Question

find the CI and A ,when p= rs.5000, R=9% ,T=2years​

Answer 1

Answer

A=P(1+R/100)^T

=5000(100+9/100)^2

=5000*109/100*109/100

=5*11881/10

=59405/10

=Rs.5940.5

CI= A-P

= 5940.5-5000

=Rs. 940.5

Answer 2

ＧＩＶＥＮ:-

• Principal = Rs. 5000
• Rate = 9%
• Time = 2 years

Ｔｏ Ｆｉｎｄ:-

• Compound Interest
• Ａｍｏｕｎｔ

ＳＯＬＵＴＩＯＮ:-

We know that,

$\large{\green{\underline{\boxed{\bf{Amount=P\left(1+\dfrac{R}{100}\right)^n}}}}}$

where,

• P is principal = Rs.5000
• R is rate = 9%
• n is time = 2 years

Putting the values,

$\large\implies{\sf{Amount=5000\left(1+\dfrac{9}{100}\right)^2}}$

$\large\implies{\sf{Amount=5000\left(\dfrac{100+9}{100}\right)^2}}$

$\large\implies{\sf{Amount=5000\times\left(\dfrac{109}{100}\right)^2}}$

$\large\implies{\sf{Amount=5000\times\left(\dfrac{109}{100}\times\dfrac{109}{100}\right)}}$

$\large\implies{\sf{Amount=5000\times\dfrac{109}{100}\times\dfrac{109}{100}}}$

$\large\implies{\sf{Amount=5\cancel{0}\cancel{00}\times\dfrac{109}{1\cancel{00}}\times\dfrac{109}{10\cancel{0}}}}$

$\large\implies{\sf{Amount=\dfrac{5\times109\times109}{10}}}$

$\large\implies{\sf{Amount=\dfrac{59405}{10}}}$

$\large\therefore\boxed{\bf{Amount=Rs.5940.5}}$

Now,

$\large{\green{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

where,

• Amount is Rs.5940.5
• Principal is Rs.5000

Putting the values,

$\large\implies{\sf{Compound\:Interest=Rs.(5940.5-5000)}}$

$\large\therefore\boxed{\bf{Compound\:Interest=Rs.940.5}}$

1)) Amount = Rs.5940.5

2)) Compound Interest = Rs.940.5