Question

P=10,000
R= 12 %
T=2 year


Find
year
the compound Interset in solution​

Answer 1

Answer:

1254.4

Step-by-step explanation:

P(1+R/100)²= compound interest for two years

so,. = 1000(1+12/100)²

=1000(112/100)²

= 1000×112/100×112/100

= 112×112/10

=11.2×112

=1254.4 ₹

Answer 2

Given:-

• P(Principal) = Rs. 10000
• R (Rate of interest) = 12%
• T (Time period) = 2 years.

To find:-

• Compound interest.

Solution:-

We know that,

$\large \boxed{ \star\sf \: Compound \: interest = Amount - Principal }$

• We do not have Amount. So, We will find Amount first by using compounded annually formula that is:

$\large \boxed{ \star \sf \: \: Amount = Principal {( 1 + \dfrac{Rate}{100} )}^{Time} }$

So,

P = Rs. 10000

R = 12%

T = 2 years.

Put all above values in compounded annually formula:

$\implies \sf 10000 \times {(1 + \dfrac{12}{100} )}^{2}$

$\implies \sf 10000 \times {( \dfrac{100 + 12}{100}) }^{2}$

$\implies \sf 10000 \times {( \dfrac{112}{100} )}^{2}$

$\implies \sf 1 \cancel{00} \cancel{00} \times \dfrac{112}{1 \cancel{00}} \times \dfrac{112}{1 \cancel{00}}$

$\implies \sf 112 \times 112$

$\implies \sf 12544$

Amount = Rs. 12544

Compound interest = Amount - Principal

= 12544 - 10000

= 2544

Therefore,

Compound interest is Rs. 2544.