Question

Rs.2000 amounts to 2226 in two years at compound interest. The rate of interest is

Answer 1

Given :

• Amount = Rs. 2226
• Principal = Rs. 2000
• Time = 2 years

To Find :

• Rate of interest = ?

Formula to be used :

• A = P(1 + r/100)ⁿ

Solution :

Let's calculate the rate of interest :

→ A = P(1 + r/100)ⁿ

Where,

• A = amount
• P = principal
• r = rate of interest
• n = time

Now, plug in the given values in above formula :

→ 2226 = 2000(1 + r/100)²

→ 2226 ÷ 2000 = (1 + r/100)²

→ 1113 ÷ 1000 = (1 + r/100)²

→ 1.113 = (1 + r/100)²

Taking square root on both the sides we get:

→ √1.113 = 1 + r/100

→ 1 + r/100 = √1.113

→ 1 + r/100 = 1.055

→ r/100 = 1.055 - 1

→ r/100 = 0.055

→ r = 0.055 × 100

→ r = 5.5 %

• Hence,the rate of interest is 5.5%.

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large\underline{\green{\bf Given :-}}$

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• Rs 2000 amounts to Rs 2226 in two years at compound interest

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

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• Rate of interest

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$\large\underline{\orange{\bf Solution :-}}$

$\:\:$

We know that , for compound interest

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➠ $\sf A = P(1 + \dfrac { r } { 100 } )^n$ ⚊⚊⚊⚊ ⓵

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Where,

$\:\:$

• A = Amount

• P = Principle amount

• r = Rate of interest

• n = Time period

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$\underline{\bold{\texttt{For this case :}}}$

$\:\:$

• A = 2226

• P = 2000

• r = r

• n = 2

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⟮ Putting these values in ⓵ ⟯

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➜ $\sf A = P(1 + \dfrac { r } { 100 } )^n$

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➜ $\sf 2226 = 2000(1 + \dfrac { r} { 100 } )^2$

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➜ $\sf 2226 = 2000(\dfrac { 100 + r } { 100 } )^2$

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➜ $\sf 2226 = 2000(\dfrac { 100 + r } { \cancel { 100} } × \dfrac { 100 + r } { 10\cancel0 } )$

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➜ $\sf 2226 = \cancel 2(\dfrac { 100 + r } { 1} × \dfrac { 100 + r } { \cancel { 10} } )$

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➜ $\sf 2226 = (100 + r) × (\dfrac { 100 + r } { 5} )$

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➜ 2226 × 5 = (100 + r)(100 + r)

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➜ 11130 = (100 + r)²

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➜ $\sf 100 + r = \sqrt { 11130 }$

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➜ 100 + r = 105.5

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➜ r = 105.5 - 100

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➨ r = 5.5

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• Hence the rate of interest is 5.5

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