Question

Find the amount and the compound interest on rupees 64000 for 1 1/2 years at 15% per annum, compunded half-yerly
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Answer 1

QUESTION......

ғɪɴᴅ ᴛʜᴇ ᴀᴍᴏᴜɴᴛ ᴀɴᴅ ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ ɪɴᴛᴇʀᴇsᴛ ᴏɴ ʀᴜᴘᴇᴇs 64000 ғᴏʀ 1 1/2 ʏᴇᴀʀs ᴀᴛ 15% ᴘᴇʀ ᴀɴɴᴜᴍ, ᴄᴏᴍᴘᴜɴᴅᴇᴅ ʜᴀʟғ-ʏᴇʀʟʏ

Sᴏʟᴜᴛɪᴏɴ........

•ᴛɪᴍᴇ (ᴛ) = 1/2 ʏᴇᴀʀs = 3/2

=ᴄᴏᴍᴘᴏᴜɴᴅ ғᴏʀ ʜᴀʟғ ʏᴇᴀʀʟʏ .

= ʀᴀᴛᴇ (ʀ) =(15/2) % [ Fᴏʀ ʜᴀʟғ ᴀ ʏᴇᴀʀ ]

= ᴛɪᴍᴇ (ᴛ) = 3/2 ×2 = 3 ʏᴇᴀʀs [ ʜᴀʟғ ʏᴇᴀʀʟʏ ]

ғᴏʀᴍᴜʟᴀ

= Aᴍᴏᴜɴᴛ = ᴘʀɪɴᴄɪᴘʟᴇ [ 1 + ʀᴀᴛᴇ / 100] t

= ᴄᴀʟᴜʟᴀᴛɪᴏɴ

= 1) ᴀᴍᴏᴜɴᴛ = 64000 [ 1 + 15/2/${100}]^{3}$

= ᴀᴍᴏᴜɴᴛ = 64000 ( 215 / ${200})^{3}$

ᴀᴍᴏᴜɴᴛ ( ᴀ )= ʀs.79507

Answer 2

QUESTION:-

$\large\bold Q.$ ғɪɴᴅ ᴛʜᴇ ᴀᴍᴏᴜɴᴛ ᴀɴᴅ ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ ɪɴᴛᴇʀᴇsᴛ ᴏɴ ʀᴜᴘᴇᴇs 64000 ғᴏʀ 1 1/2 ʏᴇᴀʀs ᴀᴛ 15% ᴘᴇʀ ᴀɴɴᴜᴍ, ᴄᴏᴍᴘᴜɴᴅᴇᴅ ʜᴀʟғ-ʏᴇʀʟʏ

ANSWER✔

$\Large\underline\bold{solution,}$

$\Large\bold{given,}$

$\sf\dashrightarrow time(T)= 1 \times \dfrac{1}{2} years = \dfrac{3}{2}$

$\sf\underline\bold{\therefore compound\:for\:half\:yearly,}$

$\sf\implies rate= \dfrac{15}{2} \% ............. \frac{1}{2} \:a\:year$

$\sf\implies time= \dfrac{3}{2} \times 2 =3\:years ............. \frac{1}{2} yearly$

$\large{\boxed{\sf{amount=principle \times \bigg[ 1+ \dfrac{rate}{100} \bigg] \times T}}}$

$\sf\underline\bold{now,}$

$\sf\implies amount=64000 \times \bigg[ \bigg( 1+ \dfrac{ \frac{15}{2}}{100} \bigg)^3 \times 100 \bigg]$

$\sf\implies amount=64000 \times \bigg[ \bigg( \dfrac{215}{200} \bigg)^3 \times 200 \bigg]$

$\sf\implies amount= \cancel {64000} \times \bigg[ \bigg( \dfrac{\cancel {215}}{ \cancel {200}} \bigg)^3 \times \cancel {200} \bigg]$

ᴀᴍᴏᴜɴᴛ ( ᴀ )= ʀs.79507

$\large{\boxed{\sf{amount=rs.79507}}}$

___________________

✯ADDITIONAL INFORMATION,

✯DERIVATION,

DERIVING, compound interest formula,

LET,

$\sf\therefore principal\:amount\:=p$

$\sf\therefore time= n\:years$

$\sf\therefore rate =R$

$\sf\large\therefore simple \:interest\:for\:firsr\:year,$

$\sf\therefore SI_1= \dfrac{P \times R \times T }{100}$

$\sf\therefore amount\:after\:1_st \:year=p+SI_1$

$\sf\implies P+ \dfrac{P times R \times T }{100}$

$\sf\therefore P \times \bigg( 1+ \dfrac{R}{100} \bigg)=P_2$

$\sf\therefore simple\:intrest\:for\:2_nd \:= \dfrac{P_2 \times R \times T }{100}$

$\sf\therefore Amount\:after\:second\:year,$

$\sf\implies P_2 +SI_2$

$\sf\implies P_2+ \dfrac{P_2 \times R \times T }{100}$

$\sf\implies P_2 \bigg( 1+ \dfrac{R}{100} \bigg)$

$\sf\implies P \bigg( 1+ \dfrac{R}{100} \bigg) \times \bigg( 1+ \dfrac{R}{100} \bigg)$

$\sf\therefore P \bigg( 1+ \dfrac{R}{100} \bigg)^2$

$\sf\therefore simlarly\:if\:we\:continued,\:for\:"n"\:years.$

THEN,

$\sf\therefore A=P \times \bigg( 1+ \dfrac{R}{100} \bigg)^n$

$\sf\large\therefore CI=A-P$

$\sf\implies p \times \bigg[ \bigg( 1+ \dfrac{R}{100} \bigg)^n \bigg]$

$\sf{\boxed{\sf{p \times \bigg[ \bigg( 1+ \dfrac{R}{100} \bigg)^n \bigg]}}}$