Question

Calculate the compound interest for the second
year on * 12,000 invested for 3 years at 10%
per year. Also, find the sum due at the end of
the third year.​

Answer 1

Given :

• Principal = Rs. 12,000

• Time = 3 years

• Rate of interest = 10% p.a.

To find :

• Compound interest for second year !

• The sum at the end of third year !

Solution :

To find the compound interest for second year :

$:\implies \sf{A = P\bigg(1 + \dfrac{R}{100}\bigg)^{n}}$

Where :-

• P = Principal
• R = Rate of interest
• n = Time period
• A = Amount

$:\implies \sf{A = 12000 \times \bigg(1 + \dfrac{10}{100}\bigg)^{3}}$

$:\implies \sf{A = 12000 \times \bigg(\dfrac{100 + 10}{100}\bigg)^{3}}$

$:\implies \sf{A = 12000 \times \bigg(\dfrac{110}{100}\bigg)^{3}}$

$:\implies \sf{A = 12000 \times \dfrac{110}{10} \times \dfrac{110}{10} \times \dfrac{110}{10}}$

$:\implies \sf{A = 15972}$

$\underline{\therefore \sf{Amount\:(A) = Rs. 15972}}$

Thus, the amount for second year is Rs 15972.

And we know that Compound Interest is the difference of the amount and the principal.

→ CI = A - P

→ CI = 15792 - 12000

→ CI = 3792

Hence, the compound interest for second year is Rs.3792

To Find the sum at the end of third :

We know that the amount for the second year is equal to the principal for the third year, hence the sum of money at the end of third year is 15972.

Answer 2

${\huge{\huge{\underline{\underline{\mathrm{\red{QuesTion:-}}}}}}}$

Calculate the compound interest for the second

year on * 12,000 invested for 3 years at 10%

per year. Also, find the sum due at the end of

the third year.

${\huge{\huge{\underline{\underline{\mathrm{\green{SoluTion:-}}}}}}}$

$\underline {\purple{\boxed{ \sf{Given:-}}}}$

✍︎Principal=12000

✍︎Rate=10%

✍︎Time=3 years

Lets understand

☞︎︎︎The money given to the borrower is called Principal.

☞︎︎︎the extra money given to the lender by borrower is called interest.

☞︎︎︎the money given to borrower for a specified time is called Time.

lets learn the formula for calculacting amount for CI(compound interest).

$\mathrm{ a = p(1 + \frac{r}{100}) {}^{n} }$

 ‎

$\underline {\green{\boxed{ \sf{Here:-}}}}$

 ‎

 ‎

A=Amount

P=Principal

r=rate

n=time

 ‎

 ‎

lets solve the formula

$\mathrm{ a = 12000( 1 + \frac{10}{100}) {}^{3} }$

 ‎

 ‎

100 will be simplified by 10

so

 ‎

 ‎

$\mathrm{ a = 12000( 1 + \frac{1}{10}) {}^{3} }$

 ‎

 ‎

now lets evaluate the formula

 ‎

 ‎

$\mathrm{ a = 12000( \frac{11}{10}) {}^{3} }$

 ‎

 ‎

$\mathrm{ a = 12000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} }$

 ‎

 ‎

all zeroes of 10s will be divided by 12,000

 ‎

 ‎

$\mathrm{ a = 12 \times \frac{11}{1} \times \frac{11}{1} \times \frac{11}{1} }$

 ‎

 ‎

there is no value of 1 in the formula so

we can write formula as

 ‎

 ‎

$\mathrm{ a = 12 \times \frac{11}{} \times \frac{11}{} \times \frac{11}{} }$

 ‎

 ‎

now lets evavulate

 ‎

 ‎

$\mathrm{ a = 12 \times \frac{11}{} \times \frac{11}{} \times \frac{11}{}=15972 }$

 ‎

 ‎

$\underline {\purple{\boxed{ \sf{Amount=15972}}}}$

 ‎

 ‎

lets find out CI(compound interest)

CI=A-P

 ‎

 ‎

CI=15972-12000

=3972

 ‎

 ‎

The amount is 15972 so the sum due at the end of third year will be 15,972.

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