# A man lends √¢‚Äö¬π4000 at 10.5% per annum, compound interest payable annually and other men lends the same sum at 10% per annum, but compound interest payable half-yearly. Who gains more at the end of 1 year and 8 months and by how much?

Given :

A man lends  ‚Çπ4000 at 10.5% per annum compound interest payable annually.

The other man lends the same amount at 10% per annum compound interest

payable half-yearly.

To do :

We have to find who gains more at the end of 1 year and 8 months and by how much.

Solution :

A man lends  ‚Çπ4000 at 10.5% per annum compound interest payable annually.

Interest $= 10.5$%

Time $= 1$ year

Compound interest can be calculated by the formula,

$$A = P(1 + \frac{r}{100})^n$$

Amount at the end of 1 year $= Rs. 4000 \times (1 + \frac{10.5}{100})^1$

$= Rs. 4000 \times (1+ \frac{2.1}{20})$

$= Rs. 200 \times 22.1$

$= Rs. 4420$

Now, principal $= Rs. 4420$

Time $= 8 months = \frac{8}{12} year = \frac{2}{3} years$

Interest $= 10.5$%

Amount at the end of 1 year 8 months $= Rs. 4420 + 4420\times \frac{2}{3} \times \frac{10.5}{100}$

$= Rs. 4420 + 309.40$

$= Rs. 4729.40$

The other man lends the same amount at 10% per annum compound interest

payable half-yearly.

Interest for year $= 10$%

Interest for 6 months $= \frac{10}{2}$% $= 5$%

$Time = 1 year 6 months = 1 year + 6 months = 1 year + \frac{6}{12} year = 1 \frac{1}{2} years$

Time periods $= \frac{\frac{3}{2}}{\frac{1}{2}} = 3$

Amount at the end of $1 \frac{1}{2} years = Rs. 4000 (1+\frac{10}{100})^3$

$= Rs. 4000\times (1.1)^3$

$= Rs. 4000 \times 1.331$

$=Rs. 5324$

Now, principal $= Rs. 5324$

$Time = 2 months = \frac{2}{12} year = \frac{1}{6} years$

Interest $= 10$%

Amount at the end of 1 year 8 months $= Rs. 5324 + 5324\times \frac{1}{6} \times \frac{10}{100}$

$= Rs. 5324 + 88.73$

$= Rs. 5412.73$

The second person gained $Rs. (5412.73-4729.40) = Rs. 683.33$ more than the first person.

Therefore, the second person gained more than the first person by ‚Çπ683.33.

Updated on: 10-Oct-2022

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