Question

a sum of Rs.1000 after 3 years at compound interest becomes a certain amount that is equal to the amount that is the result of 3 years deprication from Rs.1728 .find the difference between the rate of CI and deprication ( given rate CI is 10% per annum

Answer 1

The sum of Rs. 1000 after 3 years at CI at 10 % will become 1331.

Now as per the question, 1728 became 1331 after 3 years of depreciation at a certain unknown rate. So this entirey becomes a new question. We now need to find the rate at which 1728 became 1331 in 3 years.

This formula might be of help- - rate = [(amount/principle)^1/t] - 1 × 100

r = ([(1331/1728)^1/3] - 1) × 100

r = [(11/12) - 1 ] × 100

r = -1/12 × 100

r = 100/12 % ( WE can neglect the negative sign)

Let's find the difference now.

the rate of CI = 10 %

the rate of depreciation = 100/12% = 25/3 %

difference = 10 - 25/3 = 5/3 % Answer.. .

Answer 2

The difference between rate of CI and depreciation is 2%

Let the amount after 3 years for Rs.1000 at the rate of 10% be a . It's equation is

$a = p (1 + \frac{r}{n} ) ^{nt}$

where p is principal, r is interest rate, n is no of times interest is compounded per year and t is time in years.

Here,

$a = 1000(1 + \frac{0.10}{1} ) ^{3}$

= 1331 rs.

By given question the amount after depreciation from rs.1728 is 1331 rs. we have to find rate.

$r = ( (\frac{a}{p} )^{ \frac{1}{t} } - 1) \times 100$

$r =( ( \frac{1331}{1728} ) ^{ \frac{1}{3} } - 1) \times 100$

$r = (0.92 - 1) \times 100$

= 8% ( neglect the negative sign since it is depreciation rate)

The difference between CI rate and depreciation rate is

10 - 8 = 2%.

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