Question

Find the compound interest on rs31250 at8% per annum for 11/4 years

Answer 1

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

$\green{\tt{\therefore{Compound\:Interest=736.82\:rupees}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Principal (p) = 31250 \: rupees \\ \\ \tt: \implies Rate \%(r) = 8\% \\ \\ \tt : \implies Time(t) = \frac{11}{4}\: years \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Compound \: Interest= ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies A= p(1 + \frac{r}{100})^{t} \\ \\ \tt: \implies A= 31250( 1 + \frac{8}{100} )^{\frac{11}{4}} \\ \\ \tt: \implies A = 31250( 1+0.08 )^{\frac{11}{4}} \\ \\ \tt: \implies A = 31250\times 1.235\\ \\ \green{\tt: \implies A= 38615.82 \: rupees} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies C.I= A- p \\ \\ \tt: \implies C.I = 38615.82 - 31250 \\ \\ \green{\tt: \implies C.I = 736.82 \: rupees}$

Answer 2

$\mathtt \green{given}$

${\mathtt{\red{Principal}}}$ = ${\mathtt{\red{31250}}}$

${\mathtt{\red{Rate}}}$ = ${\mathtt{\red{8}}}$

${\mathtt{\red{Time}}}$ = 11/4 ${\mathtt{\red{months}}}$

We can write 11/4 months as :-

$⟹ \: \mathtt{ 2 \frac{3}{4} }$

${\mathtt{\green{Formula}}}$

$\mathtt{amount = p( {1 + \frac{r}{100} )}^{n} \times (1 + \frac{ \frac{b}{c} r}{100} )}$

⟹ Here b = 3 ; c = 4 ; n = 2

$\mathtt{⟹ \: 31250( {1 + \frac{8}{100}) }^{2} \times (1 + \frac{ \frac{3}{4}8 }{100} )}$

$\mathtt{⟹ \: 31250( { \frac{108}{100} }^{2} ) \times ( \frac{106}{100} )}$

$\mathtt{⟹ \: 31250( \frac{11664}{10000}) \times ( \frac{106}{100} )}$

$\mathtt{⟹ \: \: 50( 729)( \frac{106}{100} )}$

$\mathtt{⟹ \: 38637}$

Amount = 38637

C.I = 38637 - 31250

$\mathtt \red{7387}$