Question

p=12000,r=10% t=3 yrs.c.i.=???​

Answer 1

Answer:

${\huge{\pink{\underline{\underline{QuEsTiOn}}}}}$

Kala invested Rs. 12,000 in a business. She will be paid an interest at 10 % per annum compounded annually. Find the Amount and C I credited against her name after 3 years.

${\huge{\pink{\underline{\underline{Given}}}}}$

Principal(p)=Rs.12000

Rate(r)=10%per annum

Time(n)=3 years

${\huge{\pink{\underline{\underline{To\ find}}}}}$

Amount=????

C.I.=?????

${\huge{\pink{\underline{\underline{Formula\ used}}}}}$

$Amount=p \times {(1 + \frac{r}{100} )}^{n} \\</p><p>Compound\ interest(C.I.)\\</p><p>=Amount -Principal$

${\huge{\pink{\underline{\underline{SoLuTiOn}}}}}$

$Amount=12000×{(1 + \frac{10}{100} )}^{3}\\</p><p>=>12000×{(\frac{110}{100} )}^{3}\\</p><p>=>12000×{(\frac{11}{10} )}^{3}\\</p><p>=>12000× \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \\</p><p>=>12×11×11×11\\</p><p>=>Rs.15,972$

C.I.=Amount-Principal

=15,972-12000

=Rs.3,972

${\huge{\pink{\underline{\underline{Answer}}}}}$

Amount after 3 yrs.=Rs.15,972

C.I. after 3 yrs =Rs.3,972

Answer 2

Given :

p(principal) = 12,000

r(rate) = 10%

t(time) = 3 years.

To find :

• The compound interest after substituting the given values.

Solution :

• We will take out the amount using the formula and then C. I.

Procedure :

We know that :

${ \underline{ \boxed{ \blue{ \sf{A(amount) = p(1 + \frac{r}{100} {)}^{n} }}}}}$

Substituting the given values :

$\longrightarrow \sf{a = 12000(1 + \frac{1 \cancel 0}{10 \cancel 0} {)}^{3} } \\ \sf \longrightarrow a = 12000(1 + { \frac{1}{10} )}^{3} \\ \sf \longrightarrow a = 12\cancel{000} \times \frac{11}{1\cancel0} \times \frac{11}{1 \cancel0} \times \frac{11}{1\cancel0} \\ \sf \longrightarrow a =12 \times 1331\\ { \underline { \boxed{\sf { \green{\therefore a =15972}}}}}$

We get amount :

A = 15,972

Required answer :

➠ C. I. = Amount - principal

➠ C. I. = 15,972 - 12,000

➠C. I. = ₹3,972 ans.