Question

Rs.8000 is lent at 5% compound interest per year for 2 years.Find the amount and compounds

Guys please answer to this question!........
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Answer 1

$\tt\underline\purple{★Question~given:}$

Rs.8000 is lent at 5% compound interest per year for 2 years.Find the amount and compounds.

$\tt\underline\orange{★Answer}$

$\leadsto\sf{Rs.=8000}$

$\leadsto\sf{Interest=5\%}$

$\leadsto\sf{Time=2~years}$

According to Question:

$\leadsto\sf{P}[\dfrac{1+r}{100}^{t}-{1}]$

$\leadsto\sf{8000}[\dfrac{1+5²}{100}-{1}]$

$\leadsto\sf{8000}[\dfrac{25}{100}+\dfrac{10}{100}]$

$\leadsto\sf{8000×}\dfrac{25+1000}{10000}$

$\leadsto\sf{C.I.=}\dfrac{8}{10}{[1025]}=₹14.820$

__________________

P=Principal

R= Rate of interest

T= Time

CI= Compound Interest

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Answer 2

Answer:

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Step-by-step explanation:

$Rs.=8000</p><p></p><p>\leadsto\sf{Interest=5\%}⇝Interest=5%</p><p></p><p>\leadsto\sf{Time=2~years}⇝Time=2 years</p><p></p><p>$

$Time=2 years</p><p></p><p> \\ According \: to \: Question:</p><p></p><p>$

$\leadsto\sf{P}[\dfrac{1+r}{100}^{t}-{1}]</p><p>[tex]\leadsto\sf{8000}[\dfrac{1+5²}{100}-{1}</p><p></p><p> \\ \leadsto\sf{8000}[\dfrac{25}{100}+\dfrac{10}{100}]</p><p></p><p> \\ \leadsto\sf{8000×}\dfrac{25+1000}{10000}</p><p></p><p>$

$\leadsto\sf{C.I.=}\dfrac{8}{10}{[1025]}=₹14.820$

$P=Principal$

$</p><p>R= Rate \: of \: interest$

$</p><p>T= Time$

$</p><p>CI= Compound \: Interest$