Question

The difference between the compound interest and the simple interest on a certain sum for 2 years at 5% per annum is $ 40. Find the sum​

Answer 1

Step-by-step explanation:

C.I=P(1+R/100)^n

=P(1+5/100)^2-P

=P(21/20)^2-P

=441P/400-P

=41P/400

S.I=PRT/100

=P×5×2/100

=P/10

C.I-S.I=40

41P/400-P/10=40

P/400=40

P=16000

Answer 2

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}\end{gathered}$

• The difference between the compound interest and the simple interest on a certain sum is $ 40

• Time = 2 years

• Interest rate = 5% per annum

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{To find \: :}}}}}}\end{gathered}$

• The sum

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Using formula \: :}}}}}}\end{gathered}$

$\quad\dag{\underline{\boxed{\sf{Compound \: interest =(P {( 1 + r)}^{T} -P) \: }}}}†$

we're

• P = principal

• r = rate of interest

• T = time

$\quad\dag{\underline{\boxed{\sf{Simple \: interest = \dfrac{P× R × T}{100} \: }}}}†$

we're

• P = Principal

• R = rate of interest

• T = time

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}\end{gathered}$

$\sf  ➼ \: According \: to \: the \: question :-$

⇒ ( Compound interest ) - ( simple interest ) = 40

$∴ ⇒\sf \: (P {( 1 + r)}^{T} -P) - ( \dfrac{P× R × T}{100} ) = 40$

$⇒ \sf \: (P {( 1 + 5\%)}^{2} -P) - ( \dfrac{P× 5 × 2}{100} ) = 40$

$⇒ \sf(P {( 1 + \dfrac{5}{100} )}^{2} -P) - ( \dfrac{P \times 10}{100} ) = 40$

$⇒ \sf(P {( 1 + 0.05 )}^{2} -P) - ( \dfrac{10P}{100} ) = 40$

$⇒ \sf(P {(1.05 )}^{2} -P) - ( {\sf { {\dfrac{{\cancel{{10}} ^{1}}P}{{\cancel{{100}}^{10}}}}}})= 40$

$⇒ \sf \: (P \times 1.1025 -P) - ( \dfrac{P}{10}) = 40$

$⇒ \sf \: ( 1.1025P -P) - (\dfrac{P}{10} )= 40$

$⇒ \sf \: ( 0.1025P ) - (\dfrac{P}{10} )= 40$

$⇒ \sf \: ( \dfrac{1025P}{10000} ) - ( \dfrac{P}{10} ) = 40$

$⇒ \sf \: ( \dfrac{1025P - 1000P}{10000}) = 40$

$⇒ \sf \: ( \dfrac{25P}{10000}) = 40$

$⇒ \sf \: 25P = 40000$

$⇒ \sf \: P = \dfrac{40000}{25}$

$⇒ \sf \: P ={\sf { {\dfrac{{\cancel{{40000}} ^{16000}}}{{\cancel{{25}}^{1}}}}}}$

$⇒ \sf \dag{\underline{\boxed{\sf{\purple{P= 1600}}}}}†$

As we know ,

➼ Principal = Sum of money

∴ Sum of money = 16,000

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Answer \: :}}}}}}\end{gathered}$

• 16,000

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Know \: more :}}}}}}\end{gathered}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{ Simple \: Interest = \dfrac{P \times R \times T}{100}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{Amount = Principle + Interest}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{ Principle=Amount - Interest }}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{Principle = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{Principle = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\red{Time = \dfrac{Simple \: Interest \times 100}{Principle \times Rate}}}}}}$

$\longrightarrow \red{\underline{\boxed{\sf{Compound \: interest =(P {( 1 + r)}^{T} -P) \: }}}}$

$\underline{\rule{230pt}{3pt}}$