Question

a certain sum of money is invested at the rate of 10% per annum compound interest ,the interest compounded annually. if the difference between the interest of third year and first year is Rs. 1105 ,find the sum invested

Answer 1

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Answer 2

Answer:

The sum invested is Rs 4783.55 .

Step-by-step explanation:

Formula

$Amount = P(1 +\frac{r}{100})^{t}$

As

Amount = Principle + Interest

Thus

$Principle + Interest = P(1 +\frac{r}{100})^{t}$

$Interest = P(1 +\frac{r}{100})^{t}- P$

$Interest = P[(1 +\frac{r}{100})^{t}- 1]$

Where P is the principle , r is the rate of interest and t is the timein years .

As given

A certain sum of money is invested at the rate of 10% per annum compound interest ,the interest compounded annually.

If the difference between the interest of third year and first year is Rs.1105 .

Thus

1105 = Interest compounded in three years - Interest compounded in 1 years

$1105= P[(1 +\frac{10}{100})^{3}- 1]- P[(1 +\frac{10}{100})^{1}- 1]$

$1105= P[(1.1)^{3}- 1]- P[(1.1)^{1}- 1]$

$1105= P[(1.1)^{3} -(1.1)^{1}]$

$1105= P[ 1.331-1.1]$

1105 = 0.231P

$P = \frac{1105}{0.231}$

P = Rs 4783.55 (Approx)

Therefore the sum invested is Rs 4783.55 .