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3 years ago

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Safia opens a new savings account with a 2.25% simple interest rate. She deposits $3,500 and earned $3,696.88 total (with intere

st) by the time she checks the account again. How long was the money left untouched in the savings account?

1 answer:

Art [367]3 years ago

8 0

Answer:

The money was left untouched for 2 years and 6 months or just 2.5 years

Step-by-step explanation:

In this question, we are asked to calculate the time taken for an amount deposited to earn a certain amount of interest.

Mathematically, simple interest can be calculated using the formula;

I = PRT/100

where I is the interest accrued which is 3696.88 - 3,500 = 196.88

P is the principal which is the amount deposited = 3,500 according to the question

R is the rate which is 2.25% according to the question

T is the time which we are to find

we can rearrange the formula making T the subject and we obtain the following;

T = 100I/PR

substituting the values listed above we have;

T = (100 * 196.88)/(3500 * 2.25)

T = 19688/7875

T = 2.5 years or 2 years 6 months

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Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

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$P_{n-1} = \frac{109}{100}P_{n-2}$ .

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$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

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$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

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$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

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Simplifying the expression:

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The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

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