A loan of Rs 1000 is fully repaid by two installments of Rs 530 and Rs 594, paid at the end of first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is
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Official Correct Answer: C. The problem says that a loan of Rs 1000 is repaid with two installments: Rs 530 at the end of the first year and Rs 594 at the end of the second year. The interest is compounded annually, and We find the rate of interest. The options are 9, 11, 8, or 10 percent.
Hmm, compound interest usually means that the interest is calculated on the initial principal and also on the accumulated interest from previous periods. Since the installments are paid at the end of each year, the remaining principal will earn interest for the next year.
Let me recall the formula for compound interest. The amount to be paid back after 'n' years is given by:
$$
A = P \times (1 + r)^n
$$
where:
- A is the amount after n years,
- P is the principal amount,
- r is the annual interest rate (in decimal),
- n is the number of years.
But in this case, the loan is being repaid in installments, so I think I need to consider the present value of these installments equal to the principal amount.
So, the first installment of Rs 530 is paid at the end of the first year. The second installment of Rs 594 is paid at the end of the second year. Both of these payments need to be discounted back to the present value (time zero) to equal the original loan amount of Rs 1000.
Let me write that down. The present value (PV) of the first installment is:
$$
PV_1 = \frac{530}{(1 + r)^1}
$$
And the present value of the second installment is:
$$
PV_2 = \frac{594}{(1 + r)^2}
$$
Since the total present value of both installments should equal the principal amount, we can write:
$$
PV_1 + PV_2 = 1000
$$
Substituting the expressions for PV1 and PV2:
$$
\frac{530}{1 + r} + \frac{594}{(1 + r)^2} = 1000
$$
Okay, so now I have an equation in terms of r. I need to solve for r. This seems a bit tricky because it's a quadratic equation in terms of 1/(1 + r). Maybe I can let x = 1/(1 + r) to simplify it.
Let me set:
$$
x = \frac{1}{1 + r}
$$
Then, the equation becomes:
$$
530x + 594x^2 = 1000
$$
Which is a quadratic equation:
$$
594x^2 + 530x - 1000 = 0
$$
Now, I can solve this quadratic equation for x using the quadratic formula. The quadratic formula is:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Here, a = 594, b = 530, and c = -1000.
Plugging these values in:
First, calculate the discriminant:
$$
D = b^2 - 4ac = (530)^2 - 4 \times 594 \times (-1000)
$$
Calculating each part:
530 squared is 530*530. Compute that:
530 * 530:
500*500 = 250000
500*30 = 15000
30*500 = 15000
30*30 = 900
So, 250000 + 15000 + 15000 + 900 = 280,900.
Wait, actually, that's not the right way. Compute 530^2:
530^2 = (500 + 30)^2 = 500^2 + 2*500*30 + 30^2 = 250000 + 30000 + 900 = 280,900.
Okay, so D = 280,900 - 4*594*(-1000). Wait, no, it's 280,900 + 4*594*1000 because c is negative.
So, 4*594 = 2376, and 2376*1000 = 2,376,000.
So, D = 280,900 + 2,376,000 = 2,656,900.
Now, the square root of D is sqrt(2,656,900). what that is.
I know that 1600^2 = 2,560,000. 1630^2 = (1600 + 30)^2 = 1600^2 + 2*1600*30 + 30^2 = 2,560,000 + 96,000 + 900 = 2,656,900.
Wow, that's exactly D. So sqrt(D) = 1630.
So, plugging back into the quadratic formula:
x = [-530 ± 1630]/(2*594)
We can ignore the negative solution because x must be positive (since it's 1/(1 + r) and r is positive). So:
x = (-530 + 1630)/(2*594) = (1100)/(1188)
Simplify 1100/1188. Let's divide numerator and denominator by 4:
1100 ÷ 4 = 275
1188 ÷ 4 = 297
So, x = 275/297. if this can be simplified further. Both numbers are divisible by 11:
275 ÷ 11 = 25
297 ÷ 11 = 27
So, x = 25/27.
Therefore, x = 25/27. But x was defined as 1/(1 + r), so:
1/(1 + r)