Question

If the amount at the end of 2nd 3rd year are ₹6050 and ₹6655 respectively. Find the rate and the principal if the interest is compounded annually

Answer 1

Given that:

• The amount at the end of 2nd and 3rd year are ₹ 6050 and ₹ 6655 respectively.

To Find:

• The rate and the principal if the interest is compounded annually.

Formula used:

In compound interest.

• A = P(1 + R/100)ᵀ

Where,

• A = Amount
• P = Principal
• R = Rate
• T = Time

In first case:

We have,

• Amount = ₹ 6050
• Time = 2 years

Putting given values in formula.

↣ 6050 = P(1 + R/100)²

↣ P(1 + R/100)² = 6050 (i)

In second case:

We have,

• Amount = ₹ 6655
• Time = 3 years

Putting given values in formula.

↣ 6655 = P(1 + R/100)³

↣ P(1 + R/100)³ = 6655

↣ P(1 + R/100)²(1 + R/100) = 6655

From equation (i)

↣ 6050(1 + R/100) = 6655

↣ (1 + R/100) = 6655/6050

↣ (1 + R/100) = 1.1

↣ 1 + R/100 = 1 + 0.1

Cancelling 1 both sides.

↣ R/100 = 0.1

↣ R = 0.1(100)

↣ R = 10

In equation (i)

↣ P(1 + R/100)² = 6050

Putting the value of R.

↣ P(1 + 10/100)² = 6050

↣ P(1 + 0.1)² = 6050

↣ P(1.1)² = 6050

↣ P(1.21) = 6050

↣ P = 6050/1.21

↣ P = 5000

Hence,

• The rate is 10% compounded per annum.
• Principal is Rs 5000.

Answer 2

Given

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• Amount (A) at the end of 2nd and 3rd year are Rs. 6050 and Rs. 6655 respectively.

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To Find

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• Rate (R) and principal (P) if the interest is compounded annually.

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Solution

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$\qquad\footnotesize{\dag\:{\underline{\frak{Using\:Formula\:::}}}}$

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$\quad\odot\:\underline{\boxed{\bf{\blue{A = P{\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup}^{T}}}}}$

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Where,

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• A denotes Amount.

• P denotes Principal.

• R denotes Rate.

• T denotes Time.

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Solving for 1st case ::

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✭ Putting all known values in formula ::

$\\ \dashrightarrow\:\sf 6050 = P{\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup}^{2}$

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We can also write it as ::

$\\ \dashrightarrow\:\sf P{\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup}^{2} = 6050\:[1]$

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Solving for 2nd case ::

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✭ Putting all known values in formula ::

$\\ \dashrightarrow\:\sf 6655 = P{\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup}^{3}$

$\\ \dashrightarrow\:\sf P{\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup}^{3} = 6655$

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We can also write it as ::

$\\ \dashrightarrow\:\sf P{\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup}^{2} \:\times\:\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup = 6655 \:[2]$

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From [1], putting in [2] ::

$\\ \dashrightarrow\:\sf 6050\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup = 6655$

$\\ \dashrightarrow\:\sf 1 + \dfrac{R}{100} = {\cancel{ \dfrac{6655}{6050}}}$

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After cancelling 6655 with 6050, we get ::

$\\ \dashrightarrow\:\sf 1 + \dfrac{R}{100} = 1.1$

$\\ \dashrightarrow\:\sf \dfrac{R}{100} = 1.1 - 1$

$\\ \dashrightarrow\:\sf \dfrac{R}{100} = 0.1$

$\\ \dashrightarrow\:\sf R = 0.1 \:\times\:100$

$\\ \dashrightarrow\:\sf \pink{R = 10}$

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Therefore, Rate (R) is 10% compounded per annum.

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Putting value of 'R' in [1] ::

$\\ \dashrightarrow\:\sf P{\bigg\lgroup 1 + \dfrac{10}{100}\bigg\rgroup}^{2} = 6050$

$\\ \dashrightarrow\:\sf P{\big\lgroup 1 + 0.1\big\rgroup}^{2} = 6050$

$\\ \dashrightarrow\:\sf P{\big\lgroup 1.1\big\rgroup}^{2} = 6050$

$\\ \dashrightarrow\:\sf P\big\lgroup 1.21\big\rgroup = 6050$

$\\ \dashrightarrow\:\sf P\:\times\: 1.21 = 6050$

$\\ \dashrightarrow\:\sf P = {\cancel{ \dfrac{6050}{1.21}}}$

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After cancelling 6050 with 1.21, we get ::

$\\ \dashrightarrow\:\sf \green{P = 5000}$

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Therefore, Principal (P) is Rs. 5000.

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