Question

Fabina borrows Rs 12500 at 12% per annum for 3 years at simple interest and radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

Case 1:

For Fabina:

Given:

• P = ₹ 12,500
• R = 12%
• T = 3 Years

Formula used here:

• SI = P × R × T/100

Putting the values according to the given formula:

➠ 12500 × 12 × 3/100

➠ 125 × 12 × 3

➠ Rs.4,500

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Case 2:

For Radha:

Given:

• P = ₹ 12,500
• R = 10%
• T = 3 Years

Formula used here:

• Amount = P(1 + R/100)^n

Putting the values according to the given formula:

➠ 12,500(1 + 10/100)^3

➠ 12,500(1+ 1/10)^3

➠ 12,500(11/10)^3

➠ 12,500 × 11/10 × 11/10 × 11/10

➠ 25 × 1331/2

➠ Rs. 16637.50

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Now, Need To Find:

• New Compound Interest = ?

Formula used here:

• Amount - Principal

Putting the values according to the given formula:

➠ 16637.50 - 12500

➠ Rs. 4137.50

• Now, we can say that compound interest is less than simple interested it means Fabina paid more interest than Radha.

➠ 4500 - 41375

➠ Rs. 362.50 (Your Answer)

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Here:

• P is used for Principal.
• R is used for Rate of Interest.
• T is used for Time.
• SI is used for Simple Interest.

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Answer 2

Given:

→ Fabiana borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually.

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Find:

→ Who pays more interest and by how much?

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Using formula:

→ S.I = (Principal × Rate × Time)/100

→ A = Principal (1 + Rate/100)ⁿ

→ C.I = Amount - Principal

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Note:

→ (ⁿ) = Number of years or Time period.

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Calculations:

★ Finding for Radha:

→ (12500 × 12 × 3)/100

→ 450000/100

→ 4500

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★ Finding for Fabina:

→ 12500 × (1 + 10/100)³

→ 12500 × (11/10)³

→ {12500 (11 × 11 × 11)}/1000

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→ 125 × 1331/10

→ 166375/10

→ 16637.5

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★ Amount paid by Radha:

→ 16637.50 - 12500

→ 4137.5

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★ Fabia pays more:

→ 4500 - 4137.50

→ 362.5

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Hence, Fabia pays more 362.5 than Radha.