Compound interest in exam problems.

Solving math problems using basic percentage concepts.

Problems involving percentages are taught to solve from the 5th grade.

Solving problems of this type is closely related to three algorithms:

1. finding the percentage of a number,
2. finding a number by its percentage,
3. finding the percentage.

During lessons with students, they understand that a hundredth of a meter is a centimeter, a hundredth of a ruble is a penny, and a hundredth of a centner is a kilogram. People have long noticed that hundredths of quantities are convenient in practical activities. That's why a special name was invented for them - percentage.

This means one kopeck is one percent of one ruble, and one centimeter is one percent of one meter.

One percent is one hundredth of a number. Mathematical signs one percent is written as 1%.

The definition of one percent can be written as: 1% = 0.01. A

5%=0.05, 23%=0.23, 130%=1.3, etc.

How to find 1% of a number?

Since 1% is one hundredth part, you need to divide the number by 100. Division by 100 can be replaced by multiplying by 0.01. Therefore, to find 1% of a given number, you need to multiply it by 0.01. And if you need to find 5% of a number, then multiply given number by 0.05, etc.

Example. Find: 25% of 120.

1. 25% = 0,25;
2. 120 . 0,25 = 30.

Rule 1. To find a given number of percents of a number, you need to write down the percentages decimal, and then multiply the number by this decimal fraction.

Example. The turner turned 40 parts in an hour. Using a cutter made of stronger steel, he began turning 10 more parts per hour. By what percentage did the turner's labor productivity increase?

To solve this problem, we need to find out what percentage 10 parts are from 40. To do this, let’s first find what part the number 10 is from the number 40. We know that we need to divide 10 by 40. The result is 0.25. Now let's write it down as a percentage - 25%.

Answer: lathe worker productivity increased by 25%.

Rule 2. To find what percentage one number is of another, you need to divide the first number by the second and write the resulting fraction as a percentage.

Example. With a planned target of 60 cars per day, the plant produced 66 cars. To what percentage did the plant fulfill the plan?

66: 60 = 1.1 - this part is made up of manufactured cars from the number of cars according to the plan. Let's write it as a percentage = 110%.

Answer: 110%.

Example. Bronze is an alloy of tin and copper. What percentage of the alloy is copper in a piece of bronze consisting of 6 kg of tin and 34 kg of copper?

1. 6+ 34 =40 (kg) - the mass of the entire alloy.
2. 34: 40 = 0.85 = 85 (%) - the alloy is copper.

Answer: 85%.

Example. The baby elephant lost 20% weight in the spring, then gained 30% weight over the summer, lost 20% weight again in the fall, and gained 10% weight over the winter. Has his weight remained the same this year? If it has changed, by what percentage and in what direction?

1. 100 - 20 = 80 (%) - after spring.
2. 80 + 80 . 0.3 = 104 (%) - after summer.
3. 104 - 104. 0.2 = 83.2 (%) - after autumn.
4. 83.2 + 83.2. 0.1 = 91.52 (%) - after winter.

Answer: lost 8.48% weight.

Example. We left 20 kg of gooseberries for storage, the berries of which contain 99% water. The water content in the berries decreased to 98%. How many gooseberries will you get as a result?

1. 100 - 99 = 1 (%) = 0.01 - the proportion of dry matter in the gooseberries first.
2. 20 . 0.01 = 0.2 (kg) - dry matter.
3. 100 - 98 = 2 (%) = 0.02 - the proportion of dry matter in gooseberries after storage.
4. 0.2: 0.02 = 10 (kg) - became gooseberries.

Answer: 10 kg.

Example. What will happen to the price of a product if it is first increased by 25% and then decreased by 25%?

Let the price of the product be x rub., then after the increase the product costs 125% of the previous price, i.e. 1.25x, and after a reduction of 25%, its cost is 75% or 0.75 of the increased price, i.e.

0.75 .1.25x= 0.9375x,

then the price of the goods decreased by 6.25%, because

x - 0.9375x = 0.0625x;
0,0625 . 100% = 6,25%

Answer: The original price of the product decreased by 6.25%.

Rule 3. To find percentage two numbers A and B, you need to multiply the ratio of these numbers by 100%, that is, calculate (A: B). 100%.

Example. Find a number if 15% of it is 30.

1. 15% = 0,15;
2. 30: 0,15 = 200.

x - given number;
0.15. x = 300;
x = 200.

Answer: 200.

Example. Raw cotton produces 24% fiber. How much raw cotton does it take to get 480 kg of fiber?

Let's write 24% as a decimal fraction 0.24 and get the problem of finding a number from its known part (fraction).
480: 0.24= 2000 kg = 2 t

Answer: 2 t.

Example. How many kg of porcini mushrooms must be collected to obtain 1 kg of dried mushrooms, if when processing fresh mushrooms, 50% of their mass remains, and when drying, 10% of the mass of processed mushrooms remains?

1 kg of dried mushrooms is 10% or 0.01 part processed, i.e.
1 kg: 0.1=10 kg of processed mushrooms, which is 50% or 0.5 collected mushrooms, i.e.
10 kg: 0.05=20 kg.

Answer: 20 kg.

Example. Fresh mushrooms contained 90% water by weight, and dry mushrooms contained 12%. How many dried mushrooms will you get from 22 kg of fresh mushrooms?

1. 22. 0.1 = 2.2 (kg) - mushrooms by mass in fresh mushrooms; (0.1 is 10% dry matter);
2. 2.2: 0.88 = 2.5 (kg) - dry mushrooms obtained from fresh ones (the amount of dry matter has not changed, but its percentage in mushrooms and now 2.2 kg is 88% or 0.88 dry mushrooms).

Answer: 2.5 kg.

Rule 4. To find a number given its percentages, you must express the percentages as a fraction, and then divide the percentage value by this fraction.

In problems involving bank calculations, simple and compound interest are usually encountered. What is the difference between simple and compound interest growth? With simple growth, the percentage is calculated each time based on initial value, and with complex growth it is calculated from the previous value. With simple growth, 100% is the initial amount, and with complex growth, 100% is new each time and equal to the previous value.

Example. The bank pays an income of 4% per month from the deposit amount. 300 thousand rubles were deposited into the account, income is accrued every month. Calculate the amount of the deposit after 3 months.

1. 100 + 4 = 104 (%) = 1.04 - the share of the increase in the deposit compared to the previous month.
2. 300. 1.04 = 312 (thousand rubles) - the amount of the deposit after 1 month.
3. 312. 1.04 = 324.48 (thousand rubles) - the amount of the deposit after 2 months.
4. 324.48. 1.04 = 337.4592 (thousand rubles) = 337,459.2 (r) - the amount of the deposit after 3 months.

Or you can replace points 2-4 with one, repeating the concept of degree with the children: 300.1,043 = 337.4592 (thousand rubles) = 337,459.2 (r) - the amount of the contribution after 3 months.

Answer: 337,459.2 rubles

Example. Vasya read in the newspaper that over the past 3 months, food prices have increased by an average of 10% each month. By what percentage have prices increased in 3 months?

Example. Money invested in shares of a well-known company brings 20% income annually. In how many years will the invested amount double?

Let's look at a similar task plan using specific examples.

Example. (Option 1 No. 16. OGE-2016. Mathematics. Typical test. assignments_ed. Yashchenko_2016 -80s)

The sports store is holding a promotion. Any jumper costs 400 rubles. When you buy two jumpers, you get a 75% discount on the second jumper. How many rubles will you have to pay to buy two jumpers during the promotion period?

According to the conditions of the problem, it turns out that the first jumper is bought for 100% of its original cost, and the second for 100 - 75 = 25 (%), i.e. In total, the buyer must pay 100 + 25 = 125 (%) of the original cost. The solution can then be considered in three ways.

1 way.

We accept 400 rubles as 100%. Then 1% contains 400: 100 = 4 (rub.), and 125%
4 . 125 = 500 (rub.)

Method 2.

The percentage of a number is found by multiplying the number by the fraction corresponding to the percentage or multiplying the number by the given percentage and dividing by 100.
400. 1.25 = 500 or 400. 125/100 = 500.

3 way.

Applying the proportion property:
400 rub. - 100 %
x rub. - 125%, we get x = 125. 400 / 100 = 500 (rub.)

Answer: 500 rubles.

Example. (Option 4 No. 16. OGE-2016. Mathematics. Typical test. assignments_ed. Yashchenko_2016 -80s)

The average weight of boys of the same age as Gosha is 57 kg. Gosha's weight is 150% of average weight. How many kilograms does Gosha weigh?

Similar to the example discussed above, you can create a proportion:

57 kg - 100%
x kg - 150%, we get x = 57. 150 / 100 = 85.5 (kg)

Answer: 85.5 kg.

Example. (Option 7 No. 16. OGE-2016. Mathematics. Typical test. assignments_ed. Yashchenko_2016 - 80s)

After the TV was marked down, its new price was 0.52 of the old one. By what percentage did the price decrease as a result of the markdown?

1 way.

Let's first find the fraction of the price decrease. If the original price is taken to be 1, then 1 - 0.52 = 0.48 is the share of the price reduction. Then we get 0.48. 100% = 48%. Those. The price decreased by 48% as a result of markdown.

Method 2.

If the original price is taken as A, then after markdown the new price of the TV will be equal to 0.52A, i.e. it will decrease by A - 0.52A = 0.48A.

Let's make a proportion:
A - 100%
0.48A - x%, we get x = 0.48A. 100/A = 48 (%).

Answer: the price decreased by 48% as a result of markdown.

Example. (Option 9 No. 16. OGE-2016. Mathematics. Typical test. assignments_ed. Yashchenko_2016 - 80s)

The item on sale was discounted by 15%, and it now cost 680 rubles. How many rubles did the product cost before the sale?

Before the price reduction, the product was worth 100%. The price of the product after the sale decreased by 15%, i.e. became 100 - 15 = 85 (%), in rubles this value is equal to 680 rubles.

1 way.

680: 85 = 8 (rub.) - in 1%
8 . 100 = 800 (rub.) - the cost of the product before the sale.

Method 2.

This problem of finding a number by its percentage is solved by dividing the number by the corresponding percentage and by converting the resulting fraction into a percentage, multiplying by 100, or by dividing by the fraction obtained when converting from percentages.
680:85. 100 = 800 (rub.) or 680: 0.85 = 800 (rub.)

3 way.

Using proportion:
680 rub. - 85%
x rub. - 100%, we get x = 680. 100 / 85 = 800 (rub.)

Answer: The item cost 800 rubles before the sale.

Solving problems on mixtures and alloys, using the concepts of “percentage”, “concentration”, “% solution”.

The most simple tasks of this type are given below.

Example. How many kg of salt are in 10 kg of salt water if the percentage of salt is 15%.

10 . 0.15 = 1.5 (kg) salt.

Answer: 1.5 kg.

The percentage of a substance in a solution (for example, 15%) is sometimes called a % solution (for example, 15% salt solution).

Example. The alloy contains 10 kg of tin and 15 kg of zinc. What is the percentage of tin and zinc in the alloy?

The percentage of a substance in an alloy is the portion that makes up the weight of this substance from the weight of the entire alloy.

1. 10 + 15 = 25 (kg) - alloy;
2. 10:25. 100% = 40% - percentage of tin in the alloy;
3. 15:25. 100% = 60% - percentage of zinc in the alloy.

Answer: 40%, 60%.

In tasks of this type, the main concept is “concentration”. What is it?

Consider, for example, a solution of acid in water.

Let the vessel contain 10 liters of solution, which consists of 3 liters of acid and 7 liters of water. Then the relative (relative to the entire volume) acid content in the solution is equal. This number determines the concentration of acid in the solution. Sometimes they talk about the percentage of acid in a solution. In the example given, the percentage would be: . As you can see, the transition from concentration to percentage and vice versa is very simple.

So, let a mixture of mass M contain some substance of mass m.

• the concentration of a given substance in a mixture (alloy) is called a quantity;
• the percentage of a given substance is called the value c×100%;

From the last formula it follows that with known concentrations of the substance and total mass mixture (alloy), the mass of a given substance is determined by the formula m=c×M.

Problems involving mixtures (alloys) can be divided into two types:

1. For example, two mixtures (alloys) with masses m1 and m2 and with concentrations of some substance in them equal to c1 and c2, respectively, are specified. The mixtures (alloys) are drained (fused). It is required to determine the mass of this substance in the new mixture (alloy) and its new concentration. It is clear that in the new mixture (alloy) the mass of this substance is equal to c1m1 + c2m2, and the concentration.
2. A certain volume of the mixture (alloy) is specified and from this volume they begin to cast (remove) a certain amount of the mixture (alloy), and then add (add) the same or a different amount of the mixture (alloy) with the same concentration of a given substance or with a different concentration. This operation is performed several times.

When solving such problems, it is necessary to establish control over the amount of this substance and its concentration at each low tide, as well as with each addition of the mixture. As a result of such control, we obtain a resolving equation. Let's look at specific tasks.

If the concentration of a substance in a compound by mass is P%, then this means that the mass of this substance is P% of the mass of the entire compound.

Example. The silver concentration in the 300 g alloy is 87%. This means that there is 261 g of pure silver in the alloy.

300. 0.87 = 261 (g).

In this example, the concentration of the substance is expressed as a percentage.

The ratio of the volume of a pure component in a solution to the entire volume of the mixture is called the volumetric concentration of this component.

The sum of the concentrations of all components making up the mixture is equal to 1.

If the percentage of a substance is known, then its concentration is found using the formula:
K = P/100%,
where K is the concentration of the substance;
P is the percentage of the substance (in percent).

Example. (Option 8 No. 22. OGE-2016. Mathematics. Typical test. assignments_ed. Yashchenko_2016 - 80s)

Fresh fruits contain 75% water, while dried fruits contain 25%. How much fresh fruit is required to prepare 45 kg of dried fruit?

If fresh fruits contain 75% water, then the dry matter will be 100 - 75 = 25 (%), and dried fruits will contain 25%, then the dry matter will be 100 - 25 = 75 (%).

When formulating a solution to a problem, you can use the table:

Fresh fruit x 25% = 0.25 0.25. X

Dried fruits 45 75% = 0.75 0.75. 45 = 33.75

Because the mass of dry matter for fresh and dried fruits does not change, we get the equation:

0.25. x = 33.75;
x = 33.75: 0.25;
x = 135 (kg) - fresh fruit is required.

Answer: 135 kg.

Example. (Option 8 No. 11. Unified State Exam-2016. Mathematics. Typical test. ed. Yashchenko 2016 -56s)

Mixing 70% and 60% acid solutions and adding 2 kg clean water, obtained a 50% acid solution. If instead of 2 kg of water we added 2 kg of a 90% solution of the same acid, we would get a 70% acid solution. How many kilograms of 70% solution were used to obtain the mixture?

Total weight, kg | Dry matter concentration | Dry weight
I x 70% = 0.7 0.7. X
II for 60% = 0.6 0.6. at
water 2 - -
I + II + water x + y + 2 50% = 0.5 0.5. (x + y + 2)
III 2 90% = 0.9 0.9. 2 = 1.8
I + II + III x + y + 2 70% = 0.7 0.7. (x + y + 2)

Using the last column from the table, we create 2 equations:

0.7. x + 0.6. y = 0.5. (x + y + 2) and 0.7. x + 0.6. y + 1.8 = 0.7. (x + y + 2).

Combining them into a system and solving it, we get that x = 3 kg.

Answer: 3 kilograms of a 70% solution was used to obtain the mixture.

Example. (Option 2 No. 11. Unified State Exam-2016. Mathematics. Typical test. ed. Yashchenko 2016 -56s)

Three kilograms of cherries cost the same as five kilograms of cherries, and three kilograms of cherries cost the same as two kilograms of strawberries. By what percentage is a kilogram of strawberries cheaper than a kilogram of cherries?

From the first sentence of the problem we obtain the following equalities:

3h = 5v,
3v = 2k.
From which we can express: h = 5v/3, k = 3v/2.

This way you can create a proportion:
5v/3 - 100%
3v/2 - x%, we get x = (3.100.v.3)/(2.5.v), x = 90% is the cost of a kilogram of strawberries from the cost of a kilogram of cherries.

This means that 100 - 90 = 10 (%) - a kilogram of strawberries is cheaper than a kilogram of cherries.

Answer: a kilogram of strawberries is 10 percent cheaper than a kilogram of cherries.

Solving problems involving “compound” interest, using the concept of an increase (decrease) factor.

To increase positive number And by p percent, you should multiply the number A by the increase factor K = (1 + 0.01p).

To reduce a positive number A by p percent, you should multiply the number A by the reduction factor K = (1 - 0.01p).

Example. (Option 29 No. 22. OGE-2015. Mathematics. Type. exam options: 36 options / ed. Yashchenko, 2015 - 224c)

The price of the product was reduced twice by the same percentage. By what percentage did the price of the product decrease each time if its initial cost was 5,000 rubles and the final cost was 4,050 rubles?

1 way.

Because the price of the product decreased by the same number of %, let’s denote the number of % as x. Let the price of the product be reduced by x% for the first and second time, then after the first reduction the price of the product became (100 - x)%.

Let's make a proportion
5000 rub. - 100%
in rubles - (100 - x)%, we get y = 5000. (100 - x) / 100 = 50. (100 - x) rubles - the cost of the goods after the first reduction.

Let's compose new proportion already at the new price:
50 . (100 - x) rub. - 100%
z rub. - (100 - x)%, we get z = 50. (100 - x) (100 - x) / 100 = 0.5. (100 - x) 2 rubles - the cost of the goods after the second reduction.

We get the equation 0.5. (100 - x)2 = 4050. Having solved it, we find that x = 10%.

Method 2.

Because the price of the product decreased by the same number %, let's denote the number % by x, x % = 0.01 x.

Using the concept of reduction factor, we immediately obtain the equation:
5000. (1 - 0.01x)2 = 4050.

Answer: the price of the product decreased by 10% each time.

Example. (Option 30 No. 22. OGE-2015. Mathematics. Typical exam options: 36 options / edited by Yashchenko, 2015 - 224s)

The price of the goods was increased twice by the same percentage. By what percentage did the price of the product increase each time if its initial cost was 3,000 rubles and the final cost was 3,630 rubles?

Because the price of the product increased by the same number%, let's denote the number% by x, x% = 0.01 x.

Using the concept of magnification factor, we immediately obtain the equation:
3000. (1 + 0.01x)2 = 3630.

Having solved it, we find that x = 10%.

Answer: the price of the product increased by 10% each time.

Example. (Option 4 No. 11. Unified State Exam-2016. Mathematics. Typical test. ed. Yashchenko 2016 -56s)

On Thursday, the company's shares rose in price by a certain number of percent, and on Friday they fell in price by the same number of percent. As a result, they began to cost 9% cheaper than at the opening of trading on Thursday. By what percentage did the company's shares rise in price on Thursday?

Let the company's shares rise in price and fall in price by x%, x% = 0.01 x, and the initial price of the shares was A. Using all the conditions of the problem, we obtain the equation:

(1 + 0.01 x)(1 - 0.01 x)A = (1 - 0.09)A,
1 - (0.01 x)2 = 0.91,
(0.01 x)2 = (0.3)2,
0.01 x = 0.3,
x = 30%.

Answer: The company's shares rose by 30 percent on Thursday.

Solving banking problems in new version Unified State Exam 2016 in mathematics.

Example. (Option 2 No. 17. Unified State Exam-2016. Mathematics. 50 types. version ed. Yashchenko 2016)

On January 15th it is planned to take out a bank loan for 15 months. The conditions for its return are as follows:

It is known that the eighth payment amounted to 108 thousand rubles. What amount must be returned to the bank during the entire loan term?

From the 2nd to the 14th the payment is made A/15 +0.01A.

After which the amount of debt will be 1.01A - A/15 - 0.01A = 14A/15.

After 2 months we get: 1.01. 14A/15.

Second payment A/15 + 0.01. 14A/15.

Then the debt after the second payment is 13A/15.

Similarly, we find that the eighth payment will look like:

A/15 + 0.01. 8A/15 = A/15. (1 + 0.08) = 1.08A/15.

And according to the condition, it is equal to 108 thousand rubles. This means we can create and solve the equation:

1.08A/15 = 108,

A=1500 (thousand rubles) - the initial amount of debt.

2) To find the amount that needs to be returned to the bank over the entire loan period, we must find the sum of all payments on the loan.

The amount of all loan payments will be as follows:

(A/15 + 0.01A) + (A/15 + 0.01. 14A/15) + (A/15 + 0.01. 13A/15) + … + (A/15 + 0.01. A /15) = A + 0.01A/15 (15+14+13+12+11+10+9+8+7+6+5+4+3+2+1) = A + (0.01. 120A)/15 = 1.08A.

So 1.08. 1500 = 1620 (thousand rubles) = 1,620,000 rubles must be returned to the bank during the entire loan term.

Answer: 1,620,000 rubles.

Example. (Option 6 No. 17. Unified State Exam-2016. Mathematics. 50 types. version ed. Yashchenko 2016)

On January 15th it is planned to take out a bank loan for 24 months. The conditions for its return are as follows:

• On the 1st of each month, the debt increases by 1% compared to the end of the previous month;
• from the 2nd to the 14th of each month it is necessary to repay part of the debt;
• On the 15th of each month, the debt must be the same amount less than the debt on the 15th of the previous month.

It is known that in the first 12 months you need to pay the bank 177.75 thousand rubles. How much do you plan to borrow?

1) Let A be the loan amount, 1% = 0.01.

Then 1.01A debt after the first month.

From the 2nd to the 14th the payment is made A/24 +0.01A.

After which the amount of debt will be 1.01A - A/24 - 0.01A = A - A/24 = 23A/24.

With this scheme, the debt becomes the same amount less than the debt on the 15th day of the previous month.

After 2 months we get: 1.01. 23A/24.

Second payment A/24 + 0.01. 23A/24.

Then the debt after the second payment is 1.01. 23A/24 - A/24 - 0.01. 23A/24 = 23A/24(1.01 - 0.01) - A/24 = 23A/24 - A/24 = 22A/24.

Thus, we get that for the first 12 months you need to pay the bank the following amount:
A/24 +0.01A. 24/24 + A/24 + 0.01. 23A/24 + A/24 + 0.01. 22A/24 + … + A/24 + 0.01. 13A/24 =12A/24 + 0.01A/24 (24+23+22+21+20+19+18+17+16+15+14+13) = A/2 + 222A/2400 = 711A/1200 .

And according to the condition, it is equal to 177.375 thousand rubles. This means we can create and solve the equation:
711A/1200 = 177.75,
A = 300 (thousand rubles) = 300,000 rubles - it is planned to take on credit.

Answer: 300,000 rubles.

Let's talk about tasks No. 19 of the Unified State Exam

For two years now, a task has been added to the second part c economic content, i.e. problems on complex bank interest.

They say that we are dealing with “compound interest” in the case when a certain value is subject to gradual change. Moreover, each time its change is a certain number of percent of the value that this value had at the previous stage.

At the end of each stage the value changes to the same constant quantity percent –R%. Then at the endn th stage the value of a certain quantityA , the initial value of which was equal toA 0 , is determined by the formula:

With increasing and

When decreasing

Knowing that the annual interest rate of the deposit is 12%, find

its equivalent monthly interest rate.

Solution:

If you put A rubles in the bank, then after a year we get:A 1 =A 0 (1 +0,12)

If interest was calculated every month at the interest rateX , then according to the compound interest formula after a year (12 months)A n =A 0 (1 + 0.01x) 12

Equating these values, we obtain an equation whose solution will allow us to determine the monthly interest rateA(1 +0.12) = A(1 +0.01x) 12

1.12 = (1 + 0.01x) 12

x = (-1) 100% ≈ 0.9488792934583046%

Answer: The monthly interest rate is0.9488792934583046%.

From the solution to this problem you can see that the monthly interest rate is not equal to the annual rate divided by 12.

On December 31, 2013, Sergey took out 9,930,000 rubles on credit from the bank at 10% 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 10%), then Sergey transfers a certain amount of the annual payment to the bank. What should be the amount of the annual payment for Sergei to pay off the debt in three equal annual payments?

Solution:

Let the loan amount beA , the annual payment is equal toX rubles, and the annual amount is k % . Then on December 31 of each year the remaining debt amount is multiplied by the coefficient m =1+ 0,01 k . After the first payment, the amount of debt will be: A 1 = am - X. After the second payment, the amount of debt

will be:

A 2 = a 1 m – x=(at-x)t-x=a 2 -th-x=at 2 -(1+t)x

According to the condition, Sergey must repay the loan in full in three payments, therefore

where

Ata = 9930000 Andk =10 , we getT =1.1 and

Answer : 3,993,000 rubles.

Now that we have dealt with this solution proposed in all the solvers, let's look at another solution.

LetF = 9,930,000 – loan amount,x – the required amount of the annual payment.

First year:

Duty:1.1F ;

Payment:X ;

Remainder:1.1F-x .

Second year:

Duty:1.1(1.1F-x) ;

Payment:X ;

Remainder:1.1(1.1F-x)-x .

Third year:

Duty:1.1(1.1F-x)-x );

Payment:X ;

Balance: 0, because according to the condition there were only three payments.

The only equation

1.1(1.1(1.1F-x)-x)-x=0 . 1,331 F =3.31x, x=3993000

Answer: 3,993,000 rubles.

However-1 ! If we assume that the interest rate is not a beautiful 10%, but a terrible 13.66613%. The chances of dying somewhere during the multiplications or going crazy when detailing the multiplier for the amount of debt for each year have increased sharply. Let's add to this not just a small 3 years, but 25 years. This solution will not work.

On December 31, 2014, Andrey took out a certain amount on credit from the bank at 10% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (that is, increases the debt by 10%), and then Andrey transfers 3,460,600 rubles to the bank. What amount did Andrei take from the bank if he paid off the debt in three equal payments (that is, over 3 years)?

Solution.

LetA – the required quantity,k% – interest rate on the loan,X – annual payment. Then on December 31 of each year the remaining debt amount will be multiplied by the coefficientm = 1 + 0.01k . After the first payment, the amount of debt will be:A 1 = am – x . After the second payment, the amount of debt will be:

A 2 = a 1 m – x=(at-x)t-x=a 2 -th-x=at 2 -(1+t)x

After the third payment, the amount of remaining debt:

According to the terms, Andrey paid off the debt in three years,

that isA 3 = 0 , where.

Atx = 3,460,600, k% = 10% , we get:m = 1.1 And=8 606 000 (rubles).

Answer: 8,606,000 rubles.

On December 31, 2013, Igor took out 100,000 rubles on credit from the bank. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (that is, it increases the debt by a certain amount of interest), then Igor transfers the next tranche. Igor repaid the loan in two tranches, transferring 51,000 rubles the first time and 66,600 rubles the second. At what percentage did the bank issue a loan to Igor?

Solution

Letk % – the required loan rate;m = (1 + 0.01 k ) – multiplier of the remaining debt;a = 100,000 – amount borrowed from the bank;x 1 = 51 000, x 2 = 66 600 – dimensions of the first and last trenches.

After the first payment, the amount of debt will be:a 1 = ma – x 1 .

After the second payment, the amount of debt will be:a 2 = ma 1 – x 2 = a m 2 – m x 1 – x 2 . By condition,a 2 = 0 . The equation will need to be solved first with respect tom , of course, taking only positive root:

100 000m 2 – 51,000m – 66,600 = 0; 500m 2 – 255m – 333 = 0.

This is where the difficulties begin.

D = 255 2 + 4∙500∙333= 15 2 ∙ 17 2 + 15 2 ∙37∙80= 15 2 (289+ 2 960) = 15 2 ∙3249=15 2 ∙3 2 ∙19 2 .

Then.

Answer: 11%.

On December 31, 2013, Masha took out a certain amount on credit from the bank at a certain percentage per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (that is, it increases the debt by a certain amount of interest), then Masha transfers the next tranche. If she pays 2,788,425 rubles every year, she will pay off the debt in 4 years. If 4,991,625 each, then in 2 years. At what percentage did Masha take the money from the bank?

Solution

After two years of repayment, the amount of the loan taken is calculated using the formula:

After four years of repayment, the amount of the loan taken is calculated using the formula:

Where

Then.

Answer: 12.5%.

On December 31, 2013, Vanya took out 9,009,000 rubles on credit from the bank at 20% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (that is, increases the debt by 20%), then Vanya transfers the payment to the bank. Vanya paid off the entire debt in 3 equal payments. How many less rubles would he give to the bank if he could pay off the debt in 2 equal payments?

Solution

Let's use the result from problem 2.

The required differenceX 3 -X 2 =34 276 800 – 25896800= 1 036 800 rubles

Answer: 1,036,00 rubles.

On June 1, 2013, Vsevolod Yaroslavovich took out 900,000 rubles on credit from the bank. The loan repayment scheme is as follows: on the 1st of each next month, the bank charges 1 percent on the remaining amount of the debt (that is, increases the debt by 1%), then Vsevolod Yaroslavovich transfers the payment to the bank. For what minimal amount months Vsevolod Yaroslavovich can take out a loan so that monthly payments are no more than 300,000 rubles?

We need to understand simple truth– the higher the loan payment, the lower the debt. The less debt you have, the faster you will pay it off. The maximum monthly payment that the lender can afford is 300,000 rubles according to the condition. If Vsevolod Yaroslavovich pays the maximum payment, he will pay off the debt the fastest. In other words, he will be able to take out a loan for the shortest period of time, which is required by the condition.

Let's try to solve the problem head-on.

A month has passed. July 1, 2013: debt (1 + 0.01)900,000 – 300,000 = 609,000.

A month has passed. August 1, 2013: debt (1+ 0.01)609,000 – 300,000 = 315,090.

A month has passed. September 1, 2013: debt (1 +0.01)315,090 – 300,000= 18,240.9. A month has passed. October 1, 2013: debt (1 0.01)1,240.9 = 18,423.309<300 000, кредит погашен. Итого прошло 4 месяца.

Answer: 4 months.

Let's solve the problem using the standard method.

I will use the results of Problem 3 taking into account the following reasoning: the inequality of the remaining part of the debt has the forma x ≤ 0 .

Letx – the required quantity,a = 900,000 - amount borrowed from the bank,k% = 1% – loan rate,y = 300,000 - monthly payment,m = (1 + 0.01k) – monthly multiplier of the remaining debt. Then, according to the already known formula, we obtain the inequality: ≤0 ;

We got an unpleasant inequality, but a true one.

We take the integer part of the number because the number of payments cannot be a non-integer number. We take the nearest larger integer, we cannot take the smaller one (because then there will be a debt) and it is clear that the resulting logarithm is not an integer. It turns out 4 payments, 4 months.

The farmer received a loan from a bank at a certain percentage per annum. A year later, in order to repay the loan, the farmer returned to the bank the entire amount that he owed the bank by that time, and a year later, in order to fully repay the loan, he deposited into the bank an amount that was 21% greater than the amount of the loan received. What is the annual interest rate on a loan from this bank?

Solution:

The loan amount does not affect the situation. Let's take 4 rubles from the bank (divisible by 4).

In a year, the debt to the bank will increase by exactlyX times and will become equal4x rubles

Let's divide it into 4 parts and return it3x rubles and we must stayX rubles

It is known that by the end of next year we will have to pay4 1.21 rubles

It is known that the amount of debt for the year turned from the numberX in numberX 2 .

Since the farmer fully repaid the debt after two years, then

X 2 = 4 1.21 x = 2 1.1 x = 2.2

CoefficientX means that 100% turns into 220% in a year.

This means that the bank’s annual percentage is: 220% - 100%

Answer: 120%

The amount of 3,900 thousand rubles was placed in the bank at 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 accrual of interest, it turned out that the size of the deposit had increased by 725% compared to the original. What amount did the investor add to the deposit annually?

Solution:

Let the deposited amount be fixedX rubles

Then, after all operations were carried out, after the first year, the amount on the deposit became

+x

After 2 years

After3 of the year

After4 of the year

After5 of the year

Since by the end of the fifth year after accrual of interest it turned out that the size of the deposit had increased by 725% compared to the initial one, we will draw up the equation:

3900 ·8.25=3900·1.5 5 +x·(1.5 4 +1,5 3 +1,5 2 +1,5) /:1,5

3900·5.5=3900·1.5 4 +x(1.5 3 +1,5 2 +1,5+1)

Answer: 210 rubles.

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 annual interest rate by 40%. By the end of the next year, the amount accumulated was 1.44 times the initial contribution. What is the new APR percentage?

Solution:

The situation will not change depending on the deposit amount. Let's put 4 rubles in the bank (divided by 4).

In a year, the amount in the account will increase by exactlyp times and will become equal4p rubles

Let's divide it into 4 parts and take it homep rubles, we'll leave it in the bank3p rubles

It is known that by the end of the next year there were 4·1.44 = 5.76 rubles in the bank.

So the number3p turned into the number 5.76. How many times did it increase?

Thus, the second increasing coefficient has been foundx jar.

Interestingly, the product of both coefficients is 1.92:

It follows from the condition that the second coefficient is 0.4 greater than the first.

p · x = p ·( p +0,4)=1,92

Already now the coefficients can be selected: 1.2 and 1.6.

But let us continue, however, to solve the equation:

10p ·(10p+4)=192 let 10p=k

k ·(k+4)=192

k =12, i.e. p=1.2; and x=1.6

Answer: 60%

Today we will take a little break from standard logarithms, integrals, trigonometry, etc., and together we will look at a more vital problem from the Unified State Exam in mathematics, which is directly related to our backward Russian raw materials economy. To be precise, we will consider a problem about deposits, interest and loans. Because problems with percentages have recently been added to the second part of the unified state exam in mathematics. Let me make a reservation right away that for solving this problem, according to the Unified State Exam specifications, three primary points are offered at once, i.e., examiners consider this task one of the most difficult.

At the same time, to solve any of the specified problems from the Unified State Exam in mathematics, you need to know only two formulas, each of which is quite accessible to any school graduate, however, for reasons unknown to me, these formulas are completely ignored by both school teachers and the compilers of all kinds of problems for preparation to the Unified State Exam. Therefore, today I will not just tell you what these formulas are and how to apply them, but I will derive each of these formulas literally before your eyes, taking as a basis tasks from the open bank of the Unified State Exam in mathematics.

Therefore, the lesson turned out to be quite voluminous, quite informative, so make yourself comfortable, and we will begin.

We invest money in the bank

First of all, I would like to make a small digression related to finance, banks, loans and deposits, on the basis of which we will obtain the formulas that we will use to solve this problem. So, let's take a little break from exams, from upcoming school problems, and look to the future.

Let's say you have grown up and are going to buy an apartment. Let's say you are going to buy not some bad apartment on the outskirts, but a good quality apartment for 20 million rubles. At the same time, let’s also assume that you got a more or less normal job and earn 300 thousand rubles a month. In this case, you can save approximately three million rubles per year. Of course, earning 300 thousand rubles a month, in a year you will get a slightly larger amount - 3,600,000 - but let these 600,000 be spent on food, clothing and other daily household joys. The total input data is as follows: you need to earn twenty million rubles, but we only have three million rubles a year at our disposal. A natural question arises: how many years do we need to save three million each in order to get these same twenty million? This is considered elementary:

\[\frac(20)(3)=6,....\to 7\]

However, as we have already noted, you earn 300 thousand rubles a month, this means that you are smart people and will not put money “under the pillow”, but take it to the bank. And, therefore, interest will be accrued annually on those deposits that you bring to the bank. Let's say you choose a reliable, but at the same time more or less profitable bank, and therefore your deposits will grow annually by 15% per annum. In other words, we can say that the amount in your accounts will increase annually by 1.15 times. Let me remind you the formula:

Let's calculate how much money will be in your accounts after each year:

In the first year, when you just start saving money, no interest will accumulate, i.e. at the end of the year you will save three million rubles:

At the end of the second year, interest will already be accrued on the three million rubles remaining from the first year, i.e. we need to multiply by 1.15. However, during the second year you also reported another three million rubles. Of course, interest has not yet accrued on these three million, because by the end of the second year these three million had just appeared in the account:

So, third year. At the end of the third year, interest will be charged on this amount, i.e. this entire amount must be multiplied by 1.15. And again, you worked hard throughout the year and also saved three million rubles:

\[\left(3m\cdot 1.15+3m \right)\cdot 1.15+3m\]

Let's calculate another fourth year. Again, the entire amount that we had at the end of the third year is multiplied by 1.15, i.e. Interest will be charged on the entire amount. This includes interest on interest. And another three million is added to this amount, because during the fourth year you also worked and also saved money:

\[\left(\left(3m\cdot 1.15+3m \right)\cdot 1.15+3m \right)\cdot 1.15+3m\]

Now let's open the brackets and see what amount we will have by the end of the fourth year of saving money:

\[\begin(align)& \left(\left(3m\cdot 1.15+3m \right)\cdot 1.15+3m \right)\cdot 1.15+3m= \\& =\left( 3m\cdot ((1,15)^(2))+3m\cdot 1,15+3m \right)\cdot 1,15+3m= \\& =3m\cdot ((1,15)^(3 ))+3m\cdot ((1.15)^(2))+3m\cdot 1.15+3m= \\& =3m\left(((1.15)^(3))+((1 ,15)^(2))+1.15+1 \right)= \\& =3m\left(1+1.15+((1.15)^(2))+((1.15) ^(3)) \right) \\\end(align)\]

As you can see, we have elements of a geometric progression in brackets, i.e. we have the sum of the elements of a geometric progression.

Let me remind you that if a geometric progression is given by the element $((b)_(1))$, as well as the denominator $q$, then the sum of the elements will be calculated using the following formula:

This formula must be known and clearly applied.

Please note: formula n-th element sounds like this:

\[((b)_(n))=((b)_(1))\cdot ((q)^(n-1))\]

Because of this degree, many students get confused. In total we have just n for the amount n- elements, and himself n The th element has degree $n-1$. In other words, if we now try to calculate the sum of a geometric progression, then we need to consider the following:

\[\begin(align)& ((b)_(1))=1 \\& q=1.15 \\\end(align)\]

\[((S)_(4))=1\cdot \frac(((1.15)^(4))-1)(1.15-1)\]

Let's calculate the numerator separately:

\[((1.15)^(4))=((\left(((1.15)^(2)) \right))^(2))=((\left(1.3225 \right ))^(2))=1.74900625\approx 1.75\]

In total, returning to the sum of the geometric progression, we get:

\[((S)_(4))=1\cdot \frac(1.75-1)(0.15)=\frac(0.75)(0.15)=\frac(75)(15 )=5\]

As a result, we get that after four years of savings, our initial amount will increase not four times, as if we had not put the money in the bank, but five times, i.e. fifteen million. Let's write this down separately:

4 years → 5 times

Looking ahead, I will say that if we saved not for four years, but for five years, then in the end our amount of savings would increase by 6.7 times:

5 years → 6.7 times

In other words, by the end of the fifth year we would have the following amount in the account:

That is, by the end of the fifth year of savings, taking into account interest on the deposit, we would have already received over twenty million rubles. Thus, the total savings account due to bank interest would be reduced from almost seven years to five years, i.e., by almost two years.

Thus, even despite the fact that the bank charges a fairly low interest rate on our deposits (15%), after just five years this same 15% gives an increase that significantly exceeds our annual earnings. Moreover, the main multiplier effect occurs in recent years and even, rather, in the last year of savings.

Why did I write all this? Of course, this is not to encourage you to take money to the bank. Because if you really want to increase your savings, then you need to invest them not in a bank, but in a real operating business, where these same interests, i.e., profitability in the Russian economy rarely falls below 30%, i.e. twice as much bank deposits.

But what is really useful in all these discussions is a formula that allows us to find the total amount of the deposit through the amount of annual payments, as well as through the interest charged by the bank. So let's write it down:

\[\text(Vklad)=\text(platezh)\frac(((\text(%))^(n))-1)(\text(%)-1)\]

The % itself is calculated using the following formula:

This formula also needs to be known, as well as the basic formula for the deposit amount. And, in turn, the basic formula can significantly reduce calculations in those problems with percentages where it is necessary to calculate the contribution.

Why use formulas rather than tables?

Many will probably have a question: why all this complexity? Is it not possible to simply write down each year on a table, as is done in many textbooks, count each year separately, and then calculate the total amount of the contribution? Of course, you can completely forget about the sum of the geometric progression and calculate everything using classic tablets - this is done in most collections for preparing for the Unified State Exam. However, firstly, the volume of calculations increases sharply, and secondly, as a consequence, the likelihood of making an error increases.

And in general, using tables instead of this wonderful formula is the same as digging trenches at a construction site with your hands instead of using a standing excavator standing nearby and fully working.

Well, or the same thing as multiplying five by ten not using the multiplication table, but adding five to itself ten times in a row. However, I’ve already digressed, so I’ll repeat the most important idea once again: if there is some way to simplify and shorten the calculations, then this is the method that should be used.

Interest on loans

We have dealt with deposits, so we move on to the next topic, namely, interest on loans.

So, while you are saving money, carefully planning your budget, thinking about your future apartment, your classmate, and now a simple unemployed person, decided to live for today and simply took out a loan. At the same time, he will still tease and laugh at you, saying that he has a credit phone and a used car, taken on credit, and you still take the subway and use an old push-button telephone. Of course, your former classmate will have to pay dearly for all these cheap “show-offs”. How expensive it is - that’s what we’ll calculate right now.

First, a brief introductory information. Let's say your former classmate took out two million rubles on credit. Moreover, according to the agreement, he must pay x rubles per month. Let’s say that he took out a loan at an annual rate of 20%, which in current conditions looks quite decent. Additionally, let's assume that the loan term is only three months. Let's try to connect all these quantities into one formula.

So, at the very beginning, as soon as your former classmate left the bank, he has two million in his pocket, and this is his debt. Moreover, not a year has passed, and not a month, but this is only the very beginning:

Then, after one month, interest will be charged on the amount owed. As we already know, to calculate interest, it is enough to multiply the original debt by a coefficient, which is calculated using the following formula:

In our case, we are talking about a rate of 20% per annum, i.e. we can write:

This is the coefficient of the amount that will be accrued per year. However, our classmate is not very smart and he did not read the contract, and in fact, he was given a loan not at 20% per year, but at 20% per month. And by the end of the first month, interest will be accrued on this amount, and it will increase by 1.2 times. Immediately after this, the person will need to pay the agreed amount, i.e. x rubles per month:

\[\left(2m\cdot 1,2- x\right)\cdot 1,2-x\]

And again our guy makes a payment in the amount of $x$ rubles.

Then, by the end of the third month, the amount of his debt increases again by 20%:

\[\left(\left(2m\cdot 1,2- x\right)\cdot 1,2- x\right)1,2- x\]

And according to the condition, he must pay in full within three months, i.e., after making the last third payment, his amount of debt must be equal to zero. We can write the following equation:

\[\left(\left(2m\cdot 1,2- x\right)\cdot 1,2- x\right)1,2 - x=0\]

Let's decide:

\[\begin(align)& \left(2m\cdot ((1,2)^(2))- x\cdot 1,2- x\right)\cdot 1,2- x=0 \\& 2m \cdot ((1,2)^(3))- x\cdot ((1,2)^(2))- x\cdot 1,2- x=0 \\& 2m\cdot ((1,2 )^(3))=\cdot ((1,2)^(2))+\cdot 1,2+ \\& 2m\cdot ((1,2)^(3))=\left((( 1,2)^(2))+1,2+1 \right) \\\end(align)\]

Before us again is a geometric progression, or rather, the sum of three elements of a geometric progression. Let's rewrite it in ascending order of elements:

Now we need to find the sum of three elements of a geometric progression. Let's write it down:

\[\begin(align)& ((b)_(1))=1; \\& q=1,2 \\\end(align)\]

Now let's find the sum of the geometric progression:

\[((S)_(3))=1\cdot \frac(((1,2)^(3))-1)(1,2-1)\]

It should be recalled that the sum of a geometric progression with such parameters $\left(((b)_(1));q \right)$ is calculated according to the formula:

\[((S)_(n))=((b)_(1))\cdot \frac(((q)^(n))-1)(q-1)\]

This is the formula we just used. We substitute this formula into our expression:

For further calculations, we need to find out what $((1,2)^(3))$ is equal to. Unfortunately, in this case we can no longer write it as last time in the form of a double square, but we can calculate it like this:

\[\begin(align)& ((1,2)^(3))=((1,2)^(2))\cdot 1,2 \\& ((1,2)^(3)) =1.44\cdot 1.2 \\& ((1,2)^(3))=1.728 \\\end(align)\]

Let's rewrite our expression:

This is a classic linear expression. Let's go back to the following formula:

In fact, if we generalize it, we get a formula connecting interest, loans, payments and terms. The formula is as follows:

Here it is, the most important formula of today's video lesson, with the help of which at least 80% of all economic problems from the Unified State Exam in mathematics in the second part are calculated.

Most often in real tasks you will be asked for a payment, or a little less often a loan, i.e. the total amount of debt that our classmate had at the very beginning of payments. In more complex problems, you will be asked to find the percentage, but for very complex ones, which we will analyze in a separate video lesson, you will be asked to find the time frame during which, given the loan and payment parameters, our unemployed classmate will be able to fully pay off the bank.

Perhaps someone will now think that I am a fierce opponent of loans, finance and the banking system in general. So, nothing like that! On the contrary, I believe that credit instruments are very useful and extremely necessary for our economy, but only on the condition that the loan is taken for business development. As a last resort, you can take out a loan to buy a home, i.e., a mortgage or for emergency medical treatment - that’s all; there are simply no other reasons to take out a loan. And all kinds of unemployed people who take out loans to buy “show-offs” and at the same time do not think at all about the consequences in the end and become the cause of crises and problems in our economy.

Returning to the topic of today's lesson, I would like to note that knowing this formula connecting loans, payments and interest is as necessary as the sum of a geometric progression. It is with the help of these formulas that real economic problems from the Unified State Examination in mathematics are solved. Well, now that you know all this very well, when you understand what a loan is and why you shouldn’t take it, let’s move on to solving real economic problems from the Unified State Examination in mathematics.

Solving real problems from the Unified State Examination in mathematics

Example #1

So, the first task:

On December 31, 2014, Alexey took out 9,282,000 rubles on credit from the bank at 10% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (i.e., increases the debt by 10%), then Alexey transfers X rubles to the bank. What should the amount X be for Alexey to pay off the debt in four equal payments (i.e. over four years)?

So, this is a problem about credit, so we immediately write down our formula:

The loan is known to us - 9,282,000 rubles.

We'll deal with percentages now. We are talking about 10% in the problem. Therefore, we can translate them:

We can create an equation:

We have obtained an ordinary linear equation for $x$, although with rather formidable coefficients. Let's try to solve it. First, let's find the expression $((1,1)^(4))$:

$\begin(align)& ((1,1)^(4))=((\left(((1,1)^(2)) \right))^(2)) \\& 1,1 \cdot 1,1=1,21 \\& ((1,1)^(4))=1,4641 \\\end(align)$

Now let's rewrite the equation:

\[\begin(align)& 9289000\cdot 1.4641=x\cdot \frac(1.4641-1)(0.1) \\& 9282000\cdot 1.4641=x\cdot \frac(0, 4641)(0.1)|:10000 \\& 9282000\cdot \frac(14641)(10000)=x\cdot \frac(4641)(1000) \\& \frac(9282\cdot 14641)(10) =x\cdot \frac(4641)(1000)|:\frac(4641)(1000) \\& x=\frac(9282\cdot 14641)(10)\cdot \frac(1000)(4641) \\ & x=\frac(2\cdot 14641\cdot 1000)(10) \\& x=200\cdot 14641 \\& x=2928200 \\\end(align)\]\[\]

That's it, our problem with interest is solved.

Of course, this was only the simplest problem with percentages from the Unified State Examination in mathematics. Most likely, such a task will not appear in the real exam. And if it does, consider yourself very lucky. Well, for those who like to count and don’t like to take risks, let’s move on to the next more complex tasks.

Example No. 2

On December 31, 2014, Stepan took out 4,004,000 rubles on credit from the bank at 20% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (i.e., increases the debt by 20%), then Stepan makes a payment to the bank. Stepan paid off the entire debt in 3 equal payments. How many less rubles would he give to the bank if he could pay off the debt in 2 equal payments?

We have a problem about loans, so we write down our formula:

\[\]\

What do we know? First, we know the total credit. We also know the percentages. Let's find the coefficient:

As for $n$, you need to carefully read the problem statement. That is, first we need to calculate how much he paid for three years, i.e. $n=3$, and then perform the same steps again but calculate payments for two years. Let's write the equation for the case when the payment is paid over three years:

Let's solve this equation. But first, let’s find the expression $((1,2)^(3))$:

\[\begin(align)& ((1,2)^(3))=1,2\cdot ((1,2)^(2)) \\& ((1,2)^(3)) =1.44\cdot 1.2 \\& ((1,2)^(3))=1.728 \\\end(align)\]

Let's rewrite our expression:

\[\begin(align)& 4004000\cdot 1.728=x\cdot \frac(1.728-1)(0.2) \\& 4004000\cdot \frac(1728)(1000)=x\cdot \frac(728 )(200)|:\frac(728)(200) \\& x=\frac(4004\cdot 1728\cdot 200)(728) \\& x=\frac(4004\cdot 216\cdot 200)( 91) \\& x=44\cdot 216\cdot 200 \\& x=8800\cdot 216 \\& x=1900800 \\\end(align)\]

In total, our payment will be 1,900,800 rubles. However, pay attention: in the problem we were required to find not the monthly payment, but how much Stepan would pay in total for three equal payments, i.e. for the entire time of using the loan. Therefore, the resulting value must be multiplied by three again. Let's count:

In total, Stepan will pay 5,702,400 rubles for three equal payments. This is how much it will cost him to use the loan for three years.

Now let's consider the second situation, when Stepan pulled himself together, pulled himself together and paid off the entire loan not in three, but in two equal payments. We write down our same formula:

\[\begin(align)& 4004000\cdot ((1,2)^(2))=x\cdot \frac(((1,2)^(2))-1)(1,2-1) \\& 4004000\cdot \frac(144)(100)=x\cdot \frac(11)(5)|\cdot \frac(5)(11) \\& x=\frac(40040\cdot 144\ cdot 5)(11) \\& x=3640\cdot 144\cdot 5=3640\cdot 720 \\& x=2620800 \\\end(align)\]

But that’s not all, because now we have calculated only one of the two payments, so in total Stepan will pay exactly twice as much:

Great, now we are closer to the final answer. But please note: in no case have we received the final answer yet, because for three years of payments Stepan will pay 5,702,400 rubles, and for two years of payments he will pay 5,241,600 rubles, i.e. a little less. How much less? To find out, you need to subtract the second payment amount from the first payment amount:

The total final answer is 460,800 rubles. Exactly how much Stepan will save if he pays not for three years, but for two.

As you can see, the formula connecting interest, terms and payments significantly simplifies calculations compared to classical tables and, unfortunately, for unknown reasons, in most collections of problems, nevertheless, tables are still used.

Separately, I would like to draw your attention to the term for which the loan was taken and the amount of monthly payments. The fact is that this connection is not directly visible from the formulas that we wrote down, but its understanding is necessary for quickly and effectively solving real problems on the exam. In fact, this connection is very simple: the longer the loan is taken out for, the smaller the amount will be in monthly payments, but the larger the amount will accumulate over the entire period of using the loan. And vice versa: the shorter the term, the higher the monthly payment, but the final overpayment is lower and the total cost of the loan is lower.

Of course, all these statements will be equal only if the loan amount and interest rate are the same in both cases. In general, for now just remember this fact - it will be used to solve the most complex problems on this topic, but for now we will analyze a simpler problem, where we just need to find the total amount of the original loan.

Example No. 3

So, one more task for credit and also the last task in today’s video lesson.

On December 31, 2014, Vasily took out a certain amount on credit from the bank at 13% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank charges interest on the remaining amount of the debt (i.e. increases the debt by 13%), then Vasily transfers 5,107,600 rubles to the bank. What amount did Vasily take from the bank if he paid off the debt in two equal payments (over two years)?

So, first of all, this problem is again about loans, so we write down our wonderful formula:

Let's see what we know from the problem statement. Firstly, the payment is equal to 5,107,600 rubles per year. Secondly, it's a percentage, so we can find the coefficient:

In addition, according to the conditions of the problem, Vasily took out a loan from the bank for two years, i.e. paid in two equal payments, therefore $n=2$. Let's substitute everything and also note that the loan is unknown to us, i.e. the amount he took, and let's denote it by $x$. We get:

\[{{1,13}^{2}}=1,2769\]

Let's rewrite our equation taking this fact into account:

\[\begin(align)& x\cdot \frac(12769)(10000)=5107600\cdot \frac(1.2769-1)(0.13) \\& x\cdot \frac(12769)(10000) )=\frac(5107600\cdot 2769)(1300)|:\frac(12769)(10000) \\& x=\frac(51076\cdot 2769)(13)\cdot \frac(10000)(12769)\ \& x=4\cdot 213\cdot 10000 \\& x=8520000 \\\end(align)\]

That's it, this is the final answer. This is exactly the amount Vasily took out on credit at the very beginning.

Now it’s clear why in this problem we are asked to take out a loan for only two years, because there are double-digit percentages involved, namely 13%, which when squared gives a rather “brutal” number. But this is not the limit - in the next separate lesson we will look at more complex problems where we will need to find the loan term, and the rate will be one, two or three percent.

In general, learn to solve problems on deposits and loans, prepare for exams and pass them “excellently”. And if something is unclear in the materials of today’s video lesson, then do not hesitate - write, call, and I will try to help you.

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 an 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 the trial Unified State Examination in mathematics in December of this year. 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:

	on Monday morning	on Monday night	on Tuesday evening
Share price

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”.

	husband	wife	daughter	Total income
In real
Situation
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: .

The next type of problems are problems involving solutions, mixtures and alloys. They are found not only in mathematics, but also in chemistry. We will tell you about the simplest way to solve them.

Liters of water were added to a vessel containing liters of a percentage aqueous solution of a certain substance. What percentage is the concentration of the resulting solution?

A picture helps in solving such problems. 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 roughly 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 a simple system of equations:

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

Solving, we get that .

Answer: .

By mixing -percentage and -percentage solutions of acid and adding kg of pure water, we obtained -percentage acid solution. 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 equal to . 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 movement problems. They also use the formula. But there is one trick that we will tell you about.

A cyclist left a point on the circular route, and minutes later a motorcyclist followed him. Minutes after leaving, 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:

cyclist
motorcyclist

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.

cyclist
motorcyclist

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 will the minute hand line up with the hour hand for the fourth time?

This is perhaps the most difficult task of the 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 ..
But what if you have an electronic watch and 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. The minute hand will travel a distance, and the hour hand will travel one more circle. 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, is the total distance, is the 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 . The total time is .
The average speed is 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 an hour. 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.

	performance	Job
Andrey
Pasha
Volodya
Together

Andrey and Pasha painted the fence in hours. We remember that when we work together, performance adds 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.

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Slide captions:

Theory on the topic: “Solving problems with percentages.”

Type 1: Converting percentages to decimals. percentages  fraction A%  A divided by 100 Problems:20%;75%;125%;50%;40%;1%;70%;35%;80%.... Fill in the table 1% 5% 10% 20% 25% 50% 75% 100%

Type 2: Converting fractions to percentages. number  percent A  A multiplied by 100% Convert fractions to percentages: 3/4; 0.07; 2.4. (GIA, thematic assignments) Match the fractions that express fractions of a certain quantity and the corresponding percentages. A.1/4; B)3/5; B) 0.5; D) 0.05 1) 5%; 2) 25%; 3) 50%; 4) 60% Answer: A B C D

Type 3: Finding the percentage of a number. X% of A 1) X% is represented as a decimal fraction 2) Number A is multiplied by a decimal fraction. The task is a sample. Within a month, the company produced 500 devices. 20% of manufactured devices failed to pass quality control. How many devices have not passed quality control? Solution. You need to find 20% of the total number of manufactured devices (500). 20% = 0.2. 500 * 0.2 = 100. 100 of the total number of manufactured devices did not pass quality control.

Type 4: Find a number by its percentage. And this is X%: 1) X% is represented as a decimal fraction 2) And we divide by a decimal fraction. The task is a sample. In preparation for the exam, the student solved 38 problems from the self-study manual. Which is 25% of the number of all tasks in the manual. How many tasks are collected in this self-study manual? Solution. We don't know how many tasks there are in the manual. But we know that 38 tasks is 25% of their total number. 25%=0.25 38/0.25 = 152. 152 problems in this collection.

Type 5: Find the percentage of two numbers. A and B numbers. What % is B of A? 1)B/A 2) Multiply the resulting quotient by 100% Problem - sample. There are 30 students in the class. 15 of them are girls. What percentage of girls are in the class? Solution. To find out what percentage one number is of another, you need to divide the number you want to find by the total and multiply by 100%. This means 1)15 /30=0.5 2)0.5*100%=50% The problem is a sample. In 1 hour, the automatic machine produced 240 parts. After the reconstruction of this machine, he began to produce 288 of the same parts per hour. By what percentage did the productivity of the machine increase? Solution. The productivity of the machine increased by 288-240=48 parts per hour. You need to find out what percentage of 240 parts are 48 parts. In order to find out what percentage the number 48 is from the number 240, you need to divide the number 48 by 240 and multiply the result by 100%. 48/240 *100% =20% Answer: machine productivity increased by 20%

Type 6: Increase the number by a percentage. Reduce the number by a percentage. A - number; if we increase it by X%, it will increase by (1 + x/100) times. : 1) multiply the number A by 2) (1 + x /100). The task is a sample. . On last year's math exam, 140 high school students received A's. This year the number of excellent students increased by 15%. How many people got A's on their math exam this year? Solution. 140 * (1 + 15/100) = 161. A is the number; decrease by X%, then it decreased by (1 - x /100) times. : 1) multiply number A by 2) (1 - x /100). The task is a sample. A year ago, 100 children graduated from school. And this year there are 25% fewer graduates. How many graduates this year? Solution. 100 * (1 – 25/100) = 75.

Type7: Solution concentration. The task is a sample. A kilogram of salt was dissolved in 9 liters of water. What is the concentration of the resulting solution? (The mass of 1 liter of water is 1 kg) (Peterson 6 cl.) Solution 1) The mass of the dissolved substance is 1 kg 2) The mass of the entire solution 1 + 9 = 10 (kg) 9 kg is the mass of water in the solution (not to be confused with the total mass of the solution ) 3)1/10*100%=10% 10% is the concentration of the solution

Type 8: Percentage of metal in the alloy. Problem – sample 1. There is a piece of copper-tin alloy with a total mass of 12 kg, containing 45% copper. How much pure tin must be added to this piece of alloy so that the resulting alloy contains 40% copper? Solution.1)12. 0.45= 5.4 (kg) - pure copper in the first alloy; 2) 5.4: 0.4= 13.5 (kg) - the weight of the new alloy; 3) 13.5- 12 = 1.5 (kg) tin. Answer: you need 1.5 kg of tin.

Problem - sample 2. There are two alloys consisting of copper, zinc and tin. It is known that the first alloy contains 40% tin, and the second - 26% copper. The percentage of zinc in the first and second alloys is the same. Having alloyed 150 kg of the first alloy and 250 kg of the second, we obtained a new alloy, which contained 30% zinc. Determine how many kilograms of tin are contained in the resulting new alloy. Since the percentage of zinc in the first and second alloys is the same and in the third alloy it turned out to be 30%, then in the first and second alloys the percentage of zinc is 30%. 250*0.3= 75 (kg) - zinc in the second alloy; 250 * 0.26= 65 (kg) - copper in the second alloy; 250-(75+65)= 110 (kg) tin in the second alloy; 150. 0.4= 60 (kg) - tin in the first alloy; 110 + 60 = 170 (kg) - tin in the third alloy. Answer: 170 kg. 1 alloy 2 alloy New alloy (3) Copper 26% Zinc 30% 30% 30% Tin 40% ? kg weight 150 kg 250 kg 150+250=400

Type 9: For “dry matter”. Almost any product - apples, watermelons, mushrooms, potatoes, cereals, bread, etc. consists of water and dry matter. Moreover, both fresh and dried foods contain water. During the drying process, only water evaporates, and the mass of dry matter does not change. A.G. Mordkovich “Mathematics 6” Problem No. 362 Problem - sample. Fresh mushroom contains 90% water, while dried mushroom contains 15%. How many dried mushrooms will you get from 17 kg of fresh mushrooms? How many fresh mushrooms do you need to take to get 3.4 kg of dried ones? Solution. Let's make a table: Part 1 of the problem: substance Mass of the substance (kg) Percentage of water Percentage of dry matter Mass of dry matter (kg) Fresh mushroom 17kg 90% 10% 17*0.1=1.7 Dried mushroom X kg 15% 85% X*o.85 = 0.85x Since the mass of dry matter in dry and fresh mushrooms remains unchanged, we obtain the equation: 0.85x = 1.7, x = 1.7: 0.85, x = 2.

Part 2 of the problem: Substance Mass of the substance (kg) Percentage content of water Percentage content of water Dry matter mass (kg) Fresh mushroom x 90% 10% 0.1x Dried mushroom 3.4 15% 85% 3.4*0.85=2 .89 0.1x = 2.89, x = 2.89: 0.1, x = 28.9. Answer: from 17 kg of fresh mushrooms you will get 2 kg of dried ones; to get 3.4 kg of dried mushrooms, you need to take 28.9 kg of fresh ones.