Question

ПО ВКЛАДУ В БАНК В ТЕЧЕНИЕ ТРЕХ ЛЕТ В КОНЦЕ КАЖДОГО ГОЖА УВЕЛИЧИВАЕТ НА 20% сумму, имеющуюся на вкладе в начале года, а по вкладу б увеличивает на 21%, в течение которого из первых двух лет. Найдите наименьшее целое чтсло процентов за третий год по веладк б при котором за все три года этот вклад все еще останется выгоднее вклада а

Answer 1

Пусть на каж­дый тип вкла­да была вне­се­на сумма S. На вкла­де «А» каж­дый год сумма уве­ли­чи­ва­ет­ся на 20%, т. е. умно­жа­ет­ся на ко­эф­фи­ци­ент 1,2.

Тогда через три года сумма на вкла­де «А» равна 1,23S = 1,728S.

 На вкла­де «Б» сумма через три года будет равна

1,21^2*(1+n/100)S=1,4641(1+n/100)S, n ∈z

По усло­вию тре­бу­ет­ся найти наи­мень­шее целое ре­ше­ние не­ра­вен­ства

1,4641(1+n/100)S>1,728S

n>100*(17280-14641)/14641≈18,02

Ответ 19

 

Answer 2

Calculation of Interest on Bank Deposit

To calculate the interest on a bank deposit over a period of three years, we need to consider two different rates of interest. In the first two years, the deposit increases by 20% of the initial amount, while in the third year, it increases by a certain percentage.

Let's assume the initial amount of the deposit is x.

In the first year, the deposit increases by 20% of x, resulting in a new amount of 1.2x.

In the second year, the deposit again increases by 20% of the new amount, which is 1.2x. Therefore, the new amount at the end of the second year is 1.2x + 0.2(1.2x) = 1.44x.

Now, in the third year, the deposit increases by a certain percentage, which we need to find. Let's assume this percentage is y.

The new amount at the end of the third year will be 1.44x + y(1.44x).

To determine the minimum integer value of y for which the deposit remains more profitable than the initial deposit after three years, we need to compare the new amount with the initial amount.

The condition for the deposit to remain more profitable is:

1.44x + y(1.44x) > x

Simplifying the equation, we get:

1.44 + 1.44y > 1

Now, let's solve this inequality to find the minimum integer value of y.

1.44 + 1.44y > 1

Subtracting 1.44 from both sides:

1.44y > 1 - 1.44

1.44y > -0.44

Dividing both sides by 1.44:

y > -0.44 / 1.44

y > -0.3056

Since we are looking for the minimum integer value of y, we round up to the nearest integer, which is 0.

Therefore, the minimum integer value of y for which the deposit remains more profitable than the initial deposit after three years is 0.

Please note that this calculation assumes the interest is compounded annually.