At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to
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Official Correct Answer: D. First, the problem says that a certain sum amounts to Rs 13920 in 3 years at a simple interest rate. Then, the same sum amounts to Rs 18960 in 6 years and 6 months. After that, it asks for the interest earned if the same sum is invested for 2 years with interest compounded every 6 months. The options are given, and the correct answer is D, which is 3221.
Alright, let's break this down. simple interest is calculated using the formula:
$$
A = P(1 + rt)
$$
where \( A \) is the amount, \( P \) is the principal, \( r \) is the rate of interest per annum, and \( t \) is the time in years.
Given that the amount after 3 years is 13920, so:
$$
13920 = P(1 + 3r) \quad \text{(1)}
$$
And after 6 years and 6 months, which is 6.5 years, the amount is 18960:
$$
18960 = P(1 + 6.5r) \quad \text{(2)}
$$
Hmm, so I have two equations here. Maybe I can subtract equation (1) from equation (2) to eliminate \( P \). Let's try that.
Subtracting (1) from (2):
$$
18960 - 13920 = P(1 + 6.5r) - P(1 + 3r)
$$
Simplify the left side:
$$
5040 = P[(1 + 6.5r) - (1 + 3r)] = P(3.5r)
$$
So,
$$
5040 = 3.5Pr
$$
Which means,
$$
Pr = \frac{5040}{3.5} = 1440
$$
So, \( Pr = 1440 \). Now, let's plug this back into equation (1):
$$
13920 = P(1 + 3r) = P + 3Pr = P + 3*1440 = P + 4320
$$
Therefore,
$$
P = 13920 - 4320 = 9600
$$
So, the principal \( P \) is 9600 rupees. Now, since \( Pr = 1440 \), we can find \( r \):
$$
r = \frac{1440}{P} = \frac{1440}{9600} = 0.15
$$
So, the rate of interest is 15% per annum.
Now, the next part is about compound interest. It says the sum is invested for 2 years with interest compounded every 6 months. So, the formula for compound interest is:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
where \( n \) is the number of times interest is compounded per year. Since it's compounded every 6 months, \( n = 2 \). The time \( t \) is 2 years.
Plugging in the values:
$$
A = 9600 \left(1 + \frac{0.15}{2}\right)^{2*2} = 9600 \left(1 + 0.075\right)^4
$$
Calculating \( 1.075^4 \). Compute this step by step.
First, \( 1.075^2 \):
$$
1.075 * 1.075 = 1.155625
$$
Then, square that result to get \( 1.075^4 \):
$$
1.155625 * 1.155625 ≈ 1.334838867
$$
So, approximately 1.334838867.
Now, multiply this by 9600:
$$
9600 * 1.334838867 ≈ 9600 * 1.3348 ≈ Computing 9600 * 1.3348.
First, 9600 * 1 = 9600.
9600 * 0.3348 = ?
Compute 9600 * 0.3 = 2880
9600 * 0.0348 = 9600 * 0.03 = 288; 9600 * 0.0048 = 46.08
So, 288 + 46.08 = 334.08
So, 2880 + 334.08 = 3214.08
Therefore, total amount is 9600 + 3214.08 = 12814.08
Wait, that doesn't seem right. the total amount is 9600 * 1.3348 ≈ 12814.08
But wait, the interest earned would be A - P, which is 12814.08 - 9600 = 3214.08
Looking at the options, the closest is D) 3221.
Alternatively, maybe my approximation of 1.075^4 was a bit off. Let me check more accurately.
Compute 1.075^4:
First, 1.075^2 = 1.155625
Then, 1.155625^2:
Compute 1.155625 * 1.155625:
Compute 1.155625 * 1.155625:
First, 1 * 1.155625 = 1.155625
0.155625 * 1.155625:
Compute 0.1 * 1.155625 = 0.1155625
0.05 * 1.155625 = 0.05778125
0.005625 * 1