Question

Find compound interest on $7000 at 10% per annum for 2 year 5 months ,compounded annually

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Compound\:Interest=\ \:1813}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Principal(p)= \ \: 7000 \\ \\ \tt: \implies Rate\% = 10\% \\ \\ \tt: \implies Time(t) = 2 \: year \: 5 \: months \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Compound \: Interest = ?$

• According to given question :

$\tt \circ \: Time = \frac{29}{12} \: years \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies A= p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies A = 7000 \times (1 + \frac{10}{100} )^{ \frac{29}{12} } \\ \\ \tt: \implies A = 7000 \times (1 + 0.1)^{ \frac{29}{12} } \\ \\ \tt: \implies A= 7000 \times (1.1)^{ \frac{29}{12} } \\ \\ \tt: \implies A= 7000 \times 1.259 \\ \\ \green{\tt: \implies A = \ \: 8813 } \\ \\ \bold{For \: Compound \: Interest : } \\ \tt: \implies C.I = A - p \\ \\ \tt: \implies C.I=8813 - 7000 \\ \\ \green{\tt: \implies C.I =\ \: 1813}$

Answer 2

Given:-

• Principal (P) = $7000

• Rate (r) = 10 %

• Time (n) = 2 years 5 months

To find:-

The compound interest.

Solution:-

Total time = 24 months + 5 months

                 = 29 months

                 = 29/12 years

We know,

$\bold{A=P(1+\frac{r}{100})^n}$

⇒ $\bold{A=7000(1+\frac{10}{100})^\frac{29}{12} }$

⇒ $\bold{A=7000(1+\frac{1}{10})^\frac{29}{12} }$

⇒ $\bold{A=7000*1.1^\frac{29}{12} }$

⇒ $\bold{A=7000*1.259}$

⇒ A = $ 8813

Compound interest:-

C.I. = A - P

⇒C.I. = 8813 - 7000

⇒C.I. = $ 1813