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A man lends ₹4000 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 ₹4000 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 ₹4000 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 ₹683.33.
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