Find the least number of years in which the sum put at 25% rate of int...
Amount > = P*(1+25/100)n
Amount = p*(5/4)n
For n = 4, (625/256) which is greater than 2.
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Find the least number of years in which the sum put at 25% rate of int...
Amount > = P*(1+25/100)n
Amount = p*(5/4)n
For n = 4, (625/256) which is greater than 2.
Find the least number of years in which the sum put at 25% rate of int...
Solution:
Let the sum be x.
After 1 year, the amount will become $x+\frac{25}{100}x=\frac{5}{4}x$
After 2 years, the amount will become $\frac{5}{4}x+\frac{25}{100}\times\frac{5}{4}x=\frac{25}{16}x$
After 3 years, the amount will become $\frac{25}{16}x+\frac{25}{100}\times\frac{25}{16}x=\frac{15625}{1024}x$
After 4 years, the amount will become $\frac{15625}{1024}x+\frac{25}{100}\times\frac{15625}{1024}x=\frac{244140625}{16777216}x$
After 5 years, the amount will become $\frac{244140625}{16777216}x+\frac{25}{100}\times\frac{244140625}{16777216}x=\frac{152587890625}{8589934592}x$
We need to find the least number of years in which the sum put at 25% rate of interest will be more than doubled.
So, we need to find the minimum value of n for which $\left(\frac{5}{4}\right)^n>2$
Taking log on both sides, we get
$n\log\frac{5}{4}>\log 2$
$n>\frac{\log 2}{\log\frac{5}{4}}$
$n>4.3219$
The minimum value of n for which the sum put at 25% rate of interest will be more than doubled is 5.
Therefore, the correct option is (D) 5 years.