Question : The least number of years in which a sum of money on 19% p.a. compound interest will be more than double is _____.
Option 1: 3 years
Option 2: 4 years
Option 3: 5 years
Option 4: 2 years
Correct Answer: 4 years
Solution :
Let the principal be Rs. $P$.
Amount (A) = Rs. $2P$
Rate = 19%
Time = $T$ years
When compounded annually, $ A= P(1+\frac{R}{100})^{T}$
Where A is the total amount, P is the principal amount, R is the rate of interest per annum, and T is the time in years.
$⇒A = P(1 + \frac{R}{100})^T$
$⇒2P = P(1 + \frac{19}{100})^T$
$⇒2 =(\frac{119}{100})^T$
$⇒2 = (1.19) ^T$
If $T$ = 4 years, (1.19)
4
> 2
(1.19)
4
= 2.005 > 2
$\therefore$ The required time is 4 years.
Hence, the correct answer is 4 years.
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