Certain sum becomes 3 times itself at compound interest in 10 years. In how many years it becomes 9 times?
Answers
Answer:
20 Years
Step-by-step explanation:
Let Say Amount = P
Amount with compound interest = P ( 1 + r/100)^t
3P = P ( 1 + r/100)¹⁰
3 = ( 1 + r/100)¹⁰ - eq 1
Let say in T years it become 9 times
9P = P(1 + r/100)^T
=> 9 = (1 + r/100)^T
=> 3² = (1 + r/100)^T
from eq 1
=> (( 1 + r/100)¹⁰)² = (1 + r/100)^T
=> ( 1 + r/100)²⁰ = (1 + r/100)^T
Comparing both
T = 20
Time = 20 Years
in 20 years amount become 9 times
Answer:
20 years
Step-by-step explanation:
Let the sum be ₹ 100.
Then,
Principal = ₹ 100
Time = 10 years
Amount = 3 x Principal = 3 x 100 = ₹300/-
Using mathematical formula of compound interest
Amount = P [1 + r/100]ⁿ
⇒300 = 100 [1 + r/100]¹⁰
⇒3 = [1 + r/100]¹⁰
⇒3⁽¹/¹⁰⁾ = 1 + r/100
⇒1.116 = 1 +r /100
⇒0.1116 = r/100
⇒11.16 = r
Thus rate of interest = 11.16 percent per annum
Now, we need to calculate the time in which this amount will become 9 times.
Thus, Principal = ₹ 100
Amount = ₹ 900
Rate of interest = 11.16 %
substituting these values in the above equations
⇒900 = 100 [1 + 11.16/100]ⁿ
⇒9 = [1 + 11.16/100]ⁿ
⇒9 = 1.1116ⁿ
Taking logrithm on both sides
⇒log 9 = n x log 1.116
[From log value table, log 9 = 0.9542; log 1.1116 = 0.0478]
Thus, n = 0.9542/0.0478
20.01 ≈ 20 years