Find the sum of money when increases 1/10 of itself every year amount ...
This is an example of an exponential growth problem. Exponential growth can be modeled by the equation P = a(r)t, where a is the initial amount, t is the time that has passed, P is the amount after time t, and r is the rate of growth.
For your problem, t = 5 years and P = 600. The money increases by 1/10 = 10% each year, so each year, the amount of money is 110% of the previous year. That's a rate of growth of r = 1.1. So:
600 = a(1.1)5
600 = 1.61051a
372.55 = a
The sum was initially Rs. 372.55.
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Find the sum of money when increases 1/10 of itself every year amount ...
Let the initial sum of money be x.
After one year, the sum of money will become x + 1/10 x = 11/10 x.
After two years, the sum of money will become 11/10 x + 1/10 (11/10 x) = 121/100 x.
Similarly, after three years, it becomes 1331/1000 x, after four years it becomes 14641/10000 x, and after five years it becomes 161051/100000 x.
According to the question,
161051/100000 x = 600
x = 600 x 100000/161051
x = 372.95
Therefore, the initial sum of money is Rs. 372.95.
None of the given options is correct.
Find the sum of money when increases 1/10 of itself every year amount ...
The answer will be 372.95 by exponential growth P=a(r)t.
by solving further you'll obtain the answer and it's none of the above option(d)