Question

Let us write by calculating the amount on Rs.30000 at the rate of 9% compound interest per annum 3 years.​

Answer 1

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

$\green{\tt{\therefore{Compound\:Interest=8850.87\:rupees}}}$

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

$\green{\underline \bold{Given:}} \\ \tt: \implies Principal(p) = 30000 \: rupees \\ \\ \tt: \implies Rate\%(r) = 9\% \\ \\ \tt: \implies Time(t) = 3 \: years \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Compound \: Interest(C.I)= ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies A= p(1 + \frac{ {r} }{100} )^{t} \\ \\ \tt: \implies A=30000 \times (1 + \frac{9}{100} )^{ 3} \\ \\ \tt: \implies A =30000 \times (1 + 0.09)^{3} \\ \\ \tt: \implies A=30000 \times {(1.09)}^{3} \\ \\ \tt: \implies A =30000 \times 1.295029 \\ \\ \green{\tt: \implies A=38850.87 \: rupees} \\ \\ \bold{For \: compound \: interest : } \\ \tt: \implies C.I=A - p \\ \\ \tt: \implies C.I =38850.87 - 30000\\ \\ \green{\tt: \implies C.I =8850.87 \: rupees}$

Answer 2

ANSWER -

$\large{\purple{\underline{\tt{AMOUNT=38850.87\:Rs}}}}$

$\large{\red{\underline{\tt{Given-}}}}$

→Principal (P) = Rs 30, 000

→Rate = 9 % p.a.

→Time (N) = 3 years

$\large{\red{\underline{\tt{To\:Find-}}}}$

→Amount

$\large{\red{\underline{\tt{Solution-}}}}$

$\sf A = P (\dfrac{1+R}{100})^{N}$

$\sf A = 30, 000 (\dfrac{1+9}{100})^{3}$

$\sf A = 30, 000 (\dfrac{109}{100})^{3}$

$\sf A = 30, 000 (\dfrac{109\times109\times109}{100\times100\times100})$

$\sf A = 3 (\dfrac{109\times109\times109}{100})$

$\sf A = (\dfrac{327\times11881}{100})$

$\sf A = (\dfrac{3885087}{100})$

$\sf A = 38, 850.87$

$\large{\red{\underline{\tt{AMOUNT=38850.87\:Rs}}}}$

$\rule{200}2$