A and B borrowed Rs. 60000 and Rs. 50,000 respectively for a period of 3 years. A paid simple interest at a rate of $10 %$ per annum while B paid compound interest at the rate of $10 %$ per annum compounded annually. Who paid more and by how much?

A paid more by Rs. 1650

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B paid more by Rs. 1650

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A paid more by Rs. 1450

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B paid more by Rs. 1450

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Solution

The correct option is C
A paid more by Rs. 1450

For A

P = Rs. 60000

T = 3 years

R = $10 %$ per annum

I = $\frac{P R T}{100}$

= $\frac{(R s . 60000 \times 10 \times 3)}{100}$

= Rs. 18000

For B

P = Rs. 50000

T = 3 years

R = $10 %$ per annum

Amount = P ${(\frac{1 + R}{100})}^{n}$

= Rs. 50000 ${(\frac{1 + 10}{100})}^{3}$

= Rs. 66550

C.I = A -P

= Rs. 66550 - Rs. 50000

= Rs. 16550

A paid more than B by ( Rs. 18000 - Rs. 16550 ) i.e. Rs. 1450.

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A and B borrowed Rs. 60000 and Rs. 50,000 respectively for a period of 3 years. A paid simple interest at a rate of 10% per annum while B paid compound interest at the rate of 10% per annum compounded annually. Who paid more and by how much?

A and B borrowed Rs. 60000 and Rs. 50,000 respectively for a period of 3 years. A paid simple interest at a rate of $10 %$ per annum while B paid compound interest at the rate of $10 %$ per annum compounded annually. Who paid more and by how much?