Question

Find the compound interest, correct to the
nearest rupee, on 2,400 for 2whole 1/2 years at 5 %per annum​

Answer 1

Step-by-step explanation:

amount for 2 year = P(1+R/100)^n

2400×(1+5/100)^2

=2400×(105/100)×(105/100)

=2646rupees

interest on 1/2 year = (2646×5)/200

= 66.15

total interest = 246 + 66.15

= 312.15rupees

Answer 2

$\: \: \: \: \: \: \: \: \: \sf{principal = 2400}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{time = 2 \frac{1}{2} years}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{rate = 5\% \: pa}$

$\: \sf{we \: know}$

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed {\blue{ \sf{a = p(1 + \frac{r}{100} ) {}^{t} }}}$

$= > \: \: \: \sf{ a = 2400(1 + \frac{5}{100}) {}^{2 } }$

$= > \: \: \: \: \: \: \: \: \: \: \sf{a = 2400( \frac{105}{100} ) {}^{2} }$

$= > \: \sf{ a = \cancel{2400} \times \frac{ \cancel{105}}{ \cancel{100} } \times \frac{ \cancel{105}} { \cancel{100}} }$

$= > \: \: \: \: \: \: \sf{a = 6 \times 21 \times 21}$

$= > \: \: \: \: \: \: \: \: \: \: \: \sf{ a = 441 \times 6}$

$= > \: \: \: \: \: \: \: \: \: \: \: \sf{a =264 6}$

$= > \: \sf{interest =2646 - 2400 }$

$= > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ 246 \: rs}$

$\: \: \: \: \: \: \: \: \: \: \: \sf{s.i. = \frac{p \times r \times t}{100} }$

$= > \: \: \: \: \: \: \: \: \: \sf {\frac{2646 \times 5 \times \frac{1}{2} }{100} }$

$= > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 66.15$

$\sf{total \: compound \: interest \: = 246+ 66.15}$

$= > \: \: \: \: \: \: \: \: \: \: \: \: \green { \boxed{\sf \pink{ 312.15 \: rs}}}$