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In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).
- a)7.93%
- b)8.93%
- c)9.93%
- d)6.93%
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a bank, principal increases continuously at the rate of r% per year...
Let P be the principal at any time t. then,
When P = 100 and t = 0., then, c = 100, therefore, we have:
Now, let t = T, when P = 100., then;
When P = 100 and t = 0., then, c = 100, therefore, we have:
Now, let t = T, when P = 100., then;
Most Upvoted Answer
In a bank, principal increases continuously at the rate of r% per year...
To solve this problem, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A is the final amount
P is the principal amount (initial amount)
r is the annual interest rate (as a decimal)
n is the number of times the interest is compounded per year
t is the number of years
In this case, the principal doubles itself in 10 years. So, the final amount A is 2 times the principal P.
2P = P(1 + r/n)^(nt)
Since we are given that the principal increases continuously, we can take the limit as n approaches infinity. This simplifies the formula to:
2 = e^(rt)
Where e is the base of natural logarithms and is approximately equal to 2.71828.
Now, we can take the natural logarithm of both sides to solve for r:
ln(2) = rt
Dividing both sides by t:
r = ln(2)/t
Given that t = 10 years, we can substitute this value into the equation:
r = ln(2)/10
Using the given value of loge2 = 0.6931, we can calculate r:
r = 0.6931/10 = 0.06931 = 6.93%
Therefore, the value of r is 6.93%.
So, the correct answer is option D.
A = P(1 + r/n)^(nt)
Where:
A is the final amount
P is the principal amount (initial amount)
r is the annual interest rate (as a decimal)
n is the number of times the interest is compounded per year
t is the number of years
In this case, the principal doubles itself in 10 years. So, the final amount A is 2 times the principal P.
2P = P(1 + r/n)^(nt)
Since we are given that the principal increases continuously, we can take the limit as n approaches infinity. This simplifies the formula to:
2 = e^(rt)
Where e is the base of natural logarithms and is approximately equal to 2.71828.
Now, we can take the natural logarithm of both sides to solve for r:
ln(2) = rt
Dividing both sides by t:
r = ln(2)/t
Given that t = 10 years, we can substitute this value into the equation:
r = ln(2)/10
Using the given value of loge2 = 0.6931, we can calculate r:
r = 0.6931/10 = 0.06931 = 6.93%
Therefore, the value of r is 6.93%.
So, the correct answer is option D.
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In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).a)7.93%b)8.93%c)9.93%d)6.93%Correct answer is option 'D'. Can you explain this answer?
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