The price of a flat costing Rs.7000000 depreciates at the rate of 20% per annum. In how many years will its cost be reduced to Rs.3584000?

Options

A. 2years
B. 3years
C. 5years
D. 4years

Answer

To solve the problem, we will use the formula for depreciation. The formula for the amount after depreciation is given by:

\[ A = P \times \left(1 - \frac{r}{100}\right)^n \]

Where:
- \( A \) is the final amount after depreciation.
- \( P \) is the principal amount (initial cost).
- \( r \) is the rate of depreciation.
- \( n \) is the number of years.

**Step 1: Identify the values.**
- The initial cost of the flat \( P = 7000000 \) (or Rs. 70 lakh).
- The final amount \( A = 3584000 \).
- The rate of depreciation \( r = 20\% \).

**Step 2: Substitute the values into the formula.**
We need to find \( n \) such that:

\[ 3584000 = 7000000 \times \left(1 - \frac{20}{100}\right)^n \]

**Step 3: Simplify the expression.**
Calculate \( 1 - \frac{20}{100} = 0.8 \):

\[ 3584000 = 7000000 \times (0.8)^n \]

**Step 4: Divide both sides by 7000000.**
This gives us:

\[ \frac{3584000}{7000000} = (0.8)^n \]

**Step 5: Calculate the left side.**
Now, calculate \( \frac{3584000}{7000000} \):

\[ \frac{3584000}{7000000} = 0.512 \]

So, we have:

\[ 0.512 = (0.8)^n \]

**Step 6: Recognize the powers.**
We can express \( 0.512 \) and \( 0.8 \) in terms of powers:

- \( 0.512 = \left(\frac{8}{10}\right)^3 \) (since \( 8^3 = 512 \) and \( 10^3 = 1000 \)).
- \( 0.8 = \left(\frac{8}{10}\right) \).

Thus, we can write:

\[ (0.8)^n = \left(\frac{8}{10}\right)^n \]

**Step 7: Set the exponents equal.**
Since both sides are equal and have the same base, we can equate the exponents:

\[ n = 3 \]

**Step 8: Conclusion.**
The number of years it will take for the cost of the flat to reduce to Rs. 3584000 is \( n = 3 \) years.

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