Question

find the amount and the compound interest on 25000 for 3 years at 12% per annum, compounded annually​

Answer 1

• Amount = 35123.2
• Compound interest is 10123.2 .

Step-by-step explanation:

Given:-

• Principal is 25000.
• Rate of interest is 12%.
• Time period is 3 years.

To find:-

• Amount.
• Compound interest.

Solution:-

We know that,

$\boxed { \sf \star \: \bold{Amount = P ( 1 + \frac{R}{100} )^{n}}}$

In which,

• P is principal.
• R is Rate of interest.
• n is time period.

Put the values in formula :

$\longrightarrow \sf \bold{25000 \times (1 + \cfrac{12}{100})^{3} }$

$\longrightarrow \sf \bold{25000 \times ( \cfrac{100 + 12}{100}) ^{3} }$

$\longrightarrow \sf \bold{25000 \times \cfrac{112}{100} \times \cfrac{112}{100} \times \cfrac{112}{100} }$

$\longrightarrow \sf \bold{ \cfrac{351232000}{10000} }$

$\longrightarrow \sf \bold{35123.2}$

Amount = 35123.2

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

$\longrightarrow \sf \bold{35123.2 - 25000}$

$\longrightarrow \sf \bold{10123.2}$

Compound interest is 10123.2 .

Answer 2

$\huge\bold{\mathtt{Question⇒}}$

Find the amount and the compound interest on ₹ 25000 for 3 years at 12% per annum, compounded annually.

$\huge\bold{\mathtt{Given⇒}}$

Principal = ₹ 25000

Rate of interest = 12%

Time = 3

$\huge\bold{\mathtt{To\:find⇒}}$

The amount and the compound interest.

$\huge\bold{\mathtt{Solution⇒}}$

We know that:-

• $\large\boxed{\mathtt{A=P(1+{\frac{R}{100}})^{n}}}$

Where,

• A = Amount

• P = Principal

• R = Rate of interest

• n = Time

Here,

• P = ₹ 25000

• R = 12%

• n = 3

First, we have to find the amount.

$\large{\mathtt{25000×(1+{\frac{12}{100}})^{3}}}$

$\large{\mathtt{=25000×({\frac{100+12}{100}})^{3}}}$

$\large{\mathtt{=25000({\frac{112}{100}})^{3}}}$

$\large{\mathtt{=250\cancel{00}×{\frac{112}{1\cancel{00}}}×{\frac{112}{100}}×{\frac{112}{100}}}}$

$\large{\mathtt{=250×{\frac{112}{1}}×{\frac{112}{100}}×{\frac{112}{100}}}}$

$\large{\mathtt{={\frac{(250×112×112×112)}{(1×100×100)}}}}$

$\large{\mathtt{={\frac{351232000}{10000}}}}$

$\large{\mathtt{=35123.2}}$

∴ Amount = ₹ 35123.2

We know that:-

$\large\boxed{\mathtt{CI=A-P}}$

Where,

• CI = Compound Interest

• A = Amount

• P = Principal

Now, we have to find the compound interest.

$\large{\mathtt{35123.2-25000}}$

$\large{\mathtt{=10123.2}}$

∴ Compound Interest = ₹ 10123.2

$\huge\bold{\mathtt{Therefore⇒}}$

The amount is ₹ 35123.2 and compound interest is ₹ 10123.2.

$\huge\bold{\mathtt{Done࿐}}$

$\large\bold{\mathtt{Hope\:this\:helps\:you.}}$

$\large\bold{\mathtt{Have\:a\:nice\:day.}}$