Question

If P=₹1000,R= 10% and N=2yr, then calculate the compound interest ​

Answer 1

Answer:

Solution:

Given

Principal (p) = Rs50000

Rate of interest (r) = 8% p.a

Let the amount at the end of the second year = A

\boxed{Amount = p\times(1+\frac{r}{100})^{n}}

Amount=p×(1+

100

r

)

n

i)

Let the number of times interest paid in 2 years ( n)=2

A = 50000[1+8/100]²

= 50000 × (108/100)²

= 5×108×108

= Rs58320

Therefore,

The amount standing to her

credit at the end of the second

year (A) = Rs58320

ii ) Finding interest for third year:

P = Rs 58320,

R = 8%

T = 1 year

Simple Interest (I) = (PTR)/100

= ( 58320 × 1 × 8 )/100

= Rs 4665.60

••••

Answer 2

Step-by-step explanation:

TO FIND:-

If P=₹1000,R= 10% and N=2yr, then calculate the compound interest

$\huge \tt\underline{\text{A} \blue{N} \orange{S} \pink{W} \green{E} \red{R}} :) :)$

Here, Principal P = Rs. 1000

R =10% per annum and n=2 years

$\tt{∴ Amount \: after \: 2 \: years = P (1+ \frac{r}{200} {)}^{2n} }$

$\tt{= > Rs. 1000×(1+ \frac{10}{200} {)}^{2 \times 2}} \\ \tt{ = > Rs.1000×(1+ \frac{1}{20} {)}^{2}} \\ \tt{= > Rs.1000×( \frac{21}{20} {)}^{4}}$

$= Rs.1000× \frac{21}{20} × \frac{21}{20} × \frac{21}{20} × \frac{21}{20} \\ \tt{= > Rs.1215.50 }$

Hence, compound interest = Amount - Principal

=Rs.1215.50−Rs.1000

=>Rs.215.50

Hope it helpful...☃️⚡️

#Somya Here.

Mark as brainliest please... ⚡️