Theory on the topic of interest. Solving problems involving percentages in preparation for the OGE and USE in mathematics

The ability to correctly and quickly solve word problems involving percentages is necessary not only for students who will passing the Unified State Exam in basic mathematics or profile level, but also to all adults, since such tasks are constantly encountered in everyday life. Increasing prices, planning a family budget, profitable investment of funds and many other issues cannot be resolved without these skills. When preparing to take the certification test, you definitely need to repeat how to solve problems with percentages: in the Unified State Exam in mathematics, they are found at both the basic and specialized levels.

Need to remember

A percentage is a \(\frac(1)(100)\) part of a number. Denotes a share of something in relation to the whole. The written symbol is \(\%\) . When preparing for the Unified State Exam on the topic “Percentage,” schoolchildren both in Moscow and in other parts of the Russian Federation need to remember the following formula:

How to apply it?

In order to solve a simple task with percentages in the Unified State Examination in mathematics, you need:

1. Divide the given number by \(100\) .
2. Multiply the resulting value by the amount of \(\%\) that needs to be found.

For example, in order to calculate \(10\%\) of the number \(300\) , you need to find \(1\) percentage by dividing \(300:100=3\) . And the number obtained from the previous action is \(3\cdot10=30\) . Answer: \(30\).

These are the simplest tasks. 11th grade students in the Unified State Exam are faced with the need to solve complex problems involving percentages. As a rule, they refer to bank deposits or payments. You can familiarize yourself with the formulas and rules for their application by going to the “Theoretical Information” section. Here you can not only repeat the basic definitions, but also get acquainted with options for solving complex problems involving interest on a bank loan, as well as exercises from other sections of algebra, for example,

See also the video "Text problems on the Unified State Exam in mathematics".
A word problem is not only a movement and work task. There are also tasks on percentages, on solutions, alloys and mixtures, on moving in a circle and finding the average speed. We will tell you about them.

Let's start with problems involving percentages. We have already met this topic in task 1. In particular, they formulated important rule: we take as the value with which we compare.

We have also derived useful formulas:

if we increase the value by percent, we get .
if the value is reduced by percent, we get .
if the value is increased by percent and then decreased by , we get .

if we increase the value twice by percent, we get
if the value is reduced twice by percent, we get

Let's use them to solve problems.

There were people living in the city block per year. In the year, as a result of the construction of new houses, the number of residents increased by, and in the year - by compared to the year. How many people began to live in the quarter per year?

According to the condition, in a year the number of inhabitants increased by , that is, it became equal to people.

And in the year the number of residents increased by , now compared to the year. We get that in a year there were more residents living in the block.

The following problem was proposed at trial Unified State Exam in mathematics in December. It is simple, but few have mastered it.

On Monday, the company's shares rose in price by a certain amount of percent, and on Tuesday they fell in price by the same amount of percent. As a result, they became cheaper than when trading opened on Monday. By what percentage did the company's shares rise in price on Monday?

At first glance, it seems that there is an error in the condition and the stock price should not change at all. After all, they have risen in price and fallen in price by the same percentage! But let's not rush. Suppose that at the opening of trading on Monday the shares were worth rubles. By Monday evening they had risen in price and began to cost . Now this value is taken as , and by Tuesday evening the shares fell in price by this value. Let's collect the data into a table:

According to the condition, the shares eventually fell in price by .

We get that

Let's divide both sides of the equation by (after all, it is not equal to zero) and apply the abbreviated multiplication formula on the left side.

According to the meaning of the problem, the value is positive.
We get that .

The price of a refrigerator in a store decreases annually by the same percentage from the previous price. Determine by what percentage the price of a refrigerator decreased each year if, put up for sale for rubles, two years later it was sold for rubles.

This problem is also solved using one of the formulas given at the beginning of the article. The refrigerator cost rubles. Its price has decreased twice by , and now it is equal to

Four shirts are cheaper than a jacket by . What percentage are five shirts more expensive than a jacket?

Let the cost of the shirt be equal to , the cost of the jacket . As always, we take as one hundred percent the value with which we compare, that is, the price of the jacket. Then the cost of four shirts is equal to the price of the jacket, that is
.

The cost of one shirt is several times less:
,
and the cost of five shirts:

What we got was five shirts that were more expensive than the jacket.

Answer: .

The family consists of a husband, wife and their student daughter. If the husband's salary doubled, the total family income would increase by . If the daughter's scholarship were to be halved, the family's total income would be reduced by . What percentage of the total family income is the wife's salary?

Let's draw a table. We will call the situations referred to in the problem (“if the husband’s salary increased, if the daughter’s scholarship decreased...”) “situation” and “situation”.

It remains to write down the system of equations.

But what do we see? Two equations and three unknowns! We won't be able to find them separately. True, we don’t need this. It’s better to take the first equation and subtract the sum from both its sides. We get:

This means that the husband’s salary is part of the total family income.

In the second equation, we also subtract the expression from both sides, simplify and get that

This means that the daughter’s scholarship is based on the total family income. Then the wife's salary constitutes the total income.

Answer: .

Next type problems - problems on solutions, mixtures and alloys. They are found not only in mathematics, but also in chemistry. We'll tell you about the in a simple way their decisions.

In a vessel containing liters of -percentage aqueous solution some substance, added liters of water. What percentage is the concentration of the resulting solution?

In the decision similar tasks the picture helps. Let's depict a vessel with a solution schematically - as if the substance and water in it were not mixed with each other, but separated from each other, as in a cocktail. And let’s write down how many liters the vessels contain and what percentage of the substance they contain. Let's denote the concentration of the resulting solution.

The first vessel contained liters of the substance. The second vessel contained only water. This means that the third vessel contains the same number of liters of substance as the first:

.

We mixed a certain amount of a -percentage solution of a certain substance with the same amount of a -percentage solution of this substance. What percentage is the concentration of the resulting solution?

Let the mass of the first solution be equal to . The mass of the second one is the same. As a result, we obtained a solution with a mass of . Let's draw a picture.

We get:

Answer: .

Grapes contain moisture, and raisins contain moisture. How many kilograms of grapes are required to produce kilograms of raisins?

Attention! If you come across a problem “about products”, that is, one where raisins are made from grapes, apricots from apricots, crackers from bread or cottage cheese from milk - know that this is actually a problem on solutions. We can also conventionally depict grapes as a solution. It contains water and "dry matter". The “dry matter” has a complex chemical composition, and by its taste, color and smell we could understand that these are grapes and not potatoes. Raisins are produced when the water evaporates from the grapes. At the same time, the amount of “dry matter” remains constant. The grapes contained water, which means there was “dry matter”. Raisins contain water and “dry matter”. Let a kg of grapes produce a kg of raisins. Then

From from

Let's make an equation:

and we'll find it.

Answer: .

There are two alloys. The first alloy contains nickel, the second - nickel. From these two alloys, a third alloy weighing kg containing nickel was obtained. How many kilograms is the mass of the first alloy less than the mass of the second?

Let the mass of the first alloy be x, and the mass of the second alloy be y. The result was an alloy with a mass of .

Let's write it down simple system equations:

The first equation is the mass of the resulting alloy, the second is the mass of nickel.

Solving, we get that .

Answer: .

Mixing -percentage and -percentage solutions of acid and adding kg clean water, we obtained a -percentage solution of acid. If instead of kg of water a kg -percent solution of the same acid were added, we would get a -percent solution of the acid. How many kilograms of -percentage solution were used to obtain the mixture?

Let the mass of the first solution be , the mass of the second is . The mass of the resulting solution is equal to . Let's write two equations for the amount of acid.

We solve the resulting system. Let’s immediately multiply both sides of the equations by , since it is more convenient to work with integer coefficients than with fractional ones. Let's open the brackets.

Answer: .

Circular motion problems also turned out to be difficult for many students. They are solved in almost the same way as ordinary tasks for movement. They also use the formula. But there is one trick that we will tell you about.

From point circular track A cyclist rode out, and minutes later a motorcyclist followed him. Minutes after setting off, he caught up with the cyclist for the first time, and minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is km. Give your answer in km/h.

First, let's convert minutes to hours, since the speed must be found in km/h. We denote the speeds of the participants as and . For the first time, a motorcyclist overtook a cyclist minutes later, that is, an hour after the start. Up to this point, the cyclist had been on the road for minutes, that is, an hour.

Let's write this data into a table:

Both traveled the same distances, that is.

The motorcyclist then passed the cyclist a second time. This happened minutes, that is, an hour after the first overtaking.

Let's draw the second table.

And what distances did they travel? A motorcyclist overtook a cyclist. This means he drove one more lap. This is the secret of this task. One lap is the length of the track, it is equal to km. We get the second equation:

Let's solve the resulting system.

We get that . In response, we write down the speed of the motorcyclist.

Answer: .

A clock with hands shows hours minutes. In how many minutes minute hand will it be aligned with the clock for the fourth time?

This is perhaps the most difficult task from Unified State Exam options. Of course, there is a simple solution - take a watch with hands and make sure that the hands align for the fourth time in an hour, exactly at ..
What if you have electronic watch and you cannot solve the problem experimentally?

In one hour, the minute hand travels one circle, and the hour hand travels one circle. Let their speeds be (circles per hour) and (circles per hour). Start - at .. Let's find the time during which the minute hand will catch up with the hour hand for the first time.

The minute hand will travel one more circle, so the equation will be:

Having solved it, we get that hour. So, for the first time the hands will align in an hour. Let them become equal the second time after a while. Minute hand will go the distance, and the hour hand, and the minute hand will travel one circle more. Let's write the equation:

Having solved it, we get that hour. So, in an hour the hands will align for the second time, after another hour for the third time, and after another hour for the fourth time.

This means that if the start was at ., then for the fourth time the arrows will align through
hours.

The answer is completely consistent with the "experimental" solution! :-)

In your math exam, you may also be asked to find the average speed. Remember that the average speed is not equal to the arithmetic mean of the speeds. It is found using a special formula:

,
where is the average speed, - common path, - total time.

If there were two sections of the path, then

The traveler crossed the sea on a yacht at an average speed of km/h. He flew back on a sports plane at a speed of km/h. Find the traveler's average speed along the entire journey. Give your answer in km/h.

We do not know what the distance was that the traveler covered. We only know that this distance was the same on the way there and back. For simplicity, let's take this distance as (one sea). Then the time that the traveler sailed on the yacht is equal to , and the time spent on the flight is equal to . Total time equals .
Average speed equal to km/h.

Answer: .

Let's show another effective technique that helps you quickly solve the system of equations in problem 13.

Andrey and Pasha paint the fence in hours. Pasha and Volodya paint the same fence in an hour, and Volodya and Andrey - in an hour. How many hours will it take the boys to paint the fence, working together?

We have already solved work and productivity problems. The rules are the same. The only difference is that there are three people working here, and there will also be three variables. Let be Andrey’s productivity, be Pasha’s productivity, and be Volodya’s productivity. We will take the fence, that is, the amount of work, as - after all, we cannot say anything about its size.

Andrey and Pasha painted the fence in hours. We remember that when working together performances add up. Let's write the equation:

Likewise,

Then


.

You can search for , and separately, but it’s better to just add all three equations. We get that

This means that, working together, Andrey, Pasha and Volodya paint one-eighth of the fence in an hour. They will paint the entire fence in hours.

"Simple and compound interest »

Relevance of the topic.

Understanding percentages and the ability to make percentage calculations are currently necessary for every person: applied value This topic is very large and affects financial, demographic, environmental, sociological and other aspects of our lives.

The material is relevant for everyone who is in grade 11 this year.

When Yashchenko, who is directly involved in the compilation of CIMs in mathematics, came to our seminar in October, he said that all prototypes of task 19 would be posted in open jar, since the task is new.

The task is being solved for my not very strong class, and it would be possible to train for it.

A little theory...

“Interest”.

Task1

a) What is interest called? (A percentage is one hundredth of a number.)

b) What is 1% indicated? ( 1%? = 0,01 )

c) What is 1% of a hundredweight called? ( kg. ) Meter? (see) Hectare? (ar or hundredth)

d) What is called 1% interest given number A? (The percentage of a given number a is the number 0.01 a, i.e. 1% (a) = 0.01*a)

e) How to determine p% of a given number a? (find the number 0.01 p a, i.e.р% = 0.01*р*а)

f) How to convert a decimal fraction to a percentage? ( multiply by 100 ). How about percentages to decimals? (divide by one hundred, i.e. multiply by 0.01)

g) How to find a percentage of a number? (To find a part in from the number x as a percentage, you need to divide this part by the number and multiply by 100, i.e. a(%)=(w/x)*100)

e) How is a number found by its percentage?(If it is known that a% of x is equal to b, then x can be found using the formula x = (v/a)*100)

Task 2

Present these decimal fractions as percentages:

A)1; 0.5; 0.763; 1.7; 256.

b) Express the percentages in decimal fractions: 2%; 12%; 12.5%; 0.1%; 200%.

Task 3

Find the % of the number:

c) 0.1% of the number 1200?(1,2)

d) 15% of the number 2? (0.30)

Task 4

Find a number by its percentage:

e) How many centners does the bag weigh? granulated sugar, if 13% is 6.5 kg.?(50 kg.= 0.5 c.)

c) What percentage of 10 is 9?

Answers: a) 9%, b) 0.09%, c) 90%; d) 900%?

Simple and compound interest.

These terms are most often found in banking, in financial tasks.

Banks attract funds (deposits) at certain interest rates. Depending on the interest rate, income is calculated.

In practice, two approaches to assessing interest income are used - simple and compound interest.

When applying simple interest, income is calculated from the initial amount of invested funds, regardless of the investment period. In financial transactions, simple interest is used primarily for short-term financial transactions.

Let some quantity be subject to gradual change. Moreover, each time its change is certain number percent of the value that this value hadon initial stage . This is how they are calculatedsimple interest.

When applying compound interest, the accumulated amount of interest is added to the deposit at the end of the next accrual period. Moreover, each time its change is a certain number of percent of the value that this value hadat the previous stage. In this case we are dealing with “compound interest” (i.e., “interest on interest” calculations are used)

The initial amount and the interest received are collectively called the accumulated (accumulated) amount.

So, if the bank rate is 10%, and the initial amount is 100 rubles, then the accumulated amount over five years, using simple and compound interest, will look like:

Table 1. Accumulated amount using simple and compound interest.

Formulas for simple and compound interest.

I. Let a certain value A increase n times (n year) and each time by p%.

We introduce the notation: A 0 – initial value of quantity A;

r – constant quantity percent;

a interest rate; a=р/100 = 0.01*р

A n – accumulated amount for n times (by the end of the nth year) - according to the simple interest formula;

S n - the accumulated amount for n times (by the end of the nth year) - according to the compound interest formula.

Then its value A 1 for simple interest after the first increase (by the end of the first year) is calculated by the formula: A 1 = A 0 + A 0 * (0.01p) = A 0 (1 + (0.01p) = A 0 (1 + p)

At the end of the second stage A 2 = A 1 + A 0 * (0.01r) = A 0 (1 + a) + A 0 * a = A 0 (1 + 2 a).

At the end of the third stage A 3 = A 2 + A 0 * (0.01r) = A 0 (1 + 2 a) + A 0 * a = A 0 (1 + 3 a).

Then for simple interest the amount over the years is equal to:

A n = A 0 (1 + 0.01р*n) or A n = A 0 (1 + ?* n) (1)

For compound interest it looks different:

Let some quantity S 0 increases n times (n year) and each time by p%.

Then its meaning S 1 for compound interest after the first increase (by the end of the first year) is calculated using the formula:

S1 = S0 + S0 (0.01r) = S0 * (1 + 0.01r) = S0 * (1 + ?).

At the end of the second stage S 2 = S 1 + S 1 (0.01р) = S 1 * (1 + 0.01р) = S 0 (1 + ????р) 2 = S 0 (1 + ?) 2.

At the end of the third stage S 3 = S 2 + S 2 (0.01r) = S 2 * (1 +0.01r) = S 0 (1 +0.01r) 2 *(1 +0.01r)=S 0 (1 +0, 01р) 3 = S 0 (1 + a ) 3 .

Then for compound interest the amount over the years is equal to:

S n = S 0 (1 + 0.01р) n or S n = S 0 (1 + a ) n (2)

Example 1.

The bank has opened a time deposit in the amount of 50 thousand rubles. 12% for 3 years. Calculate the accumulated amount if interest:

a) simple; b) complex.

Solution 1.

Using the simple interest formula

Sn=(1+3*0.12)*50,000 = 68,000 rub. (res. 68,000 rub.)

Using the simple interest formula

Sn=(1+0.12) 3 *50,000 = 70,246 rubles. (res. 70246 rub.)

The compound interest formula connects four quantities: the initial deposit, the accumulated amount ( future value deposit), annual interest rate and time in years. Therefore, knowing three quantities, you can always find the fourth:

S n = S 0 * (1+0.01р) n

To determine the number of percent p it is necessary:

р = 100 * ((S n / S 0 ) 1/n – 1) (2.1)

The operation of finding the initial deposit S 0 , if it is known that in n years it should amount to the sum S n , is called discounting:

S 0 = S n * (1 + 0.01р) –n (2.2)

How many years is the contribution S 0 must lie in the bank at p% per annum in order to achieve the value S n.

n = (lnS n – lnS 0 ) / (ln(1 + 0.01р) (2.3)

In banking practice, interest may be accrued more than once a year. In this case, the bank rate is usually set in annual terms. The compound interest formula will look like:

S n = (1 + ?/t) n t S 0 (3)

where t is the number of interest reinvestments per year.

Example 2.

The bank has opened a time deposit in the amount of 50 thousand rubles. 12% for 3 years. Calculate the accrued amount if interest is calculated quarterly.

Solution 2.

n=3

t = 4 (per year – 4 quarters)

Using the compound interest formula

S 3 = (1+0.12/4) 3*4 *50000 = 1.03 12 *50000 = 71288 rub. Rep. RUB 71,288

As follows from examples 1 and 2, the accumulated amount will increase faster, the more often interest is accrued.

Let us present a generalization of formula (2), when the increase in the value of S at each stage is different. Let S O , the initial value of S, at the end of the first stage experiences a change by p 1 %, at the end of the second on p 2 %, and at the end of the third stage on p 3 %, etc. At the end of the nth stage, the value of S is determined by the formula

S n = S 0 (1 + 0.01р 1 )(1 + 0.01р 2 )...(1 + 0.01р n ) (4)

Example 3.

The trading base purchased a batch of goods from the manufacturer and delivered it to the store at a wholesale price, which is 30% higher more price manufacturer. The store set the retail price for the product 20% higher than the wholesale price. During the sale, the store reduced this price by 10%. How many rubles more did the buyer pay compared to the manufacturer’s price if he purchased an item at a sale for 140 rubles? 40 kopecks

Solution 3.

Let the initial price be S rub., then according to formula (4) we have:

S 0 (1 + 0,01*30)(1 + 0,01*20)***(1 – 0,01*10) = 140,4

S 0 *1.3*1.2*0.9 = S 0 *1.404 = 140.4

S 0 = 140.4: 1.404 = 100 (rub.)

Find the difference between the last and initial prices

140.4 – 100 = 40.4 Answer. 40.4 rub.

Examples of problems with solutions

Option 1

Task 1. The gas station owner increased the price of gasoline by 10%. Noticing that the number of customers had dropped sharply, he lowered the price by 10%. How did the starting price of gasoline change after this? (increased or decreased and by how many %?)

Solution: Let S 0 – starting price, S 2 – final price, x – desired number of percent change, where x = (1 - S 2 /S 0 )*100% (*)

Then according to the formula S n = S 0 (1 + 0.01р 1 )(1 + 0.01р 2 )***(1 + 0.01р n ) (4), we get

S 2 = S 0 (1 + 0.01*10 )(1 - 0.01*10) = S 0 *1.1*0.9 = 0.99*S 0.

S 2 = 0.99*S 0; 0.99 = 99%, S value 2 is 99% of the original cost, which means 100% lower - 99% = 1%.

Or using formula (*) we get: x = (1 – 0.99)*100% = 1%.

Answer: decreased by 1%.

Task 2. During the year, the enterprise increased its production output twice by the same percentage. Find this number if it is known that at the beginning of the year the company produced 600 products monthly, and at the end of the year began to produce 726 products monthly.

Solution: Let S 0 – starting price, S 2 – final price, p – constant amount of interest.

According to formula (2.1) we obtain: p = 100 * ((726/ 600 ) 1/2 – 1) = 10%.

Answer: 10%

Task 3. The price of computer equipment was increased by 44%. After this, as a result of two successive identical percentage reductions, the price of computers was 19% less than the original price. By what percentage did they reduce the price each time?

Solution: Using formula (4), we compose the equation

S 3 = S 0 (1 + 0.01*44)(1 - 0.01r)(1 - 0.01r) = S0 *1.44*(1 - 0.01r) 2 = S0 * (1-0.01*19). Solving the equation, we get 2 roots: 175 and 25, where 175 does not suit the conditions of the problem. Therefore p = 25%.

Answer: 25%

Task 4. To determine the optimal price increase regime, the company decided, starting January 1, to increase the price of the same product in two stores in two ways. In one store - at the beginning of each month (starting in February) by 2%, in another - every two months, at the beginning of the third (starting in March) by the same number of percent, and such that after six months (July 1 ) prices became the same again. By how many percent should the price of the product be increased every two months in the second store?

Solution: Let S 0 – starting price,p – constant percentage.

Then after 6 months (after six increases by 2%) in the first store the price of the product will be equal to S 0 (1 + 0,01*2) 6 , and in the second store (after three increases by p%), the price of the product will be equal to S 0 (1 + 0.01r) 3 . We get the equation S 0 (1 + 0.01*2) 6 = S 0 (1 + 0.01р) 3 . Solving it, we get

(1 + 0.01*2) 2 = (1 + 0.01r); 1.02 2 = (1 + 0.01r); p = 4.04

Answer: 4.04%

Option 2.

Task 1. A car was driving along a highway at a certain speed. Going out to country road, he reduced the speed by 20%, and then on a steep climb section, he reduced the speed by 30%. What percentage is this new speed lower than the original?

Solution: Let V 0 – initial speed,V – new speed, which is obtained after two various changes, p – the required amount of interest.

Then, using formula (4), we compose the equation V 0 (1 - 0.01*20)(1 - 0.01*30) = V 0 (1 - 0.01r). Solving it we get V 0 *0.8*0.7 = V 0 (1 - 0.01r); p = 44

Answer: 44%

Task 2. Let's assume that at room temperature water evaporates by 3% per day. How many liters of water will remain after 2 days from 100 liters? How much water will evaporate?

Solution: n=2; p=3%; S 0 = 100l. Then, according to formula (2), we get

S 2 = S 0 (1 - 0.01p) 2 = 100*(1-0.01*3) 2 = 100*0.97 2 = 94.09; S 0 – S 2 = 100 - 94.09 = 5.91

Answer: 94.09l.; 5.91l.

Task 3. The deposit placed in the bank 2 years ago reached 11,449 rubles. What was the initial contribution at 7% per annum? What is the profit?

Solution: n=2; p=7%; S2 = 11449; S0 = ?

In formula (2.2) S 0 = S n * (1 + 0.01р) –n We substitute these values, we get:

S 0 = 11449* (1 + 0.01*7) –2 = 11449/ (1.07) 2 =11449/ 1.1449 = 10000.

11449 – 10000 = 1449

Answer: 10,000 rubles; 1449 rub.

Task 4. Sberkassa accrues annually 3% of the deposit amount. In how many years will the amount double?

Solution: p=3%; S 0 – initial amount; n=?

Let's make an equation: 2*S 0 = S 0 (1 + 0.01р) n ; 2*S 0 = S 0 (1 + 0.03) n ; 2 = 1.03 n n=log 1.03 2; n ?23.

Independent work

1st level. After reconstruction, the plant increased production output by 10%, and after replacing equipment by another 30%. By what percentage did the initial output increase?

(Answer: 43%)

2nd level. The number 50 was increased three times by the same number of percent, and then decreased by the same number of percent. The result was 69.12. By what percentage did you increase and then decrease this number?

(Answer: 20%)

3rd level. The bank charges annually 7% of the deposit amount. Find smallest number years, during which the contribution grows by more than 20%.

(Answer: 3 years)

No. 1. The savings bank accrues 5.5% per annum on deposits annually. The depositor deposited 150 thousand rubles into the bank. What will be the deposit amount after 2 years?

(Answer: RUB 166,953.75)

No. 3. The bank offers two deposit options

1) at 120% with interest accrued at the end of the year;

2) at 100% with interest accrued at the end of each quarter.

Determine a more profitable option for placing deposits for one year.

Solution.

The more profitable option is considered to be the one in which the amount increased over the year will be greater. To evaluate the options, we will take the initial amount equal to 100 rubles.

According to the first option, the accumulated amount will be equal to (1+1.2)*100 rubles. = 220 rub.

Under the second option, interest is accrued quarterly. At the end of the first quarter, the accumulated amount is (1+1.0/4)*100 rubles. = 125 rub.

At the end of the 2nd quarter (1+1.0/4) 2 *100 rub. = 156 rub.

The accumulated amount for the year is (1+1.0/4) 4 *100 rub. = 244 rub.

As follows from the calculations, the second option is much more profitable (244 > 220). True, only if compound interest is used.

A selection of prototypes for task No. 19 of the Unified State Exam in mathematics 2015 at the profile level.

19. On December 31, 2012, Ekaterina took out 850,000 rubles on credit from the bank at 15% per annum. The loan repayment schedule is as follows: December 31 of each next year the bank charges interest on the remaining amount of the debt (that is, increases the debt by 15%), then Ekaterina transfers it to the bank a certain amount annual payment. What should the amount of the annual payment be for Catherine to pay off the debt in three equal annual payments?

19. A bank issues a loan to a young family at 20% per annum to purchase an apartment.

The loan repayment scheme is as follows: exactly one year after the loan is issued by the bank

charges interest on the remaining amount of the debt (that is, increases the debt by 20%),

then this family transfers a certain amount to the bank over the next year

(fixed) annual payment amount. The Ivanov family plans to repay

loan with equal payments over 4 years. How much money can he give them?

bank, if the Ivanovs are able to repay the loan 810,000 annually

rubles?

19. An 8-liter flask contains a mixture of nitrogen and oxygen containing 32% oxygen. A certain amount of the mixture was released from the flask and the same amount of nitrogen was added; then they again released the same amount of new mixture as the first time and added the same amount of nitrogen. As a result, the percentage of oxygen in the mixture was 12.5%. How many liters of mixture were released each time?

19. A deposit was made to the bank at a bank interest of 10%. A year later, the owner of the deposit withdrew 2,000 rubles from the account, and a year later again deposited 2,000 rubles. However, as a result of these actions, three years after the initial investment of the deposit, he received an amount less than planned (if there were no intermediate transactions with the deposit). How many rubles less than the planned amount did the investor receive in the end?

19. On the first working day of the month, a number of tractors rolled off the factory assembly line. Each subsequent working day, their production increased by 3 tractors daily, and the monthly plan of 55 tractors was completed ahead of schedule, and in a whole number of days. After that, 11 tractors were produced daily. Determine how many tractors were produced on the first working day, and by what percentage the monthly plan was exceeded, if it is known that there were 26 working days in the month, and the planned work lasted no less than 3 and no more than 10 days.

19. On March 8, Lenya Golubkov took out 53,680 rubles from the bank on credit for 4 years at 20% per annum to buy his wife Rita a new fur coat. The loan repayment scheme is as follows: on the morning of March 8 of the next year, the bank charges interest on the remaining amount of the debt (that is, it increases the debt by 20%), and in the evening of the same day Lenya transfers a certain amount of the annual payment to the bank (this amount is the same for all four years). What amount in excess of the borrowed 53,680 rubles will Lenya Golubkov have to pay the bank over these four years?

19. Semyon Kuznetsov planned to invest all his savings in a savings account in the Navroda bank at 500%, expecting to take A rubles in a year. However, the collapse of the Navrode Bank changed his plans, preventing a rash act. As a result, Mr. Kuznetsov put part of the money in the First Municipal bank, and the rest in a pasta jar. A year later, First Municipal increased the payment percentage by two and a half times, and Mr. Kuznetsov decided to leave the deposit for another year. As a result, the amount received at First Municipal wasAnd rubles. Determine what interest the First Municipal Bank accrued for the first year if Semyon “invested” in a can of pasta And rubles.

19. The bank plans to invest 30% of its client funds in shares of a gold mining plant for 1 year, and the remaining 70% in the construction of a shopping complex. Depending on the circumstances, the first project can bring the bank a profit of 32% to 37% per annum, and the second project - from 22% to 27% per annum. At the end of the year, the bank is obliged to return the money to customers and pay them interest at a predetermined rate, the level of which should range from 10% to 20% per annum. Determine what is the smallest and largest net profit as a percentage per annum from the total investments in the purchase of shares and the construction of a shopping complex that the bank can receive.

A percentage is a hundredth of a number.

Percentage is indicated by the symbol $%$.

To represent percentages in the form decimal, you need to divide the value by $100$.

$35%={35}/{100}=0.35$.

To find the percentage of a number, you need given number divide by $100$ and multiply by the percentage.

$n%$ of $а=(а⋅n)/(100)$

How many degrees does an angle contain if it is $5%$ of the straight angle?

The straight angle is $180°$.

Let's find $5%$ of $180°$, for this $(180°⋅5)/(100)=9°$.

Answer: $9°$.

To find a number by its specified percentage, you need to divide the given number by specified value percent, and multiply the result by $100$.

Find a number whose $20%$ is $80$.

We find a number whose $20%$ is $80$ as follows:

${80⋅100}/{20}=400$.

Answer: $400$.

Discount tasks

A discount is a reduction in the price of a product or service. Most often, the discount is indicated as a percentage.

To find the price of a product taking into account the discount you need to:

1. Subtract the discount percentage from $100%$.
2. Find the resulting percentage of the total cost of the product.

A winter jacket costs $4,500 rubles. Seasonal discount is $20%$. How much should I pay for a jacket taking into account the discount?

Let's find what percentage of the initial cost the discounted jacket price will be:

Let's calculate how much $80%$ of $4500$ rubles is. To find the percentage of a number, you need to divide the given number by $100$ and multiply by the percentage value.

$(4500·80)/(100)=$3600 - the cost of the jacket taking into account the discount.

Tasks for deposits, loans, markups

To find the amount of money taking into account the annual rate, you need to:

1. Add the annual percentage of the deposit to $100%$.
2. Find the resulting percentage of the original amount of money.

The client deposited 150,000 rubles into the bank at $12% per annum. How much can he withdraw in a year?

$100%+12%=112%$ is the percentage of the client's money after a year relative to the original amount.

Let's find $112%$ of $150,000$ rubles:

$(112⋅150000)/(100)=$168000 rubles.

Answer: $168,000$.

In some percentage problems it is convenient to use proportion, for example:

A bag of potatoes cost $200 rubles. After the price increase, it began to cost $250$ rubles. By what percentage was the price of a bag of potatoes increased?

Let’s take the initial cost of the product as $100%$ (since it is with this that we will compare the cost after the price increase):

Let $x%$ be the percentage of the new price relative to the old one.

With these data we will compose and solve the proportion:

$(100%)/(x%)=(200)/(250)$.

The product of the extreme terms of a proportion is equal to the product of the middle terms of the proportion:

$200⋅х=100⋅250$.

$х=(100⋅250)/(200)=125%$.

The new cost of a bag of potatoes is $125%$ relative to the original price.

The price has increased by $125%-100%=25%$.

Answer: $25$.

The math workbook costs $65$ rubles. How many notebooks can a student buy for $450 rubles if there is a $8%$ discount?

Let's find the percentage of the cost of the notebook taking into account the discount:

Let's find $92%$ of $65$ rubles and get the price of $1$ notebook at a discount:

${450}/{59.8}={4500}/{598}≈7.5$

We cannot buy a fractional number of notebooks; there is not enough money for eight notebooks, so the student will only be able to buy $7$ of notebooks.

Answer: $7$.

To solve some problems you need to be familiar with the term "compound interest", which is often needed to solve problems about deposits, loans, etc. In simple words, “compound interest” occurs when we compound interest on interest. Let's look at it with an example.

Let's say we deposited $X$ rubles in the bank at $N%$ per annum. And they left the money in the bank for not one, but two years. This means that at the end of the first year we could take away $X + X*(N/100) = X(1+(N/100))$ rubles, but we don’t take them, but leave them for the second year. And now, as it were, the amount of our “new” contribution for the second year at $N%$ is no longer $X$, but $X(1+(N/100))$ rubles. That is, during the second year, interest will be accrued, including on the interest accumulated during the first year. Total at the end of the second year we will be able to take $X(1+(N/100)) + X(1+(N/100))*(N/100) = X(1+(N/100))(1+ (N/100)) = X(1+(N/100))^2$.
If we made a deposit not for two, but for $Y$ years, then at the end we would receive $X(1+(N/100))^Y$ rubles.

“A good teacher must understand that no task can be exhausted to the end. He should instill this view in his students.”
D. Polya.

Introduction.

I pay special attention word problems on percentages that are often found in practice entrance exams V economic universities, but are not sufficiently addressed in school. The ability to perform percentage calculations is certainly one of the most necessary mathematical competencies. However, it is not only those who have already graduated from school a long time ago who are timid at the sight of interest. Even on the Unified State Exam, the solvability of problems involving percentages does not exceed 20%. This suggests that this type of problem should be solved not only in junior classes where this topic is studied, but also throughout all years of schooling.

1. When solving problems involving percentages, the following basic formulas are used:

1% of a is equal to a.

p% of the number a is equal to a.

If it is known that a certain number a is p% of x, then x can be found from the proportion

A− р%

X − 100%,

whence x=a.

Let there be numbers a, b, and a

Number b is 100% greater than number a.

Number a is 100% less than number b.

2. Compound interest formula.

If the deposit contains an amount of a monetary units, the bank charges p% per annum, then after n years the amount on the deposit will be

a monetary units

3. Problems involving percentages.

Task 1.

There are 45% fewer smart people than beautiful people; 36% of smart people have a beautiful appearance. What is the percentage of smart people among beautiful people?

Solution: let x be the number of beautiful people, then the number of smart people:

x − 0.45x = 0.55x.

Among smart people, 36% are beautiful people, therefore, the number of smart and at the same time beautiful people:

0.36 ·0.55x= 0.198x.

Let's make a proportion:

From here we get:

Answer: 19,8%

Students are interested in solving word problems involving percentages that are closer to real life. A special “fun” is the presentation of problems not from a problem book, but directly from a newspaper page. Here there are no thoughts about the uselessness of mathematics. And “interest journalism” is literally flourishing on the pages of newspapers in connection with the outbreak of the economic crisis.

Task 2.

Prices for tours have already increased: for example, tours to France - by 20%. Is it possible to say how many percent earlier a tour to France was cheaper?

Solution: let x be the old price and n be the new price.

1) Let's make the first proportion:

We get n=1.2x.

2) Let's make the second proportion:

x − (100-a%)

(100-a) 1.2x = 100x

Having solved the equation, we get: a ≈17%.

Answer: 17%.

4. Using the compound interest formula.

Task 3.

10 thousand rubles were deposited into the bank account. After the money had been lying around for one year, 1 thousand rubles were withdrawn from the account. A year later, there were 11 thousand rubles in the account. Determine what percentage per annum the bank charges.

Solution: Let the bank charge p% per annum.

1) The amount of 10,000 rubles deposited on a bank account at p% per annum will increase in a year to the amount

2) When 1000 rubles are withdrawn from the account, it will remain there 9000+100rub rub.

3) In another year, the latter value, due to the accrual of interest, will increase to the value

By condition, this value is equal to 11000:

Solving this equation we get: =10, =−200 - a negative root is not suitable.

Answer: 10%

Task 4. (Unified State Exam-2015)

The bank accepted a certain amount at a certain percentage. A year later, a quarter of the accumulated amount was withdrawn from the account. But the bank increased the interest rate per annum by 40%. By the end of next year, the amount accumulated 1.44 times exceeded the initial investment. What is the new APR percentage?

Solution: The situation will not change depending on the deposit amount. Let's put it in the bank 4 ruble (divided into 4 ). In a year, the amount in the account will increase exactly p times and will become equal (4p) rubles

Let's divide it by 4 parts, we'll take them home (p) rubles, we'll leave it in the bank (3p) rubles

It is known that by the end of next year the bank contained 4 1.44 = 5.76 rubles So the number (3p) turned into a number (5,76) . How many times did it increase?

Thus, the second increasing coefficient has been found k jar.

Interestingly, the product of both coefficients is equal to 1,92 :

It follows from the condition that the second coefficient on 0,4 more than the first.

Having gotten rid of the commas, let's make a replacement t = 10r:

From such an equation it is quite easy to get 12.

So p = 1.2, k = 1.6.

The deposit amount increased 1.2 times the first time, 1.6 times the second time.

It was 100%, it became 160%. The new percentage per annum is 160%-100% = 60%.

Answer: 60%.

Task 5. (Unified State Exam 2015)

Amount deposited in the bank 3900 thousand rubles under 50% per annum. At the end of each of the first four years of storage, after calculating interest, the depositor made an additional deposit of the same fixed amount into the account. By the end of the fifth year, after calculating interest, it turned out that

The size of the deposit increased compared to the initial one by 725% .

What amount did the investor add to the deposit annually?

Solution: Let x rubles be added annually by the depositor to the deposit.

50% per annum means that every year the amount in the depositor’s account increases by 1.5 times. If the investor did not add anything to the initial amount, then after a year there would be 3900·1.5, two years later - 3900·1.52 and so on.

Let's calculate how much income all four supplements brought.

x∙1.5 4 + x∙1.5 3 + x∙1.5 2 +x∙1.5

To do this, let's take out X outside the bracket and calculate the sum of the geometric progression in which b = 1.5 And q = 1.5.

It is known that the size of the deposit increased compared to the initial one by 725% .