Question

The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum​

Answer 1

GIVEN :-

• CI - SI = Rs. 120.
• Rate ( R ) = 8 %.
• Time ( n ) = 2 years.

TO FIND :-

• The Sum.

SOLUTION :-

Let the Principal be "x".

$\\ : \implies \displaystyle \sf \: SI = \dfrac{P \times R \times T}{1 00 }$

$: \implies \displaystyle \sf \: SI = \frac{x \times 8 \times 2}{100}$

$: \implies \underline{\boxed{\displaystyle \sf SI = \frac{16x}{100} }}$

Now,

$\\ \\ : \implies \displaystyle \sf \: CI = P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} -1\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg(1 + \dfrac{8}{100}\bigg)^{2} -1\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{100 + 8}{100}\bigg)^{2} -1\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{108}{100}\bigg)^{2} -1\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{108 \times 108}{100 \times 100} -1\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664}{10000} -1\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664 - 10000}{10000}\Bigg]$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{1664 }{10000}\Bigg]$

$: \implies \displaystyle \sf \: CI = x \times \frac{1664}{10000}$

$: \implies \underline{ \boxed{ \displaystyle \sf \: CI = \frac{1664x}{10000} }}$

Now According to the question,

$\\ \\ : \implies \displaystyle \sf \: CI - SI = 120$

$: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{16x}{100} = 120$

$: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{1600x}{10000} = 120$

$: \implies \displaystyle \sf \: \frac{1664x - 1600x}{10000} = 120$

$: \implies \displaystyle \sf \: \frac{64x}{10000} = 120$

$: \implies \displaystyle \sf \:64x = 120 \times 10000$

$: \implies \displaystyle \sf \:64x = 1200000$

$: \implies \displaystyle \sf \:x = \frac{1200000}{64}$

$: \implies \underline{ \boxed{ \displaystyle \sf \:x = 18750}}$

Hence the required sum is Rs. 18750.

Answer 2

Step-by-step explanation:

GIVEN :-

CI - SI = Rs. 120.

Rate ( R ) = 8 %.

Time ( n ) = 2 years.

TO FIND :-

The Sum.

SOLUTION :-

Let the Principal be "x".

\begin{gathered} \\ : \implies \displaystyle \sf \: SI = \dfrac{P \times R \times T}{1 00 } \\ \\ \\ \end{gathered}

:⟹SI=

100

P×R×T

\begin{gathered} : \implies \displaystyle \sf \: SI = \frac{x \times 8 \times 2}{100} \\ \\ \\ \end{gathered}

:⟹SI=

100

x×8×2

\begin{gathered} : \implies \underline{\boxed{\displaystyle \sf SI = \frac{16x}{100} }} \\ \\ \end{gathered}

:⟹

SI=

100

16x

Now,

\begin{gathered} \\ \\ : \implies \displaystyle \sf \: CI = P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} -1\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=P[(1+

100

R

)

n

−1]

\begin{gathered} : \implies \displaystyle \sf \: CI = x \Bigg[ \bigg(1 + \dfrac{8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[(1+

100

8

)

2

−1]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{100 + 8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[(

100

100+8

)

2

−1]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{108}{100}\bigg)^{2} -1\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[(

100

108

)

2

−1]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{108 \times 108}{100 \times 100} -1\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[

100×100

108×108

−1]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664}{10000} -1\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[

10000

11664

−1]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664 - 10000}{10000}\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[

10000

11664−10000

]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{1664 }{10000}\Bigg] \\ \\ \\ \end{gathered}

:⟹CI=x[

10000

1664

]

\begin{gathered}: \implies \displaystyle \sf \: CI = x \times \frac{1664}{10000} \\ \\ \\ \end{gathered}

:⟹CI=x×

10000

1664

\begin{gathered}: \implies \underline{ \boxed{ \displaystyle \sf \: CI = \frac{1664x}{10000} }} \\ \\ \end{gathered}

:⟹

CI=

10000

1664x

Now According to the question,

\begin{gathered} \\ \\ : \implies \displaystyle \sf \: CI - SI = 120 \\ \\ \\ \end{gathered}

:⟹CI−SI=120

\begin{gathered}: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{16x}{100} = 120 \\ \\ \\ \end{gathered}

:⟹

10000

1664x

−

100

16x

=120

\begin{gathered}: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{1600x}{10000} = 120 \\ \\ \\ \end{gathered}

:⟹

10000

1664x

−

10000

1600x

=120

\begin{gathered}: \implies \displaystyle \sf \: \frac{1664x - 1600x}{10000} = 120 \\ \\ \\ \end{gathered}

:⟹

10000

1664x−1600x

=120

\begin{gathered}: \implies \displaystyle \sf \: \frac{64x}{10000} = 120 \\ \\ \\ \end{gathered}

:⟹

10000

64x

=120

\begin{gathered}: \implies \displaystyle \sf \:64x = 120 \times 10000 \\ \\ \\ \end{gathered}

:⟹64x=120×10000

\begin{gathered}: \implies \displaystyle \sf \:64x = 1200000 \\ \\ \\ \end{gathered}

:⟹64x=1200000

\begin{gathered}: \implies \displaystyle \sf \:x = \frac{1200000}{64} \\ \\ \\ \end{gathered}

:⟹x=

64

1200000

\begin{gathered}: \implies \underline{ \boxed{ \displaystyle \sf \:x = 18750}} \\ \\ \end{gathered}

:⟹

x=18750

Hence the required sum is Rs. 18750.