Question

find the difference between compound interest and simple interest on rupees 12,000 and in 3/2 years at10% p.a compounded yearly​

Answer 1

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

$\green{\tt{\therefore{Difference=60\:rupees}}}$

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

$\green{\underline \bold{Fiven: }} \\ \tt: \implies Principal(p) = 12000 \: rupees \\ \\ \tt:\implies Time(t) = \frac{3}{2} \: years \\ \\ \tt: \implies Rate\%(r)= 10\% \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies C.I- S.I= ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies S.I= \frac{p \times r \times t}{100} \\ \\ \tt: \implies S.I= \frac{12000 \times 10 \times 3}{2 \times 100} \\ \\ \tt: \implies S.I= 60 \times 10 \times 3 \\ \\ \green{\tt: \implies S.I=1800 \: rupees} \\\\ \tt\circ\:For \:first\:year \\ \bold{As \: we \: know \: that} \\ \tt: \implies A =p(1 + \frac{r}{100} ) ^{t} \\ \\ \tt: \implies A =12000 \times (1 + \frac{10}{100} )^{ 1 } \\ \\ \tt: \implies A=12000 \times (1 + 0.1)^{1} \\ \\ \tt: \implies A =12000 \times (1.1)^{1} \\ \\ \tt: \implies A=12000 \times 1.1 \\ \\ \green{\tt: \implies A =13200 \: rupees}\\\\ \bold{For\:next\:half\:year:}\\ \tt:\implies I=\frac{13200\times 1\times 10}{2\times 100} \\\\ \tt:\implies I=660\:rupees \\ \bold{For \: compound \: interest : } \\ \tt: \implies C.I=A - p \\ \\ \tt: \implies C.I=13200+660 - 12000 \\ \\ \green{\tt: \implies C.I=1860\: rupees} \\ \\ \bold{For \: Difference : } \\ \tt: \implies C.I - S.I =1860 - 1800 \\ \\ \green{\tt: \implies C.I- S.I =60 \: rupees}$