a sum of money is invested for 3 years at a certain rate of interest. had the rate of interest been 1% more, it would have fetched rs 18 more than before. find the sum invested?
Well, well, well, looks like we've got a classic financial brain teaser here! Let's dive in, shall we?
Let's call the initial rate of interest 'x.' So, the additional 1% would be (x + 1).
Now, according to the problem, this small increment in interest would have earned an extra Rs 18 over the 3-year period. So, we can set up an equation like this:
3 * (x + 1) - 3 * x = 18
Now, let's solve this equation:
3x + 3 - 3x = 18
3 = 18
Oh, wait! Something's not right here. It seems like there's a mistake in my calculations. My apologies for the confusion caused. Unfortunately, this particular problem doesn't have enough information to find the exact sum invested.
If you have any other questions or need a good laugh, feel free to ask!
To find the sum invested, we will use the concept of compound interest.
Let's assume the original rate of interest is x%. According to the problem, if the rate of interest had been 1% more, it would have fetched Rs 18 more than before.
So, the difference in interest is Rs 18 for 3 years. Therefore, the annual interest is Rs 18/3 = Rs 6.
We can set up an equation based on compound interest formula:
A = P(1 + r/100)^t
Where:
A = Amount after interest
P = Principal/Sum invested
r = Rate of interest
t = Time period
Substituting the given values in the equation:
A1 = P(1 + x/100)^3 ...(1)
A2 = P(1 + (x+1)/100)^3 ...(2)
According to the problem, A2 - A1 = Rs 18.
P(1 + (x+1)/100)^3 - P(1 + x/100)^3 = 18
Dividing both sides by P, we get:
(1 + (x+1)/100)^3 - (1 + x/100)^3 = 18/P ...(3)
Now, let's solve equation (3) to find the value of P:
(1 + (x+1)/100)^3 - (1 + x/100)^3 = 18/P
Expanding both terms using the identity (a^3 - b^3) = (a - b)(a^2 + ab + b^2), we have:
[(1 + (x+1)/100) - (1 + x/100)][(1 + (x+1)/100)^2 + (1 + (x+1)/100)(1 + x/100) + (x/100)^2] = 18/P
Simplifying further:
[((x+1)/100) - (x/100)][(1 + (x+1)/100)^2 + (1 + (x+1)/100)(1 + x/100) + (x/100)^2] = 18/P
Cancelling out the common factors, we get:
[(1/100)][(1 + (x+1)/100)^2 + (1 + (x+1)/100)(1 + x/100) + (x/100)^2] = 18/P
[(1/100)][(1 + (x+1)/100)^2 + (1 + (x+1)/100)(1 + x/100) + (x/100)^2] = 18/P
Multiplying both sides by 100, we have:
[(1 + (x+1)/100)^2 + (1 + (x+1)/100)(1 + x/100) + (x/100)^2] = 1800/P
Expanding the square term and simplifying further:
[1 + 2(x+1)/100 + (x+1)/100^2 + (1 + (x+1)/100)(1 + x/100) + (x/100)^2] = 1800/P
Simplifying finally:
1 + 2(x+1)/100 + (x+1)/100^2 + 1 + (x+1)/100 + x/100 + (x+1)/100 * x/100 + (x/100)^2 = 1800/P
Now, we can solve this equation to find the value of x and consequently, the sum invested, P.
To find the sum invested, we need to solve this problem using algebraic equations.
Let's assume the initial rate of interest is 'r', and the sum invested is 'x'.
According to the given information, if the rate of interest had been 1% more, it would have fetched rs 18 more than before. This can be written as an equation:
(x * (r + 1/100) * 3) - (x * r * 3) = 18
Simplifying the equation, we get:
(x * 3/100) = 18
Now, solving for 'x' will give us the answer.
x = (18 * 100)/3
x = 600
Therefore, the sum invested is rs 600.
amount invested --- p
first rate of interest --- r
second rate of interest r + .01
interest at first rate = rp
interest at 2nd rate = p(r+.01)
p(r+.01) - rp = 18
rp + .01p - rp = 18
.01p = 18
p = 18/.01 = 1800