Question

le 4.
2
Mr. Singh borrowed 21600 for 3 years at 16% compounded annually while
3
Mr. Gupta borrowed the same amount at simple interest for the same period and at
the same rate of interest. Who has to pay more and by how much?​

Answer 1

Answer:

Mr. Gupta borrowed the same amount at simple interest fo

Step-by-step explanation:

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Answer 2

❍ Question -:

Mr. Singh borrowed 21600 for 3 years at 16% compounded annually while Mr. Gupta borrowed the same amount at simple interest for the same period and at the same rate of interest. Who has to pay more and by how much?

❍ Explanation -:

Case 1:

We observe here that principal is ₹ 21600 Time is 3 years and rate of interest is 16% compounded annually.

Case 2:

Principal, time and rate of interest is same at simple interest.

We are asked to calculate who has to pay more and by how much ?

Let us solve this question.

First we will calculate the interest paid by Mr. Singh

Principal = ₹ 21600

Rate of interest= 16% p.a.

Time = 3 years

We know,

$\small\boxed {\pmb{Amount = Principal} \huge( \small \sf{1 + \dfrac{R}{100} \huge {)}^{ \small{n}} }}$

Substitute the values we get

$\small\pmb{ Amount = \: } \frak{ 21600 \huge( \small{1 + \dfrac{16}{100} \huge{ {)}^{ \small{3}} }}}$

$\small \sf{Amount = 21600 \huge( \small{1 + \dfrac{4}{25} \huge{ {)}^{ \small{3}}} }}$

$\small\sf{ Amount = 21600 \huge ( \small \dfrac{29}{25} \huge{ {)}^{{ \small {3}} }}}$

$\small\sf{Amount = 21600 \: \times \dfrac{29}{25} \times \dfrac{29}{25} \: \times \dfrac{29}{25} }$

$\small\sf{Amount = 33715.35}$

$\small\boxed{ \rm{ CI = Amount - Principal}}$

CI = 33715.3536 - 21600

CI = ₹ 12,115.3536

Now we will calculate the interest paid by Mr. Gupta.

We know,

$\small\boxed { \rm{SI = \dfrac{P× R×T}{100}}}$

$\small\sf{ SI = \dfrac{216 \cancel{00} × 16 × 3}{1 \cancel{00}}}$

$\small\sf{ SI = 216 \times 16 \times 3 = 10368}$

Clearly Mr.Singh has to pay more as interest than Mr.Gupta.

Required difference= ₹(12115.3536 - 10368) = ₹1,747.3536