They were friends many years ago. - They were friends many years ago. (they are not friends now)
That museum had a great collection of paintings. - In that museum was huge collection of paintings. (currently the museum does not have a huge collection)

1. We use Past Simple to talk about old habits and repetitive actions. These actions have occurred many times in the past, but are no longer being done. Such sentences may contain adverbs often(often), sometimes(Sometimes), always(always) etc.

We took evening courses two years ago. - We walked for evening courses 2 years ago. (currently we do not take evening courses)
He always bought newspapers on Sunday. - He always bought newspapers on Sundays. (he doesn't do that now)

1. • We can also use used to to talk about old habits.
2. We use Past Simple when we tell a story or list several events that happened one after another in the past.

She came in, sat at the table, and started writing. - She came in, villages at the table and started write.
He entered the cafe, he ordered a cup of tea and a piece of cake. - He entered in a cafe, ordered a cup of tea and a piece of pie.
Where else is Past Simple found?

1. We use Past Simple to tell the details of some news or some event that happened in our lives. We report the news itself in time Present Perfect. We can use other past tenses to give details, but Past Simple is used in these cases more often than others.

I've hurt my leg. I fell off a ladder when I was repairing the roof. My telephone rank unexpectedly. - I hurt my leg. I fell from the stairs when I was fixing the roof, because suddenly rang telephone.
I've got this job. It was a hard and exhausting interview, but it turned out that I was a perfect applicant. - I got this job. This was difficult, exhausting interview, but it turned out that I am the ideal candidate.

1. Past Simple used in subordinate tenses after conjunctions after(after), before(before) when(When), until(not yet) as soon as(as soon as). In such a sentence Past Simple shows a completed action in the past.

As soon as she graduated from the university, she found a suitable job. - Howonly she graduated university, she found a suitable job.
He was astonished when I told him the news. - He was amazed When I reported him this news.

Calculating the amount of information

For completing the task correctly you will receive 1 point. It takes approximately 3 minutes.

To complete task 13 in computer science you need to know:

• To find the information volume of a message (text) I, you need to multiply the number of symbols (counts) N by the number of bits per symbol (count) K :
• Two lines of text cannot occupy 100 KB of memory
• Power of the alphabet M– is the number of characters in this alphabet
• If the alphabet has power M, then the number of all possible “words” (character chains) of length N(ignoring the meaning) is equal to Q=M N ; for binary coding (alphabet power M– 2 characters) we get the well-known formula: Q=2 N

Tasks for training

When registering in computer system Each user is given a password consisting of 10 characters. Capital letters of the Latin alphabet are used as symbols, i.e. 26 different symbols. The database allocates the same and minimum possible integer number of bytes to store each password. In this case, character-by-character encoding of passwords is used; all characters are encoded with the same and minimum possible number of bits. Determine the amount of memory (in bytes) required to store data about 50 users. In your answer, write down only an integer - the number of bytes.


Solution

1. When registering on the site, each user is given an identifier consisting of 7 characters. The characters used are uppercase and lowercase letters Latin alphabet, i.e. 26 different characters. The database allocates the same and minimum possible integer number of bytes to store each identifier. In this case, character-by-character coding of identifiers is used; all characters are encoded with the same and minimum possible number of bits. Determine the amount of memory (in bytes) required to store data about 45 users. In your answer, write down only an integer - the number of bytes.

   
   Solution

2. When registering on the site, each user is given an identifier consisting of 15 characters. Uppercase and lowercase letters of the Latin alphabet are used as symbols, i.e. 26 different symbols. The database allocates the same and minimum possible integer number of bytes to store each identifier. In this case, character-by-character coding of identifiers is used; all characters are encoded with the same and minimum possible number of bits. Determine the amount of memory (in KB) required to store data about 256 users. In your answer, write down only an integer - the number of KB.

   
   Solution

3. When registering on the site, each user is given an identifier consisting of 13 characters and a password consisting of 12 characters. The symbols used are uppercase and lowercase letters of the Latin alphabet, i.e. 26 different symbols and decimal digits. The database allocates the same and minimum possible integer number of bytes to store each ID and password. In this case, all characters are encoded with the same and minimum possible number of bits. Determine the amount of memory (in bytes) required to store data about 40 users. In your answer, write down only an integer - the number of bytes.

   
   Solution

4. When registering on the site, each user is given an identifier consisting of 6 characters and a password consisting of 10 characters. Capital letters of the Latin alphabet are used as symbols, i.e. 26 different symbols. The database allocates the same and minimum possible integer number of bytes to store each ID and password. In this case, all characters are encoded with the same and minimum possible number of bits. Determine the amount of memory (in bytes) required to store data about 25 users. In your answer, write down only an integer - the number of bytes.

   
   Solution

5. When registering on the site, each user is given an identifier consisting of 18 characters and a password consisting of 23 characters. Capital letters of the Latin alphabet are used as symbols for the identifier, i.e. 26 different characters, and decimal digits are used for the password. The database allocates the same and minimum possible integer number of bytes to store each ID and password. In this case, all characters are encoded with the same and minimum possible number of bits. Determine the amount of memory (in bytes) required to store data about 17 users. In your answer, write down only an integer - the number of bytes.

13th task: “Amount of information”
Difficulty level - increased,
Maximum score - 1,
Approximate execution time is 3 minutes.

Unified State Examination in Informatics 2017 task 13 FIPI option 1 (Krylov S.S., Churkina T.E.):

7 33 -character alphabet. The database allocates the same and smallest possible integer to store information about each user byte bit. In addition to your own password, additional information is stored in the system for each user, for which an integer number of bytes are allocated; this number is the same for all users.

To store information about 60 users required 900 byte.

How many bytes are allocated for storage additional information about one user?
In response, write down only an integer - the number of bytes.

Answer: 9

✍ Show solution:

• First, let's decide on a password. According to the formula Q = M N we get:

33 = 2 N -> N = 6 bits per character

The password consists of 7 characters:

-> 7*6 =42 bit just for the password

Since all user data is stored in bytes, let’s take the nearest number larger 42 and multiple 8 :

48/8 = 6 42 bits ~ 6 bytes

Now let’s find how many bytes are allocated to store information about one user:

900 bytes / 60 (users) = 15 bytes per user

Let's get the amount of memory to store additional information:

15 bytes (to store all information) - 6 bytes (to store the password) = 9 bytes for additional information

Unified State Examination 2017 collection by D.M. Ushakova “10 training options...” option 1:

The cable network is voting on which of four films they would like to watch that night. Cable network enjoy 2000 Human. Participated in the voting 1200 Human.
What is the amount of information ( in bytes), recorded automated system voting?

Answer: 300

✍ Show solution:

• Since the four movie numbers are stored in the computer system, we can find the number of bits needed to store the movie number:

Q = 2 k -> 4 = 2 k -> k = 2 bat

Since all 1200 people will vote for one of the films, the same amount of memory must be allocated for each vote (i.e. 2 bits).

Let's find the number of bits required to store all 1200 votes:

1200 * 2 = 2400 bits = 2400/8 bytes = 300 byte

Unified State Examination 2017 collection by D.M. Ushakova “10 training options...” option 6:

When registering in a computer system, each user is given a password consisting of 15 characters and containing only characters from 12 - character set A, B, C, D, E, F, G, H, I, K, L, M, N. The database allocates the same and smallest possible integer to store information about each user byte. In this case, character-by-character encoding of passwords is used, all characters are encoded with the same and the minimum possible number bit. In addition to the password itself, additional information is stored in the system for each user, for which 12 bytes per user.

Determine the amount of memory ( in bytes), necessary for storing information about 30 users.
In your answer, write down only an integer - the number of bytes.

Answer: 600

✍ Show solution:

Unified State Examination 2017 collection by D.M. Ushakova “10 training options...” option 10:

Taking a rehearsal exam at school 105 Human. Each of them is allocated special number, identifying him in automatic system checking answers. When registering a participant to record his number, the system uses the minimum possible number of bit, the same for each participant.

How much information is there? in bits, recorded by the device after registration 60 participants?

Answer: 420

✍ Show solution:

13 task. Demo version of the Unified State Exam 2018 computer science:

10 characters. Capital letters of the Latin alphabet are used as symbols, i.e. 26 various symbols. In the database, each password is stored in the same and smallest possible integer byte. In this case, character-by-character encoding of passwords is used, all characters are encoded with the same and the minimum possible number bit.

Determine the amount of memory ( in bytes), necessary for storing data about 50 users.
In your answer, write down only an integer - the number of bytes.

Answer: 350

✍ Show solution:

• The basic formula for solving this problem is:

Where Q— the number of character variants that can be encoded using N bit.

• To find the number of bits required to store one password, you first need to find the number of bits required to store 1 character in the password. Using the formula we get:

26 = 2 N -> N~5 bits

The password consists of 10 characters. This means you need to allocate a bit for the password:

10 * 5 = 50 bits total per password

Since password information is stored in bytes, we translate:

50 bits / 8 ~ 7 bytes (take the nearest number greater than 50 and a multiple of 8: 57/8 = 7)

Now let's find how many bytes are allocated to store information about 50 users:

7 bytes * 50 (users) = 350 byte

Solution 13 Unified State Exam assignments in computer science (diagnostic version of the examination paper, Unified State Examination simulator 2018, S.S. Krylov, D.M. Ushakov):

In some countries, the license plate consists of 7 characters. Each character can be one of 18 different letters or decimal number.

Each such number in computer program is written in the smallest possible and identical integer quantity byte, in this case character-by-character encoding is used and each character is encoded with the same and minimum possible number bit.

Determine the amount of memory in bytes, allocated by this program for recording 50 numbers.

Answer: 250

✍ Show solution:

• Since the number can use either one letter from 18 , or one digit from 10 , then just one character in the number can be used one of 28 characters:

18 + 10 = 28

Let's determine how many bits are needed to store one character in the number; for this we use the formula N=2i:

28 = 2 i => i = 5

Since the total number of characters in the number is 7 , then we get the required number of bits to store one number:

I = 7 * 5 = 35 bits

Since the same amount is allocated for storing the number byte, then convert it to bytes:

35 / 8 ~ 5 bytes

The problem asks how much memory is needed to store 50 numbers. We find:

I = 50 * 5 = 250 bytes for storing 50 numbers

Solution 13 of the Unified State Examination task in computer science (control version No. 1 of the exam paper, Simulator 2018, S.S. Krylov, D.M. Ushakov):

Passing the rehearsal exam 9 flows by 100 a person in everyone. Each of them is assigned a special code consisting of a thread number and a number in the stream. When encoding these participant numbers, the verification system uses the minimum possible number of bit, the same for each participant, separately for the thread number and the number in the stream. In this case, the minimum possible and identically integer number is used to write the code bytes.
What is the amount of information in bytes recorded by the device after registration 80 participants?
Please indicate only the number in your answer.

Answer: 160

✍ Show solution:

• The code consists of two components: 1. stream number (in bits) and 2. sequence number (in bits). Let's find the number of bits required to store them:

1. N = 2 i -> 9 = 2 i -> i = 4 bits (2 3 100 = 2 i -> i = 7 bits (2 6

Total we get 4 + 7 = 11 bits for one code. But according to the condition, an integer number of bytes are allocated to store the code. So let’s convert the resulting result into bytes:

11/ 8 ~ 2 bytes (one byte is not enough, 8

Since we need to obtain a volume of information after registration 80 participants, then we calculate:

2 * 80 = 160 byte

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 4):

Message volume – 7.5 KB. It is known that this message contains 7680 characters. What is the power of the alphabet?

Answer: 256

✍ Show solution:

• Let's use the formula:

I - message volume N - number of characters K - number of bits per character

In our case N=7680 characters highlighted I = 7.5 KB of memory. Let's find the number of bits required to store one character (first converting KB to bits):

I = 7.5 KB = 7.5 * 2 13 bits

\[ K = \frac (7.5 * 2^(13))(7680) = \frac (7.5 * 2^(13))(15 * 2^9) = \frac (7.5 * 16 )(15) = 8\]

8 bits per character allow you to encode:

2 8 = 256 various characters
(according to the formula Q = 2 N)

256 characters - that's power

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 6):

The power of the alphabet is 256 . How many KB of memory will be required to save 160 pages of text, containing on average 192 characters on every page?

Answer: 30

✍ Show solution:

• Let's find the total number of characters on all pages (for convenience, we will use powers of two):

160 * 192 = 15 * 2 11

According to the formula Q = 2n let's find the number of bits required to store one character (in our case Q=256):

256 = 2 n -> n = 8 bits per character

Let's use the formula I=N*K and find the required volume:

\[ I = (15 * 2^(11)) * 2^3 bits = \frac (15 * 2^(14))(2^(13)) KB = 30 KB \]

I = 30 KB

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 3):

The two texts contain the same number of characters. The first text is composed in the alphabet with capacity 16 characters, and the second text is in the alphabet from 256 characters.
How many times more information is in the second text than in the first?

Answer: 2

✍ Show solution:

• Formula needed Q = 2n
• Let's calculate the required number of bits to store one character for both texts:

1. 16 = 2 n -> n = 4 2. 256 = 2 n -> n = 8

Let's find how many times more information (volume) is in the second text:

8 / 4 = 2

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 17):

The database stores records containing information about dates. Each record contains three fields: year (date from 1 to 2100), month number (day from 1 to 12) and the number of the day in the month (day from 1 to 31). Each field is written separately from other fields using the smallest possible number of bits.
Determine the minimum number of bits required to encode one record.

Formula needed Q = mn.

Q - number of options m - power of the alphabet n - length

Let's compose the right side of the formula based on the given task conditions (an unknown number of letters (from five options) and three numbers (from 10 options)):

5 ... 5 10 10 10 = 5 x * 10 3

This entire result, by condition, must be no less than 100000 . Let's substitute the rest of the data into the formula:

100000

From here we find the smallest suitable x:

x = 3 : 5 3 * 1000 = 125000 (125000 > 100000)

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 58):

When registering in a computer system, each user is given a password consisting of 9 characters. The symbols are used uppercase and lowercase letters of the Latin alphabet (in it 26 characters), and also decimal digits. The database allocates the same and minimum possible integer number of bytes to store information about each user. In this case, character-by-character encoding of passwords is used; all characters are encoded with the same and minimum possible number of bits. In addition to the password itself, additional information is stored in the system for each user, for which purpose 18 bytes per user. In the computer system it is allocated 1 KB to store information about users.

About what the greatest number User information can be stored in the system? In your answer, write down only an integer - the number of users.

Answer: 40

✍ Show solution:

• Since both uppercase and lowercase letters are used, we get a total of character options for encoding:

26 + 26 + 10 = 62

From the formula Q = 2 n we obtain the number of bits required to encode 1 password character:

Q = 2 n -> 62 = 2 n -> n = 6

Since the password has 9 characters, we get the number of bits to store 1 password:

6 * 9 = 54

Let's convert it to bytes (since, by convention, passwords are stored in bytes):

54 / 8 = 7 bytes

18 bytes are allocated for storing additional information. Let's get the number of bytes to store all information for one user:

18 + 7 = 25 bytes

According to the condition, 1 KB is allocated to store information about all users. Let's convert this value into bytes:

1 KB = 1024 bytes

Let's get the possible number of users:

1024 / 25 = 40,96

Let's discard the fractional part: 40

How to solve this problem step by step?

1. Find the number of characters in the alphabet.
2. Find out how many bits of information occupies 1 character in this alphabet.
3. Find out how many bits of information 1 password takes.
4. Find out the minimum possible number of bytes that can be used to encode 1 password.
5. Calculate how many bytes are needed to store 20 passwords.
6. Subtract the number of bytes calculated in step 5 from 400 (how many were allocated).
7. Divide the result by 20, since it corresponds to 20 users.

Step 1.

The alphabet, according to the conditions of the problem, has exactly 12 characters.

Step 2.

Let's look at the minimum number of bits needed to allocate per 1 character from an alphabet consisting of 12 characters.

If we allocated 1 bit of information, it would be able to encode 1 character of an alphabet consisting of no more than 2 characters. And we have 12 of them. This means that 1 bit is not enough.

• If you select 2 bits, then you can encode a character in an alphabet of maximum 4 characters. Few.
• If you select 3 bits, then you can encode a character in an alphabet of maximum 8 characters. Few.
• If you select 4 bits, then you can encode a character in an alphabet of a maximum of 16 characters. Enough.

This means that 4 bits are enough to encode 1 character of this alphabet.

Step 3.

• 1 password consists of 15 characters.
• 1 character “weighs” 4 bits.
• This means that 15 characters will “weigh” 15x4=60 bits.

Step 4.

• 1 password is conditionally encoded minimally possible whole quantity byte.
• How many bytes are needed to store a 60-bit password?
• 7 bytes is not enough, since 7 bytes = 7x8 = 56 bits.
• 8 bytes is just right: 8 bytes = 8x8 = 64 bits.
• Therefore, 8 bytes are needed to store one password.

Step 5

• One password weighs 8 bytes.
• We have 20 users (and 20 passwords, respectively).
• Therefore, they “weigh” 8x20 = 160 bytes.

Step 6

• 400 bytes were allocated for passwords.
• Purely for storage, according to point 5, 160 bytes were used.
• So, it remains for additional information 400-160=240 bytes.

Analysis of task 13 of the Unified State Exam 2016 in computer science from the demo version. This is a task on the ability to calculate the information volume of a message (to be able to estimate the amount of memory required to store information). This is a task of an increased level of difficulty. Approximate time to complete the task is 3 minutes.

Task 13:

When registering in a computer system, each user is given a password consisting of 15 characters and containing only characters from the 12-character set: A, B, C, D, E, F, G, H, K, L, M, N. In the database The data for storing information about each user is allocated the same and the minimum possible integer number of bytes. In this case, character-by-character encoding of passwords is used; all characters are encoded with the same and minimum possible number of bits. In addition to the password itself, additional information is stored in the system for each user, for which an integer number of bytes are allocated; this number is the same for all users. To store information about 20 users, 400 bytes were required. How many bytes are allocated to store additional information about one user? In your answer, write down only an integer - the number of bytes.

Answer: ________

Analysis of task 13 of the Unified State Exam 2016:

We determine how many bits are needed for one character using the formula N=2 i , where:
N is the number of characters in the set,
i is the number of bits per character.

12=2 4
i=4

Our password consists of 15 characters, therefore the information volume of one password is equal to:

15x4 = 60 bits

Passwords are encoded with the minimum possible integer number of bytes, that is:

60:8 = 8 bytes(take the nearest integer upward).

But in addition to the password itself, additional information is stored in the system for each user. We need to find out the scope of this very information.

To store information about 20 users, 400 bytes were required, that is, 400:20 = 20 bytes allocated for one user.

Of these 20 bytes, 8 are the password, and the rest are additional information. That is, the amount of additional information about one user is equal.

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