Question

a man deposits rs 25000 for 2 years at the annual rate of 6 % in the P N B. howmuch interest will the bank pay to him? ​

Answer 1

Answer:

The bank has to pay an interest of Rs 3090

Step-by-step explanation:

Process = 1

Formula = S.I = P×R×T ÷ 100

Capital = P = Rs 25000

Rate of interest = R = 6%

Time given = 2 yrs ( annually)

Since the term means Interest receive in one year

So S. I.= P×R×T÷100 = 25000×6×1÷100

=> S.I. = Rs 1500

Now for 2nd Year ,

New Capital = P = Rs(25000+1500) =Rs 26500

So, s.i. = P×R×T÷100 = 26500×6×1÷100

=> s.i. = Rs 1590

So , total interest band will have to pay to him is

S.I. + s.i. = Rs (1500 + 1590) = Rs 3090. (Ans)

Process = 2

Formula = c.i.=

$p(1 + \frac{r}{n \times 100} ) {}^{n \times t} - p$

where p = Capital

= Capital r = rate of interest

= Capital r = rate of interest n = terms ( quarter, half, annually)

= Capital r = rate of interest n = terms ( quarter, half, annually) t =. time period

We have ,

Capital = p = Rs 25000

Rate of interest = r = 6%

n = 1 ( since annually)

Time period = t = 2 yrs

$= > c.i. = {p(1 + \frac{r}{n \times 100 }) }^{n \times t} - p\\ \\ = > c.i. = 25000(1 + \frac{6}{1 \times 100} ) {}^{2 \times 1} - p \\ \\ = > c.i. = 25000(1 + 0.06) {}^{2} - 25000 \\ \\ = > c.i. = 25000 \times 1.06 {}^{2} - 25000\\ \\ = > c.i. = 28090 - 25000 = 3090$

Thus , required interest bank has to pay is Rs 3090 ( ans)