In what time will ₹1200 amount to ₹1344 at 6% per annum?
I = Po*r*t = 1344-1200.
Po = 1200, r = 0.06, t = ?.
To calculate the time it takes for an amount to grow at a given interest rate, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount
P = the principal amount (₹1200 in this case)
r = the annual interest rate (6% or 0.06 as a decimal)
n = the number of times interest is compounded per year (usually 1 for annual compounding)
t = the time in years (what we need to find)
In this case, we need to find the time t, so let's rearrange the formula:
A/P = (1 + r/n)^(nt)
Now we can plug in the values given:
1344 / 1200 = (1 + 0.06/1)^(t * 1)
Simplifying:
1.12 = 1.06^t
Now, we need to solve for t. Take the natural logarithm (ln) of both sides:
ln(1.12) = ln(1.06^t)
Using the property of logarithms that ln(a^b) = b * ln(a):
ln(1.12) = t * ln(1.06)
Finally, solve for t by dividing both sides by ln(1.06):
t = ln(1.12) / ln(1.06)
Using a scientific calculator or an online calculator, we can find:
t ≈ 2.57 years
Therefore, ₹1200 will amount to ₹1344 in approximately 2.57 years at an annual interest rate of 6%.