Question

Mr Tiwari obtaind
a loan
of
£ 50, ooo from union Bank
of India After 3 years be paid
Rs
86, 400
. Find the rate
of
Interest,
if the interest
is compounded annually​

Answer 1

${\large{\boxed{\underline{\mathrm{\bf{\orange{Given:-}}}}}}}$

• Principal (Sum of money) = £50,000
• Time = 3 years
• Amount = £86,400

${\large{\boxed{\underline{\mathrm{\bf{\red{To \ Find:-}}}}}}}$

• Rate of Interest

${\large{\boxed{\underline{\mathrm{\bf{\pink{Formula \ Used:-}}}}}}}$

$\bigstar \ {\boxed{\tt{\green{P \ \bigg ( 1 + \dfrac{R}{100} \bigg ) ^n = A }}}} \ \bigstar$

where,

• A = Amount i.e. £86,400
• P = Principal i.e. £50,000
• n = Time period i.e. 3 years
• R = Rate of Interest

${\large{\boxed{\underline{\mathrm{\bf{\blue{Solution:-}}}}}}}$

Let, Rate of interest be R,

Given :-

• Principal (Sum of money) = £50,000
• Time = 3 years
• Amount = £86,400

According to the question by using the formula of Amount, we get,

$\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( 1 + \dfrac{R}{100} \bigg ) ^3 = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( \dfrac{100 + R}{100} \bigg ) ^3 = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. 50,000 \ \times \dfrac{(100 + R)^3}{10,00,000} = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. \ 0.05 \ \times (100 + R)^3 = Lb. \ 86,400}}$

$\mapsto {\mathsf{(100 + R)^3 = \dfrac{Lb. \ 86,400}{Lb. \ 0.05 }}}$

$\mapsto {\mathsf{(100 + R)^3 = Lb. \ 17,28,000}}$

$\mapsto {\mathsf{(100 + R)^3 = (Lb. \ 120)^3}}$

$\mapsto {\mathsf{(R)^3 = (Lb. \ 120 - 100)^3}}$

$\mapsto {\mathsf{(R)^3 = (Lb. \ 20)^3}}$

$\Longrightarrow {\mathsf{R = 20\%}}$

${\orange{\bigstar}} \ \therefore {\boxed{\underline{\mathsf{\green{Rate \% \ of \ Interest}{\pink{ \ is \ }{\blue{\bf{20\% \ p.a.}}}}}}}} \ {\orange{\bigstar}}$

${\huge{\green{\checkmark}}}$ ${\large{\boxed{\underline{\mathrm{\bf{\orange{Verification:-}}}}}}}$ ${\huge{\green{\checkmark}}}$

$\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( 1 + \dfrac{20}{100} \bigg ) ^3 = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. \ 50,000 \bigg ( \dfrac{100 + 20}{100} \bigg ) ^3 = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( \dfrac{120}{100} \bigg ) ^3 = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. 50,000 \bigg ( \dfrac{6}{5} \times \dfrac{6}{5} \times \dfrac{6}{5} \bigg ) ^3 = Lb. 86,400}}$

$\mapsto {\mathsf{Lb. \ \bigg ( 50,000 \times \dfrac{216}{125} \bigg ) = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. \ (4,000 \times 216) = Lb. \ 86,400}}$

$\mapsto {\mathsf{Lb. \ 86,400 = Lb. \ 86,400}}$

$\Longrightarrow {\mathsf{\bf{LHS = RHS}}} \Longleftarrow$

${\mathsf{Hence, Verified \ {\large{\checkmark}} }}$

Answer 2

Answer:

Let, Rate of interest be R,

Given :-

Principal (Sum of money) = £50,000

Time = 3 years

Amount = £86,400

According to the question by using the formula of Amount, we get,

\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( 1 + \dfrac{R}{100} \bigg ) ^3 = Lb. \ 86,400}}↦Lb. 50,000 (1+

100

R

)

3

=Lb. 86,400

\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( \dfrac{100 + R}{100} \bigg ) ^3 = Lb. \ 86,400}}↦Lb. 50,000 (

100

100+R

)

3

=Lb. 86,400

\mapsto {\mathsf{Lb. 50,000 \ \times \dfrac{(100 + R)^3}{10,00,000} = Lb. \ 86,400}}↦Lb.50,000 ×

10,00,000

(100+R)

3

=Lb. 86,400

\mapsto {\mathsf{Lb. \ 0.05 \ \times (100 + R)^3 = Lb. \ 86,400}}↦Lb. 0.05 ×(100+R)

3

=Lb. 86,400

\mapsto {\mathsf{(100 + R)^3 = \dfrac{Lb. \ 86,400}{Lb. \ 0.05 }}}↦(100+R)

3

=

Lb. 0.05

Lb. 86,400

\mapsto {\mathsf{(100 + R)^3 = Lb. \ 17,28,000}}↦(100+R)

3

=Lb. 17,28,000

\mapsto {\mathsf{(100 + R)^3 = (Lb. \ 120)^3}}↦(100+R)

3

=(Lb. 120)

3

\mapsto {\mathsf{(R)^3 = (Lb. \ 120 - 100)^3}}↦(R)

3

=(Lb. 120−100)

3

\mapsto {\mathsf{(R)^3 = (Lb. \ 20)^3}}↦(R)

3

=(Lb. 20)

3

\Longrightarrow {\mathsf{R = 20\%}}⟹R=20%

{\orange{\bigstar}} \ \therefore {\boxed{\underline{\mathsf{\green{Rate \% \ of \ Interest}{\pink{ \ is \ }{\blue{\bf{20\% \ p.a.}}}}}}}} \ {\orange{\bigstar}}★ ∴

Rate% of Interest is 20% p.a.

★

{\huge{\green{\checkmark}}}✓ {\large{\boxed{\underline{\mathrm{\bf{\orange{Verification:-}}}}}}}

Verification:−

{\huge{\green{\checkmark}}}✓

\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( 1 + \dfrac{20}{100} \bigg ) ^3 = Lb. \ 86,400}}↦Lb. 50,000 (1+

100

20

)

3

=Lb. 86,400

\mapsto {\mathsf{Lb. \ 50,000 \bigg ( \dfrac{100 + 20}{100} \bigg ) ^3 = Lb. \ 86,400}}↦Lb. 50,000(

100

100+20

)

3

=Lb. 86,400

\mapsto {\mathsf{Lb. \ 50,000 \ \bigg ( \dfrac{120}{100} \bigg ) ^3 = Lb. \ 86,400}}↦Lb. 50,000 (

100

120

)

3

=Lb. 86,400

\mapsto {\mathsf{Lb. 50,000 \bigg ( \dfrac{6}{5} \times \dfrac{6}{5} \times \dfrac{6}{5} \bigg ) ^3 = Lb. 86,400}}↦Lb.50,000(

5

6

×

5

6

×

5

6

)

3

=Lb.86,400

\mapsto {\mathsf{Lb. \ \bigg ( 50,000 \times \dfrac{216}{125} \bigg ) = Lb. \ 86,400}}↦Lb. (50,000×

125

216

)=Lb. 86,400

\mapsto {\mathsf{Lb. \ (4,000 \times 216) = Lb. \ 86,400}}↦Lb. (4,000×216)=Lb. 86,400

\mapsto {\mathsf{Lb. \ 86,400 = Lb. \ 86,400}}↦Lb. 86,400=Lb. 86,400

\Longrightarrow {\mathsf{\bf{LHS = RHS}}} \Longleftarrow⟹LHS=RHS⟸

{\mathsf{Hence, Verified \ {\large{\checkmark}} }}Hence,Verified ✓