The lesson is devoted to the analysis of task 8 of the exam in computer science
The 8th topic - "Programming algorithms with cycles" - is characterized as tasks of the basic level of complexity, the execution time is about 3 minutes, the maximum score is 1
Algorithmic structures with cycles
In task 8 of the exam, algorithmic structures with cycles are used. Let's consider them on the example of the Pascal language.
- For acquaintance and repetition while loop, .
- For acquaintance and repetition For loop, .
The sum of an arithmetic progression
Formula to calculate n-th element of an arithmetic progression:
a n = a 1 + d(n-1)
n members of an arithmetic progression:
- a i
- d– step (difference) of the sequence.
The sum of a geometric progression
Geometric progression property:
b n 2 = b n+1 * q n-1
Formula to calculate denominator geometric progression:
\[ q = \frac (b_(n+1))(b_n) \]
Formula to calculate n th element of a geometric progression:
b n = b 1 * q n-1
Formula to calculate denominator geometric progression:
Formula for calculating the sum of the first n members of a geometric progression:
\[ S_(n) = \frac (b_1-b_(n)*q)(1-q) \]
\[ S_(n) = b_(1) * \frac (1-q^n)(1-q) \]
- b i– i-th element of the sequence,
- q is the denominator of the sequence.
Solving tasks 8 USE in Informatics
USE in Informatics 2017 assignment FIPI option 15 (Krylov S.S., Churkina T.E.):
| 1 2 3 4 5 | var k, s: integer ; begin s:= 512 ; k:=0; while s |
vark,s:integer; begin s:=512; k:=0; while s
✍ Solution:
- In a cycle k increases by unit (k - counter). Respectively, k will be equal to the number of iterations (repetitions) of the loop. After the completion of the cycle k is displayed on the screen, i.e. this is the output of the program.
- In a cycle s increases by 64 . For simplicity of calculations, we take the initial s not 512 , a 0 . Then the loop condition will change to s< 1536 (2048 — 512 = 1536):
- The loop will run while s<1536 , а s increases by 64 , it follows that the loop iterations (steps) will be:
- Respectively, k = 24.
Result: 24
For a more detailed analysis, we suggest watching the video of the solution to this 8 task of the exam in computer science:
10 Training options for examination papers to prepare for the Unified State Examination in Informatics 2017, task 8, option 1 (Ushakov D.M.):
Determine what will be printed as a result of executing the following program fragment:
| 1 2 3 4 5 6 7 8 9 10 11 | var k, s: integer ; begin k:= 1024 ; s:=50; while s› 30 do begin s: = s- 4 ; k: = k div 2 ; end ; write (k) end . |
var k,s: integer; begink:=1024; s:=50; while s>30 do begin s:=s-4; k:=kdiv 2; end; write(k)end.
✍ Solution:
Result: 32
For a detailed solution, see the video:
USE 8.3:
For what is the smallest integer introduced number d after executing the program, the number will be printed 192 ?
| 1 2 3 4 5 6 7 8 9 10 11 12 | var k, s, d: integer ; begin readln(d) ; s:=0; k:=0; while k ‹ 200 do begin s: = s+ 64 ; k: = k + d; end ; write(s); end. |
var k,s,d: integer; begin readln(d); s:=0; k:=0; while k< 200 do begin s:=s+64; k:=k+d; end; write(s); end.
✍ Solution:
Consider the program algorithm:
- Loop depends on variable k, which every iteration of the loop increases by the value d(input). The loop will finish "work" when k equal to 200 or exceed it k >= 200).
- The result of the program is the output of the value of the variable s. In a loop s increases by 64 .
- Since, according to the assignment, it is necessary that the number be displayed 192 , then the number of repetitions of the cycle is determined as follows:
- Since in a cycle k increases by value d, and loop repetitions 3 (the cycle ends when k>=200), we write the equation:
- Since the number turned out to be non-integer, we check and 66 and 67 . If we take 66 , then:
those. the cycle after three passes will still continue to work, which does not suit us.
- For 67 :
- Given number 67 suits us, it is the smallest possible, which is required by the assignment.
Result: 67
Watch the video for a breakdown of the task:
USE in informatics task 8.4 (source: option 3, K. Polyakov)
Determine what will be printed as a result of the following program fragment:
| 1 2 3 4 5 6 7 8 9 10 | var k, s: integer ; begin s:= 3 ; k: = 1; while k ‹ 25 do begin s: = s + k; k: = k+ 2 ; end ; write(s); end. |
var k, s: integer; begin s:=3; k:=1; while k< 25 do begin s:=s+k; k:=k+2; end; write(s); end.
✍ Solution:
Let's look at the listing of the program:
- The result of the program is the output of the value s.
- In a cycle s changes by increasing k, at the initial value s = 3.
- cycle depends on k. The loop will end when k >= 25. Initial value k = 1.
- In a cycle k constantly increasing by 2 -> means you can find the number of loop iterations.
- The number of loop iterations is:
(because k originally equaled 1 , then in the last, 12th passage of the cycle, k = 25; loop condition is false)
- AT s the sum of an arithmetic progression is accumulated, the sequence of elements of which is more convenient to start with 0 (and not with 3 , as in the program). So imagine that at the beginning of the program s = 0. But let's not forget that at the end it will be necessary to add 3 to the result!
- Then the arithmetic progression will look like:
- There is a formula for calculating the sum of an arithmetic progression:
s = ((2 * a1 + d * (n - 1)) / 2) * n
where a1 is the first member of the progression,
d- difference,
n- the number of members of the progression (in our case - the number of iterations of the cycle)
- Substitute the values in the formula:
- Let's not forget that we must add to the result 3 :
- This is the meaning s, which is displayed as a result of the program.
Result: 147
The solution to this task of the exam in computer science video:
USE in computer science task 8.5 (source: option 36, K. Polyakov)
| 1 2 3 4 5 6 7 8 9 10 | var s, n: integer ; begin s := 0 ; n := 0 while 2 * s* s ‹ 123 do begin s : = s + 1 ; n := n + 2 writeln (n) end . |
var s, n: integer; begin s:= 0; n:=0; while 2*s*s< 123 do begin s:= s + 1; n:= n + 2 end; writeln(n) end.
✍ Solution:
Let's look at the listing of the program:
- variable in the loop s constantly increasing per unit(works like a counter) and the variable n in a cycle increases by 2 .
- As a result of the program, the value is displayed on the screen n.
- cycle depends on s, and the loop will end when 2 * s 2 >= 123.
- It is necessary to determine the number of loop iterations (loop iterations): to do this, we define the smallest possible s, to 2 * s 2 >= 123:
Or it would simply be necessary to find such the smallest possible even number >= 123, which, when divided by 2 would return the calculated root of the number:
S=124/2 = √62 - not suitable! s=126/2 = √63 - not suitable! s=128/2 = √64 = 8 - fits!
- So the program will do 8 loop iterations.
- Let's define n, which increases each step of the loop by 2 , means:
Result: 16
The video of this exam task is available here:
USE in informatics task 8.6 (source: option 37, K. Polyakov with reference to O.V. Gasanov)
Write the smallest and largest value of a number separated by a comma d, which must be entered so that after the execution of the program it will be printed 153 ?
| 1 2 3 4 5 6 7 8 9 10 11 | var n, s, d: integer ; begin readln(d) ; n:=33; s:=4; while s ‹ = 1725 do begin s : = s + d; n := n + 8 write (n) end . |
var n, s, d: integer; begin readln(d); n:= 33; s:= 4; while s<= 1725 do begin s:= s + d; n:= n + 8 end; write(n) end.
✍ Solution:
Let's look at the listing of the program:
- The program loop depends on the value of the variable s, which in the cycle is constantly increasing by the value d (d entered by the user at the beginning of the program).
- Also, in the loop, the variable n increases by 8 . Variable value n is displayed on the screen at the end of the program, i.e. on assignment n by the end of the program should n=153.
- It is necessary to determine the number of loop iterations (passages). Since the initial value n=33, and at the end it should become 153 , in the cycle increasing by 8 then how many times 8 "fit" in 120 (153 — 33)? :
- As we have defined, the cycle depends on s, which at the beginning of the program s = 4. For simplicity, let us assume that s = 0, then we will change the loop condition: instead of s<= 1725 сделаем s <= 1721 (1725-1721)
- Let's find d. Since the loop is running 15 times, then you need to find an integer that, when multiplied by 15 would return a number more 1721:
- 115 is the least d under which n becomes equal 153 (for 15 cycle steps).
- Let's find the largest d. To do this, you need to find a number that corresponds to the inequalities:
- Let's find:
- Maximum d= 122
Result: 115, 122
Watch the video of this 8 task of the exam:
8 task. Demo version of the exam 2018 informatics:
Write down the number that will be printed as a result of the following program.
| 1 2 3 4 5 6 7 8 9 10 11 | var s, n: integer ; begin s := 260 ; n := 0 while s › 0 do begin s : = s - 15 ; n := n + 2 writeln (n) end . |
var s, n: integer; begin s:= 260; n:=0; while s > 0 do begin s:= s - 15; n:= n + 2 writeln(n) end.
✍ Solution:
- Consider the algorithm:
- The loop depends on the value of the variable s, which is initially equal to 260 . variable in the loop s constantly changing its value, decreasing at 15.
- The loop will end when s<= 0 . So you need to count how many numbers 15 "will enter" 260 , in other words:
- This figure should correspond to the number of steps (iterations) of the loop. Since the cycle condition is strict — s > 0 , then increase the resulting number by one:
- Let's check with a simpler example. Let's say initially s=32. Two iterations of the loop will give us s = 32/15 = 2.133... Number 2 more 0 , respectively, the loop will run a third time.
- As a result of the work, the program prints the value of the variable n(desired result). variable in the loop n, initially equal to 0 , increases by 2 . Since the loop includes 18 iterations, we have:
Result: 36
For a detailed solution of this task 8 from the USE demo version of 2018, see the video:
Solution 8 of the task of the Unified State Examination in Informatics (control version No. 2 of the examination paper of 2018, S.S. Krylov, D.M. Ushakov):
Determine what will be printed as a result of executing the program:
| 1 2 3 4 5 6 7 8 9 10 11 | var s, i: integer ; begin i := 1 ; s := 105 ; while s › 5 do begin s : = s - 2 ; i := i + 1 end ; writeln (i) end . |
vars, i: integer; i:= 1; s:= 105; while s > 5 do begin s:= s - 2; i:= i + 1 end; writeln(i)end.
✍ Solution:
- Let's consider the algorithm. Loop depends on variable s, which decreases every iteration of the loop on 2.
- There is also a counter in the loop - a variable i, which will increase per unit exactly as many times as there are iterations (passes) of the loop. Those. as a result of the program execution, a value equal to the number of iterations of the loop will be printed.
- Because the loop condition depends on s, we need to calculate how many times can s decrease by 2 in a cycle. For the convenience of counting, let's change the loop condition to while s > 0 ; as we s reduced by 5 , respectively, change the 4th line to s:=100 (105-5):
- In order to calculate how many times the loop will be executed, it is necessary 100 divide by 2 , because s each loop step decreases by 2: 100 / 2 = 50 -> number of loop iterations
- In the 3rd line we see that the initial value i is an 1 , i.e. in the first iteration of the loop i = 2. Hence, we need to add to the result (50) 1 .
- 50 + 1 = 51
- This value will be displayed on the screen.
Result: 51
Solution 8 of the USE task in informatics 2018 (diagnostic version of the examination paper of 2018, S.S. Krylov, D.M. Ushakov, USE simulator):
Determine the value of a variable c after the execution of the following program fragment. Write your answer as an integer.
| 1 2 3 4 5 6 7 | a:=-5; c:=1024; while a‹ › 0 do begin c: = c div 2 ; a:= a+ 1 end ; |
a:=-5; c:=1024; while a<>0 do begin c:=c div 2; a:=a+1 end;1000 do begin s := s + n; n := n * 2 write (s) end .
varn, s: integer; beginn:= 1; s:= 0; while n<= 1000 do begin s:= s + n; n:= n * 2 end; write(s) end.
✍ Solution:
- The loop condition depends on the variable n, which changes in a cycle according to obtaining powers of two:
Consider the algorithm:
Write down the number that will be printed as a result of the following program:
Pascal:
| 1 2 3 4 5 6 7 8 9 10 11 | var s, n: integer ; begin s := 522 ; n:=400; while s - n > 0 do begin s : = s - 20 ; n := n - 15 end ; write (s) end . |
var s, n: integer; begin s:= 522; n:= 400; while s - n > 0 do begin s:= s - 20; n:= n - 15 write(s)end.
✍ Solution:
- The algorithm contains a cycle. In order to understand the algorithm, let's trace the initial iterations of the loop:
- We see that in the condition the difference between the values is 5 :
This means that on the 24th iteration of the loop, the variables s and n got such values after which the condition still remains true: 2 > 0. At the 25th step, this condition is fulfilled:
Result: 22
We offer you to watch the video of the solution of the task:
USE. Russian language. How easy is task number 8?
Task number 8 has 3 options.
1 version of task number 8
Answer algorithm:
Eliminate all words with alternating root vowel. How to determine them, read below (in this example, these are the words for MEP et and on CLONE ish)
Eliminate all words with unverifiable root vowel. How to find them, read below. (In this example, this is the word F E DERAL)
Do not confuse the letter missing in the root with the letter missing in suffix(in this example it is WRONG And flax)
Be sure to check the vowel in the root, which, in your opinion, has unstressed verifiable (approx. and ryat -m And R).
Difficult. Do not confuse an alternating vowel with a checked one. The main difference between an alternating root is that such a root always has pair with a different letter, and these are words with the same root, so the meaning of the root is approximately the same.
Compare:
at WORLD at-y MEP et is a pair of words with different letters in the root, in which the meaning of the root is approximately the same.
BUT!!! etc WORLD five friends - this root with the meaning "peace" will never be written with a vowel E.
Difficult. Find the right root. For example, in the word ON SLAZD ATS root is not messy(you could then easily write FALSE), and SLAZD, so there is no alternating root here. This is the checked vowel in the root of the words- SWEET uy.
ANSWER in this example : to reconcile enemies.
Write down the answer only when you were able to pick up a word in which this letter is under stress. Correctly found word, but spelled with an error, is incorrect unswer!
Option 2 task number 8
Answer algorithm:
Eliminate all words with alternating root vowel (voz RAST, races STEL it)
Eliminate words with verifiable root vowel (dr O beat - dr O b. unfold E waving flag-in E em)
Remember that most often words with unverifiable vowel - these are foreign words, that is, their meaning needs to be explained, it may not be clear (to O institutional). However, there are words that are not of foreign origin (def. E share)
Answer : constitutional
3 option for task number 8
Answer algorithm:
Eliminate words with verifiable root vowel (embedded E tea-built-in E cha)
Eliminate all words with unverifiable root vowel (apl O wiping, art And leria, d And rector)
alternating roots remember visually, just learn them and remember that they always have pair, root value in them about the same(with BIR at-so BER yet)
In the task, it is necessary not only to find the given words, but also to write it CORRECTLY. So teach regulations, remember exceptions.
Answer: collect
Remember alternating roots:
|
No accent - Oh gar-gor clone clan creature-creature No accent - A zar-zor plov Depends on the subsequent letter in the root rast-rasch-ros lag-lodge jump-skoch Depends on the suffix A after the root cas + A - kos Depends on value equal-even mak-mok |
Depends on the suffix A. If the suffix A is after the root, then AND is written in the root ber-beer der dir mer-world lane at ter-tir best bliss steel-steel burn-burn even-cheat Letters at the root them (occupy - occupy) in (begin - start) |
Remember exceptions and write them correctly
Let's look at examples
Example 1
Reasoning pattern
I find checked roots: op And sleigh-op And shet, fun E value - rzvl E whose, vych And fuck-h And sla
I find alternating roots: you creative yat
Answer: l E pour
Example 2
Reasoning pattern
I find checked roots(not available in this version)
I find unverifiable roots (ek O lgia, g And mnazist, with And mpatiya, at And ketka)
I find alternating roots (beginning And nay)
Example 3
|
Identify the word that is missing an unstressed checked vowel root. Write out this word by inserting the missing letter. manufacturing f..lology dist.. lay Offer E-mail |
Their writing is subject to different patterns. Read the rules, look at the examples.
To correctly complete the task, you need to be able to determine which of these three cases the examples in your version of KIM refer to.
The task involves knowing how words are spelled. For the correct answer, you need to be able to accurately identify the spelling. This action is similar to passing through passport control at the border: the border guard looked at you, looked at the photo in your passport, and the path is clear. Before you are 5 words and the wording of the task. First understand what to look for.
If the task is for checked vowels, this is one thing, if for alternating ones, then another.
Let's remember what is what.
Checked unstressed vowels in the root: goat´, goat - go´zy
In an unstressed position, the same vowel is written at the root of the word as under stress.
Why do we write a letter about, but not a in words: goats a, goat? Because - to ´ h s!
How to check?
Choose the same word but in a different form: goat´, goat - goat´ zy(plural) or a related word, that is, a word with the same root: goat´, goat - goat, goat´ zlik.
Do not confuse:
It is a mistake to select as test words:
- words with different meanings: frequency - pure, purity - often
- words with alternating vowels in the root: sunbathe - sunburn, dawn - dawns
- other kinds of verbs: to be late - to be late, to assimilate - to assimilate
Unverifiable unstressed vowels in the root: dog´ ka, caraca´ titsa, camo´ rka
This is the most stupid of all rules, because it is not a rule at all! Why? Because there is no action that would help to decide which letter to write. The only thing you can do is look up the right word in the dictionary. That's what all textbooks say. But the dictionary may not be at hand. So, these words need to be memorized. The more of these words you know, the better. Gradually you will develop intuition, and words with unchecked vowels will not create difficulties. So, despite the seeming stupidity, this rule is very useful!
If you can’t attribute the word to words with alternating vowels in the roots and pick up test words (by changing the form of the word or looking for single-root words), then you have a word with an unverifiable vowel.
Note:
After successfully passing the exam, you can return to advice number 1. It won't hurt your life.
Alternating vowels at the root
The alternation of vowels in the roots of Russian words is a systemic phenomenon. There are many roots with alternating vowels, but in school practice they study the mandatory list. It includes 20 roots, which are presented below. They need to be remembered and firmly know the whole list. This will help you avoid many annoying mistakes.
Sometimes they ask me in letters: “Why don’t you give the root -log-? After all, in fact, the alternation is not - lag-//-false-, a -lag-//-log-//-false-?»
Everything is simple. I offer the option that is presented in most school textbooks. A unified exam is being prepared by FIPI, which promises that differences in programs will not affect the result of the exam. So far, this setting has been followed. BUT - log- don't worry about it: it happens only under stress! Forgery, pledge, pretense and others. And in the tasks you will only have words with unstressed roots. Therefore, I believe that you should not distract you with this and other similar cases.
For those who want to know not a short, school list, consisting of 20 roots, but a complete list of all roots with alternating vowels, I recommend the most authoritative source: "Complete academic reference book"Edited by V. Lopatin.
And here we will remember what you all studied at school.
1. Vowel alternation in roots ber//beer, lane//peer, measures//peace, burn//burn, ter//tyre, der//dir, h em//cheat, glitter//glitter, stele//steel: erase-erase
If in a word with roots: -ber-//-bir-, -per-//-feast-, -burned-//-burn-, -measures-//-world-, -ter-//-shoot-, -der-//-dir-, -even-//-cheat-, - shine-//-blist-, -steel-//-style-
have a suffix a,
write at the root and: at choose
, shine
,
no suffix
a,
write e:ber et, glitter et.
2. Vowel alternation in roots kos//cas: touch - touch
If in a word
- have a suffix a, write at the root a: touch, touch,
- no suffix a , write about:touch, at weaving
3. Vowel alternation in roots mok//poppy: wet - dunk
If in words with roots -mok- //-poppy- meaning:
“pass liquid, absorb liquid”, then write o: you wet in the rain
,
"immerse in liquid", write a: poppy at
.
4. Vowel alternation in roots equal//equal: equalize - equalize
If in words with roots -even-//-equal- meaning:
"smooth, smooth", write o: you take the road
, behind even pits
,
equality, write a: at equal rights, equally sided.
5. Vowel alternation in roots mountains//gar, creation//creature, clone//clan: pancake burnt' l
If the roots -gar- //-mountain- , -creature- //-creative- , -clan- //-clone-
without stress, write about: mountains and´ t
, clone
, create and´ be
,
under stress, write as you hear: behind ha R
, cla ´ n yatsya
, yours R quality
, you ´ p
.
Exception:
at ´ cinder
6. Vowel alternation in roots zor//zar: dawn´
If the roots -zar- //-zor-
without stress, write a: dawn´, lightning´ tsy
under stress, write as you hear: zo´ ri.
Exception: dawn ´ t
7. Vowel alternation in roots pilaf//melt: float´ to- swimmers´(excl.)
If the roots -float- //-float-
without stress, write a: float to,
write under stress as you hear: pla ´ vat, pla ´ ing, heats in.
Exception: swimmers ´ , swimmer´ ts, swimmers´ ha, quicksand´
8. Vowel alternation in roots grew//grow//grow, lag//lie, jump // jump: the plant has grown, term - addition, jump - upstart
If root consonants:
- with, write about: grew up,
st or sch, then a: grow, grown, - well, write about: offer ,
G, then a: term , - h, write about: upstart,
to, then a: skipping rope.
Exception: growth ´ to, growth´ th, usurer´ to, you´ height, teenage´ (along with the literary normsubdro ´ stock ) and their derivatives: sprout, usurer and etc.
The spelling of word roots is, at first glance, a simple topic. Moreover, it is studied at the lessons of the Russian language already in elementary school. However, it is in the roots that students often make mistakes.
Reasons for the incorrect spelling of the roots of words:
- Ignorance of the rules for writing vowels and consonants at the root.
- The inability to choose the right word to be checked, by which it is easy to check both the vowel and the consonant.
- Mistakes in identifying roots with alternating vowels. Checking such vowels with stress, which is a gross mistake. Alternating vowels should be written only according to the rule.
- There are frequent cases when among the words with missing spellings, those are suggested in which the letter is missing in a prefix!!! Be careful not to confuse the prefix with the root (for example: d ... stoverny, O is omitted here in the prefix)
As you can see, the main reason is ignorance of the rules. You need to learn the rules of the Russian language, guys. Only then will you be able to write the words correctly.
At the exam in the Russian language in task number 8, you need to find the word from the list of words with the verifiable unstressed vowel in the root and write this word on the answer sheet. Thus, the task, in comparison with previous years, has become much more complicated. Now you need not only to find this word, but also to know very well how it is spelled. An incorrectly spelled but correctly found word will be an erroneous answer.
Learn to pick the right one test words. In them, the stress should fall on the vowel being checked:
How to complete task number 8
1. Eliminate interleaved words from the list. They are not checked by stress, but are written according to the rule.
|
Alternating letters A-O |
Alternating letters I-E |
|
gar-gor |
ber-beer |
|
clone clan |
der dir |
|
creature-creature |
mer-world |
|
zar-zor |
per-feast |
|
rast-rasch-ros |
ter-tir |
|
lag-lodge |
glitter blist |
|
plov |
steel-steel |
|
jump-skoch |
burn-burn |
|
mak-mok |
even-cheat |
|
equal-even |
|
|
kas-kos |
A (i) - them, in (occupy - occupy) |
|
(understand - understand) |
2. Exclude from the list words with an unchecked vowel in the root. These words are easy to find - these are mostly words of foreign origin:
|
|
|
 |
|
|
|
|
3. The remaining word will be the answer. Do not forget to check this word with an accent to be sure of the correct answer.
Train more, do tests, exercises. Task options No. 8 given on our website.
GOOD LUCK!
Melnikova Vera Alexandrovna
Practice 8
b..gage zap..are rushing off..to rush into..new k..ridor
burnt..took a call..twitched to listen..to kr..styanin sh..did
obscured..imaginary..to get out bl..stete comp..awning m..tafora
4. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
earthquake..seniority..report (friends)
5. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
board..natally located..live..light (with a lamp) apply..sleep d..tektiv
6. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
st..give birth to..nforka approx..ryat (dress) incompressible exp.
7. Determine the word in which the unstressed unchecked vowel of the root is missing. Write out this word by inserting the missing letter.
s..redty pl..vchikha el..ment vyt..raw prom..porridge
8. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
alm..nah d..enrichment directive
9. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
exc.
10. Determine the word in which the unstressed unchecked vowel of the root is missing. Write out this word by inserting the missing letter.
the application
11. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
attached..switching an..small exchange..whip fire..fly r..stock
12. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
round..shchenie b..lans r..spublika k..talog prisk..kat
13. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
up..to fight art..lleria note..to (friends)
14. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
see..homeland show..zatel d..stoverny locom..tiv lit..gatel
15. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
appl.. to direct.. to turn out.. d.. rector to ask ...
16. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
expensive..gie k..satelnaya r..mesleny zag..rely k..ridor
17. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
roasted
18. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
mon.. manie r.. storovschik l.. ktoriy g.. relay tr.. umf
19. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
r..dawn dist..pouring suppose..gat zh..luzi yuv..lir
20. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
g..rison por..vnu vyt..reet to..media k..side
21. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
please..fly dov..to..change prop..gandhian exchange..kick
22. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
designation r.
23. Determine the word in which the unstressed unchecked vowel of the root is missing. Write out this word by inserting the missing letter.
to..sovrotka omr..chit in..hter adj..smoking an..small
24. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
internal..drilling
25. determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
river
26. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
pon..mania..offer
27. Determine the word in which the unstressed unchecked vowel of the root is missing. Write out this word by inserting the missing letter.
por..vnu vyt..ret one..narny p..lemika count..talk
28. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
call..pour expression..fading representative..vital..stelin or..entir
29. Determine the word in which the unstressed checked vowel of the root is missing. Write out this word by inserting the missing letter.
pok..army k..sat k..companion z..rya sk..chok
30. Determine the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.
one..ravenous to..side-to-side g..roar pl..schikha note..ryat (dress)
